4.1 WORK DONE BY A CONSTANT FORCE · energy. 4.2 KINETIC ENERGY AND THE WORK–ENERGY THEOREM...

Copyright © 2003 Nelson Chapter 4 Work and Energy 233

5.

6. The shock absorber would be a spring with a large force constant to absorb lots of energy quickly, combined with a slow

air release to prevent too much bouncing. The size would be suited to fit into the length of the forks. The forces involved would mean the choice of a strong material, such as steel, for strength.

Try This Activity: Which Ball Wins? (77) (a) The ball on track Y will win because it converts its gravitational energy into kinetic energy more quickly, and will speed

up more quickly than the ball on track X. (b) The total energy of the balls remains constant down the track. The energy is just transformed from one type to another. 4.1 WORK DONE BY A CONSTANT FORCE

PRACTICE (Pages 181–182)

Understanding Concepts 1. F1 will do more work than F2 because the component of F2 in the direction of motion is smaller than F1. 2. No. The force of kinetic friction is always acting opposite to the direction of motion. Since negative work is always

opposite the direction of motion, the kinetic friction will always do negative work. 3. Yes. The force of gravity can move an object toward itself, and therefore does positive work on that object. 4. m = 2.75 kg (a) ∆d = 1.37 m

W = ?

( cos )( cos )(2.75 kg)(9.80 N/kg)(cos 0 )(1.37 m)36.9 J

W F dmg d

W

θθ

= ∆= ∆= °=

The work done to move the plant 1.37 m up is 36.9 J. (b) ∆y = 1.07 m

µK = 0.549 W = ?

First we must calculate the normal force acting on the potted plant:

N g

N g

N

0

0

(2.75 kg)(9.80 N/kg)26.95 N

y yF ma

F F

F F

mg

F

Σ = =

− =

=

===


Let FA be the applied force to move the potted plant horizontally:

A K

A K

N

A

00

(0.549)(26.95 N)14.796 N

x xF maF F

F FF

F

µ

Σ = =− =

====

To calculate the work done on the potted plant:

A( cos )

(14.796 N)(cos 0 )(1.07 m)15.8 J

W F d

W

θ= ∆= °=

The work done to move the plant 1.07 m across the shelf is 15.8 J. 5. m = 24.5 kg

F = 14.2 N [22.5° below the horizontal] ∆d = 14.8 m W = ? We only need to consider the component of force in the direction of motion:

( cos )(14.2 N)(cos 22.5 )(14.8 m)194 J

W F d

W

θ= ∆= °=

The work done by the force is 194 J. 6. T 12.5 N [19.5 above the horizontal]F = °

W = 225 J ∆d = ?

( cos )

cos225 J

(12.5 N)(cos19.5 )19.1 m

W F dWd

F

d

θ

θ

= ∆

∆ =

=°

∆ =

The toboggan moves 19.1 m. 7. (a) We will calculate the area using the formula for a rectangle

(4.0 N)(2.0 m)8.0 J

A lw

A

===

The area represents the work done on the object. (b) First calculate the area of the second portion of the graph:

( 4.0 N)(6 0 m 2.0 m)8.0 J

A lw.

A

== − −= −

The total work is 8.0 + (–8.0) = 0.0 J (c) One situation could be pushing a box across a table and pulling it back.

Applying Inquiry Skills 8. You could set the pen on the paper and pull the paper across the desk. The force of static friction between the paper and

the pen is in the direction of motion, doing positive work on the pen.


Understanding Concepts 9. Four different situations are:

• a book sitting on a desk (∆d = 0) • a student carrying a book at a constant speed (θ = 90º) • a teacher whirling a putty pat in a circle at the end of a string • a toy car travelling in a circular path

10. (a) A box being pulled by string at an angle involves the forward component doing positive work, and the vertical component doing zero work.

(b) A box being pulled up a ramp involves the parallel component of gravity doing negative work, and the perpendicular

component of gravity doing zero work.

Section 4.1 Questions (Page 183)

Understanding Concepts 1. The everyday use of the word “work” is different from the usage in physics when it is referring to employment, or a duty

to perform. “Working” as a teacher involves very little physical work. The physics definition of work means the transfer of energy to an object to move it a certain distance. Many types of employment or daily activities involve physical work. For example, the sentence “Loading the cement bags onto the truck was a lot of work,” uses the word “work” similar to the physics definition of work.

2. The centripetal force is always directed toward the centre of the circle, and is by definition perpendicular to the motion of the object. The 90º angle means that work is not done on the object by the centripetal force.

3. As you push on a wall, you are exerting a force, which involves the use of energy. Even though no physical work is being done, your muscles are still burning your body’s fuel, causing you to become tired.

4. Assuming the classroom to be 4 m tall, and the student to have a mass of 70 kg:

( )

3

( cos )cos

(70 N)(9.80 N/kg)(cos 0 )(4.0 m)

3.0 10 J

W F dmg d

W

θθ

= ∆= ∆= °

= ×

It would take about 3.0 103 J, or 3.0 kJ of work to climb the ladder. 5. AF = 75 N [22° below the horizontal]

TF = 75 N [32° above the horizontal]


(a) The work done by the boy (WB):

B T

B

( cos )(75 N)(cos32 )(13 m)826.8 J

W F d

W

θ= ∆= °=

The work done by the girl (WG):

G A

G

( cos )(75 N)(cos 22 )(13 m)904.0 J

W F d

W

θ= ∆= °=

The total work done:

total B G

3total

828.8 J 904.0 J

1.73 10 J

W W W

W

= += +

= ×

The total amount of work done on the crate is 1.7 10–3 J. (b) The crate is moving at a constant speed, so it is not gaining any energy. This means that the crate will have the same

amount of energy before and after the move, so it must have work done on it opposite the direction of motion. Therefore, the work done on the crate by the floor is –1.7 103 J.

6. W = 9.65 102 J ∆d = 45.3 m F = 24.1 N [parallel to the handle of the sleigh] θ = ?

1

1

( cos )

cos

cos

965 Jcos(24.1 N)(45.3 m)

27.9

W F dW

F dW

F d

θ

θ

θ

θ

−

−

= ∆

=∆

= ∆

=

= °

The angle between the snowy surface and the handle is 27.9º. 7. (a) ∆d = 38 m

m = 66 kg

AF = 58 N [18° above the horizontal] FN = ? µK = ? First we must calculate the normal force:

N A g

N g A

A

N

0

sin18 0

sin18

sin18(66 kg)(9.80 N/kg) (58 N)sin18628.88 N

y yF ma

F F F

F F F

mg F

F

Σ = =

+ ° − =

= − °

= − °= − °=


We then calculate the force of gravity:

A K

K A

K

0cos18 0

cos18(58 N)cos1855.161 N

x xF maF F

F F

F

Σ = =° − =

= °= °=

The coefficient of kinetic friction is:

KK

N

K

55.161 N628.88 N0.088

FF

µ

µ

=

=

=

The normal force is 6.3 102 N, and the coefficient of kinetic friction between the toboggan and the snow is 0.088. (b) W = ?

3

( cos )(55.161 N)(cos180 )(38 m)

2.1 10 J

W F d

W

θ= ∆= °

= − ×

The work done by kinetic friction is –2.1 103 J. (c) The normal force, the gravitational force, and the vertical component of the applied force do no work on the toboggan. (d) ∆d = 25 m W = ?

3

( cos )(58 N)(cos18 )(25 m)

1.4 10 J

W F d

W

θ= ∆= °

= ×

The work done by the parent is 1.4 103 J.

Applying Inquiry Skills 8. As shown below, the graph indicates that the work done on a object is positive for angles less than 90º, zero for angles

equal to 90º, and negative for angles between 90º and 180º.

Making Connections 9. Work done by friction has the effect of heating up the environment. Most forms of energy usually end up as thermal

energy. 4.2 KINETIC ENERGY AND THE WORK–ENERGY THEOREM


Understanding Concepts 1. The kinetic energy of a moving object is related to both the mass and the velocity. If the mass of the truck is large enough,

a slow moving truck can have more kinetic energy than a fast moving car.


2. The kinetic energy is proportional to the speed, so if the speed increases by (a) 2, the kinetic energy will increase by a factor of 22, or 4 (b) 3, the kinetic energy will increase by a factor of 32, or 9 (c) 37%, the kinetic energy will increase by a factor of 1.372, or 1.9 3. Assume a 75-kg student running at 8.0 m/s:

2K

2

3K

121 (75 kg)(8.0 m/s)22.4 10 J

E mv

E

=

=

= ×

The kinetic energy at maximum speed is 2.4 kJ. 4. m = 45 g = 4.5 10–2 kg

vi = 0 m/s vf = 43 m/s

(a) W = ?

( ) ( )( )

K

2 2f i

2 22

1 ( )21 (4.5 10 kg) 43 m/s 0 m/s241.6025 J, or42 J

W E

m v v

W

−

= ∆

= −

= × −

=

The work done by the club is 42 J. (b) ∆d = 2.0 cm = 2.0 10–2 m F = ?

2

3

41.6025 J2.0 10 m2.1 10 N

W F dWF

d

F

−

= ∆

=∆

=×

= ×

The average force exerted by the club is 2.1 × 103 N. 5. m = 27 g = 2.7 10–1 kg

F = 95 N ∆d = 31 cm = 3.1 10–1 m vf = ?

K

2f

2f

f

1

1

f

1 (since the initial speed is zero)22

2

2(95 N)(3.1 10 m)2.7 10 kg

47 m/s

W E

F d mv

F dvmF dvm

v

−

−

= ∆

∆ =

∆=

∆=

×=×

=

The final speed of the arrow is 47 m/s. 6. m = 4.55 104 kg

vi = 1.22 104 m/s F = 3.85 105 N ∆d = 2.45 106 m vf = ?


K

2 2f i

2 2f i

2f i

5 64 2

4

4f

1 12 2

1 12 2

2

2(3.85 10 N)(2.45 10 m) (1.22 10 m/s)4.55 10 kg

1.38 10 m/s

W E

F d mv mv

mv F d mv

F dv vm

v

= ∆

∆ = −

= ∆ +

∆= +

× ×= + ××

= ×

The final speed of the probe is 1.38 104 m/s. 7. m = 28.0 kg AF = 95.6 N [35° above the horizontal]

FK = 75.5 N vi = 0 m/s ∆d = 0.750 m vf = ?

The total work done on the box will become kinetic energy. Since the initial speed is zero:

( )

( )

( )

2f

2A K f

2f A K

f A K

f

121cos35.0 cos18022 cos35.0 cos180

2 cos35.0 cos180

2(0.750 m) (95.6 N)(0.81952) (75.5 N)( 1)20.8 kg

0.45 m/s

W mv

F d F d mv

dv F Fm

dv F Fm

v

=

°∆ + °∆ =

∆= ° + °

∆= ° + °

= + −

=

The final speed of the box is 0.45 m/s. 8. W = 1.47(cos 38º ) = 1.16 The toboggan would have increased its kinetic energy by 16%.

