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Newton’s lawsAlthough he did not know it at the time, Isaac Newton’s

work in the 17th century signalled an end to the transition in the way the world was conceived and understood. The

transition was begun by Copernicus and Galileo, but Newton was able to use mathematics to develop laws and theories that could account for the motion of the heavens. These showed the universe to be a mechanism that could readily be understood, one that was regulated by simple natural laws. The universe taught by Aristotle, and accepted up until the time of Newton, was one in which objects were classified into categories and their motion depended on the category to which they belonged.

Isaac Newton was born in rural England in 1642, the year Galileo died. He so impressed his mentors that he was made Professor of Mathematics at Cambridge University at the age of 26. His interests spanned light and optics, mathematics (he was one of the inventors of calculus), astronomy and the study of mechanics. His greatest achievement was the formulation of the law of universal gravitation. This, along with a complete explanation of the laws that govern motion, is laid out in his book Philosophiae Naturalis Principia Mathematica (Mathematical Principles of Natural Philosophy), which was published in 1687. The Principia is one of the most influential publications in natural science. Newton’s framework for understanding the universe remained intact right up to the advent of Einstein’s relativity more than 200 years later.

Newton died in 1727 a famous man. He remained self-critical and shy. That he could ‘see so far’ was only because he was ‘able to stand on the shoulders of giants’. In saying this, he was referring to the work of Galileo, Copernicus and many others who had paved the way for his discoveries.

5CHAPTER

you will have covered material from the study of movement including:

• vector techniques in two dimensions

• forces in two dimensions

• Newton’s laws of motion

• problems in mechanics including weight and friction.

by the end of this chapter

Chapter 5 Newton’s laws

5.1

143

Force as a vector

The previous chapter developed the concepts and ideas needed to describe the motion of a moving body. This branch of mechanics is called kinematics.

In this chapter, rather than simply describe the motion, we will consider the forces that cause the motion to occur. Treating motion in this way falls within the branch of mechanics called dynamics. In simple terms, a force can be thought of as a push or a pull, but forces exist in a wide variety of situations in our daily lives and are fundamental to the nature of matter and the structure of the universe. Consider each of the photographs in Figure 5.1 and identify each force—push or pull—that is acting.

In each of the situations depicted in Figure 5.1, forces are acting. Some are applied directly to an object and some act on a body without touching it. Forces that act directly on a body are called contact forces, because the body will only experience the force while contact is maintained. Forces that act on a body at a distance are non-contact forces.

Contact forces are the easiest to understand and include the simple pushes and pulls that are experienced daily in people’s lives. Examples of these include the forces between colliding billiard balls, the force that you exert on a light switch to turn it on, and the forces that act between you and your chair as you sit reading this book. Friction and drag forces are other contact forces that you should be familiar with.

Non-contact forces occur when the object causing the push or pull is physically separated from the object that experiences the force. These forces are said to ‘act at a distance’. Gravitation, magnetic and electric forces are examples of non-contact forces.

The action of a force is usually recognised through its effect on an object or body. A force may do one or more of a number of things to the object. It may change its shape, change its speed or change only the direction of its motion. The tennis racquet in Figure 5.1a has applied a force to the tennis ball, and, as a consequence, the speed of the ball changes along with its direction. The ball also changes shape while the force acts!

Figure 5.1 (a) At the moment of impact, both the tennis ball and racquet strings are distorted by the forces acting at this instant. (b) The rock climber is relying on the frictional force between his hands and feet and the rock face. (c) A continual force causes the clay to deform into the required shape. (d) The gravitational force between the Earth and the Moon is responsible for two high tides each day. (e) The globe is suspended in mid-air because of the magnetic forces of repulsion and attraction.

(c)(b)

(d)

(e)

(a)

Motion144

The amount of force acting can be measured using the SI unit called the newton, which is given the symbol N. The unit, which will be defined later in the chapter, honours Sir Isaac Newton (1642–1727), who is still considered to be one of the most significant physicists to have lived. A force of one newton, 1 N, is approximately the force you have to exert when holding a 100 g mass against the downward pull of gravity. In everyday life this is about the same as holding a small apple. Table 5.1 provides a comparison of the magnitude of some forces.

Table 5.1 A comparison of the magnitude of various forces

Force Magnitude (N)

Force on the electron in a hydrogen atom 10−7

Holding a small apple against gravity 1

Opening a door 10

Pedalling a bicycle 300

Thrust of a Boeing 747 at take-off 106

Gravitational force between the Earth and the Sun 1022

Force: a vector quantityIn Chapter 4, quantities associated with motion were classified as being either vectors or scalars. Scalar quantities such as time and mass do not have a direction. Only their size or ‘magnitude’ should be given. Quantities that require a direction as well as a magnitude are called vectors. Force is a vector quantity because the direction in which a force acts is always significant. In this text, vectors are set in bold italics.

If a question only requires the magnitude of a vector, the direction can be ignored. In this text, italics will be used to show this.

In a diagram, a force is usually shown as an arrow whose length represents the magnitude of the force and whose direction is indicated by the arrow.

Consider the case of a soccer player who kicks the ball horizontally with a force of 95 N towards the east. The horizontal forces acting on the ball can be illustrated by a vector diagram as shown in Figure 5.3.

If there are two or more forces acting on the same object, these forces can be shown on the same diagram. If one force is larger, it should be represented by a longer vector. If, for example, the soccer ball just discussed was sitting in thick mud so that a frictional force of 20 N towards the west was acting as it was kicked, this could be represented as shown in Figure 5.4.

The subsequent motion of the soccer ball will be different in the two situations described above. When there is a large frictional force acting on the ball, its speed will be significantly reduced. The muddy ground will act to make the ball travel more slowly as it leaves the boot. To analyse the horizontal motion of the ball, it is necessary to add all the horizontal forces

FORC… is measured in newtons (N) and is a vector quantity. It requires a magnitude and a direction to describe it fully.

i

Figure 5.2 The netball will only go through the hoop if a force of the right magnitude and direction is applied. Force is a vector; it can only be completely specified if both the direction and magnitude are given.

Figure 5.3 When drawing force diagrams, it is important that the force is shown to be acting at the correct location. In this example, the force on the ball acts at the point of contact between the ball and the foot.

W E

95 N

145Chapter 5 Newton’s laws

that are acting on it at this instant. The ball is simply treated as a point mass located at its centre of mass.

If more than one force acts on a body at the same time, the body behaves as if only one force—the vector sum of all the forces—is acting. The vector sum of the forces is called the resultant or net force, ΣF (shown as a double-headed arrow).

Because force is a vector quantity, the addition of a number of forces must be undertaken with the directions of the individual forces in mind. Vector addition is shown in Figure 5.5.

95 N+

20 N=

95 N

20 NΣF = 75 N

Figure 5.5 When the forces (95 N acting towards the east and 20 N acting to the west) are added, the resultant or net force is 75 N towards the east. The ball will move as though this resultant force is the only force acting on it.

If the forces that are acting are perpendicular (or any other angle) to each other, the resultant force must still be found by performing a vector addition. Consider the example of a shopping trolley that is being simultaneously pushed from behind by one person and pushed from the side by another. This situation is illustrated in Figure 5.6.

To find the magnitude of the resultant force, Pythagoras’s theorem must be used:

ΣF = √802 + 602 = √10 000 = 100 N

60 N

80 N

North

80 N + 60 N =

80 N

View from above

ΣF = 100 N 60 N

Person 1

Person 2(a)

(b)

Figure 5.6 (a) Two perpendicular forces are acting on the trolley. (b) The vector addition of these two forces gives the resultant force (ΣF) that is acting on the trolley to be 100 N at 127°T. The trolley is treated as a point mass located at its centre of mass.

The N…T FORC… acting on a body experiencing a number of forces acting simultaneously is given by the vector sum of all the individual forces acting: ΣF = F

1 + F

2 + ... + F

n

i

Figure 5.4 The two forces being considered are acting at different locations and have different strengths. The larger force is shown as a longer vector.

W E

20 N

95 N

Remember, when adding vectors, the tail of the second vector is placed at the head of the first. The resultant vector is from the tail of the first vector to the head of the second. A full explanation of one-dimensional and two-dimensional vector addition is included in Appendix A.

Physics file

Motion146

To find the direction of the resultant force, trigonometry must be used:

tan θ = 6080

= 0.75

θ = 37°This is a direction of 37° south of east, which is equivalent to a bearing

of 127°T.Hence, the net force acting on the trolley is 100 N at a bearing of 127°T.

Worked example 5.1A A motorboat is being driven west along the Yarra River. The engine is providing a driving force of 560 N towards the west. A frictional force of 180 N from the water and a drag force of 60 N from the air are acting towards the east as the boat travels along.a Draw a force diagram showing the horizontal forces of this situation.b Determine the resultant force acting on the motorboat.

Solution a

b For the vector addition, treat the boat as a point mass located at its centre of mass.

The resultant or net force acting on the motorboat is ΣF = 320 N due west.

Worked example 5.1B While playing at the beach, Sally and Ken kick a stationary beachball simultaneously with forces of 100 N south and 150 N west respectively. The ball moves as if it were only subjected to the net force. In what direction will it travel, and what is the magnitude of the net force on the ball?

Solution The net force is found by treating the beachball as a point mass and is given by: ΣF = F

Sally + F

Ken

Figure 5.7 The golf ball moves in the direction of the applied force and is in the direction of the line joining the centre of the club-head with the centre of the ball. The force will be very large.

60 N

180 N 560 N

W E

F 560 N= + 180 N 60 N+

560 N

180 N 60 N F = 320 N=

Σ

Σ

100 N +150 N = Σ F

100 N

150 N

There are two methods for describing the direction of a vector in a two-dimensional plane. In each case, the direction has to be referenced to a known direction.

A full circle bearing describes north as ‘zero degrees true’—written as 0°T. In this convention, all directions are given as a clockwise angle from north. 90°T is 90° clockwise from north, i.e. due east. A force acting in a direction 220°T is acting in a direction of 220° clockwise from north. This is the method most commonly used in industry.

An alternative method is to provide a quadrant bearing, where all angles are between 0° and 90° and so lie within one quadrant. The particular quadrant is identified using two cardinal directions, the first being either north or south. In this method, 220°T becomes S40°W, literally ‘40° west of south’.

220°Tor S40°W

F

N

W E

S

Figure 5.8 The direction 220°T lies 220° clockwise from north. This direction can also be written as S40°W meaning 40° west of south.

Physics file


ΣF = √1002 + 1502 = 180 N

tan θ = 150100

= 1.5

θ = 56°This is a quadrant bearing of 56° west of south, which is equivalent to a true bearing of 236°T. Hence, the net force acting on the beachball is 180 N in the direction 236°T.

Vector components

It is often helpful to divide a force acting in a two-dimensional plane into two vectors. These two vectors are called the components of the force. This can be done because the force can be considered to act in each of the two directions at once. Consider, for example, the pulling force of 45 N acting on the cart shown in Figure 5.9.

This pulling force is acting through the rope and is known as tension or a tensile force. The force is acting at an angle of 20° to the horizontal, so it has some effect in the horizontal direction and some effect in the vertical direction. The amounts of force acting in each direction are the components of the force.

It is usual to construct a right-angled triangle around the force vector. The force vector is the hypotenuse of the triangle, and the adjacent and opposite sides become the components of the force. The horizontal and vertical components of the pulling force can then be determined using trigonometry. It is important to remember that there is only one pulling force acting on the cart, but this force can be treated as two component forces.

So, the cart will move as though a horizontal force of 42 N pulling the cart along and a vertical pulling force of 15 N upwards were acting on it simultaneously. When the components are added together, the original 45 N force is the resultant force.

Is this the most effective way of using a 45 N force to move the cart forwards? No, it would be slightly more effective if the 45 N force was acting in the horizontal direction. This would make the cart travel faster, but it may be impractical or inconvenient to apply the force in this way.

Worked example 5.1C A stationary hockey ball is struck with a force of 100 N in the direction N30°W. What are the northerly and westerly components of this force?

Solution

Figure 5.9 The pulling force acting on the cart has a component in the horizontal direction and a component in the vertical direction.

Figure 5.10 The magnitudes of the vector components F

h and F

v can be calculated using

trigonometry.

Fv = 45 sin 20° = 15 N

Fh = 45 cos 20° = 42 N

20°

45 N

30

30

N

S

E

W

westerly component ofthe force = 50 N

northerlycomponent ofthe force = 87 Nforce from

the hockey stick = 100 N

F = 45 N

20°

Motion148

FW

= 100sin30° = 50 NF

N = 100cos30° = 87 N

The ball moves as though forces of 50 N west and 87 N north were acting on it simultaneously.

Worked example 5.1D When walking, a person’s foot pushes backwards and downwards at the same time. While playing basketball, Kate’s foot pushes back along the court with a force of 400 N, and down with a force of 600 N. What is the actual force applied by Kate’s foot?

Solution 400 N horizontally and 600 N vertically downwards are the components of the force supplied by Kate’s foot. Therefore, the force she supplies will be F = F

horizontal + F

vertical and a

vector diagram is needed.Using Pythagoras’ theorem:

F = √Fh

2 + Fv

2 = √4002 + 6002 = √520 000 = 721 N

θ = tan−1 600400

= tan−11.5 = 56°

So Kate supplies a force of 721 N backwards at 56° down from the horizontal.

Fv = 600 N downF

Fh = 400 N

• A force is a push or a pull. Some forces act on contact while others can act at a distance.

• Force is a vector quantity whose SI unit is the newton (N).

• A vector can be represented by a directed line seg ment whose length represents the magnitude of the vector and whose arrowhead gives the direction of the vector.

• The net force acting on a body that experiences a number of forces acting simultaneously is given by the vector sum of all the individual forces acting:

ΣF = F1 + F2 + … + Fn

• A vector addition may be calculated using a sketch vector diagram that can be solved using trigonometry.

• A force F acting at an angle θ to a given direction will have components F cos θ parallel to the reference direction, and F sin θ perpendicular to that reference direction.

Force as a vector5.1 summary


1 1 Which one or more of the following quantities are vectors?

A mass B velocity C C temperature D force

2 2 Calculate the resultant force in each of the following vector additions:a 200 N up and 50 N downb 65 N west and 25 N eastc 10 N north and 10 N southd 10 N north and 10 N west

3 3 If the force you have to exert when holding a small apple is about 1 N and holding a kilogram of sugar is 10 N, estimate the force required for:a using a staplerb kicking a beachballc lifting your school bagd doing a chin-up exercise.

4 4 Which one or more of the following directions are identical?A 40°T and S40°E B 140°T and S40°E C 200°T and S20°W D 280°T and N80°W

5 5 Convert the following into full circle bearings (i.e. °T).a N60°E b N40°Wc S60°W d SEe NNE

6 6 Use the vectors below to determine the forces rep-resented in the following situations.

Scale: 1 cm represents 20 N

7 7 Use trigonometry if necessary to add the following forces:a 3 N east and 4 N westb 60 N east and 80 N south

8 8 A small car is pulled by two people using ropes. Each person supplies a force of 400 N at an angle of 40° to the direction in which the car travels. What is the total force applied to the car?

40°40°

400 N

400 N

9 9 Resolve the following forces into their perpendicular components around the north–south line. In part d, use the horizontal and vertical directions.a 100 N south 60° eastb 60 N northc 300 N 160°Td 3.0 × 105 N 30° upward from the horizontal

10 10 What are the horizontal and vertical components of a 300 N force that is applied along a rope at 60° to the horizontal used to drag a Christmas tree across the backyard?

Force as a vector5.1 questions

a

F

b

F

c

F

Worked Solutions

Motion150

Newton’s fi rst law of motion5.2

Aristotle and GalileoThe first attempt to explain why bodies move as they do was made more than 2000 years ago by the Greek philosopher Aristotle. As discussed in Chapter 4, Aristotle and his followers felt that there was a natural state for matter and that all matter would always tend towards its natural place where it would be at rest. Aristotle’s thesis was based on the everyday observation that a moving body will always slow down and come to rest unless a force is continually applied. Try giving this book a (gentle) push along a table top and see what happens.

Aristotle’s ideas were an attempt to explain the motion of a body as it was seen, but they do not help to explain why a body moves as it does. It was not until the early 17th century that Galileo Galilei was able to explain things more fully. Galileo performed experiments that led him to conclude that the natural state of a moving body is not at rest. Significantly, Galileo introduced the idea that friction was a force that, like other forces, could be added to other forces. A generation later, Newton developed Galileo’s ideas further to produce what we now call the first law of motion.

To understand Newton’s first law, follow the logic of this thought experiment, similar to one used by Galileo. Consider a steel ball and a smooth length of track. In Figure 5.11a, the ball is held at one end of an elevated track; the other end of the track is also elevated. When the ball is released, it will roll downhill, then along the horizontal section and then up the elevated section. In reality, it will not quite reach the height it started at, due to friction.

Now, imagine there were no friction. The ball, in this ideal case, would slow down as it rolled uphill and finally come to rest when it reached the height at which it started.

Now consider what would happen if the angle of the elevated section was made smaller as in Figure 5.11b. The ball would now roll further along the track before coming to rest because the track is not as steep. If we could again imagine zero friction, the ball would again slow down as it rolls uphill and reach the same height before stopping, but travelling further in the process.

What would happen if we made the angle of this elevated section progressively smaller? Logically, we would expect that, if we had enough track, the ball would travel even further before stopping.

Now consider Figure 5.11c. Here the end of the track is not elevated at all. As the ball rolls along, there is nothing to slow it down because it is not going uphill. If we can ignore friction, what will the ball do as it rolls along the horizontal track? Galileo reasoned that it would not speed up, nor would it slow down. Ideally, the ball should keep travelling horizontally with constant speed and never reach its starting height. According to Galileo, the natural state of a body was to keep doing what it was doing. This tendency of objects to maintain their original motion is known as inertia.

As the ball travels along the horizontal track, there is no driving force, nor is there any retarding force acting. The net force on the ball is zero and so it keeps moving with a constant velocity. This is the breakthrough in

Figure 5.11 Galileo used a thought experiment to derive his law of inertia. This became Newton’s first law and stated that the natural state of bodies was to maintain their original motion. This contradicted Aristotle’s idea that the natural state of bodies was at rest.

Ideally you would expect the ball to reach this height.

Ball ideally would roll furtherand reach this position.

The ball ideally will keepmoving with a constant velocity.

(a)

(b)

(c)

Interactive


understanding that Newton was able to make. Any body will continue with constant velocity if zero net force (ΣF = 0) acts upon it.

A good example of inertia and Newton’s first law is illustrated by the air-track. With the air turned on, give a glider a gentle push along the track. It will travel along the track with a constant velocity as described by Newton’s first law. There are no driving or retarding forces acting on the glider, so it simply maintains its original motion. Aristotle’s laws would not be able to explain the motion of the glider.

The motion of a spacecraft cruising in deep space is another good example of a body moving with constant velocity as required by Newton’s first law. As there is no gravitational force, and no air in space to retard its motion, the spacecraft will continue with constant speed in a straight line. The absence of air explains why there is no need to make a space probe aerodynamic in shape.

N…WTON’S FIRST LAW OF MOTION states that a body will either remain at rest or continue with constant speed in a straight line (i.e. constant velocity) unless it is acted on by an unbalanced force.

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At the time of the Roman Empire some 2000 years ago, it cost as much money to have a bag of wheat moved 100 km across land as it did to transport it across the whole Mediterranean Sea. One of the reasons for this stemmed from the enormous friction that acted between the wheel and the axle in the cart of the day. Some animal fats were used as crude lubricants, but the effect was minimal. It has only been during the last century that engineering has provided a mechanical solution.

Today’s wheels are connected to the axle by a wheel bearing. An outer ring is attached to the wheel, and an inner ring is attached to the axle. Separating the rings are a number of small ball bearings, which are able to roll freely between the rings. In this way, the area of contact and the friction between the wheel and axle is reduced dramatically.

Figure 5.13 Ball bearings reduce friction and enable wheels to work very efficiently.

Physics file

Several decades before Newton, Galileo Galilei had concluded that objects tended to maintain their state of motion. He called this tendency inertia, so this conclusion is also known as Galileo’s law of inertia. Inertia is not a force; it simply describes the property of bodies to continue their motion.

Physics file

Figure 5.14 Two Voyager spacecraft were launched from Cape Canaveral in 1977 with the mission to investigate the outer planets of the solar system at close hand. Both craft completed the mission successfully, passing Saturn in 1981, Uranus in 1986 and Neptune in 1989. Voyager 1 and 2 have now left the solar system and since they have effectively zero net force acting on them, they continue to travel away from the Earth with a constant velocity.

Figure 5.12 An air-track glider moves with a constant velocity because there is zero net force acting on it. This is an example of inertia.

Motion152

Forces in equilibriumNewton’s first law states that a body will travel with a constant velocity (or remain at rest) when the vector sum of all the forces acting on it is zero, i.e. when the net force is zero. When the net force is zero, the forces are said to be in equilibrium or balance.• If a body is at rest and zero net force acts on it, it will remain at rest. This

applies to any stationary object such as a parked car or a book resting on a desk, as shown in Figure 5.15. In these cases, the velocity is zero and it is constant. The forces that are acting are balanced and so the net force is zero.

• If a body is moving with a constant velocity and zero net force acts on it, it will continue to move with the same constant velocity. An example of this is a spacecraft with its engines off travelling through deep space. If gravitation is ignored, there is nothing to slow the craft down or to speed it up, and so it will continue with a constant velocity. The net force acting on the spacecraft is zero, so it will move with a constant velocity. Similarly, if a bus is travelling along a road with a constant velocity, the vector sum of the forces acting on the bus must be zero. The driving forces must balance the retarding forces, i.e. ΣF = 0.

Worked example 5.2A During a car accident, a passenger travelling without a fastened seatbelt may fly through the windscreen and land on the road. Using Newton’s first law of motion, explain how this will occur.

Solution During the accident, the car is brought to rest suddenly. Any occupants of the car will continue to travel with the original speed of the car until a force acts to slow them down. If the seatbelt were fastened, this would provide the necessary force to slow the passenger within the car. In the absence of an opposing force, the passenger continues to move—often crashing through the windscreen.(The injuries received as a consequence of not wearing a seatbelt are usually far more serious than those received if the person were fixed in the car during an accident. This is why laws require seatbelts to be worn.)

Figure 5.15 The forces acting on this book are balanced—i.e. ΣF = 0—so the velocity of the book will not change; the book will continue to stay at rest. That is, the book’s velocity will be constant at 0 m s−1.

Figure 5.17 (a) There are no forces acting on the spacecraft. The net force is zero and so it continues to move with constant velocity. (b) The forces acting on the bus are in equilibrium, so the net force is zero. The bus continues to move with constant velocity.

FT

Fg

Σ F = 0 velocity doesn’t

change

no forces acting

deep space(a) v = 50 km h–1Fground

Fgravity

Fdrag

Fdriving

forces are balanced

velocity doesn’t change

(b)

F = 0

velocity doesn’t changeF = 0

Friction had a lot to do with Australia’s Steven Bradbury winning a surprise gold medal at the 2002 Winter Olympics. His opponents did not experience enough friction as they skated around the home turn. Their inertia, in the absence of any horizontal forces, sent them crashing into the side wall. Steven stayed upright, his skates cutting into the ice and producing enough friction to allow him to turn the corner and win the gold medal.