Applying Inquiry Skills 9. W F d= ∆

2

2

2

N mkg m m

skg m

sW

= ⋅⋅= ⋅

⋅=

2K

2

2

K 2

12

mkgs

kg ms

E mv

E

=

= ⋅ ⋅=

The base units for both are the same.

Making Connections 10. m = 6.85 103 kg

vA = 2.81 103 m/s vB = 8.38 103 m/s

W = ?


(a) The work done is equal to the change in kinetic energy

( )

K

2 21B A2

2 2B A

3 3 2 3 2

11

121 ( )21 (6.85 10 kg) (8.38 10 m/s) (2.81 10 m/s)22.13 10 J

W E

mv mv

m v v

W

= ∆

= −

= −

= × × − ×

= ×

(b) The work done by Earth to move the satellite from A to B is 2.13 × 1011 J.

( )

K

2 2A B

2 2A B

3 3 2 3 2

11

1 12 21 ( )21 (6.85 10 kg) (2.81 10 m/s) (8.38 10 m/s)22.13 10 J

W E

mv mv

m v v

W

= ∆

= −

= −

= × × − ×

= − ×

The work done by Earth to move the satellite from B to A is –2.13 × 1011 J.


Understanding Concepts 1. The first doubling will require much less energy than the second doubling of the speed. This can clearly be shown using:

2 2f i

2 2f i

1 ( )2

W m v v

W v v

= −

∝ −

To go from v to 2v:

2 2f i

2 2

2 2

2

(2 ) ( )

4

3

W v v

v v

v v

W v

= −

= −

= −

=

To go from 2v to 4v:

2 2f i

2 2

2 2

2

(4 ) (2 )

16 4

11

W v v

v v

v v

W v

= −

= −

= −

=

The first doubling of speed will require work proportional to 3 times the square of the original speed. The second doubling will require work proportional to 11 times the square of the original speed.


2. m = 1.50 103 kg v = 18.0 m/s [E] EK = ?

2K

3 2

5K

121 (1.5 10 kg)(18.0 m/s)22.43 10 J

E mv

E

=

= ×

= ×

The kinetic energy of the car is 2.43 105 J. 3. (a) v = 1.150 × 18.0 = 20.7 m/s EK = ?

2K

3 2

5K

121 (1.5 10 kg)(20.7 m/s)23.21 10 J

E mv

E

=

= ×

= ×

The new kinetic energy of the car is 3.21 105 J. (b) The increase in EK is:

5

53.21 10 J 1.322.43 10 J

× =×

We can verify this with (1.15)2 = 1.32 This represents an increase in the kinetic energy of 32%. (c) W = ?

K

5 5

4

3.21 10 J 2.43 10 J

7.8 10 J

W E

W

= ∆

= × − ×

= ×

The work done to speed up the car was 7.8 104 J. 4. m = 55 kg

EK = 3.3 103 J v = ?

2K

K

3

12

2

2(3.3 10 J)55 kg

11 m/s

E mv

Evm

v

=

=

×=

=

The speed of the sprinter is 11 m/s. 5. v = 12 m/s

EK = 43 J m = ?

2K

K2

2

122

2(43 J)(12 m/s)0.60 kg

E mv

Emv

m

=

=

=

=

The mass of the basketball is 0.60 kg.


6. m = 0.353 kg ∆d = 89.3 cm = 0.893 m

(a) W = ?

( cos )(0.353 kg)(9.80 N/kg)(cos 0 )(0.893 m)3.0892 J

W mg d

W

θ= ∆= °=

The work done by gravity is 3.09 J. (b) Using the work energy theorem, KW E= ∆ . Since vi = 0:

2f

f

f

12

2

2(3.0892 J)0.353 kg

4.18 m/s

W mv

Wvm

v

=

=

=

=

The speed of the plate just before it hits the floor is 4.18 m/s. 7. m = 61 kg

θ = 23° FK = 72 N vi = 3.5 m/s ∆d = 62 m vf = ?

The component of gravity along the slope is mg sin 23º. Using the work energy theorem:

2 2K f i

2 2f K i

2 2Kf i

2Kf i

2 2

f

1 1sin 23 (cos 0 ) (cos180 )2 2

1 1sin 23 (1) ( 1)2 2

22 sin 23

22 sin 23

2(72 N)(62 m)2(9.8 m/s )(sin 23 )(62 m) (3.5 m/s)61 kg

18 m/s

mg d F d mv mv

mv mg d F d mv

F dv g d vmF dv g d v

m

v

° ° ∆ + ° ∆ = −

= ° ∆ + − ∆ +

∆= °∆ − +

∆= °∆ − +

= ° − +

=

The speed of the skier after travelling 62 m downhill is 18 m/s. 8. m = 55.2 kg

∆d = 4.18 m µK = 0.27 vi = ?


Using the FBD to calculate FN,

N

N

N

0

0

(55.2 kg)(9.80 N/kg)540.96 N

y yF ma

F mgF mg

F

Σ = =

− ====

To calculate FK:

K K N

K

(0.27)(540.96 N)146.06 N

F F

F

µ===

Using the work-energy theorem:

K

2 2K f i

1(cos180 ) ( )2

W E

F d m v v

= ∆

° ∆ = −

Since vf = 0,

2K i

Ki

i

1(cos180 )2

2 (cos180 )

2(146.06 N)( 1)(4.18 m)55.2 kg

4.7 m/s

F d mv

F dvm

v

° ∆ = −

− ° ∆=

− −=

=

The initial speed of the skater was 4.7 m/s.

Applying Inquiry Skills 9. (a)

Car Speed (m/s) Car Energy (J) Truck Speed (m/s) Truck Energy (J) 10.0

6.0 × 104 10.0

2.5 × 107

20.0

2.4 × 105 20.0

1.0 × 108

30.0

5.4 × 105 30.0

2.2 × 108

40.0

9.6 × 105 40.0

4.0 × 108

To calculate the energy of the car and the truck, use the equation 2K

12

E mv= .

To convert tonnes to kilograms, multiply by 1000: m = 1.2 t = 1.2 103 kg (for the car) m = 5.0 102 t = 5.0 105 kg (for the truck)

(b)


(c) A vehicle with a much larger mass has a much larger kinetic energy. As the speed of a vehicle increase, the kinetic energy increases proportional to the square of the speed.

Making Connections 10. (a) The kinetic energy is used to permanently deform the object as energy is transformed into heat. (b) The kinetic energy of a vehicle is transmitted (at least in part) to the occupants of the vehicle, and anything it contacts.

The high kinetic energy of a fast moving vehicle is more than sufficient to damage the human body beyond its limit, causing death.

4.3 GRAVITATIONAL POTENTIAL ENERGY AT EARTH’S SURFACE

PRACTICE (Page 191)

Understanding Concepts 1. The total work done by gravity is zero. The work on the way down is positive and is equal to the work done on the way

up, which is negative. At the end, the pen has not lost or gained gravitational potential energy. Alternatively, you could argue that because the ∆d = 0 (25 cm – 25 cm), the work done must be equal to zero.

2. m = 62.5 kg ∆y = 346 m (using the ground as y = 0) Eg = ?

g

5g

(62.5 kg)(9.80 N/kg)(346 m)

2.12 10 J

E mg y

E

∆ = ∆

=

∆ = ×

Relative to the ground, the gravitational potential energy is 2.12 105 J. 3. m = 58.2 g = 5.82 10–2 kg

∆y = 1.55 m (a) ∆Eg = ? At 1.55 m above the court:

g

2

g

(5.82 10 m)(9.80 N/kg)(1.55 m)0.884 J

E mg y

E

−

∆ = ∆

= ×∆ =

At the court height:

g

2

g

(5.82 10 m)(9.80 N/kg)(0.00 m)0.00 J

E mg y

E

−

∆ = ∆

= ×∆ =

The gravitational potential energy when the ball is above the court is 0.884 J, and as it strikes the court surface is 0.00 J.

(b) W = ?

2

( cos )( cos )

(5.82 10 kg)(9.80 N/kg)(cos 0 )(1.55 m)0.884 J

W F dmg d

W

θθ

−

= ∆= ∆

= × °=

At the instant the ball strikes the court surface, the force of gravity has done 0.884 J of work on the ball. (c) The work done in (b) is equal to the change in kinetic energy of the ball. 4. m = 68.5 kg

m = 2.56 km = 2.56 103 m θ = 13.9° Eg = ?


First, calculate the vertical lift:

3

3

sin13.92.56 10 m2.56 10 m(sin13.9 )614.98 m

y

yy

−

∆° =×

∆ = × °∆ =

Then calculate the gravitational potential energy:

g

5g

(68.5 kg)(9.80 N/kg)(614.98 m)

4.13 10 J

E mg y

E

∆ = ∆

=

∆ = ×

The skier’s gravitational potential energy at the top of the mountain is 4.13 105 J. 5. ∆y = 2.36 m

∆Eg = –1.65 103 J m = ? The ground is 2.36 m down from the pole, therefore:

g

g

31.65 10 J(9.80 N/kg)( 2.36 m)71.3 kg

E mg y

Em

g y

m

∆ = ∆

∆=

∆

− ×=−

=

The mass of the jumper is 71.3 kg. 6. (a) The first coin does not need to be lifted, so no work is done on it. Each successive coin will need to be raised one more

coin thickness, t, than the previous. The thickness of each coin will be ytN

= . The work done on each coin will be

equal to its increase in gravitational potential energy.

T 1 2 N

1 2 N

1 2 N

T

( )(0 2 )(0 1 2 )

(1 2 )

W W W Wmg y mg y mg ymg y y ymg t t Ntmgt NmgyW NN

= + + += ∆ + ∆ + + ∆= ∆ + ∆ + + ∆= + + += + + +

= + + +

The sum of an arithmetic series is:

1

2n

nt t

S n+ =

For the series (1 + 2 + ··· + N):

12nNS N + =


Substituting in the original equation to find the amount of work that must be done on the last coin:

T

T

12

12

mgy NW NN

NW mgy

+ = + =

(b) The energy stored is equal to the work done, therefore:

g1

2NE mgy + ∆ =

Applying Inquiry Skills 7. The units for gravitational potential energy:

( ) 2

2 2

mkg ms

kg m /s

mg y ∆ = = ⋅

The units for work:

( )( )

( )2

2 2

N m

mkg ms

kg m /s

F d∆ =

= = ⋅

The units for kinetic energy:

( )2

2

2 2

1 mkg2 s

kg m /s

mv = = ⋅

Therefore, all three units are the same.

Making Connections 8. ∆Eg = 6.1 109 J (a) ∆y = ?

Assume 920 students in the school. Assume an average mass of 70.0 kg per student. m = 70.0 kg 920 = 6.44 104 kg

g

g

9

4

4

6.1 10 J(6.44 10 kg)(9.80 N/kg)

9.472 10 m

E mg y

Ey

mg

y

∆ = ∆

∆∆ =

×=×

∆ = ×

The energy from one barrel of oil could raise the students 9.472 104 m above ground level. (b) There are 158.987 L in a barrel of oil

9

76.1 10 J 3.8 10 J/L158.987 L

× = ×

There are 3.8 107 J stored in each litre of oil.


PRACTICE (Page 193)

Understanding Concepts 9. (a) The Sun’s radiant energy is converted to thermal energy in the water, which is then converted into gravitational

potential energy as the water rises. The gravitational energy converts into kinetic energy as the water falls and turns the turbine. The kinetic energy of the turbine is converted into electrical energy by the generator.

(b) A run-of-the-river generating station does not dam the water to create a large vertical drop in a short area, but rather uses the natural drop of the land over a certain distance, and diverts part of the water to flow down this path.

Making Connections (c) The main difference students will find is that most run-of-the-river generating stations in Canada are much smaller. (d) (i) rainfall, rivers, and glacier/mountain snow melting

(ii) Bhutan relies heavily on its environment for exports from logging and energy. (e) possible points: cost, limited suitable locations, low population density, long-term effects


Understanding Concepts 1. As the construction worker raises the wood, the wood’s gravitational energy increases. 2. m = 63 kg ∆y = 3.4 m (a) On Earth, g = 9.80 N/kg ∆Eg = ?

g

3g

(63 kg)(9.80 N/kg)(3.4 m)

2.1 10 J

E mg y

E

∆ = ∆

=

∆ = ×

The astronaut’s gravitational potential energy is 2.1 103 J. (b) On the Moon, g = 1.6 N/kg ∆Eg = ?

2

(63 kg)(1.6 N/kg)(3.4 m)

3.4 10 J

g

g

E mg y

E

∆ = ∆

=

∆ = ×

The astronaut’s gravitational potential energy is 3.4 102 J. 3. m = 125 g = 0.125 kg

∆y = 3.50 m (a) ∆Eg = ? (of the pear relative to the ground) The pear on the branch:

g

g

(0.125 kg)(9.80 N/kg)(3.50 m)4.29 J

E mg y

E

∆ = ∆

=∆ =

The pear at ground level:

g

g

(0.125 kg)(9.80 N/kg)(0.00 m)0.00 J

E mg y

E

∆ = ∆

=∆ =

The gravitational potential energy of the pear on the branch relative to the ground is 4.29 J. The gravitational potential energy of the pear on the ground relative to the ground is 0.00 J.


(b) ∆Eg = ? (of the pear relative to the branch) The pear on the branch:

g

g

(0.125 kg)(9.80 N/kg)(0.00 m)0.00 J

E mg y

E

∆ = ∆

=∆ =

The pear on the ground:

g

g

(0.125 kg)(9.80 N/kg)( 3.50 m)4.29 J

E mg y

E

∆ = ∆

= −∆ = −

The gravitational potential energy of the pear on the branch relative to the branch is 0.00 J. The gravitational potential energy of the pear on the ground relative to the branch is –4.29 J.

4. m = 0.15 kg ∆Eg = 22 J ∆y = ?

g

g

22 J(0.15 kg)(9.80 N/kg)15 m

E mg y

Ey

mg

y

∆ = ∆

∆∆ =

=

∆ =

The ball’s maximum height is 15 m above the point where it was hit. 5. Let the subscript g represent gravity, and W represent the weightlifter.

m = 15 kg ∆y = 66 cm = 0.66

(a) Wg = ?

g

g

( cos )

( cos )(15 kg)( 9.80 N/m)(cos180 )(0.66 m)

97 J

W F d

mg d

W

θ

θ

= ∆

= ∆= − °= −

The amount of work done by gravity on the mass is –97 J. (b) WW = ?

W

W

( cos )( cos )(15 kg)(9.80 N/kg)(cos 0 )(0.66 m)97 J

W F dmg d

W

θθ

= ∆= ∆= °=

The amount of work done by the weightlifter on the mass is 97 J. (c) ∆Eg = ?

g

g

(15 kg)(9.80 N/kg)(0.66 m)97 J

E mg y

E

∆ = ∆

=∆ =

The gravitational potential energy of the mass increases by 97 J.


Applying Inquiry Skills 6.

Making Connections 7. (a) V = 32.8 km3 = 3.28 × 1010 m3

∆y = 23.1 m ρ = 1.00 103 kg/m3 ∆Eg = ?

First we must determine the mass of the water:

3 3 10 3

13

(1.00 10 kg/m )(3.28 10 m )

3.28 10 kg

m V

m

ρ=

= × ×

= ×

To calculate the gravitational potential energy:

g

13

15g

(3.28 10 kg)(9.80 N/kg)(23.1 m)

7.43 10 J

E mg y

E

∆ = ∆

= ×

∆ = ×

The gravitational potential energy of the lake relative to the turbines is 7.43 1015 J.

(b) This value is approximately 15

157.43 10 J 6.521.14 10 J

× =×

times the annual energy output of the Chukha plant in Bhutan.

4.4 THE LAW OF CONSERVATION OF ENERGY

PRACTICE (7)

Understanding Concepts 1. The force has done negative work on the ball. The speed of the ball has decreased, which corresponds to a decrease in

kinetic energy. This can only be done with negative work. 2. If the losses of energy are the same, the only source of kinetic energy is the conversion of gravitational potential. Mass

doesn’t matter (it cancels out in the equation), so they will both acquire the same final velocity. 3. ∆y = 59.4 m

vi = 0.0 m/s vf = ?



Making Connections 7. (a) V = 32.8 km3 = 3.28 × 1010 m3

∆y = 23.1 m ρ = 1.00 103 kg/m3 ∆Eg = ?

First we must determine the mass of the water:

3 3 10 3

13

(1.00 10 kg/m )(3.28 10 m )

3.28 10 kg

m V

m

ρ=

= × ×

= ×

To calculate the gravitational potential energy:

g

13

15g

(3.28 10 kg)(9.80 N/kg)(23.1 m)

7.43 10 J

E mg y

E

∆ = ∆

= ×

∆ = ×

The gravitational potential energy of the lake relative to the turbines is 7.43 1015 J.

(b) This value is approximately 15

157.43 10 J 6.521.14 10 J

× =×

times the annual energy output of the Chukha plant in Bhutan.

4.4 THE LAW OF CONSERVATION OF ENERGY

PRACTICE (Page 197)

Understanding Concepts 1. The force has done negative work on the ball. The speed of the ball has decreased, which corresponds to a decrease in

kinetic energy. This can only be done with negative work. 2. If the losses of energy are the same, the only source of kinetic energy is the conversion of gravitational potential. Mass

doesn’t matter (it cancels out in the equation), so they will both acquire the same final velocity. 3. ∆y = 59.4 m

vi = 0.0 m/s vf = ?


Using conservation of energy, we will have no kinetic energy at the top, and no gravitational energy at the bottom.

T1 T2

21 f

f 1

2

f

12

2

2(9.80 m/s )(59.4 m)34.1 m/s

E E

mgy mv

v gy

v

=

=/ /

=

==

To convert to km/h:

m 3600 s 1 km34.1 123 km/hs h 1000 m

=

The roller coaster reaches a maximum speed of 123 km/h at the bottom of the hill. 4. vi = v1 = 9.7 m/s

∆y = 4.2 m vf = v2 = ? We will use y = 0 at the point of contact on the hill.

T1 T2

2 211 1 2 22

2 21 1 2 2

2 22 1 1 2

12

2 2

2 2

E E

mv mgy mv mgy

v gy v gy

v v gy gy

=

+ = +

+ = +

= + −

Since y2 = 0, 2gy2 = 0, therefore:

22 1 1

2 2

2

2

(9.7 m/s) 2(9.80 m/s )(4.2 m)13 m/s

v v gy

v

= +

= +=

The skier’s speed upon touching the hillside is 13 m/s. 5. ∆y = 4.4 102 m v2 = 93 m/s

v1 = ?

T1 T2

2 21 1 2 2

2 21 1 2 2

2 21 2 2 1

21 2 2 1

2 2

1

1 12 2

2 2

2 2

2 ( )

(93 m/s) 2(9.8 m/s )(0 440 m)5.0 m/s

E E

mv mgy mv mgy

v gy v gy

v v gy gy

v v g y y

v

=

+ = +

+ = +

= + −

= + −

= + −=

The speed of the water at the top of the waterfall is 5.0 m/s. 6. v1 = 9.7 m/s

∆y = 4.7 m v2 = ?


T1 T2

2 21 1 2 2

2 21 1 2 2

2 22 1 1 2

22 1 1 2

2 2

2

1 12 2

2 2

2 2

2 ( )

(9.7 m/s) 2(9.8 m/s )(0 4.7 m)1.4 m/s

E E

mv mgy mv mgy

v gy v gy

v v gy gy

v v g y y

v

=

+ = +

+ = +

= + −

= + −

= + −=

The cyclist crests the hill at a speed of 1.4 m/s. 7. v2 = ?

First determine how high the pendulum is vertically raised:

2 2 21

2 21

1

24.5 cm 85.5 cm

85.5 cm 24.5 cm81.915 cm

h

hh

+ =

= −=

1 2

2 1

2

85.5 cm85.5 cm85.5 cm 81.915 cm3.585 cm

h hh h

h

+ == −= −=

Using the value h2 = 3.585 cm, or 0.03585 m, calculate the maximum speed:

T1 T2

2 21 1 2 2

12

1 2 222 1 2

2 1 2

2

2

1 12 2

since 0

2 2

2 2

2 ( )

2(9.80 m/s )(0.03585 m 0)0.838 m/s

E E

mv mgy mv mgy

v

gy v gy

v gy gy

v g y y

v

=

+ = +

=

= +

= −

= −

= −=

The maximum speed of the pendulum bob is 0.838 m/s.



∆∆∆∆y (m) ∆∆∆∆Eg (J) EK (J) ET (J) 8.00 392 0.00 392 6.00 294 98.0 392 4.00 196 196 392 2.00 98.0 294 392 0.00 0.00 392 392

Making Connections 9. A wrecking ball works as a pendulum that is slowly pulled back, increasing its gravitational energy. When it is released,

the gravitational potential energy is converted into kinetic energy which is used to destroy buildings.

PRACTICE (Page 200)

Understanding Concepts 10. (a) The energy supplied becomes sound and thermal energy through friction. (b) The energy supplied still produces sound and thermal energy, but some is also converted into kinetic energy. 11. FK = 67 N ∆d = 3.5 m (a) W = ?

2

( cos )(67 N)(cos180 )(3.5 m)

2.3 10 J

W F d

W

θ= ∆= °

= − ×

The amount of work done by friction is –2.3 102 J. (b) Eth = ?

th K

2th

(67 N)(3.5 m)

2.3 10 J

E F d

E

= ∆=

= ×

The amount of thermal energy produced is 2.3 102 J. 12. Eth = 0.620 J

FK = 0.83 N ∆d = ?