Figure 5.16 Steven Bradbury won the gold medal as his opponents learned first-hand about inertia and friction (or rather a lack of friction!).

Physics file


Worked example 5.2B A cyclist keeps her bicycle travelling with a constant velocity of 8.0 m s−1 east on a horizontal surface by continuing to pedal. A force (due to air resistance) of 60 N acts against the motion. How large is the driving force that is provided by the rear tyre at the point of contact with the ground?

v = 8 m s–1

Fdrag = 60 N

Fapplied = 60 N

ΣF = Fapplied + Fdrag = 0

Solution If the cyclist is to continue at a constant 8.0 m s−1 east, then the forces that act on the bicycle must be in equilibrium, i.e. ΣF = 0. This means that the forces due to air resistance are exactly balanced by the pedalling force. A force of 60 N east must be produced at the rear wheel.(The cyclist will actually have to produce more than 60 N as the gearing of the bike is designed to increase speed, not reduce the force that has to be applied.)

Prac 21

Figure 5.18 In these examples, the forces acting on the objects are in equilibrium—the net force is zero. The body will either remain at rest, like the picture hanging on the wall, or continue with constant velocity, like the aircraft. A more complex situation involves the groundsman pushing the heavy roller with constant velocity. The horizontal component of the force he applies along the handle exactly balances the frictional force that opposes the motion of the roller in the horizontal direction. The vertical component of the applied force acts downwards, and adds to the weight of the roller, but these two downward forces are balanced by an upward force provided by the ground.

lift

weight, Fg

Fg

dragFthrust

Fthrust

Fdrag

Flift

F2

F1

F1

Fg

Fg

appliedforce

weight of rollerfriction

upwardsforce fromground onroller

applied push

upward forcefrom ground

weight of roller

friction

Fg + F1 + F2 = 0 Fthrust + Fg + Fdrag + Flift = 0

F2

SPARKlab

Motion154

Galileo Galilei was born into an academic family in Pisa, Italy, in 1564. Galileo made significant contributions to physics, mathematics and the scientific method through intellectual rigour and the quality of his experimental design. But more than this, Galileo helped to change the way in which the universe was understood.

Galileo’s most significant contributions were in astronomy. Through his development of the refracting telescope he discovered sunspots, lunar mountains and valleys, the four largest moons of Jupiter (now called the Galilean moons) and the phases of Venus. In mechanics, he demonstrated that projectiles moved with a parabolic path and that different masses fall at the same rate (the law of falling bodies).

These developments were most important because they changed the framework within which mechanics was understood. This framework had been in place since Aristotle had constructed it in the 4th century BCE. By the 16th century, the work of the Greek philosophers had become entrenched, and it was widely supported in the universities. It was also supported at a political level. In Italy at that time, government was controlled by the Catholic Church. Today one would think that Galileo would have been praised by his peers for making such progress, but so ingrained and supported was the Aristotelian view that Galileo actually lost his job as a professor of mathematics in Pisa in 1592.

Galileo was not without supporters, though, and he was able to move from Pisa to Padua where he continued teaching mathematics. At Padua, Galileo began to use measurements from carefully constructed experiments to strengthen his ideas. He entered into vigorous debate in which his ideas (founded as they were on observation) were pitted against the philosophy of the past and the politics of the day. The most divisive debate involved the motion of the planets. The ancient Greek view, formalised by Ptolemy in the 2nd century AD, was that the Earth was at the centre of the solar system and that all the planets, the Moon and the Sun were in orbit around it. This view was taught by the Church and was also supported by common sense. As such, it was accepted as the establishment view. In 1630 Galileo published a book in which he debated the Ptolemaic view and the new

Sun-centred model proposed by Copernicus. On the basis of his own observations, Galileo supported the Copernican view of the universe. However, despite the book having been passed by the censors of the day, Galileo was summoned to Rome to face the Inquisition for heresy. The finding went against Galileo, and all copies of his book had to be burned and he was sentenced to permanent house arrest for the term of his life.

Galileo died in 1642 in a village near Florence. He had become an influential thinker across Europe and the scientific revolution he had helped start accelerated in the freer Protestant countries in northern Europe. For its part, the Catholic Church under Pope John Paul II began an investigation in 1979 into Galileo’s trial, and in 1992 a papal commission reversed the Church’s condemnation of him.

Galileo Galilei—revolutionaryPhysics in action

Figure 5.19 Galileo Galilei (1564–1642) was a short, active man with red hair. Galileo made significant contributions to our understanding of the forces that act on moving bodies. In his book the Principia, Newton was quick to acknowledge his debt to Galileo’s genius. This portrait was drawn 8 years before Galileo’s trial.

• Aristotle theorised that the natural state of matter was to be at rest in its natural place.

• Galileo performed experiments and from these developed the idea of inertia.

• Newton developed Galileo’s ideas further and devised the first law of motion, stated as ‘A body

will either remain at rest or continue with constant velocity unless it is acted on by a non-zero net force (or an unbalanced force).’

• Where the net force on a body is zero, i.e. ΣF = 0, the forces are said to be balanced and are in equilibrium.

Newton’s first law of motion5.2 summary


1 1 In just a few sentences, distinguish between the understandings held by Aristotle and Newton about the natural state of matter. Describe an experiment that might help support each of these views.

2 2 A billiard ball is rolling freely across a smooth horizontal surface. Ignore drag and frictional forces when answering these questions.a Which of the following force diagrams shows the

horizontal forces acting on the ball according to the theories of Aristotle?

b Which of the following force diagrams is correct for the ball according to the theories of Newton?

c Which force diagram correctly describes this situation?

3 3 If a person is standing up in a moving bus that stops suddenly, the person will seem to fall forwards. Has a force acted to push the person forwards? Use Newton’s first law of motion to explain what is happening.

4 4 What horizontal force has to be applied to a wheelie bin if it is to be wheeled to the street on a horizontal path against a retarding force of 20 N at a constant 1.5 m s−1?

5 5 When flying at constant speed at a constant altitude, a light aircraft has a weight of 50 kN down, and the thrust produced by the engines is 12 kN to the east. What is the lift force required by the wings of the plane, and how large is the drag force that is acting?

6 6 A young boy is using a horizontal rope to pull his go-kart at a constant velocity. A frictional force of 25 N also acts on the go-kart.a What force must the boy apply to the rope?b The boy’s father then attaches a longer rope to the

kart because the short rope is uncomfortable to use. The rope now makes an angle of 30° to the horizontal. What is the horizontal component of the force that the boy needs to apply in order to move the kart with constant velocity?

c What is the tension force acting along the rope that must be supplied by the boy?

7 7 Passengers on commercial flights are required to be seated and have their seatbelts done up when their plane is coming in to land. What would happen to a person who was standing in the aisle as the plane travelled along the runway during landing?

8 8 Consider the following situations, and name the force that causes each object not to move in a straight line.a The Earth moves in a circle around the Sun with

constant speed.b An electron orbits the nucleus with constant

speed.c A cyclist turns a corner at constant speed.d An athlete swings a hammer in a circle with

constant speed.

9 9 A magician performs a trick in which a cloth is pulled quickly from under a glass filled with water without the glass falling over or the water spilling out. a Explain the physics principles underlying this

trick.b Does using a full glass make the trick easier or

more difficult? Explain.

10 10 Which of these objects would find it most difficult to come to a stop: a cyclist travelling at 50 km h−1, a car travelling at 50 km h−1 or a fully laden semitrailer travelling at 50 km h−1? Explain.

Newton’s first law of motion5.2 questions

F F

FF

A B

C D

Worked Solutions

Motion156

Newton’s second law of motion5.3

Newton’s first law of motion states that when all the forces on a body are balanced, the body can only remain at rest or continue with constant velocity. Newton’s second law of motion deals with situations in which a body is acted on by a non-zero net force, i.e. ΣF ≠ 0; in other words, when the forces are unbalanced.

When there is a non-zero net force acting on a body, the body will accelerate in the direction of the net force. Newton explained that the rate of this acceleration will depend on both the size of the net force and the mass of the body. Experiments show that the acceleration produced is directly proportional to the size of the net force acting:

a ∝ ΣFExperiments also show that the acceleration produced by a given net

force depends on the mass of the body. We know that a greater mass has a greater inertia, so it will be more difficult to accelerate. Not surprisingly, experiments reveal that the acceleration produced by a particular force is inversely proportional to the mass of the body:

a ∝ 1m

If the two relationships are combined, we get:

a ∝ ΣF × 1m

ora ∝

ΣFm

The relationship can be converted into an equality by including a constant of proportionality, so:

a = k × ΣFm

By definition, 1 newton is the force needed to accelerate a mass of 1 kg at 1 m s−2. In the SI system of units, this makes the constant k equal to 1. The relationship is therefore simplified to ΣF = ma, a mathematical statement of Newton’s second law of motion.

The SI unit, the newton (N), will be required for force when the mass of the accelerating body is given in kilograms (kg) and its acceleration is provided in metres per second squared (m s−2).

ΣF = ma is a vector equation in which the direction of the acceleration is in the same direction as the net force. If only one force acts, the acceleration will be in the direction of that force.

Worked example 5.3A Determine the size of the force required to accelerate an 80 kg athlete from rest to 12 m s−1 in a westerly direction in 5.0 s.

N…WTON’S S…COND LAW OF MOTION states that the acceleration of a body, a, is directly proportional to the net force acting on it, ΣF, and inversely proportional to its mass, m:

ΣF = ma

i

Figure 5.20 This sprinter is about to leave the starting blocks. The starting blocks stop his foot from slipping backwards, increasing the size of the forward force acting on him and increasing his forward acceleration.

From Newton’s second law, it can be seen that the unit of a newton (N) is equivalent to the product of the mass unit (kg) and the acceleration unit (m s−2). In other words: N = kg m s−2.

When writing the value of a force, either unit is correct; but newton is the SI unit and is obviously more convenient to use!

Physics file


Solution First, determine the acceleration of the athlete:v = u + at

a = v − u

t

= 12 − 0

5.0a = 2.4 m s−2 westThe net force can now be found using Newton’s second law:ΣF = ma

= 80 × 2.4= 190 N west

If more than one force acts on a body, the acceleration will be in the direction of the net force, i.e. the vector sum of all of the forces.

Worked example 5.3B A swimmer whose mass is 75 kg applies a force of 50 N as he starts a lap. The water opposes his efforts to accelerate with a drag force of 20 N. What is his initial acceleration?

Solution The net force on the swimmer in the horizontal direction will be:ΣF = F

applied + F

drag

A vector addition gives ΣF = 30 N forwards.So,

a = ΣFm

= 3075

= 0.40 m s−2 in the direction of the applied force

(It is worth noting that the drag applied by the water will increase with the swimmer’s speed.)

Worked example 5.3C A 150 g hockey ball is simultaneously struck by two hockey sticks. If the sticks supply a force of 15 N north and 20 N east respectively, determine the acceleration of the ball, and the direction in which it will travel.

Solution Remember to work in kilograms. Calculate the net force acting on the ball by performing a vector addition:ΣF = F

1 + F

2

ΣF = √F1

2 + F2

2

= √152 + 202 = √225 + 400 = √625 = 25 N

a = ΣFm

= 25

0.15

= 170 m s−2

Looking at the vector diagram showing the addition of the forces, we can see that θ will be given by:

tan θ = F

2

F1

= 2015

= 1.33

so θ = tan−1 1.33 = 53°The ball will travel in the direction N53°E or (053ºT).

Prac 22

50 N

20 N

30 N

+

Fapplied

Fdrag

ΣF

F1 = 15 N north

F2 = 20 N east

F2

F1

N

S

W EΣF

θ

Motion158

Mass and weightMass of a body

To this point, the idea of the mass of an object has been taken for granted. However, the concept of a body’s mass is rather subtle and, importantly in physics, the mass of a body is a fundamentally different quantity from its weight—even though people (even physics teachers) tend to use these expressions interchangeably in everyday life.

In earlier science courses, mass may have been defined as ‘the amount of matter in an object’. To understand what mass really is, this description says very little. The international standard for the kilogram is not very helpful either. Since the time of the French Revolution (late 1700s), the kilogram has been defined in terms of an amount of a standard material. At first, 1 litre of water at 4°C was used to define the kilogram. More recently an international mass standard has been introduced. This is a 1 kg cylinder of platinum–iridium alloy that is kept in Paris. Copies are made from the standard and sent around the world.

Newton’s second law can help to provide a better understanding of mass through the effect of a force on a massive body. Think about a mass resting on a frictionless surface. If a force is applied to the mass in the

horizontal direction, an acceleration is produced that is given by a = ΣFm

.

The greater the mass, the smaller will be the acceleration. If the mass is reduced, the acceleration will increase. Here the mass can be seen as the property of the body resisting the force. Mass is the closest quantity in physics to the concept of inertia.

If the above experiment is repeated at another location, the same net force acting on the body will give the same acceleration regardless of where the experiment is performed. This is because—on Earth, on the Moon, in space—the mass of the body remains the same. Mass is a property of the body, and is not affected by its environment. In fact, for any situation at this level in physics, the mass of a body will be a constant value. As discussed in the adjacent Physics file, Albert Einstein was able to show in his theory of relativity that the mass of an object does change as its speed changes.

The (inertial) MASS of a body is its ability to resist acceleration when the body is acted on by a net force. Mass is a constant property of the body.

i

Mass is usually considered to be an unchanging property of an object. This is true in Newtonian mechanics where the speed with which an object is considered to travel matches everyday experience. However, at very high speeds, Newton’s laws of motion do not apply, and the theory of relativity must be used. In 1905, Albert Einstein showed that a body with a rest mass m

0 (i.e. mass

when stationary) will experience an increase in mass as it gets faster. This increase is usually undetectable except when the object nears the speed of light. At these very high speeds, the mass will become greater and greater, tending to infinity as the speed approaches the speed of light.

Figure 5.21 Scientists working at the Australian synchrotron in Clayton have to take account of these relativistic effects on the electrons they accelerate to extreme speeds.

Table 5.2 gives the mass of a 1 kg block if it were to travel at speeds of 0.1c (10% of the speed of light or 3 × 107 m s−1), 0.8c and 0.99c.

Table 5.2 The mass of a 1 kg block at different speeds

Speed Mass

0.1c 1.0050 kg (i.e. 5 g increase)

0.8c 1.6667 kg (667 g increase)

0.99c 7.1 kg (over 700% increase)

Relativistic mass increase provides the reason why no object can travel at the speed of light. To do so would require an infinite quantity of energy, since the mass of the body would itself be infinite. Only objects with no rest mass (such as light ‘particles’) can travel at the speed of light.

Physics file

Figure 5.22 Regardless of the external conditions, the inertial qualities of a mass remain the same. A net force of 1 N will always produce an acceleration of 1 m s−2 for a 1 kg mass. In this way, mass can be understood as the resistance to a force. The greater the mass of the body, the smaller the acceleration caused by the force.

1 kg

on Earth in deep space

a = 1 m s–2

a = 1 m s–2

F = 1 N

F = 1 N1 kg


Weight of a body

In the late 1500s, Galileo was able to show that all objects dropped near the surface of the Earth accelerate at the same rate, g, towards the centre of the Earth. The force that produces this acceleration is the force of gravity. In physics, the force on a body due to gravity is called the weight of a body, Fg or W.

Consider a pumpkin of mass 10 kg and a banana of mass 0.10 kg that are dropped together from several metres above the surface of the Earth.

The pumpkin and the banana will fall with a uniform acceleration that is equal to 9.8 m s−2. The acceleration of a freely falling object (i.e. one on which the only force that is acting is gravity) does not depend on the mass of the object.

If we took the pumpkin and the banana to the Moon and dropped them, they would fall with an acceleration of just 1.6 m s−2. Gravity is weaker on the Moon because it is much less massive than Earth.

The acceleration of a freely falling object due to gravity is known as g.i

Prac 23

Figure 5.23 If air resistance is ignored, the pumpkin and banana will fall side by side with an acceleration of 9.8 m s−2 towards the Earth.

Figure 5.24 The pumpkin and banana will fall side by side with a uniform acceleration of 1.6 m s−2 towards the Moon.

a = 9.8 m s–2 a = 9.8 m s–2

a = 1.6 m s–2 a = 1.6 m s–2

SPARKlab

Motion160

We will now use Newton’s second law to analyse the motion of the pumpkin as it falls to the Earth. The only force acting on the pumpkin is the force of gravity or weight, Fg. Hence:

ΣF = Fg

ma = Fg

The acceleration of the pumpkin is 9.8 m s−2 or g, so Fg = mg.

The acceleration of a mass due to gravity is numerically identical to the gravitational field strength, g. These two quantities have different names and different units but are numerically equal. It can be shown that 1 m s−2 is equal to 1 N kg−1.

As a consequence of this, the weight of a body will change as it is placed in different gravitational fields. On the Earth a 10 kg pumpkin will have a weight of 10 × 9.8 = 98 N downwards. On the Moon, the gravitational field strength is lower at 1.6 N kg−1, and so the pumpkin will be easier to lift as its weight is now only 10 × 1.6 = 16 N. In deep space, far from any stars or planets, where g = 0, the pumpkin would be truly weightless, although its mass would still be 10 kg.

Why do heavy and light objects fall with equal acceleration?

Although Galileo was able to show that heavy and light objects fell at the same rate, he was not able to explain why. Newton, however, after stating his laws of motion, was able to show why this happens. We can use Newton’s second law to analyse the motion of the pumpkin and the banana as they fall towards Earth.

The W…IGHT of a body, W or Fg, is defi ned as the force of attraction on a body

due to gravity:W = F

g = mg

where m is the mass of the body (kg) g is the acceleration due to gravity (m s−2)g is also known as the GRAVITATIONAL FI…LD STR…NGTH. The unit of the gravitational fi eld strength is newton/kg or N kg−1.

i

Figure 5.26 The force dragging the pumpkin to the ground is much greater than the force that is acting on the banana. However, the mass and inertia of the pumpkin is much greater than the mass and inertia of the banana. The acceleration that results in both cases is identical: 9.8 m s–2 down.

Figure 5.25 The pumpkin is in free-fall. The only force acting on it is gravity, F

g, and it

accelerates at 9.8 m s−2 towards the ground.

a = 9.8 m s–2

Fg

Fg = 98 N Fg = 0.98 N

10 kg0.10 kg


The force of gravity acting on the 10 kg pumpkin is: W = Fg = mg = 10 × 9.8 = 98 N down

This is the only force acting on the pumpkin so the net force, ΣF, is also 98 N down.

The acceleration of the pumpkin can be calculated:

a = ΣFm

= 9810

= 9.8 m s−2 down

The force of gravity acting on the 0.10 kg banana is:W = Fg = mg = 0.10 × 9.8 = 0.98 N down

This is the only force acting on the banana so the net force, ΣF, is also 0.98 N down.

The acceleration of the banana can be calculated:

a = ΣFm

= 0.980.10

= 9.8 m s−2 down

The pumpkin has a large force dragging it towards the ground, but this large force is acting on a large mass, which has more inertia. The banana has a small force acting on it, but this small force is moving a small mass. The acceleration produced in each case is exactly the same: 9.8 m s−2 down.

Worked example 5.3D A 1.5 kg trolley cart is connected by a cord to a 2.5 kg mass as shown. The cord is placed over a pulley and allowed to fall under the influence of gravity. a Assuming that the cart can move over the table unhindered by friction, determine the

acceleration of the cart.b If a frictional force of 8.5 N acts against the cart, what is the magnitude of the acceleration

now?

Solution a The cart and mass experience a net force equal to the weight of the falling mass. So

ΣF = Fg = mg = 2.5 × 9.8 = 24.5 N down. This force has to accelerate not only the cart

but also the falling mass, and so the total mass to be accelerated is 1.5 + 2.5 = 4.0 kg.

a = ΣFm

= 24.54.0

= 6.1 m s−2 to the right

b In analysing the forces that now act on the cart, the net force is:

ΣF = 24.5 − 8.5 = 16.0 N to the right,

and a = ΣFm

= 16.04.0

= 4.0 m s−2 to the right.

Figure 5.27 The weight of this boulder is the force it experiences due to gravity given by F

g = mg. This is approximately 2.5 × 105 N directed

to the centre of the Earth. The mass of the boulder is approximately 25 000 kg. If the boulder were taken to outer space where the gravitational field strength is zero, the boulder would still have the same mass but no weight.

1.5 kg

2.5 kg

8.5 N

24.5 N

Motion162

Galileo was able to show more than 400 years ago that the mass of a body does not affect the rate at which it falls towards the ground. However, our common experience is that not all objects behave in this way. A light object, such as a feather or a balloon, does not accelerate at 9.8 m s−2 as it falls. It drifts slowly to the ground, far slower than other dropped objects. Parachutists and skydivers also eventually fall with a constant speed. Why is this so?

Skydivers, BASE-jumpers and air-surfers are able to use the force of air resistance to their advantage. As a jumper first steps off, the forces acting on him are drag (air resistance),

a, and gravity,

g. Since his speed is low, the drag force is

small (Figure 5.28a). There is a large net force downwards, so he experiences a large downwards acceleration of just less than 9.8 m s−2, causing him to speed up. This causes the drag force to increase because he is colliding harder with the air molecules. In fact, the drag force increases in proportion to the square of the speed:

a ∝ v2. This results in a smaller net

force downwards (Figure 5.28b). His downwards acceleration is therefore reduced. It is important to remember that he is still speeding up, but at a reduced rate.

As his speed continues to increase, so too does the magnitude of the drag force. Eventually, the drag force becomes as large as the weight force (Figure 5.28c). When this happens, the net force is zero and the jumper will fall with a constant velocity. Since the velocity is now constant, the drag

force will also remain constant and the motion of the jumper will not change (Figure 5.28d). This velocity is commonly known as the terminal velocity.

For skydivers, the terminal velocity is usually around 200 km h−1. Opening the parachute greatly increases the air resistance force that is acting, resulting in a lower terminal velocity. This is typically around 70 km h−1.

Terminal velocityPhysics in action

Figure 5.28 As the jumper falls, the force of gravity does not change, but the drag force increases as he travels faster. Eventually these two forces will be in equilibrium and the jumper will fall with a constant or terminal velocity.

ΣF

ΣF

Fa

Fg

Fa

Fg

v

v v

v

Fa

Fg

Fa

Fg

ΣF = 0

ΣF = 0

(a)

(b)

(c)

(d)

Figure 5.29 Skydivers can change their speed by changing their body profile. If they assume a tuck position they will fall faster and if they spreadeagle they will fall slower. This enables them to meet and form spectacular patterns as they fall.


• Newton’s second law of motion states that the acceleration a body experiences is directly pro-portional to the net force acting on it, and inversely proportional to its mass:

ΣF = ma where m is measured in kilograms (kg), a is

measured in metres per second squared (m s−2), and ΣF in newtons (N).

• The mass of a body can be considered to be its ability to resist a force. Mass is a constant property of the body and is not affected by its environment or location.

• The weight of a body W or Fg is defined as the force of attraction on a body due to gravity. This will be given by W = Fg = mg where m is the mass of the body and g is the strength of the gravitational field.

Newton’s second law of motion5.3 summary

Use g = 9.8 m s−2 when answering these questions.