The 0.620 J of energy comes from the work done by friction, therefore:

th K

th

K

0.620 J0.83 N0.75 m

E F dE

dF

d

= ∆

∆ =

=

∆ =

The plate slides 0.75 m. 13. m = 22.0 kg

F = 98 N FK = 87 N vi = 0.0 m/s ∆d = 1.2 m vf = ?

th k

2K 2

2 K2

K2

2

1( cos )2

( cos )12

( cos )0.5

(98 N)(cos 0 )(1.2 m) 87 N(1.2 m)0.5(22.0 kg)

1.1 m/s

W E E

F d F d mv

F d F dvm

F d F dvm

v

θ

θ

θ

= +

∆ = ∆ +

∆ − ∆=

∆ − ∆=

° −=

=

The speed of the cabinet after it moves 1.2 m is 1.1 m/s. 14. m = 0.057 kg

∆d = 25 cm = 0.25 m vf = 5.7 cm/s = 5.7 10–2 m/s FK = 0.15 N vi = ?

K1 th K2

2 21 K 2

2K 2

21

2K 2

1

2 2

1

1 12 2

12

12

12

0.51(0.15 N)(0.25 m) (0.057 kg)(5.7 10 m/s)2

0.5(0.057 kg)1.1 m/s

E E E

mv F d mv

F d mvv

m

F d mvv

m

v

−

= +

= ∆ +

∆ +=

∆ +=

+ ×=

=

The initial speed of the pen is 1.1 m/s.

Applying Inquiry Skills 15. (a) The law of conservation of energy could be verified by observing how close to its original position the pendulum

would return after a swing.


(b) Some sources of error would be loss of energy due to friction at the point of attachment and air friction of the moving pendulum. Some error may be observed if the string is somewhat elastic.

Making Connections 16. The oil circulation system is an attempt to minimize the frictional forces between the moving parts of the engine, and

thereby reduce the loss (and damage) due to the thermal energy caused by friction. The water (coolant) circulation is used to absorb thermal energy from the engine and dissipate it rapidly into the air to prevent damage from overheating.

Section 4.4 Questions (Pages 201–202)

Understanding Concepts 1. Roller coasters are gravity rides that have an initial input of gravitational potential energy that is converted to kinetic (and

back into gravitational) energy throughout the ride. To give them this initial energy, they must be pulled up the largest hill at the beginning.

2. m = 0.052 kg ∆y = 11 cm = 0.11 m y = 0

(a) ∆Eg = ?

g

2

g

(0.052 kg)(9.80 m/s )(0.11 m)0.056 J

E mg y

E

∆ = ∆

=∆ =

The initial gravitational potential energy of the egg’s contents is 0.056 J. (b) ∆Eg = ?

g

2

g

(0.052 kg)(9.80 m/s )(0.0 m)0.0 J

E mg y

E

∆ = ∆

=∆ =

The final gravitational potential energy of the egg’s contents is 0.0 J. (c) ∆Eg = 0.0 – 0.056 = –0.056 J The change in gravitational potential energy as the egg’s contents fall is –0.056 J. (d) EK = ?

v = ? The kinetic energy will be equal to the loss of gravitational potential, so EK = 0.056 J.

2K

K

12

2

2(0.056 J)0.052 kg

1.5 m/s

E mv

Evm

v

=

=

=

=

The kinetic energy is 0.056 J. The speed of the egg’s contents just before hitting the pan is 1.5 m/s. 3. ∆y = 1.2 m

vf = 9.9 m/s vi = ?


T1 T2

2 21 1 2 2

2 21 1 2 2

2 21 2 2 1

21 2 2 1

2 2

1

1 12 2

2 2

2 2

2 ( )

(9.9 m/s) 2(9.8 m/s )(1.2 m 0 m)11 m/s

E E

mv mgy mv mgy

v gy v gy

v v gy gy

v v g y y

v

=

+ = +

+ = +

= + −

= + −

= + −=

The initial speed of the ball was 11 m/s. 4. v1= 3.74 m/s

∆y = 8.74 m m = 7.12 104 kg

(a) ∆Eg = ?

g

4 2

6g

(7.12 10 kg)(9.80 m/s )(8.74 m)

6.10 10 J

E mg y

E

∆ = ∆

= ×

∆ = ×

The gravitational potential energy of the mass of water at the top of the waterfall is 6.10 106 J. (b) v2 = ?

T1 T2

2 21 1 2 2

2 21 1 2 2

2 22 1 1 2

22 1 1 2

2 2

2

1 12 2

2 2

2 2

2 ( )

(3.74 m/s) 2(9.80 m/s )(8.74 m 0 m)13.6 m/s

E E

mv mgy mv mgy

v gy v gy

v v gy gy

v v g y y

v

=

+ = +

+ = +

= + −

= + −

= + −=

The speed of the water at the bottom of the waterfall is 13.6 m/s. 5. (a)

First, determine the vertical height above the bottom of the swing:

( )

1

1

1

cos 483.7 m3.7 m cos 482.476 m

h

hh

° =

= °=


1 2

2 1

2

3.7 m3.7 m3.7 m 2.476 m1.224 m

h hh h

h

+ == −= −=

Now we can calculate the acrobat’s speed:

T1 T2

2 21 1 2 2

1 12 2

E E

mv mgy mv mgy

=

+ = +

Since v1 = 0:

21 2 222 1 2

2 1 2

2

2

2 2

2 2

2 ( )

2(9.80 m/s )(1.224 m 0 m)4.9 m/s

gy v gy

v gy gy

v g y y

v

= +

= −

= −

= −=

The acrobat’s speed at the bottom of the swing is 4.9 m/s. (b) Due to conservation of energy, the maximum height on the other side is equal to the starting height of the acrobat. 6. m = 55 kg ∆d = 3.7 m

FK = 41.5 N v1 = 65.7 cm/s = 0.657 m/s v2 = 7.19 m/s φ = ?

Relating φ to h:

sin

11.711.7sin

h

h

φ

φ

=

=

Using conservation of energy:

( ) ( )( )

T1 T2

K1 g K2 th

2 21 2 K

2 22 K 1

2 22 1 K

2 2

2

1 12 2

1 12 21 ( )2

1 (55.0 kg) 7.19 m/s 0.657 m/s (41.5 N)(11.7 m)211.7sin

(55.0 kg)(9.80 m/s )sin 0.30054

17.5

E EE E E E

mv mgh mv F d

mgh mv F d mv

m v v F dh

mg

φ

φφ

=+ ∆ = +

+ = + ∆

= + ∆ −

− + ∆=

− +=

== °

The angle is 17.5°.


7. v1 = 0.0 m/s v2 = 6.8 m/s Since the skateboarder starts even with the height of the centre of the circle, the total vertical drop at the bottom will be

equal to the radius of the circle. r = y1 = ?

T1 T2

2 21 1 2 2

2 21 1 2 2

1 12 2

2 2

E E

mv mgy mv mgy

v gy v gy

=

+ = +

+ = +

Since v1 = 0, and y2 = 0:

21 2

22

1

2

2

1

2

2

6.8 m/s2(9.8 m/s )2.4 m

gy v

vyg

y

=

=

=

=

The radius of the half-pipe is 2.4 m. 8. m = 55 g = 5.5 10–2 kg

v1 = 1.9 m/s ∆d = 54 cm = 0.54 m v2 = 0.0 m/s

(a) µK = ? First, solve for FK:

T1 T2

K th

2K

2

K

2 2

K

12

2(5.5 10 kg)(1.9 m/s)

2(0.54 m)0.1838 N

E EE E

mv F d

mvFd

F

−

==

= ∆

=∆

×=

=

Now calculate the normal force:

N

N2 2

N

0

0

(5.5 10 kg)(9.80 m/s )0.539 N

y yF ma

F mgF mg

F

−

Σ = =

− ==

= ×=

Solve for µK:

KK

N

0.1838 N0.539 N

0.34K

FF

µ

µ

=

=

=

The coefficient of kinetic friction is 0.34.


(b) µK =? First, solve for the acceleration:

( ) ( )

2 2f i

2 2f i

2 2

2

2

21.9 m/s 0 m/s

2(0.54 m)

3.3426 m/s

v v a d

v vad

a

= + ∆

−=

∆−

=

=

Using the FBD, calculate the magnitude of kinetic friction:

K2 2

K

(5.5 10 kg)(3.3426 m/s )0.1838 N

x x

x

F maF ma

F

−

Σ ==

= ×=

Calculate the normal force:

N

N2 2

N

0

0

(5.5 10 kg)(9.80 m/s )0.539 N

y yF ma

F mgF mg

F

−

Σ = =

− ==

= ×=

Calculate the coefficient of kinetic friction:

KK

N

K

0.1838 N0.539 N

0.34

FF

µ

µ

=

=

=

The coefficient of kinetic friction is 0.34. (c) The kinetic energy is converted into thermal energy. 9. vi = 85 km/h = 23.61 m/s

vf = 0.0 m/s ∆d = 47 m FK = 7.4 103 N

(a) Eth = ?

th K

3

5th

(7.4 10 N)(47 m)

3.5 10 J

E F d

E

= ∆

= ×

= ×

The amount of thermal energy produced is 3.5 105 J. (b) Before the skid, the thermal energy was kinetic energy.


(c) m = ?

( )

T1 T2

K th

2K

K2

3

2

3

12

2

2(7.4 10 N)(47 m)23.61 m/s

1.2 10 kg

E EE E

mv F d

F dmv

m

==

= ∆

∆=

×=

= ×

The mass of the car is 1.2 103 kg. (d) µK = ? First calculate the normal force:

N

N3 2

N

0

0

(1.248 10 kg)(9.80 m/s )12 228

y yF ma

F mgF mg

F

Σ = =

− ==

= ×=

To calculate the coefficient of kinetic friction:

KK

N3

K

7.4 10 J12228 J

0.61

FF

µ

µ

=

×=

=

The coefficient of kinetic friction is 0.61. 10. m = 22 kg

∆d = 2.5 m θ = 44° FK = 79 N

(a) W = ?

K

2

( cos )(79 N)(cos180 )(2.5 m)

2.0 10 J

W F d

W

θ= ∆= °

= − ×

The work done by friction is –2.0 102 J. (b) EK = ?


First, determine the vertical drop:

sin 442.52.5sin 441.737 m

h

hh

° =

= °=

To calculate the final kinetic energy:

T1 T2

g K th

K K

K K

2K

(9.8 N/kg)(22 kg)(1.737 m) (79 N)(2.5 m)

1.8 10 J

E EE E E

mg y E F dE mg y F d

E

=∆ = +

∆ = + ∆= ∆ − ∆= −

= ×

The box’s final kinetic energy is 1.8 102 J. (c) Eth = ?

th K

2th

(79 N)(2.5 m)

2.0 10 J

E F d

E

= ∆=

= ×

The thermal energy produced is 2.0 102 J.

Applying Inquiry Skills 11. (a) The loss of gravitational potential energy from the first lift to the second is equal to the amount of thermal energy

produced. (b) The equation needed would be:

T1 T2

g1 g2 th

1 2 th

th 1 2

th 1 2( )

E EE E E

mgy mgy EE mgy mgyE mg y y

=∆ = ∆ +

= += −= −

(c) You could determine the height at each point by standing a known distance from the base of the ride and measure the angle up for each height. Using simple trigonometric ratios, you could estimate the amount of thermal energy produced.