1 1 State whether the forces are balanced (B) or unbalanced (U) for each of these situations.a a netball falling towards the groundb a stationary busc a tram travelling at a constant speed of 50 km h–1

d a cyclist slowing down at a traffic light

2 2 During a tennis serve, a ball of mass 0.060 kg is accelerated to 30 m s–1 from rest in just 7.0 ms.a Calculate the average acceleration of the ball.b What is the average net force acting on the ball

during the serve?

3 3 Use Newton’s laws to explain why a 1.0 kg shot-put can be thrown further than a 1.5 kg shotput.

4 4 In a game of soccer, the ball is simultaneously kicked by two players who impart horizontal forces of 100 N east and 125 N south on the ball. Determine:a the net force acting on the ballb the direction in which the ball will travelc the acceleration of the ball if its mass is 750 g.

5 5 When travelling at 100 km h−1 along a horizontal road, a car has to overcome a drag force due to air resistance of 800 N. If the car has a mass of 900 kg, determine the average driving force that the motor needs to provide if the car is to accelerate at 2.0 m s−2.

6 6 Mary is paddling a canoe. The paddles are providing a constant driving force of 45 N south and the drag forces total 25 N north. The mass of the canoe is 15 kg and Mary has a mass of 50 kg.

a What is Mary’s mass?b Calculate Mary’s weight.

c Find the net horizontal force acting on the canoe.d Calculate the magnitude of the acceleration of the

canoe.

7 7 On the surface of the Earth, a geological hammer has a mass of 1.5 kg. Determine its mass and weight on Mars where g = 3.6 m s−2.

8 8 What is the average braking force required of a 1200 kg car in order for it to come to rest from 60 km h−1 in a distance of 25 m?

9 9 Consider a 70 kg parachutist leaping from an aircraft and taking the time to reach terminal velocity before activating the parachute. Draw a sketch graph of the net force against time for the parachutist in the period from the start of the jump until terminal velocity has been reached. Explain your reasoning.

10 10 A 0.50 kg metal block is attached by a piece of string to a dynamics cart as shown below. The block is allowed to fall from rest, dragging the cart along. The mass of the cart is 2.5 kg.

2.5 kg

0.50 kg

a If friction is ignored, what is the acceleration of the block as it falls?

b How fast will the block be travelling after 0.50 s?c If a frictional force of 4.3 N acts on the cart, what is

its acceleration?

Newton’s second law of motion5.3 questions

Worked Solutions

Motion164

Newton’s third law of motion5.4

Newton’s first two laws of motion describe the motion of a body resulting from the forces that act on that body. The third law is easily stated, and seems to be widely known by students, but is often misunderstood and misused! It is a very important law in physics, and assists our understanding of the origin and nature of forces.

Newton realised that all forces exist in pairs and that each force in the pair acts on a different object. Consider the example of a hammer gently tapping a nail. Both the hammer and nail experience forces during this. The nail experiences a small downwards force as the hammer hits it and this moves it a small distance into the wood. The hammer experiences a small upwards force as it hits the nail causing the hammer to stop. These forces are known as an action–reaction pair and are shown in Figure 5.31a.

Now consider what will happen if the hammer is smashed into the nail.The nail now experiences a large downwards force as the hammer smashes into it and this moves it a larger distance into the wood. The hammer itself experiences a large upwards force as it hits the nail, again causing the hammer to stop. You should notice that the forces acting on the hammer and nail are both larger, as shown in Figure 5.31b. This is what Newton realised. If the hammer exerted a downwards force of 25 N on the nail, the nail would exert an upwards force of 25 N on the hammer.

It is important to recognise that the action force and the reaction force in Newton’s third law act on different objects and so should never be added together. Figure 5.32 shows some examples of action–reaction forces.

It is also important to understand that even though action–reaction forces are always equal in size, the effect of these forces may be very different. A good example of this is the collision between the bus and the car shown in 5.32c. Because of the car’s small mass, the force acting on

N…WTON’S THIRD LAW OF MOTION states that for every action force (object A on B), there is an equal and opposite reaction force (object B on A):

F (A on B) = −F (B on A) When body A exerts a force F on body B, body B will exert an equal and opposite force –F on body A.

i

Figure 5.30 A hammer hitting a nail is a good example of an action–reaction pair and Newton’s third law.

Figure 5.31 (a) As the hammer gently taps the nail, both the hammer and nail experience small forces. (b) When the hammer smashes into the nail, both the hammer and nail experience large forces.

force that nail exerts on hammer force that

hammer exertson nail

force that nail exerts on hammer force that

hammer exertson nail

(b)(a)


the car will cause the car to undergo a large acceleration backwards. The occupants may be seriously injured as a result of this. The force acting on the bus is equal in size, but is acting on a much larger mass. The bus will have a relatively small acceleration as result and the occupants will not be as seriously affected.

Worked example 5.4A In each of the following diagrams, one of the forces is given.

i Describe each given force using the phrase ‘force that exerts on ’. ii Describe the reaction pair to the given force using the phrase ‘force that

exerts on ’. iii Draw each reaction force on the diagram, carefully showing its size and location.

Figure 5.32 Some action–reaction force pairs. Notice that these can be contact or non-contact forces. (a) The girl exerts a downwards force on the floor and the floor exerts an equal upwards force on the girl. (b) The foot exerts a forwards force on the ball and the ball exerts an equal size backwards force on the foot. (c) The bus exerts a backwards force on the car and the car exerts an equal size forwards force on the bus. (d) The Earth exerts a downwards gravitational force on the brick and the brick exerts an equal size upwards force on the Earth.

Figure 5.33 The soccer ball and the player’s head exert equal forces on each other during this collision, but only the player will experience pain!

force that foot exerts on ball

force that floorexerts on girl

force that girlexerts on floor

force that busexerts on car

gravitational forcethat Earth exerts on brick (Fg)

force that carexerts on bus

force that ballexerts on foot

gravitational forcethat brick exertson Earth

(a) (b)

(c) (d)

(a) (b) (c)

Motion166

Solution a i force that bat exerts on ball ii force that ball exerts on bat iii see diagram at right

b i force that ball exerts on floor ii force that floor exerts on ball iii see diagram at right

c iii gravitational force that Earth exerts on astronaut ii gravitational force that astronaut exerts on Earth iii see diagram at right

Motion explainedNewton’s third law also explains how we are able to move around. In fact, the third law is needed to explain all locomotion. Consider walking. Your leg pushes backwards with each step. This is an action force acting on the ground. As shown in Figure 5.34, a component of the force acts downwards and another component pushes backwards horizontally along the surface of the Earth. The force is transmitted because there is friction between your shoe and the Earth’s surface. In response, the ground then pushes forwards on your leg. This forwards component of the reaction force enables you to move forwards. In other words, it is the ground pushing forwards on your leg that moves you forwards. It is important to remember that in Newton’s second law, ∑F = ma, the net force ∑F is the sum of the forces acting on the body. This does not include forces that are exerted by the body on other objects. When you push back on the ground, this force is acting on the ground and may affect the ground’s motion. If the ground is firm, this effect is usually not noticed, but if you run along a sandy beach, the sand is clearly pushed back by your feet.

The act of walking relies on there being some friction between your shoe and the ground. Without it, there is no grip and it is impossible to supply the action force to the ground. Consequently, the ground cannot supply the reaction force needed to enable forward motion. Walking on smooth ice is a good example of this, and so mountaineers will use crampons (basically a rack of nails) attached to the soles of their boots in order to gain purchase in icy conditions.

The situation outlined above is fundamental to all motion.

Figure 5.34 Walking relies on an action and reaction force pair in which the foot will push down and backwards with an action force. In response, the ground will push upwards and forwards. The forward component of the reaction force is actually friction. This is responsible for the body moving forward as a whole, while the back foot remains at rest.

θ

θ

θ

Fup

Fforwards

F (reaction)

F (reaction)

F (action)


Table 5.3 All motion can be explained in terms of action and reaction force pairs

Motion Action force Reaction force

Swimming Hand pushes back on water Water pushes forwards on hand

Jumping Legs push down on Earth Earth pushes up on legs

Bicycle/car Tyre pushes back on ground Ground pushes forwards on tyre

Jet aircraft and rockets

Hot gas is forced backwards out of engine

Gases push craft forwards

Skydiving Force of gravitation on the skydiver from Earth

Force of gravitation on Earth from skydiver

Worked example 5.4B A front-wheel drive car of mass 1.2 tonne accelerates from traffic lights at 2.5 m s–2. a Discuss and identify the horizontal force that enables the car to accelerate forwards. b Draw this force on the diagram at right. Label it force A.c Carefully draw the reaction pair to this force as described in Newton’s third law. Label it

force B.d Discuss and identify the reaction force that you have drawn.

Solution a This is the forwards force that the road exerts on the front tyre and could be called a

frictional force.

b c

d This is the backwards force that the front tyre exerts on the road. If the road surface was ice, both of the forces in (b) and (c) would be very small and the car would not be able to drive forwards.

The normal force When an object, say a rubbish bin, is allowed to fall under the influence of gravity, it is easy to see the effect of the force of gravity. As shown in Figure 5.35a, the only force acting is the weight, so the net force is the weight, and the bin therefore accelerates at g.

When the bin is at rest on a table, the force of gravity (Fg = W = mg) is still acting. Since the bin is at rest, there must be another force (equal in magnitude and opposite in direction) acting to balance the weight. This upwards force is provided by the table. Because of the weight of the bin, the table is deformed a little, and being elastic, it will push upwards. The elastic force it provides is perpendicular to its surface and is called a normal reaction force FN or N, (often abbreviated to the normal force).

The NORMAL FORC… is a reaction force supplied by a surface at 90° to its plane.i

1200 kg

force A force B

Motion168

This means that there are two forces that act on the bin which will completely balance each other so that the net force is zero. In Figure 5.35b, the bin is empty so its weight is small and the normal force is also small. However, when the bin fills up, its weight increases and so too does the normal force.

The normal force and the weight force in these examples are equal and opposite. However, this does not mean that they are an action–reaction pair as described by Newton’s third law! The weight force and the normal force act on the same body (the bin) so they cannot be an action–reaction force pair. Remember that in Newton’s third law, one force acts on one object and the other force acts on the other object. Let us identify the reaction pair to each of the forces shown in Figure 5.36.

In Figure 5.36a the action force shown is the force of gravity Fg on the bin. This is the attractive force that the Earth exerts on the bin. The reaction force, therefore, is the attractive force that the bin exerts on the Earth. This is shown as FG. In Figure 5.36b, the action force shown is the normal force on the bin, FN. This is the force that the table exerts on the bin. So the reaction force is the force that the bin exerts on the table. This is shown as FT.

Worked example 5.4C An 8.0 kg computer rests on a table.a Identify the forces that act on the computer.b Determine the magnitude and direction of the force that the computer exerts on the

table.c If a 3.0 kg monitor is placed on the computer box, determine the new normal force acting

on the computer.

Solution a The weight of the computer will be F

g = mg = 8.0 × 9.8 = 78 N down, so that if the net force

on the computer is zero, the normal force supplied by the table must be 78 N upwards: ΣF = F

g + F

N = 0.

Figure 5.35 (a) When the bin is in mid-air, there is an unbalanced force acting on it so it accelerates towards the ground. (b) When the empty bin is resting on the table, there is a small upwards force from the table acting on it. The bin remains at rest, so the forces are balanced. (c) The forces acting on the full bin are also balanced. The weight of the bin is greater, so the normal force exerted by the table is larger than for the empty bin.

Figure 5.36 (a) The reaction force pair to the weight of the bin is the gravitational force of attraction that the bin exerts on the Earth. (b) The reaction force pair to the normal force on the bin is the force that the bin exerts on the table. The force pairs are equal and opposite but they do not cancel out because they are not acting on the same object.

FN

FN Fg

Fg

Fg

(a) (b) (b)

FN =

FG =

Fg =

FT =

gravitational forceEarth exerts on bin

force tableexerts on bin

gravitational force binexerts on Earth

force bin exerts on table

(b)(a)

FN

Fg

ΣF = Fg + FN = 0


b Since the force that the table exerts on the computer has been found to be 78 N up, then, according to Newton’s third law, the force that the computer exerts on the table must be 78 N down.

c If a 3.0 kg monitor is placed on top of the computer, the table must supply a further 3.0 × 9.8 = 29 N, bringing the total normal force to 107 N. (The computer will also have to provide a normal force of 29 N upwards to balance the weight of the monitor.)

The inclined planeThe Guinness Book of Records identifies the steepest road in the world as being at an angle of 20° to the horizontal. It is located in Dunedin, New Zealand. Living on such a road requires the residents to ensure that the handbrake in their car is always in good repair! To determine the force required by the handbrake of a car parked on this steep road, the physics of forces acting on a body on an inclined plane must be used.

Start by thinking of a body at rest on a horizontal surface. Two forces act on the body: the weight of the body Fg and the normal force FN supplied by the surface. The weight force always acts downwards and is given by Fg = mg. The normal force is supplied by the surface and will vary depending on the situation, but it will always act upwards and perpendicular to the surface. This means that the net force on the body will be the sum of these two forces, and in this case it has to be zero since the body does not move.

If the surface is tilted so that it makes an angle to the horizontal, the weight force remains the same: Fg = mg. However, the normal force continues to act at right angles to the surface and will change in magnitude, getting smaller as the angle increases. If there is no friction between the body and the surface, the two forces will not balance and a non-zero net force will be directed down the incline as shown in Figure 5.37.

From the vector diagram of the forces:ΣF = Fg + FN = Fg sin θ = mgg sin θ

From Newton’s second law, the net force is:ΣF = ma

so, ma = mgg sin θ or a = gg sin θ

This means that the acceleration down an incline is a function of the angle of the incline alone, and not the mass of the body. Ignoring any friction, a car rolling down the steep street in Dunedin will accelerate at a = gg sin θ = 9.8 × sin 20° = 3.4 m s−2—quite a rate!

You may have observed that the mass of an object can be used in two different contexts. First, mass is a measure of the ability of an object to resist being accelerated by a force. This mass can be determined from Newton’s second law and is known as inertial mass. Second, mass can give an indication of the degree to which an object experiences a gravitational force when in the presence of a gravitational field. This mass is known as gravitational mass. Some very accurate experiments have shown that the inertial and gravitational masses are equal, although there is no theoretical reason why this should be the case.

Physics file

Figure 5.37 (a) Where the surface is perpendicular to the weight force, the normal force will act directly upwards and cancel the weight force. (b) On an inclined plane, F

N is at an angle to

Fg and as long as no friction acts, there will be a net

force down the incline. The body will accelerate.

ΣF = Fg + FN

= 0

ΣF = Fg + FN

ΣF

θ

θ

FN

FN

FN

FN

Fg

Fg

Fg

Fg

body remainsat rest

(a) (b)

Prac 24 SPARKlab

Motion170

80 m

ice patch

35°

35°

Fg FN

Fg

FN

ΣF

Worked example 5.4D Driver error allows a 5 tonne truck to roll down a steep ramp inclined at 30° to the horizontal. As it is a high-technology vehicle, there is negligible friction between the wheels of the truck and the wheel bearings. Find the acceleration of the truck if the acceleration due to gravity is taken as 9.8 m s−2.

Solution ΣF = F

g + F

N

From the vector diagram:ΣF = F

g sin θ

So, ma = mg sin θ a = g sin 30° = 9.8 × sin 30° = 4.9 m s−2 down the ramp

If friction exists between a body on an inclined plane and the surface, its direction will be along the incline but against the motion. If a frictional force is great enough to balance the sum of the normal force and the weight of the body, the net force is zero and the body will either travel with a constant velocity or remain stationary. Worked example 5.4E illustrates this.

Worked example 5.4E Kristie is a 60 kg skier. At the start of a ski slope that is at 35° to the horizontal, she crouches into a tuck. The surface is very icy, so there is no friction between her skis and the ice. Ignore air resistance when answering these questions.a If Kristie starts from rest, what is her speed after travelling a distance of 80 m on the ice?b The snow conditions change at the end of the ice patch so that Kristie continues down

the slope with a constant velocity. What is the force due to friction that must be acting between Kristie’s skis and the snow?

Solution a The net force on Kristie will be ΣF = F

g + F

N. This is a vector addition. From the vector

diagram: ΣF = F

g sin θ

So, a = g sin 35° = 9.8 sin 35° = 5.6 m s−2

In February 2003, a train driver pulled into Broadmeadows station and went for a toilet break. Unfortunately, he forgot to put on the handbrake. When he returned, the train was rolling away from the platform, heading for Jacana. It is a slight downhill incline from Broadmeadows to Jacana and the train simply rolled off down the hill, accelerating gradually. After Glenroy, the incline of the track is greater and the train’s acceleration increased. It is estimated that it reached speeds of around 100 km h–1 at times. Fortunately, no-one was injured as it hurtled through level crossings and stations on its way into the city. At Spencer Street station, it crashed into a stationary V/Line train at about 60 km h−1. The express trip from Broadmeadows to Spencer Street took about 16 minutes.

Jacana

Glenroy

5 km

Oak ParkPascoe Vale

Strathmore

Moonee Ponds

Glenbervie

Broadmeadows

Essendon

Ascot Vale

Newmarket

KensingtonNorth Melbourne

Spencer Street Station

0

Figure 5.38 An empty three-carriage train took 16 minutes to roll downhill from Broadmeadows to Spencer Street (now Southern Cross) station. The track was like a long inclined plane and the train accelerated along it after the driver forgot to put the handbrake on!

Physics file

30

30

F = Fg + FN

FN

FN

Fg

Fg


If this acceleration continues over 80 m, Kristie’s final speed would be: v2 = u2 + 2axso v2 = 0 + 2 × 5.6 × 80 = 900 v = 30 m s−1 (110 km h−1)

b Kristie is travelling with a constant velocity, so ΣF = 0, i.e. the force of friction Ff would

balance the component of the weight force parallel to the incline. So F

f = F

g sin 35 = 340 N up the incline.

Tension Another force that is experienced in everyday life is the tension force that is found in stretched ropes, wires, cables and rubber bands. If you stretch a rubber band, it will exert a restoring force on both your hands. This force is known as a tensile force and is present in any material that has been stretched.

Consider the situation in which a person hangs from a cable that is tied to a beam as shown in Figure 5.39a. We will assume that the mass of the cable is negligible.

At the top end of the cable, the tension force FT pulls down on the beam. At the other end, the cable exerts an upwards force FT on the person, holding them in mid-air. In other words, the same size tension force acts at both ends of the cable, but in opposite directions. If a second identical person also hung from the cable, the tension acting at both ends of the cable would double and the cable would become tauter (and more likely to snap!)

If the mass of the person in Figure 5.39a is known, the size of the tension can be determined. If the mass of the person is 50 kg, then the forces acting on them are, as shown in Figure 5.39b, an upwards tension force and the downwards pull of gravity of 490 N. If the person is at rest, the forces acting are balanced and so the upwards tensile force acting from the cable must also be 490 N.

Calculations involving tension are illustrated in the following example. A naughty monkey of mass 15 kg has escaped backstage in a circus. Nearby is a rope threaded through an ideal (frictionless and massless) pulley. Attached to one end of the rope is a 10 kg bag of sand. The monkey climbs a ladder and jumps onto the free end of the rope.

Figure 5.39 (a) The stretched cable exerts an upwards force on the person and an equal size downwards force on the beam. (b) The forces acting on the person are balanced, so the tension force is equal in magnitude to the weight force (490 N).

tension T

tension T

T = 490 N

g = 490 N

= 50 kg Σ = 0

(a) (b)

Motion172

The system of the rope and the monkey is now subjected to a net force of:ΣF = 15 × 9.8 − 10 × 9.8 = 49 N down on the side of the monkey.

As a consequence, both objects and the rope will accelerate at:

a = ΣFΣm

= 4925

= 2.0 m s−2

While all of this is occurring, the rope is under tension. To find the amount, we look at the forces acting on one of the masses (Figure 5.40c). Take the monkey: the net force on the monkey will be ΣF = Fg + FT.So, FT = ΣF − Fg = 15 × 2.0 down − 15 × 9.8 down = 30 N down + 147 N up = 117 N up

To check, find the tension acting on the sandbag.Again, ΣF = Fg + FT

So, FT = ΣF − Fg = 10 × 2.0 up − 10 × 9.8 down = 20 N up + 98 N up = 118 N up

Note that the small difference between the two results is due to rounding error. The tension is equal on both sides of the pulley, regardless of how the masses are arranged (provided the pulley is frictionless).

Intuitively, you might have thought that the tension would have been (15 + 10) × 9.8 = 245 N since the two weights are pulling in opposite directions. This is not the case because the system is allowed to accelerate, causing a reduction in tension.

Figure 5.40 The monkey has a greater weight than the sandbag, and so the rope will accelerate in the direction of the monkey. The tension in the rope is found by considering the forces that act on each weight.

98 N

10 kg

15 kg

a

Fg =147 N Fg = 98 N

ΣF = 49 N

ΣF = 15 × 2.0ΣF = 10 × 2.0

FT = 117 N FT = 118 N

= 30 N= 20 N

(b)(a) (c)

147 N


Friction is a force that opposes movement. Suppose you want to push your textbook along the table. This simple experiment can reveal a significant amount of information about the nature of friction. As you start to push the book, you find that, at first, the book does not move. You then increase the force that you apply, and suddenly, at a certain critical value, the book starts to move relatively freely.

The maximum frictional force resists the onset of sliding. This force is called the static friction force,

s. Once the book

has begun to slide, a much lower force than s is needed to

keep the book moving. This force is called the kinetic friction force, and is represented by

k.

This phenomenon can be understood when it is realised that even the smoothest surfaces are quite jagged at the microscopic level. When the book is resting on the table, the jagged points of its bottom surface have settled into the valleys of the surface of the table, and this helps to resist attempts to try to slide the book. Once the book is moving, the surfaces do not have any time to settle into each other, and so less force is required keep it moving.

Another fact that helps to explain friction arises from the forces of attraction between the atoms and molecules of the two different surfaces that are in contact. These produce weak bonding between the particles within each material; before one surface can move across the other, these bonds must be broken. This extra effort adds to the static friction force. Once there is relative motion between the surfaces, the bonds cannot re-form.

In everyday life, there are situations in which friction is desirable (e.g. walking) and others in which it is a definite problem. Consider the moving parts within the engine of a

car. Friction can rob an engine of its fuel economy and cause it to wear out. Special oils and lubricants are introduced in order to prevent moving metal surfaces from touching. If the moving surfaces actually moved over each other they would quickly wear, producing metal filings that could damage the engine. Instead, both metal surfaces are separated by a thin layer of oil. The oils are chosen on the basis of their viscosity (thickness). For example, low viscosity oils can be used in the engine while heavier oils are needed in the gear box and differential of the car where greater forces are applied to the moving parts.

At other times, we want friction to act. When driving to the snowfields, if there is any ice on the road, drivers are required to fit chains to their cars. When driving over a patch of ice, the chain will break through the ice, and the car is again able to grip the road. Similarly, friction is definitely required within the car’s brakes when the driver wants to slow down. In fact modern brake-pads are specially designed to maximise the friction between the pads and the brake drum or disk.

When a car is braking in a controlled fashion, the brake-pads grip a disk that is attached to the wheel of the car. The retarding force, applied through friction, slows the disk and hence the car will come to rest. If the brakes are applied too strongly they may grab the disk, locking up the wheels in the process. The car then slides over the road, with two undesirable consequences. First, the car usually takes about 20% longer to come to rest. This is because the car is relying on the kinetic frictional force between the tyres and the road to stop. As was seen when pushing the book over the desk, this force is less than the static friction force. The other consequence is that the car has lost its grip with the road, and so the driver can no longer steer the car. Most cars now employ anti-lock braking systems (ABS) to overcome the possibility of skidding. This is achieved by using feedback systems that automatically reduce the pressure applied by the brake-pads regardless of the pressure applied by the driver to the brake pedal.

Frictional forcesPhysics in action

Figure 5.41 To get things moving, the static friction between an object and the surface must be overcome. This requires a larger force than that needed to maintain constant velocity.

maximumstatic friction

kinetic friction

Applied force

Time

overcoming friction constant velocity

motion

Fs

Fs

Fk

Fk

PHYSICS11

PHYSICS11

Figure 5.42 This magnetic levitation train in China rides 1 cm above the track, so the frictional forces are negligible. The train is propelled by a magnetic force to a cruising speed of about 430 km h−1.

Motion174

• Newton’s third law of motion recognises that forces exist in pairs and states that for every action force, there is an equal and opposite reaction force:

F(A on B) = −F(B on A)• Action and reaction forces act on different objects

and so should never be added together.• Whenever a force acts against a fixed surface, the

surface provides a normal force, FN or N, at right angles to the surface. The size of the normal force depends on the orientation of the surface to the contact force.

• All locomotion is made possible through the existence of action and reaction force pairs.

• On smooth inclined planes at an angle of θ to the horizontal, objects will move with an acceleration of a = g sin θ.

• Materials that have been stretched, such as ropes, cables and rubber bands, exert tensile forces on the objects to which they are attached. These forces are equal and opposite.

Newton’s third law of motion5.4 summary

1 1 Determine the action and reaction forces involved when:a a tennis ball is hit with a racquetb a pine cone falls from the top of a tree towards the

groundc a pine cone lands on the groundd the Earth orbits the Sun.

2 2 A 70 kg fisherman is quietly fishing in a 40 kg dinghy at rest on a still lake when, suddenly, he is attacked by a swarm of wasps. To escape, he leaps into the water and exerts a horizontal force of 140 N on the boat.a What force does the boat exert on the fisherman?b With what acceleration will the boat move

initially?c If the force on the fisherman lasted for 0.50 s,

determine the initial speed attained by both the man and the boat.

3 3 A 100 kg astronaut (including the space suit) becomes untethered during a space walk and drifts to a distance of 10 m from the mother ship. To get back to the ship, he throws his 2.5 kg tool kit away with an acceleration of 8.0 m s−2 that acts over 0.50 s. a How does throwing the tool kit away help the

astronaut in this situation?b How large is the force that acts on the tool kit and

the astronaut?c With what speed will the astronaut drift to the

mother ship?d How long will it take for the astronaut to reach the

ship?

4 4 A 2.0 kg bowl strikes the stationary ‘jack’, which has a mass of 1.0 kg, during a game of bowls. It is a head-on collision, and the acceleration of the jack is found to be 25 m s−2 north. What is the acceleration of the bowl?

5 5 a A ball rolls down an incline as shown in diagram (a). Which one of the following best describes the speed and acceleration of the ball?

(a) (b)

A The speed and acceleration both increase.B The speed increases and the acceleration is

constant.C The speed is constant and the acceleration is

zero.D The speed and acceleration are both constant.

b A ball rolls down the slope shown in diagram (b). Which one of the following best describes its speed and acceleration?A Its speed and acceleration both increase.B Its speed and acceleration both decrease.C Its speed increases and its acceleration

decreases.D Its speed decreases and its acceleration

increases.

Newton’s third law of motion5.4 questions


6 6 During the Winter Olympics, a 65 kg competitor in the women’s luge has to accelerate down a course that is inclined at 50° to the horizontal.a Name the forces acting on the competitor.b Ignoring friction (because it’s an icy slope), deter-

mine the magnitude and direction of the forces that act.

c Determine the magnitude of the net force on the competitor.

d What acceleration will the competitor experience?

7 7 A cyclist is coasting down a hill that is inclined at 15° to the horizontal. The mass of the cyclist and her bike is 110 kg, and for the purposes of the problem, no air resistance or other forces are acting. After accelerating to the speed limit, she applies the brakes a little. What braking force is needed for her to be able to travel with a constant velocity down the hill?

8 8 Discuss and compare the size of the normal force that acts on the skater as he travels down the half-pipe from A to B to C as shown.

A

B

Cx

x

x

9 9 Two students, James and Tania, are discussing the forces acting on a lunchbox that is sitting on the laboratory bench. James states that a weight force and a normal force are acting on the lunchbox and that since these forces are equal in magnitude but opposite in direction, they comprise a Newton’s third law action–reaction pair. Tania disagrees saying that these forces are not an action–reaction pair. Who is correct and why?

10 10 A rope is allowed to move freely over a ‘frictionless’ pulley backstage at a theatre. A 30 kg sandbag, which is at rest on the ground, is attached at one end. A 50 kg work-experience student, standing on a ladder, grabs onto the other end of the rope to lower himself. a When the student steps off

the ladder, what is the net external force on the rope?

b With what acceleration will the system move?

c What is the tension in the rope?

1 A boxer receives a punch to the head during a training session. His opponent is wearing boxing gloves. Which of the following is correct?

A The force that the glove exerts to the head is greater than the force that the head exerts on the glove.

B The force that the glove exerts to the head is less than the force that the head exerts on the glove.

C The force that the glove exerts to the head is always equal to the force that the head exerts on the glove.

D None of the above is correct. 2 When pushing a shopping trolley along a horizontal path, James

has to continue to provide a force of 30 N just to maintain his speed. If the trolley (and shopping) has a mass of 35 kg, what is the total horizontal force that he will have to provide to accelerate the cart at 0.50 m s−2?

3 A force of 25 N is applied to a 6.0 kg ten-pin bowling ball for 1.2 s. If the ball was initially at rest, what is its final speed?

4 A parachutist of mass 60 kg is falling vertically towards the ground with a constant speed of 100 km h–1.

a Calculate the size of the drag force acting on the parachutist at this speed.

b The parachutist pulls the ripcord and the chute opens, slowing the parachutist to 20 km h–1. How does the drag force compare to the weight force during this period?

c During the final stage of the skydive, the parachutist continues towards the ground at 20 km h–1. How does the drag force compare to the weight force during this period?

Chapter reviewNewton’s laws

30 kg

50 kg

Continued on next page

Worked Solutions

Motion176

5 Jane has a mass of 55 kg. She steps into a lift which goes up to the second floor. The lift accelerates upward at 2.0 m s−2 for 2.5 s, then travels with constant speed.

a What is the maximum speed that the lift attains as it travels between floors?

b What is Jane’s weight: i when the lift is stationary? ii when the lift is accelerating upwards? 6 a What is the mass of an 85 kg astronaut on the surface of

Earth where g = 9.8 m s−2?

b What is the mass of an 85 kg astronaut on the surface of the Moon where g = 1.6 m s−2?

c What is the weight of an 85 kg astronaut on the surface of Mars where g = 3.6 m s−2?

7 The series of photographs shows a stack of smooth blocks in a tall pile. One of the blocks in the pile is struck by a hammer and the blocks above it fall onto the block below, and the pile remains standing. Explain this in terms of Newton’s laws of motion.

8 A force of 120 N is used to push a 20 kg shopping trolley along the line of its handle—at 20° down from the horizontal. This is enough to cause the trolley to travel with constant velocity to the north along a horizontal path.

a Determine the horizontal and vertical components of the force applied to the trolley.

b What is the value of the frictional force acting against the trolley?

c How large is the normal force that is supplied by the ground on which the trolley is pushed?

d Why is it often easier to pull rather than push a trolley? 9 A 100 g glider is at rest on a horizontal air track, and a force is

applied to it as shown in the following graph. What will be its speed at the end of the time interval?

10 The following diagrams show force vectors on a puck travelling

at constant speed across an air table in a games arcade. The puck experiences no friction as it moves across its cushion of air. Which diagram A–D correctly shows the forces that act on the puck?

Appliedforce

(N)0.5

0.2

021

t (s)

A

B

C

D

Fg

FN

Fdrag

F

Fg

FN

F

Fg

FN

Fg

Newton’s laws (continued)


11 Two shopping trolleys with masses 30 kg and 50 kg stand together. A force of 120 N is applied to the 30 kg trolley.

a With what acceleration will the trolleys move?b Calculate the size of the contact force that the 30 kg trolley

exerts on the 50 kg trolley. 12 A young girl of mass 40 kg leaps horizontally off her stationary

10 kg skateboard. Assuming that no frictional forces are involved, determine the following ratios:

a horizontal force on the girl

horizontal force on the skateboard

b acceleration of the girl

acceleration of the skateboard

c final velocity of the girl

final velocity of the skateboard 13 A rope has a breaking tension of 100 N. How can a full bucket

of mass 12 kg be lowered using the rope, without the rope breaking?

14 The force diagram below shows the forces acting on a full water tank.

FN

Fg

a Are these forces an action–reaction pair as described by Newton’s third law?

b Justify your answer to part a.

15 A car begins to roll down a steep road that has a grade of 1 in 5 (i.e. a 1 m drop for every 5 m in length). If retarding forces are ignored, determine the speed of the car in km h−1 after it has travelled a distance of 100 m if it begins its journey at rest.

16 A small boy’s racing set includes an inclined track along which a car accelerates at 1

2 g (i.e. 4.9 m s−2). At what angle is the track to the horizontal?

17 When skiing down an incline, Eddie found that there was a frictional force of 250 N acting up the incline of the mountainside due to slushy snow. The slope was at 45° to the horizontal, so if Eddie had a mass of 70 kg, what was his acceleration?

18 On a sketch, draw vectors to indicate the forces that act on a tennis ball:

a at the instant it is struckb an instant after it has been struck.

19 Two masses, 5.0 kg and 10.0 kg, are suspended from the ends of a rope that passes over a frictionless pulley. The masses are released and allowed to accelerate under the influence of gravity. What is the acceleration of the system, and what is the tension in the rope?

20 Tim and Julia are discussing a couple of physics problems over dinner.

a First, they discuss a collision between a marble and a billiard ball. Tim argues that since the billiard ball is much heavier than the marble, it will exert a larger force on the marble than the marble exerts on it. Julia thinks that the marble and billiard ball will exert equal forces on each other as they collide. Who is correct? Explain.

b Then they discuss a basketball as it bounces on a concrete floor. Tim claims that the ball must exert a smaller force on the floor than the floor exerts on it, otherwise the ball would not rebound. Julia thinks that the ball and the floor will exert forces on each other that are equal in magnitude. Who is correct this time? Explain.

120 N30 kg 50 kg

Chapter QuizWorked Solutions

Momentum, energy, work and power

6CHAPTER

Is skydiving on your list of things to do in your future? Base-jumping? Mountain boarding? Are you a person who would love to experience the exhilaration of taking that leap out of a plane, or do

you question why someone would choose to jump out of a perfectly safe aeroplane in mid-flight? When an aeroplane is climbing to the required height for the thrill-seekers, we say that the aeroplane’s engines are doing work against gravity. When the parachutist takes the jump, we say that gravity is doing work on the parachutist. The search for the ultimate extreme-sport thrill is often about ‘taking on gravity’. Whether it is snowboarding, BASE-jumping or bungee jumping, the participant is experimenting with the conversion of gravitational potential energy into kinetic energy.

Throughout this chapter you will be able to see the common thread of energy conversion that is present in so many of our activities. In our everyday lives we try to harness and transform various forms of energy as efficiently and cleverly as possible. We burn up our own personal energy stores as we climb up the steps to the classroom. We make sure the tennis ball hits the ‘sweet spot’ on our racquet when we hit it back to our opponent. In more thrill-seeking adventures we also make the most of our understanding of energy transformations. Converting lots of gravitational potential energy into kinetic energy can be an extreme experience and great fun, as shown in the photograph. Don’t be deceived though, the laws of physics cannot be switched off!

you will have covered material from the study of movement including:

• momentum and impulse

• work done as a change in energy

• Hooke’s law

• kinetic, gravitational and elastic potential energy

• energy transfers

• power.

by the end of this chapter

Chapter 6 Momentum, energy, work and power 179

6.1 The relationship between momentum and force

Consider a collision between two footballers on the football field. From Newton’s second law, each force can be expressed as:

Σ net = mand using the relationship for acceleration:

= − Δt

Σ = m( − )

Δtwhere a is the acceleration during the collision (m s−2)

Δt is the time of contact (s) u is the velocity of either one of the footballers before the collision (m s−1)v is the velocity of the footballer after the collision (m s−1).

Rearranging:Σ Δt = m( − )

or:Σ Δt = mΔ

This relationship introduces two important ideas.• The product of net force and the contact time is referred to as impulse,

I. The idea of impulse is commonly applied to objects during collisions when the time of contact is small. This concept will be further explained later.

• The product of the mass of an object and its velocity is referred to as momentum:

= where is momentum (kg m s−1)

m is the mass of the object (kg) is the velocity of the object (m s−1).

MomentumMomentum can be thought of as the tendency of an object to keep moving with the same speed in the same direction. It is a property of any moving object. As it is the product of a scalar quantity (mass) and a vector quantity (velocity), momentum is a vector quantity. The direction of the momentum of an object is the same as the direction of the velocity of that object. The unit for momentum is kg m s−1, which is readily determined from the product of the units for mass and velocity.

Momentum often indicates the difficulty a moving object has in stopping. A fast-moving car has more momentum than a slower car of the same mass; equally so, an elephant will have more momentum than a person travelling at the same speed (just as a greater force is needed to cause the same acceleration). The more momentum an object gains as its velocity increases, the more it has to lose to stop and the greater the effect it will have if involved in a collision. A football player is more likely to be knocked over if tackled by a heavy follower than a light rover, since the product p = mv will be larger for the heavy follower.

Although he used different language, Newton understood this idea, and his second law of motion can be stated in terms of momentum.

Figure 6.1 When two footballers collide, they exert an equal and opposite force on each other. The effect this force will have on the velocity of each footballer can be investigated using the concept of momentum.

Figure 6.2 The enormous mass of a large ship endows it with very large momentum despite its relatively slow speed. After turning off its engines the ship can continue against the resistance of the water for more than 4 km if no other braking is applied.

Interactive

Motion180

That is:ΣF =

ΔpΔt

where ΣF is the average net force applied to the object during the collision, in newtons (N)Δp is the change in momentum during contact for a time Δt.

An unbalanced net force is required to change the momentum of an object—to increase it, decrease it or change its direction. This force might result from a collision or an interaction with another object. The change in momentum (Δp) of the object will be given by:

Worked example 6.1A A footballer trying to take a mark collided with a goal post and came to rest. The footballer has a mass of 80 kg and was travelling at 8.2 m s−1 at the time of the collision.a What was the change in momentum during the collision for the footballer?b Estimate the average force the footballer experienced in this collision.

Solution a Prior to the collision the footballer’s momentum was given by:

p = mv= 80 × 8.2= 656 kg m s−1 towards the pole

After the collision the momentum was zero since the footballer stopped moving. So: Δp = 0 − 656 = −656 kg m s−1 towards the pole or: Δp = 6.6 × 102 kg m s−1 away from the poleb The negative value for the change in momentum indicates that the direction of the

momentum, and hence the force applied to the footballer, is opposite to the direction in which the footballer was travelling. The time that the footballer took to stop has not been given but a reasonable estimate of the force can be made by estimating the stopping time. Keeping to magnitudes of 10 for easy working, it would be reasonable to assume that the stopping time in this sort of collision would be less than 1 s (100) and greater than 0.01 s (10−2). Something in the order of 0.1–0.5 s (10−1) would make sense on the basis of observations of similar situations.

Then using:

F = ΔpΔt

F = 65610−1

= 6560 ≈ 7 × 103 N away from the pole, i.e. a retarding force.

The change in momentum of a body is proportional to the net force applied to it:Δp ∝ ΣF

i

Change in momentum = fi nal momentum − initial momentumΔp = p

f − p

i

i

181Chapter 6 Momentum, energy, work and power

ImpulseThink about what it feels like to fall onto a concrete floor. Even from a small height it can hurt. A fall from the same height onto a tumbling mat is barely felt. Your speed is the same, your mass hasn’t changed and gravity is still providing the same acceleration. So what is different about the fall onto the mat that reduces the force you experience?

Remember that, according to Newton’s second law of motion, the velocity of an object only changes when a net force is applied to that object. A larger net force will be more effective in creating a change in the velocity of the object. The faster the change occurs (i.e. a smaller time interval Δt), the greater the net force that is needed to produce that change. Landing on a concrete floor changes the velocity very quickly as you are brought to an abrupt stop. When landing on a tumble mat the change occurs over a much greater time. The force needed to produce the change is smaller.

Another illustration of this could be a tennis player striking a ball with a racquet. At the instant the ball comes in contact with the racquet the applied force will be small. As the strings distort and the ball compresses, the force will increase until the ball has been stopped. The force will then decrease as the ball accelerates away from the racquet. A graph of force against time will look like that in Figure 6.3.

The impulse affecting the ball at any time will be the product of applied force and time, i.e. I = FavΔt. The total impulse during the time the ball is in contact will be I = Fav × t, where Fav is the average force applied during the collision and t is the total time the ball is in contact with the racquet. This is equivalent to the total area under the force–time graph. The total impulse for any collision can be found in this way.

The concept of impulse is appropriate when dealing with forces during any collision since it links force and contact time, for example a person hitting the ground, as described above, or a ball being hit by a bat or racquet. If applied to situations where contact is over an extended time, the average net force involved is used since the forces are generally changing (as the ball deforms for example). The average net applied force can be found directly from the formula for impulse. The instantaneous applied force at any particular time during the collision must be determined from a graph of the force against time.

The relationship between impulse and momentumFrom the derivation earlier in the chapter, the impulse is also equal to the change of momentum for an object. Previously we had:

F = mΔΔt

The IMPULS… affecting an object during a collision is the product of the net average applied force and the time of contact and is equivalent to the area under a force–time graph.

i

Figure 6.3 (a) When a tennis player hits a ball, an unbalanced force is applied to the ball, producing a change in its momentum; hence an impulse is applied to the ball. The magnitude of the force will change over time. (b) The impulse can be found from the area under the force–time graph since area = x-axis × y-axis = F

av × Δt = impulse.

(a)

t (s)0 0.150.120.090.060.03

Impulse = Fav × t = area under graph

F (N

)

(b)

Motion182

Multiplying by the time interval Δt:FavΔt = mΔ

or:I = FavΔt = Δp

The units for impulse (N s) and the units for momentum (kg m s−1) have each been introduced separately. Since we have just seen that impulse is equal to the change in momentum:

1 N s = 1 kg m s−1

Worked example 6.1B A tennis racquet applies a force to a tennis ball for a period of 0.15 s, bringing the ball (momentarily) to a halt. The tennis ball has a mass of 58 g and was originally travelling towards the racquet at 55 m s−1.a Find the change in momentum as the ball is momentarily brought to a halt by the

racquet.b Find the magnitude of the impulse during this part of the collision.c Find the average force applied during the time it takes to stop the ball.

Solution a Initial momentum: p

i = mu = 0.058 × 55 = 3.19 kg m s−1

Final momentum: pf = mv = 0.058 × 0 = 0 kg m s−1

Change in momentum: Δp = 0 − 3.19 = −3.19 kg m s−1 in the direction of travel, i.e. ∼3.2 kg m s−1 in the opposite direction.

b Impulse = change in momentum: I = 3.19 ≈ 3.2 N sc Using I = F

avΔt

then Fav

= I

Δt

so Fav

= 3.190.15

= 21.27 N ≈ 21 N in the opposite direction to the ball’s travel.

Try this simple experiment. Grab a can of fruit or similar relatively soft non-corrugated steel can. Place your finger flat on a bench top and, carefully avoiding the can’s seam, bring the side of the can crashing down on your finger. (We take no responsibility for you using the wrong part of the can!)Were you actually game to try it? If you did, how much did it hurt? Not nearly as much as you expected, right? Why?

Bringing a rigid hammer down on your finger in similar circumstances would have caused considerable damage to the finger. Yet the can crumpled in around your finger and, even though it had a similar mass to the hammer and travelled at a reasonable velocity, it caused no damage to the finger and little pain. This observation can be explained with the concept of impulse.

By assuming a mass of 500 g for both hammer and can and an impact speed of 20 m s−1, the magnitude of the change in momentum, and hence impulse, can be estimated.

Forces during collisionsPhysics in action

Figure 6.4 A simple example of the effect on the applied force of the stopping time during a collision can be achieved with nothing more complicated than a hammer, a can of fruit and your finger. (a) A rigid object, such as the hammer, will stop quickly. The applied force will be large. (b) A can will experience the same change in momentum, but, having a simple crumple zone, will stop more slowly, thus reducing the applied force to a tolerable amount.

(a) (b)


I = Δp = mΔv= 0.5 × 20 = 10 N s

The hammer, being rigid, will quickly come to a stop. This time can be estimated at about 0.1 s, so:

F = I

Δt =

100.1

= 100 N

—a considerable force that could be expected to do some damage to the finger!

The can is able to compress and so the stopping time will be somewhat longer, say around 0.5 s. The average force will be:

F = I

Δt =

100.5

= 20 N

Increasing the stopping time by five times has reduced the average force applied to the finger to one-fifth that applied by a rigid object to something which, while perhaps not totally pain-free, is quite tolerable and will do no real damage. The applied force is inversely proportional to the stopping time. Increase the stopping time and the applied force is decreased. Try it on a friend and see if you can prove this bit of physics to them!

This simple idea is the basis upon which the absorbency systems of sports shoes, crash helmets, airbags and crumple zones of cars, and other safety devices are designed.

Walking, running and sports shoe designAs athletes walk or run, they experience action–reaction forces due to gravity, the surface of the track and the air around them. These forces have been investigated in some detail in the previous chapter.

The force the ground exerts on a runner creates a change in momentum as the runner’s feet strike the ground. This force can be quite large and cause considerable damage to the runner’s ankles, shins, knees and hips as it is transmitted up the bones of the leg. Jogging in bare feet can increase the forces experienced to nearly three times that applied when simply standing still. Table 6.1 lists the relative size of some forces associated with common movements in sport.

An understanding of the forces generated and the elastic properties of materials are used by designers in the development of athletics tracks, playing surfaces and sports shoes. Elastic materials can reduce the forces developed between foot and track by increasing the stopping time. Based on an understanding of impulse, sports shoes are designed with soles that include gels, air and cushioning grids, which extend the stopping time and thus reduce the force applied to the runner’s body. Sophisticated modern sports shoes, properly fitted to suit the wearer, have substantially reduced the size and effect of forces on the runner, with consequential benefits for the runner’s knees and hips.