Making Connections 12. A trebuchet is very accurate. The raised mass supplies a fixed amount of gravitational potential energy. This potential

energy is converted into the kinetic energy of the projectile.

4.5 ELASTIC POTENTIAL ENERGY AND SIMPLE HARMONIC MOTION


Understanding Concepts 1. The higher the spring constant, the more force is needed to stretch the spring the same amount. For this reason, spring A

would be more difficult to stretch than spring B. 2. The spring would exert a southward force on you. 3. k = 25 N/m (a) x = 16 cm = 0.16 m

(25 N/m)(0.16 m)4.0 N

x

x

F kx

F

===


x = 32 cm = 0.32 m

(25 n/m)(0.32 m)8.0 N

x

x

F kx

F

===

The magnitude of force would be 4.0 N for a stretch of 16 cm, and 8.0 N for a stretch of 32 cm. (b) x = 16 cm = 0.16m

(25 N/m)(0.16 m)4.0 N

x

x

F kx

F

= −= −= −

x = 32 cm = 0.32 m

(25 N/m)(0.32 m)8.0 N

x

x

F kx

F

= −= −= −

The magnitudes of the forces are 4.0 N and 8.0 N, respectively. 4. k = 3.2 102 N/m x = 2.0 cm = 2.0 10–2 m

2 2(3.2 10 N/m)(2.0 10 m)6.4 N

x

x

F kx

F

−

= −

= − × ×= −

The magnitude of the force applied by the air is 6.4 N. 5. m = 1.37 kg

k = 5.20 102 N/m (a) x = ?

2

0

0

(1.37 kg)(9.80 N/kg)5.20 10 N/m

0.0258 m

y

x

F ma

F mgkx mg

mgxk

x

Σ = =

− ==

=

=×

=

The spring stretches 0.0258 m. (b) x = 1.59 cm = 0.0159 m yFΣ = ?

2(5.20 10 N/m)(0.0159 m) (1.37 kg)(9.80 N/kg)5.16 N

y x

y

F F mg

kx mg

F

Σ = −

= −

= × −Σ = −

The net force on the fish is 5.16 N [down]. (c) x = 2.05 cm = 0.0205 m a = ?

2

2

(5.20 10 N/m)(0.0205 m) (1.37 kg)(9.80 N/kg)1.37 kg

2.02 m/s

y

x

F ma

F mg makx mga

m

a

Σ =

− =−=

× −=

= −

The acceleration of the fish is 2.02 m/s2 [down]


Applying Inquiry Skills 6. (a)

(b) The slope of the line is negative.

Making Connections 7. (a)

Mass (kg) Stretch (m) 1.00 0.122 2.00 0.245 3.00 0.368 4.00 0.490 5.00 0.612 6.00 0.735 7.00 0.858 8.00 0.980

(b)

(c) The mass value may not be correct if the value of g is different then where it was calibrated. The weight value would

be accurate anywhere (even on the Moon).


PRACTICE (Page 211)

Understanding Concepts 8. (a) The graph shown is the force applied by the spring. Since the spring is being stretched to the right (positive x), the force

exerted by the spring will be to the left (negative). This is what is shown on the graph. (b) The force constant is the slope of the graph. Since the equation Fx = kx relates the force exerted by the spring, we must

change the direction (i.e., the sign) of the force,

( 15 N)0.40 m

38 N/m

xFk

x

k

−=

− −=

=

The force constant of the spring is 38 N/m. (c) x = 35 cm = 0.35 m

The energy stored is the area between the curve and the x-axis.

e

e

12

1 (0.35 m)(13 N)22.3 J

E A bh

E

= =

=

=

The elastic potential energy is 2.3 J. 9. k = 9.0 103 N/m (a) x = 1.0 cm = 0.010 m Ee = ?

2e

3 2

e

121 (9.0 10 N/m)(0.010 m)20.45 J

E kx

E

=

= ×

=

The elastic potential energy stored by the spring is 0.45 J. (b) x = –2.0 cm = –0.020 m Ee = ?

2e

3 2

e

121 (9.0 10 N/m)( 0.020 m)21.8 J

E kx

E

=

= × −

=

The elastic potential energy stored in the spring is 1.8 J. 10. m = 7.8 g = 0.0078 kg

k = 3.5 102 N/m x = –4.5 cm = –0.045 m

(a) Ee = ?

2e

2 2

e

121 (3.5 10 N/m)( 0.045 kg)20.35 J

E kx

E

=

= × −

=

The elastic potential energy of the spring is 0.35 J.


(b) v = ?

T1 T2

e K

2 2

2

2 2

1 12 2

(3.5 10 N/m)( 0.045 m)0.0078 kg

9.5 m/s

E EE E

kx mv

kxvm

v

==

=

=

× −=

=

The speed of the dart as it leaves the toy is 9.5 m/s. 11. m = 3.5 10–3 kg k = 9.5 N/m (a) ∆y = 5.7 cm = 0.057 m x = ?

T1 T2

e g

2

3

12

2

2(3.5 10 kg)(9.8 N/kg)(0.057 m)9.5 N/m

0.020 m

E EE E

kx mg y

mg yxk

x

−

== ∆

= ∆

∆=

×=

=

The spring must be compressed by 0.020 m, or 2.0 cm. (b) If friction was not negligible, the amount of compression would need to be increased. The energy supplied by the

compressed spring would now become both gravitational potential and thermal energy. In order to have the same final gravitational energy, the spring would need to be compressed more.

12. m = 0.20 kg k = 55 N/m

(a) ∆y = 1.5 cm = 0.015 m v = ?

T1 T2

g K e

2 2

2 2

2

2

1 12 22

2

2(0.20 kg)(9.8 N/kg)(0.015 m) (55 N/m)(0.015 m)0.20 kg

0.48 m/s

E EE E E

mg y mv kx

mv mg y kx

mg y kxvm

v

=∆ = +

∆ = +

= ∆ −

∆ −=

−=

=

The speed of the mass is 0.48 m/s.


(b) For the fall, ∆y = x. At maximum fall, there will be no kinetic energy.

( )( ) ( )

T1 T2

g e

2

2

2

2

12

2 0

55 2(0.20)(9.8) 0

55 3.92 0(55 3.92) 0

55 3.92 0 or 03.9255

0.071m or 0 m

E EE E

mg y kx

kx mg y

x x

x xx x

x x

x

x x

=∆ =

∆ =

− ∆ =

− =

− =− =− = =

=

= =

The value x = 0 refers to the moment of release, so the maximum stretch will be 0.071 m. 13. k = 12 N/m

∆y = 93.0 cm = 0.93 m m = 8.3 10–3 kg x = 4.0 cm = 0.040 m ∆d = ? Analyze as a projectile motion question. First, determine horizontal speed at launch:

T1 T2

e K

2 2

2

2

3

1 12 2

(12 N/m)( 0.040 m)8.3 10 kg

1.521 m/s

E EE E

kx mv

kxvm

v

−

==

=

=

−=×

=

Choosing down as positive, the time for the marble to drop 0.93 m vertically is:

2i

1 ( )2

y v t a t∆ = ∆ + ∆

Since vi = 0:

2

2

1 ( )2

2

2(0.93 m)9.8 m/s

0.4357 s

y a t

yta

t

∆ = ∆

∆∆ =

=

∆ =

The horizontal distance travelled during this time is:

(1.521 m/s)(0.4357 s)0.66 m

d v t

d

∆ = ∆=

∆ =

The marble travels 0.66 m horizontally before hitting the floor.


Applying Inquiry Skills 14. (a) The measurement needed would be the mass of the largest friend. (b) Assuming a largest mass of 115 kg,

3

0

0

(115 kg)(9.80 N/kg)0.75 m

1.5 10 N/m

y

x

F ma

F mgkx mg

mgkx

k

Σ = =

− ==

=

=

= ×

The approximate force constant is 1.5 103 N/m.

Making Connections 15. m = 2.0 102 µg = 2.0 10–7 kg

∆y = 65 mm = 0.065 m x = 75 cm = 0.75 m Ee = ?

T1 T2

e g

7

8e

( )

( )

0.75 m(2.0 10 kg)(9.8 N/kg)(0.065 m)

9.6 10 J

E EE x E

x mg y

E

−

−

== ∆

= ∆

= ×

= ×

The initial quantity of elastic potential energy is 9.6 10–8 J.


Understanding Concepts 16. (a) The maximum displacement from the rest position will be at the top and bottom of the bounce. (b) The speed is a maximum at the rest position. (c) The speed will be zero at the top and the bottom of the bounce. (d) The acceleration will be a maximum at the top and the bottom of the bounce. (e) The acceleration will be zero at the rest position. 17. T = ? f = ? (a) number of vibrations = 12

t = 48 s

total timenumber of complete vibrations48124.0 s

T

T

=

=

=

number of complete vibrationstotal time

12480.25 Hz

f

f

=

=

=


Alternatively, you could calculate frequency by using the equation:

1

14.0 s0.25 Hz

fT

f

=

=

=

The period is 4.0 s, and the frequency is 0.25 Hz. (b) number of vibrations = 210 t = 1 min = 60 s

total timenumber of complete vibrations60

2100.29 s

T

T

=

=

=


21060

3.5 Hz

f

f

=

=

=

The period is 0.29 s, and the frequency is 3.5 Hz. (c) number of vibrations = 2200 t = 5.0 s

3

total timenumber of complete vibrations5.0

22002.3 10 s

T

T −

=

=

= ×

2


22005.0

4.4 10 Hz

f

f

=

=

= ×

The period is 2.3 10–3 s, and the frequency is 4.4 102 Hz. 18. m = 0.25 kg

A = 8.5 cm k = 1.4 102 N/m

(a) d = ? During each cycle, the mass moves 4 amplitudes, or 4A. In 5 cycles, the mass will move:

2

5 45 4(8.5 cm)

1.7 10 cm

d A

d

= ×= ×

= ×

The mass moves 1.7 102 cm in the first five cycles. (b) T = ?

2

2

0.25 kg21.4 10 N/m

0.27 s

mTk

T

π

π

=

=×

=

The period of vibration is 0.27 s.


19. m = 0.10 kg f = 2.5 Hz k = ?

2 2

2 2

12

2

4

4 (2.5 Hz) (0.10 kg)25 N/m

kfm

kfm

k f m

k

π

π

π

π

=

=

=

==

The force constant of the spring is 25 N/m. 20. k = 1.4 102 N/m

T = 0.85 s m = ?

2

2

2 2

2

2

2

4(0.85 s) (1.4 10 N/m)

42.6 kg

mTk

T mk

T km

m

π

π

π

π

=

=

=

×=

=

The mass would have to be 2.6 kg.

Applying Inquiry Skills 21. Examining the base SI units for each:

2

2

mms

s

s

xa

xa

=

=

=

2

2

kgNm

kg m=kg m

s

s

s

mk

mk

=

⋅⋅

=

=

Therefore, they are dimensionally equivalent.