The running surface can also be designed to minimise the forces and produce fewer injuries. Cushioned surfaces reduce the impact considerably. Grass is actually quite effective but can sometimes be too spongy. The extra time spent rebounding from the surface slows the runner down. The response must be quick if good running times are to be achieved. As a result artificial surfaces such as polyurethane, ‘AstroTurf’ and ‘Rebound Ace’ have become popular.

They offer good cushioning but are more responsive, allowing faster take-offs than grass.

Vehicle safety designDesigning a successful car is a complex task. A vehicle must be reliable, economical, powerful, visually appealing, secure and safe. Public perception of the relative importance of these issues varies. Magazines and newspapers concentrate on appearance, price and performance. The introduction of air-bag technology into most cars has altered the focus towards safety. Vehicle safety is primarily about crash avoidance. Research shows potential accidents are avoided 99% of the time. The success of accident avoidance is primarily attributable to accident avoidance systems such as antilock brakes. When a collision does happen, passive safety features come into operation, for example the air bag. Understanding the theory behind accidents involves an understanding primarily of impulse and force.

Table 6.1 Relative size of forces associated with some common movements in sport

Movement Footwear Ratio of normal to weight force

Standing still Barefoot or shoes 1.0

Walking Barefoot 1.6

Jogging Barefoot 2.9

Jogging Running shoes 2.2

Sprinting Barefoot 3.8

Fast bowling Cricket spikes 4.1

Long jump take-off Athletic spikes 7.8

Figure 6.5 The forces developed between track and foot can do considerable damage to a runner’s body unless they are reduced by increasing the stopping time through cushioning of the foot. Modern track shoes incorporate sophisticated design principles to increase stopping time and thus decrease the forces generated.


Motion184

The air bagSeat belts save lives. They also cause injuries. Strangely, the number of people surviving an accident but with serious injury has increased since the introduction of safety belts. Previously many of today’s survivors would have died instantly in the accident. A further safety device is required to minimise these injuries.

The air bag in a car is designed to inflate within a few milliseconds of a collision to reduce secondary injuries during the collision. It is designed to inflate only when the vehicle experiences an 18–20 km h−1 or greater impact with a solid object. The required deceleration must be high or accidental nudges with another car would cause the air bag to inflate. The car’s computer control makes a decision in a few milliseconds to detonate the gas cylinders that inflate the air bag. The propellant detonates and inflates the air bag while the driver collapses towards the dashboard. As the body lunges forwards into the air bag, the bag deflates, allowing the body to slow in a longer time as it moves towards the dashboard. Injury is thus minimised.

Calculating exactly when the air bag should inflate, and for how long, is a difficult task. Many cars have been crash tested and the results painstakingly analysed. High-speed film demonstrates precisely why the air bag is so effective. During a collision the arms, legs and heads of the occupants are restrained only by the joints and muscles. Enormous forces are involved because of the large deceleration. The shoulders and hips can, in most cases, sustain the large forces for the short duration. However, the neck is the weak link. Victims of road accidents regularly receive neck and spinal injuries. An air bag reduces the enormous forces the neck must

withstand by extending the duration of the collision, a direct application of the concept of impulse.

The extent of injuries during a collision is not only dependent on the size of the force but also the duration and deflection resulting from the applied force. An increase in localised pressure will result in a greater compression or deflection of the skull. The air bag reduces the localised pressure by increasing the contact surface area and decreasing the force. The effect can be seen by the relationship:

P = FA

where P is the pressure (N m−2)

F is the force (N)

A is the contact area (m2).

An air bag has a contact surface area of about 0.2 m2 compared with 0.05 m2 for a seat belt. This reduces injuries caused by seat belts, such as bruising and broken ribs and collar bones, since it increases the stopping time. It also supports the head and chest, preventing high neck loads caused by the seat belt restraining the upper torso. Most importantly, it prevents the high forces caused by contact of the head with the steering wheel. The air bag ensures that the main thrust of the expansion is directed outwards instead of towards the driver. The deflation rate, governed by the size of the holes in the rear of the air bag, provides the optimum deceleration of the head for a large range of impact speeds.

The air bag is not the answer to all safety concerns associated with a collision, but it is one of many safety features that form a chain of defence in a collision.

Forces during collisions (continued)A

pp

lied

forc

e (N

)

Time (ms)0 20 40 60 80 100 120 140

no air bag

with air bag

(a)

(b)

Figure 6.6 (a) The air bag is one of a number of passive safety features incorporated into the design of modern cars. It extends the stopping time, significantly reducing the forces on the head and neck during a collision. It also distributes the force required to decelerate the mass of the driver or passenger over a larger area than a seat belt. (b) The deflation rate of the bag is governed by the size of holes in the rear of the air bag, and is designed to provide the optimum deceleration of the head for a large range of impacts.


• The momentum of a moving object is the product of its mass and its velocity:

p = mv where p is in kg m s−1

m is in kg v is in m s−1

Momentum is a vector quantity.• The change in momentum (Δp) = final momentum

(pf) − initial momentum (pi).• Impulse is the product of the net force during a

collision and the time interval Δt during which the

force acts: I = FavΔt. It can also be found from the area under a force–time graph and is measured in newton seconds (N s). Impulse is equal to the change in momentum, Δp, caused by the action of the net applied force:

FavΔt = Δp

• Extending the time over which a collision occurs will decrease the average net force applied since Fav ∝ 1

Δt.

This is the principle behind many safety designs.

The relationship between momentum and force6.1 summary

Use g = 9.8 m s−2 where required.

1 1 What is the momentum, in kg m s−1, of a 20 kg cart travelling at:a 5.0 m s−1?b 5.0 cm s−1?c 5.0 km h−1?

2 2 The velocity of an object of mass 8.0 kg increases from an initial 3.0 m s−1 to 8.0 m s−1 when a force acts on it for 5.0 s.a What is the initial momentum?b What is the momentum after the action of the

force?c How much momentum is the object gaining each

second when the force is acting?d What impulse does the object experience?e What is the magnitude of the force?

3 3 Which object has the greater momentum—a medicine ball of mass 4.5 kg travelling at 3.5 m s−1 or one of mass 2.5 kg travelling at 6.8 m s−1?

4 4 Calculate the momentum of an object:a of mass 4.5 kg and velocity 9.1 m s−1

b of mass 250 g and velocity 3.5 km h−1

c that has fallen freely from rest for 15 s and has a mass of 3.4 kg

d that experiences a net force of magnitude 45 N, if the net force is applied for 3.5 s.

5 5 A tennis ball may leave the racquet of a top player with a speed of 61 m s−1 when served. If the mass of

the ball is 65 g and it is actually in contact with the racquet for 0.032 s:a what momentum does the ball have on leaving

the racquet?b what is the average force applied by the racquet

on the ball?

6 6 A 200 g cricket ball (at rest) is struck by a cricket bat. The ball and bat are in contact for 0.05 s, during which time the ball is accelerated to a speed of 45 m s−1.a What is the magnitude of the impulse the ball

experiences?b What is the net average force acting on the ball

during the contact time?c What is the net average force acting on the bat

during the contact time?

7 7 The following graph shows the net vertical force generated as an athlete’s foot strikes an asphalt running track.

Forc

e (N

)

140012001000800600400200

0 10 20 30 40 50 60 70Time (ms)

a Estimate the maximum force acting on the athlete’s foot during the contact time.

The relationship between momentum and force6.1 questions


Motion186

b Estimate the total impulse during the contact time.

8 8 A 25 g arrow buries its head 2 cm into a target on striking it. The arrow was travelling at 50 m s−1 just before impact.a What change in momentum does the arrow

experience as it comes to rest?b What is the impulse experienced by the arrow?c What is the average force that acts on the arrow

during the period of deceleration after it hits the target?

9 9 Crash helmets are designed to reduce the force of impact on the head during a collision.a Explain how their design reduces the net force on

the head.b Would a rigid ‘shell’ be as successful? Explain.

1010 Describe, with the aid of diagrams, a simple collision involving one moving object and one fixed in position. Estimate, by making reasonable estimates of the magnitudes of the mass and velocity of the moving object, the net force acting on the objects during the collision.

The relationship between momentum and force (continued)

Worked Solutions


Conservation of momentum6.2

The most significant feature of momentum is that it is conserved. This means that the total momentum in any complete system will be constant. For this reason momentum is very useful in investigating the forces experienced by two colliding objects—as long as they are unaffected by outside forces. The law of conservation of momentum, as it is known, is derived from Newton’s third law.

From Newton’s third law, the force applied by each object in a collision will be of the same magnitude but opposite in direction:

F1 = −F2

From Newton’s second law, ΣF = ma, so the forces could be expressed as:m1a1 = −m2a2

and using a = v − u

Δt we get:

m1

v1 − u1

Δt = −m2

v2 − u2

Δtwhere a1 and a2 are the respective accelerations of the two objects during the collision in m s−2

Δt is the time of contact (s) u1 and u2 are the velocities of the objects prior to collision (m s−1) v1 and v2 are the velocities after collision (m s−1)

Since the time that each object is in contact with the other will be the same, Δt will cancel out:

m1(v1 − u1) = −m2(v2 − u2)or:

m1u1 + m2u2 = m1v1 + m2v2

In other words, the total momentum before colliding is the same as the total momentum after the collision.

It is most important to realise that momentum is only conserved in an isolated system; that is, a system in which no external forces affect the objects involved. The only forces involved are the action–reaction forces on the objects in the collision. Consider two skaters coming together on a near-frictionless ice rink. In this near-ideal situation it is realistic to apply the law of conservation of momentum. The only significant horizontal forces between the two skaters are those of the action–reaction pair as the two skaters collide.

If the skaters were to skate through a puddle of water as they come together then friction would become noticeable. This force is an external force since it is not acting between the two skaters. The interaction between

TH… LAW OF CONS…RVATION OF MOM…NTUM states that, in any collision or interaction between two or more objects in an isolated system, the total momentum of the system will remain constant; that is, the total initial momentum will equal the total fi nal momentum:

Σpi = Σp

for

m1u

1 + m

2u

2 = m

1v

1 + m

2v

2

i

The principle of conservation of momentum was responsible for the interpretation of investigations that led to the discovery of the neutron. Neutral in charge, the neutron could not be investigated through the interactions of charged particles that had led to the discovery of the proton and electron. In 1932 Chadwick found that in collisions between alpha particles and the element beryllium, the principle of conservation of momentum only held true if it could be assumed that there was an additional particle within the atom, which had close to the same mass as a proton but no electric charge. Subsequent investigations confirmed his experiments and led to the naming of this particle as the neutron.

Physics file

If you release an inflated rubber balloon with its neck open, it will fly off around the room. In the diagram below, the momentum of the air to the left is moving the balloon to the right. Momentum is conserved.

air balloon

This is the principle upon which rockets and jet engines are based. Both rockets and jet engines employ a high-velocity stream of hot gases that are vented after the combustion of a fuel–air mixture. The hot exhaust gases have a very large momentum as a result of the high velocities involved, and can accelerate rockets and jets to high velocities as they acquire an equal momentum in the opposite direction. Rockets destined for space carry their own oxygen supply, while jet engines use the surrounding air supply.

Physics file

Motion188

a skater and the puddle would constitute a separate isolated system (as would that between puddle and ice, ice and ground etc.). Momentum would still be conserved within this other separate system. It is virtually impossible to find a perfectly isolated system here on Earth because of the presence of gravitational, frictional and air-resistance forces. Only where any external forces are insignificant in comparison to the collision forces is it reasonable to apply the law of conservation of momentum.

Also important is that in any collision involving the ground, Earth itself must be part of the system. Theoretically, any calculation based on conservation of momentum should include the Earth as one of the objects, momentum only being conserved when all the objects in the system are considered. In practice, the very large mass of the Earth in relation to the other objects involved means that there is a negligible change in the Earth’s velocity and it can be ignored in most calculations. Of course, a collision with a fast-moving asteroid would be another matter!

Worked example 6.2A Skater 1 in Figure 6.7, with mass 80 kg, was skating in a straight line with a velocity of 6.0 m s−1 while the skater 2, of mass 70 kg, was skating in the opposite direction, also with a speed of 6.0 m s−1.a The two skaters collide and skater 1 comes to rest. Assuming that friction can be ignored,

what will happen to the skater 2 after the collision?b What would happen if the two skaters had hung on to each other and stayed together

after the collision?

Solution a From conservation of momentum: Σp

i = Σp

f

or m1u

1 + m

2u

2 = m

1v

1 + m

2v

2

and m1 = 80 kg, m

2 = 70 kg

As both velocity and momentum are vector quantities, a positive direction should be established and taken into consideration. Adopting the direction of motion of skater 1 as the positive direction:

u1 = 6.0 m s−1, u

2 = −6.0 m s−1, v

1 = 0 m s−1, v

2 = ?

Substituting into equation: 80 × 6.0 + 70 × (−6.0) = 80 × 0 + 70 × v

2

and v2 = +0.86 m s−1.

The 70 kg skater bounces back in the opposite direction with a speed of 0.86 m s−1.b Treating the two skaters as one mass after the collision: 80 × 6.0 + 70 × (−6.0) = (80 + 70) × v

2

and now v2 = +0.4 m s−1.

A different outcome after the collision results in a different velocity for each skater. There is no unique answer when applying the idea of conservation of momentum. The final velocity of any object depends on what happens to all the objects involved in the collision.

The law of conservation of momentum can be extended to any number of colliding objects. The total initial momentum is found by calculating the vector sum of the initial momentum of every object involved. The total final momentum will then also be the vector sum of each separate momentum involved. Separation into two or more parts after the ‘collision’ (interaction is a better word since it does not have to be destructive), for example the firing of a bullet, can also be dealt with in the same manner.

Figure 6.7 When two skaters collide on a near-frictionless skating rink they exert equal and opposite forces on each other. The total momentum of the two skaters before the collision will equal the total momentum of the two skaters after the collision. Because no other large horizontal forces are involved other than those in the collision, this can be considered as an isolated system.

Figure 6.8 Newton’s cradle, or Newton’s balls to some, is an instructive ‘executive toy’ based on the principle of conservation of momentum extended over a number of objects.

6.0 m s–1 6.0 m s–1

Prac 25


A car is designed to keep its occupants safe. Unfortunately, however, very little can be done to protect a pedestrian from the onslaught of a 1400 kg car travelling at 60 km h−1. Bull bars in residential areas are currently under review because of the enormous damage they inflict on a pedestrian. The effect on the pedestrian will depend on the person’s height and mass, the height of the front of the oncoming vehicle, and the speed, mass and shape of the vehicle.

Consider the following possibilities:• a pedestrian being struck by a truck moving at 30 km h−1

• a pedestrian being struck by a car moving at 30 km h−1

• a pedestrian being struck by a cyclist moving at 30 km h−1.The injuries to the pedestrian are due largely to the

change in the pedestrian’s momentum. Being hit by the cyclist will obviously result in the least injury to the pedestrian because of the lower momentum of the cycle and its rider. The mass of the other vehicles is such that they have a far larger momentum to impart to the pedestrian.

Consider the following situation. A 1400 kg car is travelling at 60 km h−1 when it strikes a stationary 70 kg pedestrian. The pedestrian lands on the bonnet of the car and travels with the car until it finally comes to a halt. Assuming that frictional forces are minimal:

total momentum before the collision = total momentum after the collision

Before the collision:

car = m

= 1400 × (60 × 1000

3600)

= 2.3 × 104 kg m s−1

pedestrian

= 0 (pedestrian is stationary)

After the collision:

total

= 2.3 × 104 kg m s−1 = (m

car + m

pedestrian)

so 1470 = 2.3 × 104 kg m s−1

and

total ≈ 16 m s−1 or 57 km h−1

This means that the pedestrian accelerates from rest to a speed of 57 km h−1 in the short duration of the collision. A similar collision between the pedestrian and a cyclist travelling at 30 km h−1 would result in a final speed of 5.2 m s−1 (19 km h−1). The speed of the car changes very little. The speed of the cyclist is almost halved.

Antilock brakes, excellent road handling and reduced speed limits in some areas reduce the likelihood of a vehicle striking a pedestrian. Unfortunately accidents can still happen.

There are essentially two possibilities that can occur when a pedestrian is struck by a car.1.1. The pedestrian bounces off the front of the car

and is projected through the air. This type of motion tends to happen when the vehicle is travelling relatively slowly. The pedestrian is rapidly accelerated forwards to near the velocity of the vehicle. Injuries occur to the pedestrian when the car strikes and again when they land on the

ground. Leg and hip injuries are general in this form of pedestrian–vehicle collision.

Head injuries result from the pedestrian colliding with the road. Very little of a car’s design will alter the severity of head injuries received.

Vehicles can be designed with a low, energy-absorbing bumper bar to reduce knee and hip damage. If the pedestrian’s knee strikes the bumper bar, knee damage is very likely. Knees do not heal as well as broken legs. A lower, energy-absorbing bar is, for this reason, preferable.

2.2. When a vehicle is moving very fast at the point of impact, the pedestrian’s inertia acts against rapid acceleration. The pedestrian does not initially move forward with the same velocity as the vehicle. If the pedestrian remains in approximately the same place, he or she will go either over or under the car. The pedestrian will be either run over or run under (i.e. the car goes under the pedestrian).

Being run over usually results in serious injury or fatality. Massive head injuries occur as the pedestrian’s head strikes the ground. The relative height of the vehicle’s bumper bar and the height of the pedestrian determines whether they will be run over or under. Most bumper bars are below adult waist level. Small children, however, have much more chance of being run over as the height of the bumper bar is relatively much higher.

If the pedestrian is run under, ‘passive’ safety features of modern car design come into play. Removal of protruding hood ornaments is essential since they can easily penetrate the body of a person and cause enormous injuries. The bonnet of a car acts as a good impact absorber, particularly in comparison with the hard surface of the road. Bull bars, however, can block the path of the pedestrian, making it more likely that they be run over. Further, they have little impact-absorbing ability. It is, therefore, logical to ban bull bars in residential areas.

Collisions and pedestriansPhysics in action

Figure 6.9 This Nissan car has a pop-up bonnet that has been designed to result in less damage to a pedestrian in a collision, by making space between the bonnet and the engine.

Motion190

• The law of conservation of momentum states that in any collision or interaction between two or more objects in an isolated system the total momentum of the system will remain constant. The total initial momentum will equal the total final momentum:

Σpi = Σpf

• Conservation of momentum can be extended to any number of colliding objects within an isolated system.

Conservation of momentum6.2 summary

1 1 A white billiard ball of mass 100 g travelling at 2.0 m s−1 across a low-friction billiard table has a head-on collision with a black ball of the same mass initially at rest. The white ball stops while the black ball moves off. What is the velocity of the black ball?

2 2 A girl with mass 50 kg running at 5 m s−1 jumps onto a 4 kg skateboard travelling in the same direction at 1.0 m s−1. What is their new common velocity?

3 3 A man of mass 70 kg steps forward out of a boat and onto the nearby river bank with a velocity, when he leaves the boat, of 2.5 m s−1 relative to the ground. The boat has a mass of 400 kg and was initially at rest. With what velocity relative to the ground does the boat begin to move?

4 4 A railway car of mass 2 tonnes moving along a horizontal track at 2 m s−1 runs into a stationary train and is coupled to it. After the collision the train and car move off at a slow 0.3 m s−1. What is the mass of the train alone?

5 5 A trolley of mass 4.0 kg and moving at 4.5 m s−1 collides with, and sticks to, a stationary trolley of mass 2.0 kg. Their combined speed in m s−1 after the collision is:A 2.0 B 3.0 C 4.5 D 9.0

6 6 A superhero stops a truck simply by blocking it with his outreached arm.

a a Is this consistent with the law of conservation of momentum? Explain.

b Using reasonable estimates for the initial speed and mass of the truck and the superhero, demonstrate what will happen. Use appropriate physics concepts.

7 7 A car of mass 1100 kg has a head-on collision with a large four-wheel drive vehicle of mass 2200 kg, immediately after which both vehicles are stationary. The four-wheel drive vehicle was travelling at 50 km h−1 prior to the collision in an area where the speed limit was 70 km h−1. Was the car breaking the speed limit?

8 8 A 100 g apple is balanced on the head of young master Tell. William, the boy’s father, fires an arrow with a mass of 80 g at the apple. It reaches the apple with a velocity of 35 m s−1. The arrow passes right through the apple and goes on with a velocity of 25 m s−1. With what speed will the apple fly off the boy’s head? (Assume there is no friction between apple and head.)

9 9 A space shuttle of mass 10 000 kg, initially at rest, burns 5.0 kg of fuel and oxygen in its rockets to produce exhaust gases ejected at a velocity of 6000 m s−1. Calculate the velocity that this exchange will give to the space shuttle.

10 10 A small research rocket of mass 250 kg is launched vertically as part of a weather study. It sends out 50 kg of burnt fuel and exhaust gases with a velocity of 180 m s−1 in a 2 s initial acceleration period.a What is the velocity of the rocket after this initial

acceleration?b What upward force does this apply to the rocket?c What is the net upward acceleration acting on the

rocket? (Use g = 10 m s−2 if required.)

Conservation of momentum6.2 questions

Worked Solutions


Work6.3

Aristotle, Galileo and Newton each made significant contributions to our developing understanding of the relationships between the forces that are applied to objects and their resultant behaviour. Aristotle and those who followed him were often locked into philosophical views involving, for example, the natural resting places of objects. Unlike his predecessors, Galileo, over 400 years ago, based his proposed theories largely on the observations that he made. Although he could not be completely free from the influences of his era, observational scientific experimentation had been born through him. The implications of this change in approach were immeasurable.

In this chapter our study of the interactions between forces and objects focuses on the resultant displacement that objects experience, rather than the resulting velocities and accelerations discussed earlier in our study of Newton’s laws. This leads to the examination of the concepts of work and energy. The notion of energy was not developed until a relatively short time ago, and was only fully understood in the early 1800s. Today, the concept has become one of the most fundamental in science. We will see that in physics an object is said to have energy if it can cause particular changes to occur. Energy is a conserved quantity and is useful not only in the study of motion, but in all areas of the physical sciences. Before discussing energy, it is necessary to first examine the concept of work.

In common usage the term ‘work’ has a variety of meanings. Most convey the idea of something being done. At the end of a long, tiring day we might say that we have done a lot of work. This could also be said because the person feels that their reserves of energy have been used up. Imagine lifting a heavy book up onto a high shelf. The heavier the book, the more force must be applied to overcome its weight. The higher the shelf, the greater the displacement over which the force must be applied. A very heavy book lifted to a high shelf will require a considerably greater effort than moving a few pieces of paper from floor to table. Thus there are two features that constitute the amount of work done: the amount of effort required and the displacement involved.

In physics work is done on an object by the action of a force or forces. The object is often referred to as the load. Many interactions are complex and there is often more than one force present. As work can only be done in the presence of a force, it is imperative that any time the work done in a particular situation is being discussed, the relevant force, forces or net force should be clearly stated. For clarity, the item upon which the work is done, the load, should also be specified. Clearly specified examples of work are:• the work done by gravity on a diver as she falls• the work done by arm muscles on a schoolbag lifted to your shoulder• the work done by the heart muscle on a volume of blood during a

contraction• the work done by the net force acting on a cyclist climbing a hill.

Always being clear about the particular forces and objects examined will prevent considerable confusion in this area of study.

For work to be done on a body, the energy of the body must change. Thus the work done is measured in joules, which is also the unit of energy.

Figure 6.10 In each situation involving work, a load can clearly be specified.

Motion192

The different forms of energy are discussed later in this chapter. A test to decide whether work has been done on a particular object involves deter-mining whether the object’s energy has altered. If its energy is unaltered, no net work has been done on it, even though forces clearly may have been acting.

Work done by a constant forceIf the net force acting on an object in a particular situation has a constant value, or if it is appropriate to utilise an average force value, then:

From the definition of work, it can be seen that if a person pushes a load a horizontal distance of 5 m by exerting a horizontal force of 30 N on the load, then the person does 150 J of work on the load. This is straightforward. A displacement in the direction of the force is achieved so work is done. If an applied force does not produce any displacement of the object (x = 0), then we say that no work is done on the object.

Situations also occur in which a constant force acts at an angle θ to the direction of motion. A force acting at an angle will be less effective than the force acting solely in the direction of the displacement. The component of the force in the direction of the displacement, F cos θ, is used in calculating the work done in the required direction.

The net WORK done on an object is defi ned as the product of the net force on the object and its displacement in the direction of the net force.When the force and displacement are in the same direction, the work done by the stated force is given by:

W = Fxwhere W is the work done by the stated force in joules (J) F is the magnitude of the stated force in newtons (N) x is the magnitude of the displacement in metres (m)Work is the area under a force–displacement graph.

i

One JOUL… of work is done on an object when the application of a net force of 1 newton moves an object through a distance of 1 metre in the direction of the net force.

i

The symbol W is a little over-used in this area of physics. In the area of motion and mechanics it can stand for work or the abbreviation of watt. Be careful to read the context when you come across the symbol!

Physics file

The convention for naming units in physics is to use small letters when writing the unit in full (e.g. joule, newton, metre). A capital letter is used for the symbol only when the unit is named in recognition of a scientist’s contributions, otherwise the symbol is lower case (e.g. J for joule, N for newton, but m for metre).

Physics file

Figure 6.12 During the fall the force due to gravity does work on the person and produces a displacement.

Figure 6.11 (a) No work is done on the crate since its energy is not altered. (b) The energy of the crate is changing, so work is being done on the crate.

I’m doing no workon the crate sincex = 0W = Fx = 0

I’m doing nowork on thecrate sincex = 0

I’m doing work!

crate speeding up

work is doneon the loadW = Fx

(a) (b)

The unit for work, the joule (J), is used for all forms of energy in honour of James Prescott Joule, an English brewer and physicist, who pioneered work on energy in the 19th century.

Physics file


Work and frictionIf an object is forced to move across a surface by the application of a force, its motion may be slowed by friction. In this case the applied force is doing work on the object and the frictional force can be considered to be doing ‘negative’ work on the object. In Figure 6.14 an applied force of 300 N across a displacement of 5 m does 1500 J of work on the object. If a 100 N frictional force occurs, we can state that work done by the frictional force is:

−100 × 5 = −500 JHence the net work done on the object is 1000 J. An alternative approach

would involve first calculating the net force, ΣF, on the object to be 200 N. The net work done on the load (by the net force) is therefore:

ΣF × x = 200 × 5 = 1000 JIf work is done against a frictional force as a load continues to move,

then some of the energy expended by the person pushing is converted into heat and sound energy, and transferred to the ground and the load. As the surfaces slide past one another friction would cause them to heat up slightly and make some noise. Keep in mind that on a frictionless surface the load would accelerate, increasing its energy.

Figure 6.15 shows a situation in which the size of the frictional force is not large enough to prevent motion, but it is large enough to balance the applied force. As a result the object moves at a steady speed. Although the person is doing work on the object, this is opposed by friction and the net work on the object is zero. This is consistent with our earlier discussion, which stated that if the energy of the object is not altered, then no net work has been done on the object.

W = Fx cos θwhere θ is the angle between the applied force and the direction of motion.

i

Figure 6.14 The object slides across a displacement of 5 m. Due to friction the net work done on the object is less than the work done by the person on the object.

Figure 6.13 (a) If a force is applied in the direction of motion of the cart, then the force is at its most effective in moving the cart. (b) When the force is applied at an angle θ to the direction of motion of the cart, the force is less effective. The component of the force in the direction of the displacement, F cos θ, is used to calculate the work. (c) When the angle at which the force is acting is increased to a right angle (θ = 90°), then the component of the force in the direction of the intended displacement is zero and it does no work on the cart—provided of course that it doesn’t lift the cart, in which case work would also be done against gravity.

Figure 6.15 Due to friction the net work done on the object is zero since the object has no increase in kinetic energy.

Direction of motion

FF

F

θ

F cos θ

Direction of motion Direction of motion

(a) (b) (b)

frictional force Ff= 100 N

ΣF = 200 N

Fapplied= 300 N

Net workdone on crate = 1000 J

frictional force Ff= 100 N

Fapplied= 100 N

steady speed

object’s energy is unchanged

Motion194

Worked example 6.3A Calculate the work done against gravity by an athlete of mass 60 kg competing in the Empire State Building Run-up illustrated in Figure 6.16. Use g = 9.8 m s−2. Assume the athlete climbs at a constant speed.

Solution Only the weight force needs to be considered in this example as the work in the vertical direction is all that is required.m = 60 kg, g = 9.8 m s−2, x = Δh = 320 mForce applied = weight = mg = 60 × 9.8 = 588 NW = Fx

= 588 × 320= 188 160 N m = 1.9 × 105 J

Worked example 6.3B The girl in Figure 6.13 pulls the cart by applying a force of 50 N at an angle of 30° to the horizontal. Assuming a force due to friction of 10 N is also acting on the wheels of the cart, calculate the net work done on the cart if the cart is moved 10 m along the ground in a straight line.

Solution F

applied = 50 N, θ = 30°, F

f = 10 N, x = 10 m

ΣF = Fapplied

× cos 30° − Ff

= 50 × 0.866 − 10 = 33.3 N Now W = ΣFx = 33.3 × 10 = 330 J

Figure 6.16 The Empire State Building Run-up is one of a number of races to the tops of tall buildings around the world. The current record for the 320 m (vertical) race is 9 minutes and 33 seconds.


Upward force does no workA more difficult idea to comprehend is that in the apparent absence of friction, a force can be exerted on an object yet do no work on it. For example, when a person carries an armload of books horizontally the upward force does no work on the books since the direction of the applied force (i.e. up) is at right angles to the displacement (i.e. horizontal). An examination of the definition of work—W = Fx cos θ—confirms this finding since the value of θ is 90° and, hence, the value of cos θ is 0.

Similarly if a person is holding a heavy item, such as a TV, stationary, they may be exerting great effort. However, since the upward force applied to support the object does not produce any vertical (nor indeed horizontal) displacement, x = 0 and there is no work done by this upward applied force on the object.

Force–displacement graphsA graphical approach can also be used to understand the action of a force and the work expended in the direction of motion. This is particularly useful in situations in which the force is changing with displacement. The area under a graph of force against displacement always represents the work produced by the force, even in situations when the force is changing, such as during a collision. The area can be shown to be equivalent to work as follows. From Figure 6.17:

the area enclosed by the graph = Fav × xWork = Fav × x

When the force is changing, a good estimate of the area can be found by dividing the area into small squares and counting the number or by dividing it into thin segments. The segments can be considered to be rectangles with an area equal to the work for that small part of the displacement. The total work will be the sum of the areas of all the separate rectangles.

Whenever the net force is perpendicular to the direction of motion no (net) work is done on the object.

i

Figure 6.17 The area under a force–displacement graph is equivalent to the work done by a force acting in the direction of the displacement. Where the net applied force is changing, the area can be found by counting squares or by dividing the area into segments. The area of each segment then equals the work done by a constant force during that small displacement and the total area will represent the total work.

Units for area = N m = N m = Ji.e. area = work done

Forc

e (N

)

Displacement (m)

The area under a force–displacement graph can also be found by using calculus if the equation of the graph is known. In most instances a good estimate by counting squares or segments is sufficient.

Physics file

Motion196

Worked example 6.3C The force–displacement graph on the left represents the work done on the sole of a sports shoe as it compresses against the surface of a rigid track. The displacement shown represents the amount of compression the sole undergoes. Find the work done on the shoe by the compressive forces.

Solution This is a simple case of working out the area represented by each square and then counting the total number of squares to find the total work done. Be careful to consider the scale of each axis in your working.Area of one square = 10 N × 0.001 m = 0.01 JTotal number of squares (part squares can be added to give whole squares) = 33Work = 33 × 0.01 = 0.33 J

Impulse and workThe concepts of impulse and work seem quite similar and, when solving problems, can easily be confused. Actually, problems focusing on forces in collisions may be solved using either concept, but it should be understood that each is derived from a different idea. Impulse comes from an understanding of the action of a force on an object over time and is equal to the change in momentum the force produces. Work is related to the action of a force on an object as it moves the object, or part of it, through some displacement. This equals the change in the object’s energy, ΔE.

Summarising:• Impulse is equal to F × Δt, is equivalent to Δp, has the units newton

seconds (N s), and can be determined from the area under a force–time graph.

• Work is equal to F × x, is equivalent to ΔE, has the units joules (J), and can be determined from the area under a force–displacement graph.

• When a force does work on an object, a change occurs in the displacement and energy of the object.

• The work done on an object, W in joules (J), is the product of the net applied force and its displacement in the direction of the force:

W = Fx• The work done by a force acting at an angle to the

displacement is given by Fx cos θ where θ is the angle between the force and the direction of the

displacement. When the force is at right angles to the direction of the displacement, no work is done in that direction.

• The area under a force–displacement graph is equivalent to the work done. The area under the graph for a variable force can be found by counting squares or narrow segments.

Work6.3 summary

Forc

e (N

)

Displacement (m)

100

90

80

70

60

50

40

30

20

10

0.00

1

0.00

2

0.00

3

0.00

4

0.00

5

0.00

6

0.00

7

0.00

8


Where appropriate use g = 9.8 m s−2.

1 1 How much work is done on an object of 4.5 kg when it is lifted vertically at a constant speed through a displacement of 6.0 m?

2 2 A bushwalker climbs a hill 250 m high. If her mass is 50 kg and her pack has an additional mass of 10 kg, calculate the work she needs to do in climbing to the top of the hill.

3 3 Is the quantity you calculated in Question 2 the only work that the bushwalker has done? Explain.

4 4 The work done by a force is: i i calculated by multiplying the force by the

distance moved ii ii measured in joules iii iii not affected by the angle at which the force acts.

Which statement/s is/are correct?A i, ii, iiiB i, iiC ii, iiiD iiE iii

5 5 A removalist is loading five boxes onto a truck. Each has a mass of 10 kg and a height of 30 cm. The tray of the truck is 1.5 m above the ground and the removalist is placing each box on top of the previous one.a How much work does the removalist do in lifting

the first box onto the truck tray?b How much energy has the removalist used in

lifting this first box?c What is the total work done on the boxes in lifting

all the boxes onto the truck as described?

6 6 If a lift of mass 500 kg is raised through a height of 15 m by an electric motor:

i i the weight of the lift is 4900 N ii ii the useful work done on the lift is 73 500 J iii iii the useful work done is the only energy used by

the motor.Which statement/s is/are correct?A i, ii, iiiB i, iiC ii, iiiD iiE iii

7 7 The diagram shows the position of a student’s arm as the weight of a sandbag is measured using a spring balance. The balance is held still to take the reading. What net work is done on the sandbag while the measurement is being made?

sandbag

springbalance

8 8 A weightlifter raises a 100 kg mass 2.4 m above the ground in a weightlifting competition. After holding it for 3.0 s he places it back on the ground.a How much work has been done by the weightlifter

in raising the mass?b How much additional work is done during the

3.0 s he holds it steady?

9 9 A rope that is at 35° to the horizontal is used to pull a 10.0 kg crate across a rough floor. The crate is initially at rest and is dragged for a distance of 4.00 m. The tension in the rope is 60.0 N and the frictional force opposing the motion is 10.0 N.a Draw a diagram illustrating the direction of all

relevant forces.b Calculate the work done on the crate by the

tension in the rope.c Find the total work done on the crate.d Determine the energy lost from the system as heat

and sound due to the frictional force.

1010 The graph represents the size of a variable force, F, as a rubber band is stretched from a resting length of 5 cm to 25 cm. Estimate the total work done on the rubber band by the force.

Forc

e (N

)

Extension (cm)10 20

10

5

Work6.3 questions

Worked Solutions

Motion198

Mechanical energy6.4

Although the concept of energy is quite abstract, each of us, from an early age, will have begun to develop an understanding of its meaning. We are increasingly aware of our reliance upon the energy resources that allow our vehicles and computers to run, that keep our homes warm and fuel our bodies. Energy can take on many forms.

In this section we look at the forms of energy specifically related to motion. Mechanical energy is defined as the energy that a body possesses due to its position or motion. Kinetic energy, gravitational potential energy and elastic potential energy are all forms of mechanical energy. Recall our earlier assertion that work is done when a force is applied that results in the displacement of an object in the direction of the applied force. When work is done the energy of an object will change. We will analyse situations that result in a change in the kinetic and/or gravitational potential and/or elastic potential energy of an object.

A hockey puck gains energy when hit because work has been done by the stick on the puck. The amount of work done on the puck equals the puck’s change in kinetic energy. A tennis ball at the point of impact is compressed against the tennis racquet. It has gained elastic potential energy. Work has been done in compressing the tennis ball. However, the idea of work may be applied to many forms of energy. The common thread is that, regardless of the form of energy, whenever work is done there is a change in energy from one form to another. In order for any energy transformation to occur, say from motion to heat, work must be done.

We observe many different forms of energy each day. We have come to take for granted the availability of light, heat, sound and electrical energy whenever we require it. We rely upon the chemical potential energies that are available when petrol, diesel and LPG are burnt to run our vehicles, and food to fuel our bodies. Whenever work is done, energy is expended.

Some comparative energy transformations are included in Table 6.2.

Table 6.2 Comparison of various energy transformations

Energy use Amount of energy

Household in 1 day 150 MJ

Fan heater in 1 hour 8.6 MJ

Adult food intake in 1 day 12 MJ

Making 1 Big Mac 2.1 MJ

Climbing a flight of stairs 5 kJ

Lifting 10 kg to a height of 2 m 200 J

Kinetic energyAn object in motion has the ability to do work and therefore is said to possess energy. This energy carried by a moving object is called kinetic energy (from the Greek word ‘kinesis’, literally meaning ‘motion’).

…N…RGY is the ability to do work.i

Figure 6.18 Mechanical energy exists in many forms.

Interactive


If a moving object of mass, m, and initial velocity, u, experiences a constant net force, F, for time, t, then a uniform acceleration results. The velocity will increase to a final value, v. Work will have been done during the time the force is applied. Since work is equivalent to the change in kinetic energy of the object, there should be a relationship linking the two quantities. This can be found from the definition for work when the net applied force is in the direction of the displacement:

W = ΣFxNow substituting Newton’s second law F = ma we get:

W = max . . . . . . (i)Using one of the earlier equations of motion: v2 = u2 + 2ax

and rearranging: x = v2 − u2

2aSubstitute this for x in equation (i): W = ma ×

v2 − u2

2aRearranging gives: W = 1

2mv2 − 1

2mu2

but W = ΔEIf it is accepted that the work done results in a change in kinetic energy,

then an object of mass m with a speed v has kinetic energy equal to 1

2mv2.

Like all forms of energy, kinetic energy is a scalar quantity and is measured in joules (J). There is no direction associated with it. The kinetic energy of an object depends solely on its mass and velocity. The approximate kinetic energy of various moving objects is given in Table 6.3.

Table 6.3 Kinetic energy of moving objects

Object Mass (kg)Average speed (m s–1)

…k (J)

Earth in orbit 6 × 1024 3 × 104 2.7 × 1033

Orbiting satellite 100 8 × 103 3 × 109

Large car 1400 28 5.5 × 105

Netball player 60 8 1900

Footballer 90 8 2900

The KIN…TIC …N…RGY, …k, of a body of mass m and speed v is:

…k = 1

2 mv2i

Figure 6.19 The kinetic energy of any object depends on its mass and the square of its speed. Doubling the velocity will increase the kinetic energy by a factor of four.

The derivation described for kinetic energy is actually that for translational kinetic energy, the movement of a body along a path. A body can also have rotational kinetic energy if it is spinning, as does the Earth. A different relationship is required to calculate the kinetic energy of rotation.

Physics file

Motion200

Electron in a TV tube 9 × 10−31 7 × 107 2.2 × 10−15

Note that the relationship between work and energy, which was discussed earlier, has now been quantified for kinetic energy changes.

When an applied force results in the change in kinetic energy of an object, the work done in joules (J) can be calculated using: W = Δ…

k = …

k(fi nal) − …

k(initial)

= 1

2 mv2 − 1

2 mu2

where m is the mass of the object in kilograms (kg) v is the fi nal speed of the object in metres per second (m s−1) u is the initial speed of the object in metres per second (m s−1)As the mass of the object is generally unaltered, often this can be simplifi ed to:

W = Δ…k = 1

2 m(v 2 – u2)

i

Therefore, if an object undergoes a known change in kinetic energy during an interaction, the work done on the object by the net force is known. Hence the average net force exerted on the object during this interaction can be calculated by assuming that ΔEk = W = Favx.

Worked example 6.4A Calculate the kinetic energy of an athlete of mass 60 kg running at a speed of 8.0 m s−1.

Solution m = 60 kg, v = 8.0 m s−1

Using …k = 1

2 mv2: …

k = 1

2 × 60 × 8.02

≈ 1900 J.

Worked example 6.4B Blood is pumped by the heart into the aorta at an average speed of 0.15 m s−1. If 100 g of blood is pumped by each beat of an adult human’s heart find:a the amount of work done by the heart during each contractionb the energy used by the heart each day in pumping blood through the aorta (use an

adult’s average resting rate of 70 beats per minute). Assume that there are no other energy losses.

Solution a The work done by the heart is equal to the kinetic energy the blood gains as it is pumped

into the aorta. m = 0.10 kg, v = 0.15 m s−1, u = 0 m s−1

Using W = Δ…k = 1

2 m(v2 − u2) W = 1

2 × 0.10 × (0.152 − 02) W = 1.125 × 10−3 J = 1.1 mJb If there are 70 beats each minute then the amount of energy transferred: …

k per minute = 1.125 × 10−3 × 70 = 0.07875 J per minute

…k per day = 0.07875 × 60 min per hour × 24-hour day

…k = 113.4 J per day ≈ 110 J per day

Interactive


Potential energyAn object can have energy not only because of its motion, but also as a result of its shape or position. This is called potential energy. A gymnast, crouched ready to jump, has potential energy. During the jump, work is being done by the force exerted by the gymnast, and potential energy is converted into kinetic energy from the stores of chemical energy in the muscles of the gymnast’s body.

There are many different forms of potential energy: chemical, grav-ita tional, elastic etc. Potential energy is a stored energy giving the body potential to do work or produce a force that creates motion. In this particular study we are mainly concerned with gravitational and elastic potential energy which, for the present, we will denote U.

Gravitational potential energy

An athlete at the top of a high-jump has gravitational potential energy because of his position. As he falls, work is done (Figure 6.20). Recall that in this case the work done is given by:

Work done = ΣFxThe force acting on the body is simply the force due to gravity also called

the person’s weight:Weight = mg

The displacement that occurs is in a vertical direction and can be described as a change in height, Δh. Replacing F and x with these equivalent terms gives:

W = mgΔhSimilarly, the work done in raising the athlete against a gravitational field

is stored as gravitational potential energy; hence, the athlete has a change in potential energy:

ΔUg = mgΔh

Figure 6.20 The energy gained or lost due to a change in height within a gravitational field is called gravitational potential energy. An increase in height will require the transformation of energy from other sources. A decrease will usually increase the kinetic energy of the body.

Figure 6.21 This photograph of a pole-vaulter illustrates that elastic potential energy is stored in the pole. This energy is largely converted to kinetic energy and then the gravitational potential energy of the athlete.

Motion202

The ΔUg of a body depends only on the vertical height of the object above some reference point, in this case the ground. It does not depend on the path taken since it is based on the direction of the gravitational field. It is the work done against or by the force of gravity that leads to changes in gravitational potential energy. Similarly, the work it can do when falling does not depend on whether the object falls vertically or by some other path, but only on the vertical change in height, Δh.

The reference level from which the height is measured does not matter as long as the same reference level is used throughout a given problem solution. It is only changes in potential energy that are important. For example, the height of the high jumper is best referenced to the ground she jumped from, and commonly it is her centre of gravity that is analysed. The height of a luggage locker in an aircraft makes a lot more sense when referenced to the floor of the aircraft than it would referenced to the ground.

The need for considering a change in height in comparison to a reference level is also made apparent when considering a person standing at ground level. If the person is standing beside a hole and his centre of gravity is considered, he will have gravitational potential energy with reference to the bottom of the hole. It is also quite justifiable to suggest that even with reference to the ground he has gravitational potential energy; he could fall over! Gravity would do work on him and his gravitational potential energy would change. The change in height would be with reference to the person’s centre of mass.

Worked example 6.4C A ranger with a mass of 60 kg, checking the surface of Uluru for erosion, walks along a path that takes her past points A, B and C.

C (0 m)

A (260 m)B (389 m)

a What is her gravitational potential energy at points B and C relative to A?b What is the change in the ranger’s potential energy as she walks from B to C?c If the ranger was to walk from B to C via A would it alter your answer to part b? Explain.

Solution a In this question heights are being referenced to point A. The person would have had zero

gravitational potential energy at A using this reference. m = 60 kg, g = 9.8 m s−2, h

A = 260 m, h

B = 389 m, h

C = 0 m

Potential energy change from A to B: ΔU

g = mg(h

B − h

A)

= 60 × 9.8 × (389 − 260)= 7.6 × 104 J

The change in gravitational potential energy is due to the work done against a gravitational fi eld and is given by:

ΔUg = mgΔh

where ΔUg is the change in gravitational potential energy measured in joules (J)

m is the mass of the body (kg) Δh is the change in height (m) g is the acceleration due to gravity (m s–2)

iThe relationship for gravitational potential energy used here is only appropriate when the weight force due to gravity is constant. This will only be the case when the change in height is relatively small. As the distance from the Earth’s surface changes so will the strength of the gravitational field according to the relationship

g ∝ 1

r2

where r is the distance (in metres) from the centre of the Earth to the body’s position. The area under a force–displacement graph can then be used to find the change in potential energy due to a change in position and the varying weight force.

For the purposes of this study only relatively small changes in height close to the Earth’s surface will be considered for which the weight force can be considered constant.

Physics file


Potential energy change from A to C: ΔU

g = mg(h

C − h

A)

= 60 × 9.8 × (0 − 260)= −1.5 × 105 J

b There is no need to calculate ΔUg from B to C separately as the difference between the

two previous results can be used. Potential energy change from B to C: ΔU

g = ΔU

g (A to C) − ΔU

g (A to B)

= −1.5 × 105 − 7.6 × 104

= −2.3 × 105 Jc It makes no difference what path is taken to achieve the change in height. Potential

energy change throughout this example is being determined relative to an initial height. In general, if an object is originally at a height h

0, then the change in potential energy as

it moves to a different height, h, is: ΔU

g = mgh − mgh

0

= mg(h − h0)

In general terms, the change in potential energy of an object when it is moved between two heights is equal to the work needed to take it from one point to another.