Making Connections 22. number of vibrations = 6.0 t = 8.0 s (a) k = ? First we must calculate the frequency:


6.08.0 s0.75 Hz

f

f

=

=

=


Using a mass of 75 kg:

2 2

2 2

3

12

2

4

4 (0.75 Hz) (75 kg)

1.7 10 N/m

kfm

kfm

k f m

k

π

π

π

π

=

=

=

=

= ×

The force constant is 1.7 103 N/m. (b) No, you are not undergoing SHM. When you leave the trampoline, there is a period of time when it is not exerting any

force on you. SHM requires that the force be proportional to the displacement.


Understanding Concepts 23. (a) The speed is zero at lengths of 12 cm and 38 cm. (b) The maximum speed will be at the rest position. The rest position will be halfway between the minimum and maximum

extensions:

12 38 25 cm2+ =

(c) The amplitude is 38 – 25 = 13 cm. 24. Ee = 5.64 J

m = 0.128 kg k = 244 N/m

(a) x = ? The maximum energy is constant, and all elastic potential at either end of the system, at the maximum amplitude.

2e

e

12

2

2(5.64 J)244 N/m

0.215 m

E kx

Ex

k

x

=

=

=

=

The amplitude of the vibration is 0.215 m. (b) v = ? Approach 1: All of the energy will be kinetic as it passes through the rest position.

2K

K2

2(5.64 J)0.128 kg

9.39 m/s

E mv

Evm

v

=

=

=

=


Approach 2: All of the energy stored in the spring at maximum compression will be converted to kinetic energy.

T1 T2

2 2

2

2

1 12 2

(244 N/m)(0.215 m)0.128 kg

9.39 m/s

E E

kx mv

kxvm

v

=

=

=

=

=

The speed is 9.39 m/s regardless of the approach used. (c) x2 = 15.5 cm = 0.155 m v = ?

( ) ( )( )

T1 T2

2 2 2max 2

2 22 max 2

2 2max 2

2 2

1 1 12 2 2

( )

(244 N/m) 0.215 m 0.155 m

0.128 kg6.51 m/s

E E

kx mv kx

kx kxvm

k x xv

m

v

=

= +

−=

−=

−=

=

The speed of the mass is 6.51 m/s. 25. x = 0.18 m

m = 58 g = 0.058 kg k = 36 N/m

(a) Ee = ? v = ? Maximum energy during maximum stretch/compression of the spring:

2max

2

max

121 (36 N/m)(0.18 m)20.58 J

E kx

E

=

=

=

This will all be kinetic energy at the rest position.

2K

K

12

2

2(0.58 J)0.058 kg

4.5 m/s

E mv

Evm

v

=

=

=

=

The maximum energy of the system is 0.58 J. The maximum speed of the mass is 4.5 m/s.


(b) x = ? Double the energy would be:

max e

2

2

max

2122

(36 N/m)(0.18 m)1.1664 J

E E

kx

E

′ = ×

= ×

=′ =

This is all stored as elastic potential energy at the full amplitude.

2e

e

12

2

2(1.1664 J)36 N/m

0.25 m

E kx

Ex

k

x

=

=

=

=

The amplitude of vibration required would be 0.25 m. (c) v = ?

2K

K

12

2

2(1.1664 J)0.058 kg

6.3 m/s

E mv

Evm

v

=

=

=

=

The maximum speed of the mass is 6.3 m/s. 26. For the maximum speed, all of the elastic energy will be converted to kinetic:

T1 T2

2 2

2

1 12 2

E E

kx mv

kxvmkv xm

=

=

=

=

For a SHM system, x = A, therefore:

12

2

kfm

kfm

π

π

=

=

Substituting above:

(2 )2

kv xm

A fv fA

ππ

=

==


Applying Inquiry Skills

27. (a) 2 2 2 2

Nm (m m )kg

k (A x )m

− = −

22

2

2

kg mm s (m )

kg

ms

m/s

⋅ ⋅ =

=

=

The dimensions are m/s. (b) This expression is to calculate the speed of an object at a location during SHM.

Making Connections 28. (a) Tuning fork prongs have slow damping to produce a long tone. (b) The voltmeter needle has fast damping to stabilize the reading quickly. (c) A guitar string has slow damping to play the note as long as possible. (d) Saloon doors have medium damping to keep the doors from swinging too much, but they still swing back and forth. (e) The string of the bow has fast damping to prevent dangerous vibration.

Section 4.5 Questions (Pages 218–219)

Understanding Concepts 1. When the two students pull on either end of the spring, it will not stretch as much as when it is pulled by both students

while attached to the wall. When they both pull on it from the same side, the wall pulls back with equal force. When they pull from opposite ends, the force will be half as much, and the stretch will also be half as much.

2. The elastic potential energy is the same when the spring is stretched or compressed 2.0 cm. The amount of energy stored only depends on the magnitude of the distortion, not the direction.

3. Harmonic means that it is regularly repeated, symmetrical motion.

4. (a) Period is inversely proportional to the frequency, 1Tf

∝ .

(b) The acceleration is directly proportional to the displacement, a x∝ .

(c) The period is inversely proportional to square root of the force constant, 1Tk

∝ .

(d) The maximum speed is directly proportional to the amplitude, v A∝ . 5. m = 62 kg

k = 2.4 103 N/m x = ?

3

0

6 06

6(62 kg)(9.8 N/kg)6(2.4 10 N/m)

0.042 m

y

x

F ma

F mgkx mg

mgxk

x

Σ = =

− ==

=

=×

=

The compression of each spring is 0.042 m.


6. k = 78 N/m x = 2.3 cm = 0.023 m Fx = ?

(78 N/m)(0.023 m)1.8 N

x

x

F kx

F

===

The magnitude of force is 1.8 N. 7. x1 = 1.85 cm

Fx1 = 85.5 N x2 = 4.95 cm = 0.0495 m Fx2 = ? First we must calculate k:

1 1

1

1

85.5 N0.0185 m4621.6 N/m

x

x

F kxF

kx

k

=

=

=

=

Solve for x2:

2 2

2

(4621.6 N/m)(0.0495 m)229 N

x

x

F kx

F

===

The force required is 229 N. 8. m = 97 kg

k = 2.2 103 N/m a = 0.45 m/s2

x = ?

Ignoring friction to calculate the applied force:

A2

A

(97 kg)(0.45 m/s )43.65 N

xF maF ma

F

Σ ==

==

This force is the force exerted on the spring:

343.65 N

2.2 10 N/m0.020 m

x

x

F kxF

xk

x

=

=

=×

=

The spring stretches 0.020 m, or 2.0 10–2 m. 9. m = 289 g = 0.289 kg

k = 18.7 N/m (a) x = 10.0 cm = 0.100 m yFΣ = ? a = ?


To calculate the net force:

(18.7 N/m)(0.100 m) (0.289 kg)(9.80 N/kg)

0.962 N

y x

y

F F mg

kx mg

F

Σ = −

= −= −

Σ = −

To calculate acceleration:

2

0.962 N0.289 kg

3.33 m/s

y

y

F ma

Fa

m

a

Σ =

Σ=

−=

= −

The net force is 0.962 N [down], and the acceleration is 3.33 m/s2 [down]. (b) x = ?

0

0

(0.289 kg)(9.80 N/kg)18.7 N/m

0.151 m

y

x

F ma

F mgkx mg

mgxk

x

Σ = =

− ==

=

=

=

The spring will be stretched by 0.151 m. 10. m = 64.5 kg

∆y = 48.0 m –12.5 m = 35.5 m k = 65.5 N/m x = 35.5 m – 10.1 m = 25.4 m. v = ?

g e K

2 2

2 2

2

2

1 12 22

2

2(64.5 kg)(9.80 N/kg)(35.5 m) (65.5 N/m)(25.4 m)64.5 kg

6.37 m/s

E E E

mg y kx mv

mv mg y kx

mg y kxvm

v

∆ = +

∆ = +

= ∆ −

∆ −=

−=

=

The jumper’s speed at a height of 12.5 m above the water is 6.37 m/s. 11. Fx = 8.6 N x = 9.4 cm = 0.094 m (a) k = ?

8.6 N

0.094 m91 N/m

x

x

F kxF

kx

k

=

=

=

=

The force constant of the spring is 91 N/m.


(b) Ee = ?

2e

2

e

121 (91.49 N/m)(0.094 m)20.40 J

E kx

E

=

=

=

The maximum energy of the spring is 0.40 J. 12. x = 1.0 10–7 m

Ee = 1.0 10–13 J k = ?

2e

e2

13

7 2

1

122

2(1.0 10 J)(1.0 10 m)

2.0 10 N/m

E kx

Ek

x

k

−

−

=

=

×=×

= ×

The force constant is 2.0 101 N/m. 13. m = 22 kg

θ = 29° k = 8.9 102 N/m x = 0.30 m d = ? To calculate the total vertical drop, ∆y:

T1 T2

g e

2

2

2 2

12

2

(8.9 10 N/m)(0.30 m)2(22 kg)(9.8 N/kg)

0.1858 m

E EE E

mg y kx

kxymg

y

=∆ =

∆ =

∆ =

×=

∆ =

To calculate the distance:

sin

sin0.1858 m

sin 290.38 m

yd

yd

d

θ

θ

∆=

∆=

=°

=

The crate slides 0.38 m along the ramp. 14. m = 0.20 kg

k = 28 N/m ∆y = ?


For fall, ∆y = x. At maximum fall, there will be no kinetic energy.

T1 T2

g e

2

2

2

2

12

2 0

28 2(0.20)(9.8) 0

28 3.92 0(28 3.92) 0

28 3.92 0 or 03.9228

0.14 m or 0

E EE E

mg y kx

kx mg y

x x

x xx x

x x

x

x x

=∆ =

∆ =

− ∆ =

− =

− =− =− = =

=

= =

Since x = 0 refers to the moment of release, the maximum stretch will be 0.14 m.

Applying Inquiry Skills 15. (a) (i) Line A (total energy is constant)

(ii) Line B (no kinetic energy at maximum stretch) (iii) Line C (maximum elastic potential energy a maximum stretch)

(b) By observation from the graph, A = 10.0 cm = 0.100 m (c) At the end, the total energy of 5.0 J is all elastic potential energy,

2e

e2

2

3

122

2(5.0 J)(0.100 m)

1.0 10 N/m

E kx

Ek

x

k

=

=

=

= ×

The force constant of the spring is 1.0 103 N/m. (d) Maximum kinetic energy is 5.0 J when there is no elastic potential energy.

2K

K

12

2

2(5.0 J)0.12 m

9.1 m/s

E mv

Evm

v

=

=

=

=

The maximum speed of the mass is 9.1 m/s. 16. (a) If the spring were cut in two, it would take more force to stretch the spring the same amount because each coil would

need to be moved twice as far. This will cause the force constant to double for the remaining two half springs. If a mass hung from two identical springs attached together caused them to stretch 36 cm, than each of them would stretch half of that amount. Since each bears the same weight, the force constant of each would only allow them to stretch 18 cm under the same force, indicating a force constant that is twice as great.