Elastic materials and elastic potential energyThe third aspect of mechanical energy that we will study is elastic potential energy. Like gravitational potential energy it occurs in situations where energy can be considered to be stored temporarily so that, when this energy is released, work may be done on an object. Elastic potential energy is stored when a spring is stretched, a rubber ball is squeezed, air is compressed in a tyre or a bungee jumper’s rope is extended during a fall. Since each object possesses energy due to its position or motion, these all suit our earlier definition of mechanical energy. We will see that when some materials are manipulated we can think of this as work being done to store energy. This energy is often released or utilised via work being done on another object.

Materials that have the ability to store elastic potential energy when work is done on them and then release this energy are called elastic materials. Metal springs are common examples, but also realise that many materials are at least partially elastic. If their shape is manipulated, items such as our skin, metal hair clips and wooden rulers all have the ability to restore themselves to their original shape once released—within limits of course! Materials that do not return mechanical energy when their shape is distorted are referred to as plastic materials. Plasticine is an example of a very plastic material.

Ideal springs obey Hooke’s law Springs are very useful items in our everyday life due to the consistent way in which many of them respond to forces and store energy. When a spring is stretched or compressed by an applied force we say that elastic potential energy is being stored. In order to store this energy work must be done on the spring. Recall that in section 6.3 we have discussed that if a force of a constant value is applied to an object (and a displacement occurs in the direction of that force) then the quantity of work done can be calculated using W = Fx. This formula can therefore be used when a set force, F, has been applied to a spring and a given compression or extension, Δx, occurs.

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However, we are usually interested in examining how a spring will behave in a range of conditions.

Consider the situation in which a spring is stretched by the application of a steadily increasing force. As the force increases, the extension of the spring, Δx, can be graphed against the applied force, F. You can imagine hanging a spring vertically and gradually adding more and more weight to it so that it stretches. Many items, such as well-designed springs, will (at least for a small load) extend in proportion to the applied force. For example, if a 10 newton force produced an extension of 6 cm, then a 20 newton force would produce an extension of 12 cm. These items are called ideal springs. The resulting graph of applied force versus extension would be linear as in Figure 6.22.

Note that the gradient of this graph tells us the force, in newton, required to produce each unit of extension. The gradient of the graph is called the spring constant, k, measured in N m−1. The gradient therefore indicates the stiffness of the spring, and for an ideal spring this gradient has a set value (i.e. the F vs Δx graph is a straight line). A very stiff spring that is difficult to stretch would have a steep gradient; that is, a large value of k. Although k is usually called the spring constant, it is sometimes called the stiffness constant or force constant of a spring. A spring constant of k = 1500 N m−1 indicates that for every metre that the spring is stretched or compressed, a force of 1500 N is required. This does not necessarily mean that the spring can be stretched by 1 m, but it tells us that the force and the change in length are in this proportion.

The relationship between the applied force and the subsequent extension or compression of an ideal spring is known as Hooke’s law. Since for ideal springs F ∝ Δx, we can say F = kΔx. However, as we are often interested in using the energy stored by stretched or compressed springs we tend to refer to the force that the distorted spring is able to exert (rather than the force that was applied to it). Newton’s third law tells us that an extended or compressed spring in equilibrium is able to exert a restorative force equal in size but opposite in direction to the force that is being applied to it. Therefore Hooke’s law is often written in the form shown below.

Calculating elastic potential energy

Work must be done in order to store elastic potential energy in any elastic material. Essentially the energy is stored within the atomic bonds of the material as it is compressed or stretched. The amount of elastic potential energy stored is given by the area under the force–extension graph for the item.

For materials that obey Hooke’s law (such as the material shown in Figure 6.22), an expression can be derived for the area under the F–x graph.

HOOK…’S LAW states that the force applied by a spring is directly proportional, but opposite in direction, to the spring’s extension or compression. That is:

F = −kΔxwhere F is the force applied by the ideal spring (N) k is the spring constant (N m−1) (also called force constant or stiffness constant) Δx is the amount of extension or compression of the ideal spring (m)

i

Prac 26

Figure 6.22 Ideal materials obey Hooke’s law: F ∝ kΔx.

Ap

plie

d fo

rce

(N)

Extension (m)


Work done = area under F–x graph = area of a triangle = 1

2 × base × height = 1

2 × Δx × F But since F = kΔx: Work done = 1

2 × Δx × kΔx W = 1

2 kΔx2

= the elastic potential energy stored during the extension/compression

Although many materials (at least for a small load) extend in proportion to the applied force, many materials have force–extension graphs more like that shown in Figure 6.23. For these materials the area under the F–Δx graph must be used to determine the elastic potential energy stored.

Worked example 6.4D Three different springs A, B and C are exposed to a range of forces and the subsequent extension measured. The data collected for each spring has been graphed in the F–Δx graph at right. a Justify the statement that spring B is the only ideal spring shown.b Calculate the stiffness constant of spring B.c Calculate the work done in extending spring B by 25 mm. Assume the process of storing

energy is 100% efficient.d Estimate the work done in extending spring C by 25 mm.e If all springs are extended such that Δx = 40 mm, which spring will have stored the most

elastic potential energy? Justify your choice.

Solution a Ideal springs produce an extension that is consistently proportional to the applied

force. Since spring B has an F–Δx graph which is a straight line emerging from the origin, it is behaving ideally and obeying Hooke’s law until an extension of ∼30 mm is reached.

Spring C does not obey Hooke’s law, that is, F is not directly proportional to Δx, since the graph is not a straight line.

On close inspection it can be seen that the initial application of a small force did not produce any extension in spring A. (This is a common behaviour of real springs where a certain minimum amount of force must be applied before any extension will occur). This means that spring A has not obeyed Hooke’s law and therefore is not an ideal spring.

The elastic (or spring) potential energy, Us, stored in an item is always given

by the area below the force–extension graph for the item. The unit of Us is the

joule, J.In the case of an ideal material that obeys Hooke’s law, the elastic potential energy is given by the expression:

Us = 1

2kΔx2

where Us is the elastic potential energy stored during compression or

extension (J) k is the stiffness constant of the material (N m−1) Δx is the extension or compression (m)

i

Figure 6.23 Elastic potential energy is a form of mechanical energy. Work is done as elastic potential energy is stored, as indicated by the area under the F–Δx graph.

Extension (m)

area under graph = work done

Forc

e ap

plie

d (N

)

AB

C

10 20 30 40 50 60

20

40

60

80

100

120

F ap

plie

d (N

)

x (mm)

Take care! The two forms of potential energy, elastic (or spring) potential energy, U

s, and gravitational potential

energy, Ug, have been introduced. Take

extra care when analysing situations like pole-vaulting, since at some stages in this event both forms of potential energy are present at the same time!

Physics file

Motion206

b k = gradient of F–Δx graph = rise/run = 90/0.030 = 3.0 × 103 N m−1 c Since spring B obeys Hooke’s law, the equation U

s = 1

2kΔx2 can be applied.

W = ΔUs = U

s[final] − U

s[initial]

As there is initially zero energy stored: W = 1

2kΔx2

= 1

2 × 3.0 × 103 × 0.0252

= 0.94 Jd Spring C does not obey Hooke’s law, so the work done must be calculated using the area

under the F–Δx graph: 1 square of area is equal to (0.005 × 10) or 0.05 joules. There are approximately 7.5 squares of area. Therefore, W ≈ 7.5 × 0.05 ≈ 0.38 Je Elastic potential energy is given by the area under the F–Δx graph. At an extension of

40 mm spring A will have the greatest area under the graph, i.e. it will have stored the most elastic potential energy.

By the mid-19th century, several scientists had begun to write of the heating process as an energy change from work (mechanical energy) to heat. It was eventually realised that all forms of energy were equivalent and that when a particular form of energy seemed to disappear, the process was always associated with the appearance of the same amount of energy in other forms. This led to the development of the principle of conservation of energy. At this same time, James Joule conducted a series of experiments fundamental to our present understanding of heat.

Joule noticed that stirring water could cause a rise in temperature. He designed a way of measuring the relationship between the energy used in stirring the water and the change in temperature. A metal paddle wheel was rotated by falling masses and this churned water around in an insulated can. The amount of work done was calculated by multiplying the weight of the falling masses by the distance they fell. The heat generated was calculated from the mass of the water and the temperature rise. Joule found that exactly the same quantity of heat was always produced by exactly the same amount of work. Heat was simply another form of energy, and 4.18 joules of work was equivalent to 1 calorie of heat.

Joule’s work led to some unusual conclusions for his day. He stated that as a container of cold water is stirred, the mechanical energy is being transformed into thermal energy, heating the water. Theoretically, this means that a cup of water stirred long enough and fast enough will boil—a novel, if laborious, way of making a cup of coffee. Of course, the rate at which we can normally add energy by stirring is less than the transfer of energy to the surrounding environment. For the cup of water to boil it would need to be very well insulated.

As a result of Joule’s investigations and other experiments of the time, we now interpret the process of heating or cooling as a transfer of energy. When heat ‘flows’ from a hot object to a cold one, energy is being transferred from the hot to the cold.

James JoulePhysics in action

Figure 6.24 James Prescott Joule.


Figure 6.25 Joule’s original apparatus for investigating the mechanical work equivalent of heat energy. The falling weights caused the paddle to turn. The friction between the wheel and the water created heat energy in the water. For the first time, heat energy could be measured and related to other forms of energy.

falling masses

water

paddle

pulley

Figure 6.26 As a result of the considerable amount of mechanical work being done on the water of a waterfall, the temperature of the water at the bottom of the falls is usually 1°C or 2°C higher than at the top.

• Energy is the ability to do work. Whenever work is done, energy is transformed from one form to another.

• Kinetic energy is the energy a body has because of its motion. Ek = 1

2mv2 where Ek is the kinetic energy in

joules (J), m is the mass in kilograms (kg) and v is the speed in metres per second (m s−1).

• Potential energy is stored energy with the potential to allow work to be done. It may take many forms including chemical, elastic and gravitational.

• Gravitational potential energy is the energy a body has because of its position within a gravitational field: ΔUg = mgΔh where Ug is the gravitational potential energy in joules (J), m is the mass in

kilograms (kg) and Δh is the change in height from a reference height in metres (m).

• Ideal materials extend or compress in proportion to the applied force; that is, they obey Hooke’s law:

F = −kΔx• The elastic potential energy, Us, stored in an item is

given by the area below the force–extension graph for that item, or Us = 1

2kx2 for an ideal spring that

obeys Hooke’s law.• When work is done to store energy, one or more of

the following may be applied: W = ΔUg or W = ΔEk or W = ΔUs

or W = area under F–Δx graph

Mechanical energy6.4 summary

Motion208

Where appropriate use g = 9.8 m s−2.

1 1 Calculate the kinetic energy of a:a 1.0 kg mechanics trolley with a velocity of 2.5 m s−1

b 5.0 g bullet travelling with a velocity of 400 m s−1

c 1200 kg car travelling at 75 km h−1.

2 2 Calculate the gravitational potential energy relative to the ground when a:a mass of 1.0 kg is 5 m above the groundb bird of mass 105 g is 400 m above the groundc 1200 kg car has travelled a vertical height of 10 m

up a slope.

3 3 A 100 g rubber ball falls from a height of 2.5 m onto the ground and rebounds to a height of 1.8 m. What is the gravitational potential energy of the ball relative to the ground at its:a original position?b final position?c final position relative to its original position?

4 4 Which object has the greatest amount of energy?A a spring with a spring constant k = 40 000 N m−1

compressed by 5.0 cmB a cricket ball of mass 150 g stuck on the roof of a

grandstand 14 m above the groundC a cricket ball of mass 150 g travelling at 10 m s−1 at

a height of 10 m above the ground

5 5 What net braking force must be applied to stop a car within a straight-line distance of 50 m, if the car has a mass of 900 kg and was initially travelling at a velocity of 100 km h−1?

6 6 A small steel ball with a mass of 80 g is released from a resting height of 1.25 m above a rigid metal plate. Calculate the:a change in gravitational potential energyb kinetic energy of the ball just before impactc velocity of the ball just before impact.

The following information relates to questions 7 and 8. The force–extension graphs for three different springs are shown below.

A

B

C

0.05 0.10 0.15 0.25

100

200

300

F ap

plie

d (N

)

x (m)

7 Calculate the spring constant for each spring and determine which is the stiffest spring.

8 8 Each spring has a force of 100 N applied to it. Calculate the elastic potential energy stored by each spring.

9 9 A piece of gymnasium equipment is operated by compressing a spring that has a spring constant of 2500 N m−1. How much elastic potential energy is stored in the spring when it is compressed by:a 5.00 cm?b 10.0 cm?c 15.0 cm?

10 10 The gymnasium equipment described in Question 9 is adjusted so that its spring constant is now 3000 N m−1. If 12.5 J of energy is now stored in the spring, by what distance has the spring been compressed?

Mechanical energy6.4 questions

Worked Solutions


Energy transformation and power6.5

Besides the mechanical energy discussed in section 6.4, other forms of energy exist, for example nuclear, heat, electrical, chemical and sound energy. Atomic theory has led to each of these other forms being understood as either kinetic energy or potential energy at the molecular level. Energy stored in food or fuel and oxygen can be considered as potential energy stored as a result of the electrical forces in the molecules.

Despite the apparently different nature of the various forms of energy, any energy can be transformed from one form to another. The connecting factor is that all forms can do work on a body and therefore can be measured and compared in this way. A stone dropped from some height loses gravitational potential energy as its height decreases; at the same time its kinetic energy will increase as its speed increases.

Transformation of energyEnergy transfers or transformations enable people and machines to do work, and processes and changes to occur. Elastic potential energy stored in a diving board must be transformed into the kinetic energy of the diver at the pool. Contracting a muscle converts chemical potential energy stored in the muscle to the kinetic energy of a person’s motion. In each example, the transformation of energy means work is being done.

In many cases a transformation of energy produces an unwanted consequence—a substantial amount of the energy is ‘lost’ as heat energy. Of a typical adult’s daily food intake of about 12 MJ at least 80% is converted into heat energy during normal activity. Such transfers can be depicted by an energy-conversion flow diagram.

h

gravitationalpotential energy

kinetic energy

sound heat

heat

workdone

workdone

workdone

workdone

(b)(a)

Figure 6.27 Whenever work is done, energy is transformed from one form to another. (a) As a body falls, gravitational potential energy is transformed to kinetic energy and heat, from the friction with the air. Once the body lands, further energy transformations will take place. (b) An energy-conversion flow diagram can be useful in visualising the transformations that take place.

A simple, although infinitely unlikely, example is shown in Figure 6.27. As the body falls to the ground there will be a number of energy transformations. An energy flow diagram illustrates these changes.

Work is done whenever energy is transformed from one form to another.i

Interactive

Motion210

While the body falls, work will be done on the body by the gravitational field, and gravitational potential energy becomes kinetic energy, the energy of movement. There will also be some energy converted into heat by the action of air resistance. When the body hits the ground, the kinetic energy is converted into elastic potential energy by the compression of the body, and to other forms, particularly heat but also to sound and kinetic energy. Each transformation requires a force to do work on the body.

The efficiency of energy transformationsThe percentage of energy that is transformed to a useful form by a device is known as the efficiency of that device. All practical energy transformations ‘lose’ some energy as heat. The effectiveness of a transfer from one energy form to another is expressed as:

efficiency (%) = useful energy transferred × 100

total energy supplied =

useful output × 100total input

Table 6.4 Efficiencies of some common energy transfers

Device Desired energy transfer Efficiency (%)

Large electric motor Electric to kinetic 90

Gas heater or boiler Chemical to heat in water 75

Steam turbine Heat to kinetic 45

High-efficiency solar cell Radiation to electric 25

Coal-fired electric generator Chemical to electric 30

Compact low-energy fluorescent light

Electric to light 25

Human body Chemical to kinetic 25

Car engine Chemical to kinetic 25

Open fireplace Chemical to heat 15

Filament lamp Electric to light 5

In all the energy transformations included in Table 6.4, the energy lost in the transfer process is mainly converted into heat. Most losses are caused by the inefficiencies involved in the process of converting heat into motion. In the real world, energy must be constantly provided for a device to continue operating. A device operating at 45% efficiency is converting 45% of the supplied energy into the new form required. The other 55% is lost to the surroundings, mainly as heat but also some as sound.

Figure 6.28 In each of these situations below an energy transformation is taking place. Can you identify the forms involved in each transformation?


Conservation of energyNo matter what energy transformation occurs overall, no energy is gained or lost in the process. It is a fundamental law of nature that energy is conserved.

Consider the example of a diver as depicted in Figure 6.31. When the diver is at the top, his or her gravitational potential energy will be at a maximum. As the diver free falls the gravitational potential energy decreases but the kinetic energy will increase with the increased velocity. Some of the energy—a small amount—will be converted into heat due to contact with the air. The moment before the diver reaches the reference level all the gravitational potential energy has been converted into other energy forms, mostly kinetic energy. The total at this point will be exactly

Air, like water, is a fluid and therefore there is a force between the particles of the air and the surface of any object moving through it. This force is called air resistance or drag. Drag is due to the air particles that can be thought of as obstacles in the path of a moving object. There will also be frictional forces as air particles slide past the object.

At low speeds the effect of air resistance is slight. However, it has been found that air resistance is proportional to the square of the velocity (i.e. F

a ∝ v2). A doubling of speed

will increase air resistance approximately four times. At the racing speeds that Olympic cyclists reach of 50 km h−1 or more, 90% of the cyclist’s energy is required just to push the bicycle and rider through the surrounding air. The remaining 10% is needed to overcome frictional forces between the wheels and the ground.

Air resistance is affected by the frontal area of cross-section and the shape of the bicycle. Designers have tried to reduce the frontal area of racing bikes and their riders by dropping the handle bars and raising the position of the pedals. This allows the rider to race bent forward, reducing the area presented to the air.

Streamlining the bike helps still further. Some shapes move through fluids more easily than others. Streamlined bicycles, such as those used by the Australian Cycling Team in international competition, have low-profile frames with relatively smaller front wheels. Brake and gear cables are run through the frame rather than left loose to create drag. Moulded three-spoke and solid-disk wheels help still further.

The riders’ clothing and helmets have also been streamlined. Cyclists wear pointed shoes, streamlined helmets and skin-tight one-piece lycra bodysuits. Together these reduce the air resistance on a rider by as much as 10% at higher speeds. At race time riders shave their legs to reduce energy losses just that little bit more.

Air resistance in sportPhysics in action

Figure 6.29 The efficiency of a cyclist is affected by the velocity of the bike. As the velocity increases, the drag or air resistance can use as much as 90% of the energy the cyclist’s input.

0 5 9.0 13.5

20

10

Velocity (m s−1)

Dra

g (N

)

Figure 6.30 Technological advances in bike design, pioneered by RMIT University, have led to efficient designs such as that used by Australian riders in recent international competition.

Motion212

equal to the potential energy at the top. The total energy in the system remains the same. Energy has been conserved.

This applies to any situation involving energy transfer or transformations in an isolated system. In this particular case the sum of the gravitational potential energy and the kinetic energy at any point is called the total mechanical energy.

Here, the total mechanical energy remains constant. As an object falls, gravitational potential energy decreases but kinetic energy increases to compensate, so that the total remains constant. At any point during an object’s free fall:

total mechanical energy = 1

2mv2 + mgh

There are many examples of this conservation of energy. In athletics, the pole-vaulters and high-jumpers base their techniques on this principle. Throwing a ball in the air is another example. When the ball leaves the hand, its kinetic energy is at a maximum. As it rises, its velocity decreases, reducing the kinetic energy, and its potential energy increases by the same amount. At any point Ek + Ep will equal the initial kinetic energy. At the top of the throw, the ball will have a vertical velocity of zero and, in a vertical direction, the energy will be totally gravitational potential energy (any horizontal motion will be represented by a remaining amount of Ek). The transformation will reverse as the ball falls. Gravitational potential energy will decrease as the ball returns towards its original height and, with its speed increasing, the kinetic energy will increase once more.

A more complex example is provided by the interactions as a gymnast repeatedly bounces on a trampoline. Figure 6.32 is a series of frames from a video of a gymnast carrying out a routine on a trampoline. Kinetic energy and gravitational potential energy changes are shown in the graph below the frames. Despite the complexity of the motion, the total energy of the gymnast remains the same during each airborne phase, as illustrated in the graph. On landing on the bed of the trampoline, the energy is transferred to elastic potential energy within the trampoline and both kinetic and gravitational potential energy fall. On take-off some of this energy will be permanently transferred to the trampoline and its surrounds, thus lowering the total available to the gymnast. This is represented by the reduced total energy for each successive jump. Were the gymnast to flex his legs then additional energy would be added to the mechanical energy available and this total could be maintained or even increased until the gymnast finally ran out of available energy himself.

The TOTAL …N…RGY in an isolated system is neither increased nor decreased by any transformation. Energy can be transformed from one kind to another, but the total amount stays the same.

i

Kinetic energy + potential energy = total mechanical energyi

Figure 6.31 A diver loses gravitational potential energy but gains kinetic energy during the fall.

All Ep

12 —Ep

12 —Ek

All Ek

Prac 27 SPARKlab


Worked example 6.5A Calculate the initial velocity required for a high jumper to pass over a high bar. Assume the jumper’s centre of gravity rises through a height of 1.5 m and passes over the bar with a horizontal velocity of 1.2 m s−1, and that all of his initial horizontal kinetic energy is transferred into U

g and …

k.

Use g = 9.8 m s−2.

Solution As the total mechanical energy is assumed to be conserved after landing, the initial horizontal kinetic energy equals total mechanical energy at the peak height:1

2mu2 = …

k + U

g (at peak height)

= 1

2mv2 + mgΔh

The mass cancels out, giving an expression independent of the mass of the athlete. The same speed at take-off will be required for a light person as a heavy one. (If this doesn’t seem to make sense remember that all objects fall at the same rate regardless of their mass.)1

2u2 = 1

2v2 + gΔh

Substituting the values from the question:1

2u2 = 1

2 × 1.22 + 9.8 × 1.5

1

2u2 = 15.42

and u = √2 × 15.42 = 5.6 m s−1.In reality the take-off speed will need to be a little greater since there will be some losses to friction.

Figure 6.32 During each airborne stage of a gymnast’s trampoline routine (indicated on the graph by shading), mechanical energy is conserved. The graph shows the relationship between total energy and gravitational potential energy and kinetic energy. Each time the gymnast lands, energy is transferred to the trampoline. The energy returning from the springs after each landing allows the routine to continue.

Ene

rgy

(J)

20 40 60 80 100 120 140 160 180 200 220Number of frames

on bed on bedflight flight

gravitationalpotential energyplus kinetic energy

gravitationalpotential energy

kinetic energy

2043

64

81102

122125

147

159174

185194

207218

flight

1000

2000

3000

4000

Motion214

Worked example 6.5B A climber abseiling down a cliff uses friction between the climbing rope and specialised metal fittings to slow down. If a climber of mass 75 kg abseiling down a cliff of height 45 m reaches a velocity of 3.2 m s−1 by the time the ground is reached, calculate the average frictional force applied. Use g = 9.8 m s−2.