(b) Answers will vary.

Making Connections 17. m = 5.5 102 kg

number of vibrations = 6.0 t = 3.5 s

k = ?


Calculate the frequency:


6.03.5 s1.7143 Hz

f

f

=

=

=

To calculate the force constant:

2 2

2 2 2

4

12

2

4

4 (1.7143 Hz) (5.5 10 kg)

6.4 10 N/m

kfm

kfm

k f m

k

π

π

π

π

=

=

=

= ×

= ×

The force constant of each spring is 6.4 104 N/m. 18. a = 25g

f = 8.9 Hz A = ?

2 2

2

2 2

2

1 2

12

4

25(9.8 m/s )4 (8.9 Hz)0.078 m

ATaA

f a

Af a

aAf

A

π

π

π

π

π

=

=

=

=

=

=

The minimum amplitude is 0.078 m, or 7.8 10–2 m. CHAPTER 4 LAB ACTIVITIES

Activity 4.4.1: Applying the Law of Conservation of Energy (Page 220) (a) Energy conversions would be gravitational potential to kinetic to move the hands. (b) Different types include water clocks, hour-glass clocks, and pendulum clocks. (c) – (f) Answers will vary based on the students’ choice of design.

Investigation 4.5.1: Testing Real Springs (Pages 220–221)

Questions (i) By graphing the stretch as a function of the applied force we can learn the relationship is linear. (ii) The force constant of two combined springs is always less than the force constant for either spring individually.


Use conservation of energy to determine the amount of stretch required in the spring. Assume the spring mass is 18 g.

T1 T2

e g

2 2

2

2

1 12 2

(0.018 kg)(5.2 m/s)27.5 N/m

0.13 m

E EE E

kx mv

mvxk

x

== ∆

=

=

=

=

2. The height could be determined by analyzing the vertical component of velocity.

2 2f i

2 2f i

2 2

2

2

2(0 m/s) (5.2 m/s)

2( 9.8 m/s )1.4 m

v v a d

v vda

d

= + ∆

−∆ =

−=−

∆ =

3. The three forms of friction would use up some of the elastic potential energy intended to launch the spring. To compensate, the spring would need to be stretched an extra amount.

4. The energy to stretch the string would come from food you ate. The energy stored in the spring would be converted to kinetic energy as it left. The kinetic energy would be partially converted into gravitational potential as it rises, and than reconverted back to kinetic as it lands. As it lands on the floor and slides to a stop, the kinetic energy is converted into sound and thermal energy.

5. The spring would cause a mass to move up and down with decreasing amplitude until it came to rest. 6. The concepts and equations can be used to design and manufacture vehicles, beds, and a wide variety of other products. CHAPTER 4 SELF QUIZ (Page 225)

True/False 1. T 2. F The work done by gravity is zero. 3. F The work you do on the backpack is negative. 4. F The gravitational potential energy decreases in proportion to the distance fallen. 5. F This does not refute the law of conservation of energy because some energy is converted into other forms, such as heat

(thermal energy) in the ball and the floor. 6. T 7. T 8. F Maximum speed occurs at the equilibrium position, but elastic potential energy is at a minimum at the point of

minimum extension of the spring. 9. F A long damping time would not be appropriate for a bathroom scale. It would be appropriate for a “jolly-jumper” toy.

Multiple Choice 10. (c) 11. (c) 12. (e) 13. (d) 14. (e) 15. (a) 16. (d) ( cos )W F dθ= ∆

( sin )(cos180 )

sinmg L

W mgLθ

θ= °= −


CHAPTER 4 REVIEW (Pages 226–229)

Understanding Concepts 1. (a) No work is done because the force is perpendicular to the displacement. (b) The work is negative because the student is exerting a force opposite the direction of motion. (c) The work is negative because gravity is exerting a force opposite the direction of motion. (d) Assuming a level roadway, the work done is zero because the force is perpendicular to the displacement. (e) No work is done because the electrical force is perpendicular to the displacement. (f) No work is done because the tension is perpendicular to the displacement. 2. The force must be applied perpendicular to the object for it to do no work on the object. 3. The normal force can do work on an object. For example, when you jump, you push down on the ground and the normal

force pushes up on you and accelerates you up, giving you kinetic energy. 4. (a) No work is being done on the swimmer because the balancing forces forward and backward produce no motion. (b) Work is done on the student to speed him up, but after that, there is no work being done on the student. Once the

student reaches the speed of the current, the only force is the upward buoyant force, which is perpendicular to the waters surface. Technically, a small amount of work is being done by gravity as they whole river/student system is pulled closer to the earth.

5. (a) The velocity will be changing because Newton’s second law states that if a net force acts on an object, it will accelerate (i.e., change its velocity).

(b) It is possible the speed is constant if the particle is travelling in a circle. (c) The kinetic energy is proportional to the square of the speed and the mass. Since both can remain constant under these

conditions, the kinetic energy may be constant. 6. Agree. In the absence of friction (including air resistance), all of the gravitational potential energy will be converted into

kinetic energy. The mass will cancel in each term. 7. (a) Damped vibrations are useful in the suspension of a vehicle. (b) Damped vibrations are not useful in a pendulum clock. 8. It is not possible to have a motion that is not damped. Such a device would be a perpetual motion machine that cannot

exist. The force of friction within the system cannot be avoided. 9. m = 0.425 kg

∆y = 11.8 m (a) W = ?

2

( cos )( cos )

(0.425 kg)(9.80 m/s )(cos180 )(11.8 m)49.1 J

W F dmg y

W

θθ

= ∆= ∆

= °= −

The work gravity does on the ball on the way up is –49.1 J. (b) W = ?

2

( cos )( cos )

(0.425 kg)(9.80 m/s )(cos 0 )(11.8 m)49.1 J

W F dmg y

W

θθ

= ∆= ∆

= °=

The work gravity does on the ball on the way down is 49.1 J. 10. F = 9.3 N

W = 87 J ∆d = 11 m θ = ?


1

1

( cos )

cos

cos

87 Jcos(9.3 N)(11 m)

32

W F dW

F dW

F d

θ

θ

θ

θ

−

−

= ∆

=∆

= ∆

=

= °

The angle between the applied force and the horizontal is 32°. 11. Let the subscript C represent the child, and TO represent the toboggan.

mC = 25.6 kg mTO = 4.81 kg ∆y = 27.3 m

(a) WC = ?

First, find the actual distance:

27.3sin

27.3sin

d

d

φ

φ

=∆

∆ =

The applied force, FA, will be equal to the component of gravity down the hill, mgsin φ

C

3C

( cos )( sin )(cos )

27.3( sin )(cos )sin

( cos )(27.3)(4.81 kg)(9.80 N/kg)(cos 0 )(27.3)

1.29 10 J

W F dmg d

mg

mg

W

θφ θ

φ θφ

θ

= ∆= ∆

=

== °

= ×

The total work done by the child is 1.29 103 J. (b) From part (a), the angle doesn’t matter, therefore W = 1.29 × 103 J. (c) The total work on the child and toboggan during the slide will be equal to the work done to take them up the hill. Using

the equation derived in part (a),

T

3T

( cos )(27.3)(25.6 kg 4.81 kg)(9.80 N/kg)(cos 0 )(27.3)

8.14 10 J

W mg

W

θ== + °

= ×

The total work on the child and the toboggan is 8.14 103 J. 12. m = 73 kg

θ = 9.3° vi = 4.2 m/s

∆d = ?


The relation between the distance along the slope and the vertical height is:

sin 9.3

sin 9.3

yd

y d

∆° =∆

∆ = ∆ °

To calculate the change in distance:

T1 T2

2

2

2

2

12

2 ( sin 9.3 )

2 (sin 9.3 )

(4.2 m/s)2(9.8 m/s)(sin 9.3 )5.6 m

E E

mv mg y

v g d

vdg

d

=

= ∆

= ∆ °

∆ =°

=°

∆ =

The skier would travel 5.6 m along the hill before stopping. 13. An increase of 50% is equivalent of multiplying by 1.5:

22

K2

2K11

2K1 2

2K1 1

21

21

2 21 1 1

21 1

2

1

1

1212

1.50

( 2.00)1.5

1.5 4 4

0.5 4 4 0

( 4) ( 4) 4(0.5)( 4)2(0.5)

4 4.8991

8.90 m/s, or 0 899 m/s

mvEE mv

E vE v

vv

v v v

v v

v

v .

=

=

+=

= + +

− − =

− − ± − − −=

±=

= −

Since the negative value is not admissible, the original speed of the object was 8.90 m/s. 14. m = 7.0 109 kg

∆y = 36 m (a) ∆Eg = ?

g

9

12g

(7.0 10 kg)(9.8 N/kg)(36 m)

2.5 10 J

E mg y

E

∆ = ∆

= ×

∆ = ×

The gravitational potential energy is 2.5 1012 J. (b) The work done by on the pyramid by one person in 20 years is:

( ) ( )6

85 10 J 40 d0.20 20 a 8 10 J/persond a

× = ×

To calculate the total number of people:

12

38

2.5 10 3 10 people8 10

× = ××

There were 3 103 workers involved.


15. m = 45 kg ∆d = 66 cm = 0.66 m (a) W = ?

2

( cos )( cos )(45 kg)(9.80 N/kg)(cos180 )(0.66 m)

2.9 10 J

W F dmg d

W

θθ

= ∆= ∆= °

= − ×

The work done by gravity on the mass is –2.9 102 J. (b) W = ?

2

( cos )( cos )(45 kg)(9.80 N/kg)(cos 0 )(0.66 m)

2.9 10 J

W F dmg d

W

θθ

= ∆= ∆= °

= ×

The work done by the weightlifter on the mass is 2.9 102 J. (c) ∆Eg = ?

g

2g

(45 kg)(9.80 N/kg)(0.66 m)

2.9 10 J

E mg y

E

∆ = ∆

=

∆ = ×

The change in gravitational potential energy is 2.9 102 J. 16. m = 47 g = 0.047 kg (a) ∆y = 4.3 m v2 = ?

T1 T2

2 21 1 2 2

1 12 2

E E

mv mgy mv mgy

=

+ = +

Since v1 = 0:

21 2 222 1 2

2 1 2

2

2

2 2

2 2

2 ( )

2(9.8 m/s )(4.3 m)9.2 m/s

gy v gy

v gy gy

v g y y

v

= +

= −

= −

==

The speed of the stick just before it hits the ground is 9.2 m/s. (b) If air resistance was included, the answer in part (a) would be slightly smaller. The energy from gravity would be

shared between the thermal energy and the kinetic energy. 17. y1 – 27 m

v1 = 18 m/s θ = 37°

(a) v2 = ?