Solution m = 75 kg, u = 0 m s−1, v = 3.2 m s−1, h = 45 mGravitational potential energy at the top of the cliff:…

p = mgh = 75 × 9.8 × 45 = 33 075 J

Kinetic energy at ground level:…

k = 1

2mv2 = 1

2 × 75 × 3.22 = 384 J

Total energy transformed to forms other than gravitational potential and kinetic energy:Δ… = …

p − …

k = 33 075 − 384 = 32 691 J

This change in energy will be equivalent to the work done by the frictional force; that is:Work = Δ… = F

f × x

and:F

f =

Δ…x

= 32 691

45 = 726 N ≈ 730 N

PowerWhy is it that running up a flight of stairs can leave you more tired than walking up if both require the same amount of energy to overcome the force of gravity?

The answer lies in the rate at which the energy is used. When horses were first replaced by steam engines, the engine was rated by how fast it could perform a given task compared with a horse. An engine that could complete a task in the same time as one horse was given a rating of one horsepower. More formally, power is defined as the rate at which energy is transformed or the rate at which work is done.

Determining the power developed is fairly straightforward when mechanical work is done; but consider a situation in which a person pushes a lawnmower, say, at constant speed. Here, there is no increase in kinetic energy, but energy is being transformed to overcome the frictional forces acting against the lawnmower.

POW…R = work donetime taken =

energy transformedtime taken

or

P = WΔt =

Δ…Δt

where P is the power developed in watts (W) resulting from an energy transformation Δ… occurring in time Δt. Δ… is measured in joules (J), time is measured in seconds (s).

i

The British Imperial unit for power is the horsepower, hp, dating from the time of the Industrial Revolution when the performance of steam engines was compared with that of the horses they were replacing. 1 hp = 746 W. The SI unit for power honours the inventor of the steam engine, James Watt.

Physics file


This is useful when finding the power required to produce a constant speed against a frictional or gravitational force.

The rate of energy use is as much a limiting factor of the work a person can do as the total energy required. A person may be able to walk or climb a long distance before having to stop because all available energy is used. The same person will fall over exhausted after a much shorter time if the same journey is attempted at a run. Power is the limiting factor, the rate at which a person’s body can transform chemical energy into mechanical energy. Few humans can maintain one horsepower, about 750 W, for any length of time. Table 6.5 includes comparative figures for the power developed in various activities and devices.

Table 6.5 Average power ratings for various human activities and machines

Activity or machine Power rating (W)

Sleeping adult 100

Walking adult 300

Cycling (not racing) 500

Standard light globe 60

Television 200

Fast-boil kettle 2400

Family car 150 000

Worked example 6.5C The fastest woman to scale the Rialto building stairs in the Great Rialto Stair Trek in a particular year climbed the 1222 steps, which are a total of 242 m high, in 7 min 58 s. Given that her mass was 60 kg, at what rate was she using energy to overcome the gravitational force alone? Use g = 9.8 m s−2.

Solution The work is against gravity so:

P = Ug

Δt =

mgΔhΔt

m = 60 kg, g = 9.8 m s−2, Δh = 242 m Δt = (7 min × 60) + 58 s = 478 s

P = 60 × 9.8 × 242

478 = 3.0 × 102 W

In the special case of an applied force opposing friction or gravity and doing work with no increase in the speed of the object, we can say:As W = Fx then:

P = FxΔt

and as v = x

Δt then:

P = Fav

vav

where P is power developed (W) F

av is average applied force (N)

vav

is average speed (m s−1)

i

Figure 6.33 It is not the amount of energy required that stops the rest of us from winning the 400 m sprint, but the rate at which we can effectively convert it to useful work.

Motion216

• Whenever work is done, energy is converted from one form into another.

• The efficiency of an energy transfer from one form to the required form is:

efficiency (%) = energy outputenergy input

× 100

• Whenever energy is transformed, the total energy in the system remains constant. This con serv ation of energy is a fundamental natural principle.

• The total mechanical energy will remain constant in an isolated system. That is:

Ek + Us + Ug = constant• Power, P (in watts, W), is the rate at which work is

done or energy transformed:

P = WΔt

= ΔEΔt

• In the particular case of work being done to overcome friction, with no resulting increase in speed:

P = Favvav

Energy transformation and power6.5 summary


1 1 Describe the energy transformations that take place when:a a car slows to restb a gymnast uses a springboard to propel themselves

into the airc an archer draws back and then releases an arrow

vertically upwardd an athlete’s foot hits a track.

2 2 Draw an energy transformation flow chart for a swimmer diving off a diving board and into a pool of water.

3 3 A boy of mass 46 kg runs up a 12 m high flight of stairs in 12 s.a What is the gain in gravitational potential energy

for the boy?b What is the average power he develops?

4 4 A coach is stacking shot-puts, from the shot-put event, onto a shelf 1.0 m high following an athletics meeting. Each shot-put has a mass of 500 g and all are being lifted from the ground. The coach stacks 15 shot-puts, at the same level, in 2.0 minutes.a How much useful work has been done in lifting

all the shot-puts?b What is the total gravitational potential energy of

all the shot-puts on the shelf?c What was the coach’s average power output in

performing this task?d The actual power output would be considerably

greater than the answer to part c. Suggest two possible reasons for this difference.

5 5 One of the shot-puts in Question 4 rolls off the shelf just after the coach has finished.a What is the gravitational potential energy of the

shot-put when it is halfway to the ground?b What is the kinetic energy of the shot-put when it

is halfway to the ground?c What happens to the kinetic energy of the shot-

put when it hits the ground?

6 6 Tarzan is running at his fastest speed (9.2 m s−1) and grabs a vine hanging vertically from a tall tree in the jungle.a How high will he swing upwards while hanging

on to the end of the vine?b What other factors that have not been considered

may affect your answer?

7 7 In high jumping, the kinetic energy of an athlete is transformed into gravitational potential energy. With what minimum speed must the athlete leave the ground in order to lift his centre of gravity 1.80 m high with a remaining horizontal velocity of 0.50 m s−1?

8 8 A 100 g apple falls from a branch 5 m above the ground.a With what speed would it hit the ground if air

resistance could be ignored?b If the apple actually hits the ground with a speed

of 3.0 m s−1, what was the average force of air resistance exerted on it?

9 9 A 150 g ball is rolled onto the end of an ideal spring whose spring constant is 1000 N m−1. The spring is temporarily compressed.

Energy transformation and power6.5 questions


a The ball compresses the spring by a maximum distance of 10 cm. How much elastic potential energy is stored in the spring at this compression?

b How fast must the ball have been travelling just before it began to compress the spring? Ignore any frictional effects.

c If in another trial the ball reached a speed of 5.0 m s−1 before compressing the spring, how far would the spring be compressed?

10 10 As a 30 kg child compressed the spring of a pogo stick, it stored 150 J of elastic potential energy. Assuming the spring is 50% efficient:a how much kinetic energy will the child be given

as the spring rebounds?b with what speed will the child rebound?c ignoring air resistance, what gain in height will

the child achieve?


The following information relates to questions 1–4. A ball of mass 50 g strikes a brick wall. It compresses a maximum distance of 2.0 cm. The force extension properties of the ball are shown below.

800700600500400300200100

Forc

e (N

)

0.01 0.02Compression (m)

1 What work does the wall do on the ball in bringing it to a stop?

2 How much …p is stored in the ball at its point of maximum

compression?

3 If the ball–wall system is 50% efficient, what is the rebound speed of the ball?

4 At the instant that the ball had only been compressed by 1.0 cm, had the wall done half of the work required to stop the ball? Explain.

The following information relates to questions 5 and 6. An arrow with a mass of 80 g is travelling at 80 m s−1 when it reaches its target. It penetrates the target board a distance of 24 cm before stopping.

5 Calculate the arrow’s kinetic energy just before impact.

6 Calculate the average net force between arrow and target.

7 A 70 kg bungee jumper jumps from a platform that is 35 m above the ground. Assume that the person, the rope and the Earth form an isolated system.

a Calculate the initial total mechanical energy of this system.b Write a flow chart displaying the energy transformations

that are occurring during the first fall.

c The person can fall a distance of 10 m before the rope attached to her feet begins to extend. How much kinetic energy will the person have at this moment?

d After a fall of a further 15 m, the person momentarily stops and the rope reaches its maximum extension. How much elastic potential energy is stored in the rope at this moment?

e Since the bungee jumper bounces and eventually comes to rest, this is not a truly isolated system. Explain.

8 A stone of mass 3 kg is dropped from a height of 5 m. Neglecting air resistance, what will the kinetic energy of the stone be in joules just before the stone hits the ground?

A 3B 5C 15D 147E 150

The following information relates to questions 9 and 10. An object of mass 2 kg is fired vertically upwards with an initial kinetic energy of 100 J. Assume no air resistance.

9 What is the speed of the object in m s−1 when it first leaves the ground?

A 5B 10C 20D 100E 200

10 Which of A–E in Question 9 is the maximum height in metres that the object will reach?

Chapter reviewMomentum, energy, work and power


Worked Solutions

Motion218

The following information relates to questions 11–14.

Casey and Mandeep go bobsledding in their holidays. Figure A shows the bobsled stationary, with the handbrake engaged. Figure B shows the bobsled moving down the slope at a constant velocity. The total mass of the bobsled and occupants is 350 kg.

CaseyMandeep

CaseyMandeep

30° 30°

A B

11 For the situation shown in Figure A, what is the net force acting on the sled?

12 For the situation shown in Figure A, calculate the magnitude of the support force of the slope on the bobsled.

13 For the situation shown in Figure A, calculate the magnitude of the resistive force that is preventing the bobsled from sliding down the slope.

14 The two students attempt to explain the situation shown in Figure B.

• Casey says that as the sled is moving down the slope there must be more force in that direction. She says that friction can’t be as big as the ‘pull down the slope’.

• Mandeep says that friction is the same size as the ‘pull down the slope’ but the sled can still keep moving.

Which student is correct? Explain why you think that student is correct, making reference to any of Newton’s laws of motion.

The following information relates to questions 15 and 16.

Using one smooth lifting action, a weightlifter lifts a 240 kg barbell from the ground to above his head. The lowest point of the barbell is lifted to 1.70 m above the floor and held stationary for 3.00 seconds before being dropped back to the ground.

15 How much gravitational potential energy did the barbell possess while being held above the weightlifter’s head?

16 The complete lifting action was carried out in 0.40 seconds. What power was developed during the lift?

The following information relates to questions 17–20. A roller-coaster is shown in the following diagram. Assume no friction.

30 m

A

B

C

D25 m

12 m

17 Calculate the speed at points B, C and D, assuming an initial speed of 4.0 m s−1 at point A.

18 Draw a graph of potential energy and kinetic energy against vertical displacement for this motion. Use separate lines for each form of energy and draw in a third line to represent the total mechanical energy, assuming no frictional losses.

It is found that the roller-coaster actually just reaches point C with no remaining speed.

19 What are the energy losses due to friction and air resistance between A and C?

20 With what efficiency is the roller-coaster operating over this section of track?

21 Two players collide during a game of netball. Just before impact one player of mass 55 kg was running at 5.0 m s−1 while the other player, of mass 70 kg, was stationary. After the collision they fall over together. What is the velocity as they fall, assuming that momentum is conserved?

22 A 300 kg marshalling boat for a rowing event is floating at 2.0 m s−1 north. A starting cannon is fired from its bow, launching a 500 g ball, travelling at 100 m s−1 south as it leaves the gun. What is the final velocity of the marshalling boat?

23 A 150 g ice puck collides head on with a smaller 100 g ice puck, initially stationary, on a smooth, frictionless surface. The initial speed of the 150 g puck is 3 m s−1. After the collision the 100 g ice puck moves with a speed of 1.2 m s−1 in the same direction. What is the final velocity of the 150 g ice puck?

24 ‘When I jump, the Earth moves’. Is this true? Using reasonable estimates and appropriate physics relationships explain your answer.

The following information relates to questions 25–27. In a horrific car crash, a car skids 85 m before striking a parked car in the rear with a velocity of 15 m s−1. The cars become locked together and skid a further 5.2 m before finally coming to rest. The mass of the first car, including its occupants, is 1350 kg. The parked car has a mass of 1520 kg.

25 What is the velocity of the two cars just after impact?

26 What is the impulse on each car during the collision?

27 What is the average size of the frictional force between road and car that finally brings them to rest?

Momentum, energy, work and power (continued)

Chapter QuizWorked Solutions

219Area of study review

The following information applies to questions 1–3. The acceleration due to gravity may be taken as g = 9.8 m s−2 and the effects of air resistance can be ignored. An Olympic archery competitor tests a bow by firing an arrow of mass 25 g vertically into the air. The arrow leaves the bow with an initial vertical velocity of 100 m s−1.

1 1 At what time will the arrow reach its maximum height?

2 2 What is the maximum vertical distance that this arrow reaches?

3 3 What is the acceleration of the arrow when it reaches its maximum height?

4 4 Two students drop a lead weight from a tower and time its fall at 2.0 s. How far does the weight travel during the 2nd second, compared with the first second?

The following information applies to questions 5–7. A car with good brakes, but smooth tyres, has a maximum retardation of 4.0 m s−2 on a wet road. The driver has a reaction time of 0.50 s. The driver is travelling at 72 km h−1 when she sees a danger and reacts by braking.

5 5 How far does the car travel during the reaction time?

6 6 Assuming maximum retardation, calculate the braking time.

7 7 Determine the total distance travelled by the car from the time the driver realises the danger to the time the car finally stops.

The following information applies to questions 8–11. Two physics students, Helen and Emily, conduct the following experiment from a skyscraper. Helen drops a platinum sphere from a vertical height of 122 m while at exactly the same time Emily throws a lead sphere with an initial downward vertical velocity of 10.0 m s−1 from a vertical height of 140 m. Assume g = 9.80 m s−2 and ignore friction.

8 8 Determine the time taken by the platinum sphere to strike the ground.

9 9 Calculate the time taken by the lead sphere to strike the ground.

10 10 Determine the average velocity of each sphere over their respective distances.

11 11 In reality, will the diameters of the respective spheres affect the outcome of the experiment?

The following information applies to questions 12–14. During a physics experiment a student sets a multi-flash timer at a frequency of 10 Hz. A nickel marble is rolled across a horizontal table. The diagram shows the position of the marble for the first four flashes: A, B, C and D.

Assume that when flash A occurred t = 0, at which time the marble was at rest.

1.0 cm 3.0 cm

A B C D

5.0 cm

12 12 Determine the average speed of the marble for the following distance intervals:

a A to Bb b B to Cc c C to D

13 13 Determine the instantaneous speeds of the marble for the following times:

a t = 0.05 sb t = 0.15 sc c t = 0.25 s

1414 Describe the motion of the marble.

The following information applies to questions 15–18. A tow-truck, pulling a car of mass 1000 kg along a straight road, causes its velocity to increase from 5.00 m s−1 west to 10.0 m s−1 west in a distance of 100 m. A constant frictional force of 200 N acts on the car.

15 15 Calculate the acceleration of the car.

16 16 What is the resultant force acting on the car during this 100 m?

17 17 Calculate the force exerted on the car by the tow-truck.

18 18 What force does the car exert on the tow-truck?

19 19 A car that is initially at rest begins to roll down a steep road that makes an angle of 11.3° with the horizontal. Ignoring friction, determine the speed of the car in km h−1 after it has travelled a distance of 100 m (g = 9.8 m s−2).

The following information applies to questions 20–23. A 100 kg trolley is being pushed up a rough 30° incline by a constant force F. The frictional force F

f between the incline and the trolley is 110 N.

F

g = 9.8 m s–2

30˚

20 20 Determine the value of F that will move the trolley up the incline at a constant velocity of 5.0 m s−1.

21 21 Determine the value of F that will accelerate the trolley up the incline at a value of 2.0 m s−2.

22 22 Calculate the acceleration of the trolley if F = 1000 N.

23 23 What is the value of F if the trolley accelerates up the incline at 10 m s−2?

AREA OF STUDY REVIEW Motion


Motion220

AREA OF STUDY REVIEW (continued)

24 24 Two masses, 10 kg and 20 kg, are attached via a steel cable to a frictionless pulley, as shown in the following diagram.

10 kg

20 kg

g = 9.8 m s–2a

a

FT

FT

a Determine the acceleration of each mass.b What is the magnitude of the tension in the cable?

25 25 An 800 N force is applied as shown to a 20.0 kg mass, initially at rest on a horizontal surface. During its subsequent motion the mass encounters a constant frictional force of 100 N while moving through a horizontal distance of 10 m.

20.0 kg

60°

F = 800 N

Ff = 100 N

a Determine the resultant horizontal force acting on the 20.0 kg mass.

b Calculate the work done by the horizontal component of the 800 N force.

c Calculate the work done by the frictional force. d Calculate the work done by the resultant horizontal force.e Determine the change in kinetic energy of the mass.f What is the final speed of the mass?g Describe the effect of the frictional force on the moving

mass. 26 26 The figure shows the velocity–time graph for a car of mass

2000 kg. The engine of the car is providing a constant driving force. During the 5.0 s interval the car encounters a constant frictional force of 400 N. At t = 5.0 s, v = 40.0 m s−1.

50403020100

v (m

s–1

)

t (s)0 2 4 6

a How much kinetic energy (in MJ) does the car have at t = 5.0 s?

b What is the resultant force acting on the car?c What force is provided by the car’s engine during the 5.0 s

interval?

d How much work is done on the car during the 5.0 s interval?e Determine the power output of the car’s engine during the

5.0 s interval.f How much heat energy is produced due to friction during the

5.0 s interval?The following information applies to questions 27–30. The following diagram shows the trajectory of a 2.0 kg shotput recorded by a physics student during a practical investigation. The sphere is projected at a vertical height of 2.0 m above the ground with initial speed v = 10 m s−1. The maximum vertical height of the shotput is 5.0 m. (Ignore friction and assume g = 9.8 N kg−1.)

2.0 m

3.0 m5.0 mA

B

C

27 27 What is the total energy of the shotput just after it is released at point A?

28 28 What is the kinetic energy of the shotput at point B?

29 29 What is the minimum speed of the shotput during its flight?

30 30 What is the total energy of the shotput at point C?

31 31 A 5.0 kg trolley approaches a spring that is fixed to a wall. During the collision, the spring undergoes a compression, Δx, and the trolley is momentarily brought to rest, before bouncing back at 10 m s−1. Following is the force–compression graph for the spring. (Ignore friction.)

vspring

5.0 kg

12.010.08.06.04.02.00.0

Forc

e (k

N)

Compression (cm)0.0 1.0 2.0 3.0 4.0 5.0 6.0

a Calculate the elastic potential energy stored in the spring when its compression is equal to 2.0 cm.

b What is the elastic potential energy stored in the spring when the trolley momentarily comes to rest?

c At what compression will the trolley come to rest?d Explain why the trolley starts moving again.

221Area of study review

e What property of a spring accounts for the situation described above?

f Describe a situation in which the property of the spring in this example could be used in a practical situation.

32 32 A nickel cube of mass 200 g is sliding across a horizontal surface. One section of the surface is frictionless while the other is rough. The graph shows the kinetic energy, …

k, of the cube versus

distance, x, along the surface.

5.04.03.02.01.00.0

0.0 1.0 2.0 3.0 4.0 5.0 6.0Kin

etic

ene

rgy

(J)

x (cm)

a Which section of the surface is rough? Justify your answer.b Determine the speed of the cube during the first 2.0 cm.c How much kinetic energy is lost by the cube between

x = 2.0 cm and x = 5.0 cm? d What has happened to the kinetic energy that has been lost

by the cube?e Calculate the value of the average frictional force acting on

the cube as it is travelling over the rough surface.The following information applies to questions 33 and 34. The diagram is an idealised velocity–time graph for the motion of an Olympic sprinter.

Time (s)1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Velocity(m s–1)

109876543210

33 33 What distance was this race?

34 34 Determine the average speed of the sprinter:

a while she is racing to the finish lineb for the total time that she is moving.

The following information applies to questions 35–37. The diagram gives the position–time graph of the motion of a boy on a bicycle. The boy initially travels in a northerly direction.

A B C D E F G

Position(m)

5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80Time (s)

80

70

60

50

40

30

20

10

0

35 35 During which of the section(s) (A–G) is the boy:

a travelling towards the north?b stationary?c travelling towards the south?d speeding up?e slowing down?

36 36 For the boy’s 80 s ride, calculate:

a the total distance coveredb the average speed.

37 37 Determine the velocity of the boy:

a when t = 10 sb during section Bc when t = 60 s.

The following information applies to questions 38–40. A mass of 0.40 kg hangs from a string 1.5 m long. The string is kept taut and the mass is drawn aside a vertical distance of 0.30 m, as shown in the diagram below. A pencil is fixed in a clamp so that when the mass is released it will swing down and break the pencil. The mass swings on but now only moves through a vertical distance of 0.14 m. (Assume g = 9.8 m s−2.)

1.50 m

0.30 m0.14 m


Motion222

AREA OF STUDY REVIEW (continued)

38 38 Calculate the velocity of the mass the instant before it strikes the pencil.

39 39 Calculate the work required to break the pencil.

40 40 Can you account for the loss in energy?

The following information applies to questions 41–44. A small car is found to slow down from 90 km h−1 to 60 km h−1 in 12 seconds when the engine is switched off and the car is allowed to coast on level ground. The car has a mass of 830 kg.

41 41 What is the car’s deceleration (in m s−2) during the 12 s interval?

42 42 What was the average braking force acting on the car during the time interval?

43 43 Determine the distance that the car travels during the 12 second interval.

44 44 Explain what happens to all the initial kinetic energy of the car.

45 45 A spaceship with a mass of 20 tonnes (2.0 × 104 kg) is launched from the surface of Earth, where g has a value of 9.8 N kg−1 downwards, to land on the Moon, where g is 1.6 N kg−1

downwards. What is the weight of the spaceship when it is on Earth and when it is on the Moon?

46 46 a Describe the motion of your chair when you stand up and push it back from your desk.

b How would the chair behave if it were on castors?c Explain how your answers to parts a and b are illustrations

of Newton’s laws. 47 47 An engine pulls a line of train cars along a flat track with a steady

force, but instead of accelerating, the whole train travels at a constant velocity. How can this be consistent with Newton’s first and second laws of motion?

48 48 Students were conducting an experiment to investigate the behaviour of springs. Increasing masses (m) were hung from a vertically suspended spring and the resulting force–extension graph was plotted as shown.

10.08.06.04.02.00.0

Forc

e (N

)

Extension (mm)0 20 40 60 80 100 120

a Estimate the value of the spring constant.b Use your answer to part a to calculate the elastic potential

energy stored in the spring when the extension is 100 cm.c What other method could you have used to estimate the

energy stored in the spring when the extension is 100 cm?

Jordy is playing softball and hits a ball with her softball bat. The force versus time graph for this interaction is shown below. The ball has a mass of 170 g. Assume that the bat and ball form an isolated system during the interaction.

4949 Determine the magnitude of the change in momentum of the ball.

5050 Determine the magnitude of the change in momentum of the bat.

5151 Determine the magnitude of the change in velocity of the ball.

500

Force (N)

0.015 0.040

Time (s)

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