T1 T2

2 21 1 2 2

2 21 1 2 2

2 22 1 1 2

22 1 1 2

2 2

2

1 12 2

2 2

2 2

2 ( )

(18 m/s) 2(9.8 m/s )(27 m 0 m)29 m/s

E E

mv mgy mv mgy

v gy v gy

v v gy gy

v v g y y

v

=

+ = +

+ = +

= + −

= + −

= + −=

The speed of the stick just before it hits the ground is 29 m/s. (b) The angle the stick is thrown at does not appear in the equation, so the answer will still be 29 m/s.


18. F = 1.5 102 N [22° below the horizontal] m = 18 kg

∆d = 1.6 m µk = 0.55

(a) FN = ? FK = ? To calculate the normal force:

N A

N A2

2N

0

sin 22 0sin 22

(1.5 10 N)(sin 22 ) (18 kg)(9.8 N/kg)232.6 N

2.3 10 N

yF

F F mgF F mg

F

Σ =

− ° − == ° +

= × ° +=

= ×

To calculate the force of friction:

K K N

2K

(0.55)(232.6 N)127.9

1.3 10 N

F F

F

µ===

= ×

The normal force on the box is 2.3 102 N. The force of friction on the box is 1.3 102 N. (b) v = ? If the box is starting from rest, there is no initial kinetic energy.

th K

2K

2K

K

2

1( cos )2

2( cos ) 2

2( cos ) 2

2(1.5 10 N)(cos 22 )(cos 0 )(1.6 m) 2(127.9 N)(1.6 m)18 kg

1.4 m/s

W E E

F d F d mv

mv F d F d

F d F dvm

v

θ

θ

θ

= +

∆ = ∆ +

= ∆ − ∆

∆ − ∆=

× ° ° −=

=

The final speed of the box is 1.4 m/s. (c) Eth = ?

th K

2th

(127.9 N)(1.6 m)

2.0 10 J

E F d

E

= ∆=

= ×

The amount of thermal energy produced is 2.0 102 J.


19. m = 1.2 103 kg v1 = 9.5 103 m/s F = 9.2 104 N ∆d = 86 km = 86 103 m v2 = ?

K

2 2f i

2 2f i

2f i

4 33 2

3

4f

1 12 2

1 12 2

2

2(9.2 10 N)(86 10 m) (9.5 10 m/s)1.2 10 kg

1.0 10 m/s

W E

F d mv mv

mv F d mv

F dv vm

v

= ∆

∆ = −

= ∆ +

∆= +

× ×= + ××

= ×

The final speed of the probe is 1.0 104 m/s. 20. y1 = 1.15 m

y2 = 4.75 m v1 = ?

T1 T2

K1 g1 g2

21 1 2

21 2 1

1 2 1

2

1

12

2( )

2 ( )

2(9.80 m/s )(4.75 m 1.15 m)8.40 m/s

E EE E E

mv mgy mgy

v gy gy

v g y y

v

=+ =

+ =

= −

= −

= −=

The speed with which the gymnast leaves the trampoline is 8.40 m/s. 21. The work in the area under the graph. Consider the area in three parts:

triangle 1 rectangle triangle 2

1 1 2 21 12 21 1(1.0 m)(12.0 N) (2.0 m 1.0 m)(12.0 N) (6.0 m 2.0 m)(12.0 N)2 242 J

W A A A

b h lw b h

W

= + +

= + +

= + − + −

=

The person does 42 J of work. 22. x = 0.418 m

F = 1.00 102 N (a) k = ?

21.00 10 N

0.418 m239 N/m

x

x

F kxF

kx

k

=

=

×=

=

The force constant of the spring is 239 N/m. (b) x = 0.150 m

Fx = ?


(239 N/m)(0.150 m)35.9 N

x

x

F kx

F

===

The force required to stretch the spring is 35.9 N. (c) To stretch it 0.150 m:

2e

2

e

121 (239 N/m)(0.150 m)22.69 J

E kx

E

=

=

=

To compress it 0.300 m:

2e

2

e

121 (239 N/m)( 0.300 m)210.8 J

E kx

E

=

= −

=

The work required is 2.69 J to stretch it, and 10.8 J to compress it. 23. k = 22 N/m

m = 7.5 10–3 kg FK = 4.2 10–2 N x = 3.5 cm = 0.035 m ∆d = ?

T1 T2

e th

2K

2

K2

2

12

2

(22 N/m)(0.035 m)2(4.2 10 N)

0.32 m

E EE E

kx F d

kxdF

d

−

==

= ∆

∆ =

=×

∆ =

The eraser will slide 0.32 m along the desk. 24. k = 75 N/m

A = 0.15 m v = 1.7 m/s x = 0.12 m m = ?

For SHM, all of the energy is Ee when at the maximum amplitude, A:

( ) ( )( )

T1 T2

e max e K

2 2 2

2 2 2

2 2

2

2 2

2

1 1 12 2 2

( )

75 N/m 0.15 m 0.12 m

(1.7 m/s)0.21 kg

E EE E E

kA kx mv

mv kA kx

k A xmv

m

== +

= +

= −

−=

−=

=

The mass of the block is 0.21 kg.


25. m = 0.42 kg k = 38 N/m A = 5.3 cm = 0.053 m

(a) Ee = ? Maximum energy occurs at full amplitude.

2e

2

e

121 (38 N/m)(0.053 m)20.053 J

E kx

E

=

=

=

The maximum energy of the mass-spring system is 0.053 J. (b) v = ? Maximum speed occurs when there is no elastic potential energy:

T1 T2

e max K

2 2

2 2

2

2

1 12 2

38 N/m(0.053 m)0.42 kg

0.50 m/s

E EE E

kA mv

mv kA

kAvm

v

==

=

=

=

=

=

The maximum speed of the mass is 0.50 m/s. (c) x = 4.0 cm = 0.040 m v = ?

( ) ( )( )

T1 T2

e max e K

2 2 2

2 2 2

2 2

2 2

1 1 12 2 2

( )

38 N/m 0.053 m 0.040 m

0.42 kg0.33 m/s

E EE E E

kA kx mv

mv kA kx

k A xvm

v

== +

= +

= −

−=

−=

=

The speed of the mass is 0.33 m/s. (d) x = 4.0 cm = 0.040 m ET = ?

T e K

2 2

2 2

T

1 12 21 1(38 N/m)(0.040 m) (0.42 kg)(0.33 m/s)2 20.053 J

E E E

kx mv

E

= +

= +

= +

=

The total energy is 0.053 J. The results are the same.


Applying Inquiry Skills 26. (a) One text weighs 2.0 kg and has a thickness of 3.6 cm. The first one doesn’t need to be raised at all.

g1 g2 g3 g4 g5

1 2 3 4 5

1 2 3 4 5( )(2.0 kg)(9.8 N/kg)(0 m 0.036 m 0.072 m 0.108 m 0.144 m)7.1 J

W E E E E E

mgy mgy mgy mgy mgymg y y y y y

W

= + + + +

= + + + += + + + += + + + +=

You would have to do 7.1 J of work. (b) Some errors would be determining the mass of the text. Also, the thickness may compress the books on the bottom.

Not all texts may have the same mass. 27. The calculator would need to know the distance the box was pushed. 28.

29. The tractor seat would need to have a strong spring to absorb large bumps, and a shock absorber to prevent “launching”

the driver. There would have to be smaller springs on top of that to absorb small vibrations. Damping would be important to prevent resonance.

Making Connections 30. Roller coasters are often shut down in high winds because of the loss of energy that may occur due to increased air

resistance. Cold can cause parts to shrink and increase frictional forces beyond safe limits. 31. (a) The ball on track B will arrive first. Shortly after the start, the vertical drop of the ball on track B causes an increase in

the speed, which it will enjoy for the majority of the race. At the end, it will slow down to the same speed that the ball on track A has just accelerated to.

(b) Racing cyclists use this in a variety of ways. The sprinters stay high on the track until ready to make a break for it, converting all of their stored gravitational energy into kinetic. Team racers use the change to minimize the work of the rider. When the leader is ready to give up his spot, he rides up the hill a bit, converting some of his kinetic energy into gravitational potential energy, reducing his speed. Once the last team mate has passed, he can drop down again, gaining the gravitational potential energy back as kinetic energy without needing to supply it from his own body power.

32. (a) v1 = 0 m/s v2 = ?

T1 T2

2 21 1 2 2

2 21 1 2 2

22 1 2

2 1 2

2

2

1 12 2

2 2

2 2

2 ( )

2(9.80 m/s )(37.8 m 17.8 m)19.8 m/s

E E

mv mgy mv mgy

v gy v gy

v gy gy

v g y y

v

=

+ = +

+ = +

= −

= −

= −=

The speed of the coaster at position C is 19.8 m/s.


(b) v1 = 5.00 m/s v2 = ?

T1 T2

2 21 1 2 2

2 21 1 2 2

2 22 1 1 2

22 1 1 2

2 2

2

1 12 2

2 2

2 2

2 ( )

(5.00 m/s) 2(9.80 m/s )(37.8 m 17.8 m)20.4 m/s

E E

mv mgy mv mgy

v gy v gy

v v gy gy

v v g y y

v

=

+ = +

+ = +

= + −

= + −

= + −=

The speed of the coaster at position C is 20.4 m/s. 33. (a) The mass is not necessary because it cancels out of the mathematical equations. (b) You might expect the speed would be 5.0 m/s more at the second point, but the reality is that the small kinetic energy

supplied by having a speed going over the first hill is negligible compared with the large amount of gravitational potential energy.

34. a = 12 g [upward] (a) FA = ?

A

A

A

( )(0.12 )

1.12

yF ma

F mg maF ma mg

m a gm g g

F mg

Σ =

− == += += +=

The force required is 1.12mg. (b) W = ?

( cos )(1.12 )(cos 0 )( )1.12

W F dmg y

W mg y

θ= ∆= ° ∆= ∆

The work done is 1.12mg∆y. 35. m = 1.5 kg

k = 2.1 103 N/m ∆y = 0.37 m x = ?


T1 T2

2

3 2

2

2

121(1.5)(9.8)(0.37 ) (2.1 10 )2

5.439 14.7 1050

1050 14.7 5.439 0

E E

mg y kx

x x

x x

x x

=

∆ =

+ = ×

+ =

− − =

Using the quadratic equation:

2( 14.7) ( 14.7) 4(1050)( 5.439)2(1050)

14.7 151.92100

0.079 m or 0.065 m (negative value inadmissable)0.079 m

x

xx

− − ± − − −=

±=

= = −=

The maximum distance the spring is compressed is 0.079 m. 36. m = 0.55 g = 5.5 10–4 kg

∆d = 95 cm = 0.95 m ∆dx = 3.7 m Ee = ? First we must calculate the time required for the vertical drop (vi = 0):

2i

2

2

1 ( )2

1 ( )2

2

2( 0.95 m)9.8 m/s

0.44 s

d v t a t

d a t

dta

t

∆ = ∆ + ∆

∆ = ∆

∆∆ =

−=−

∆ =

To calculate the horizontal speed:

3.7 m0.44 s8.4 m/s

x x

xx

x

d v td

vt

v

∆ = ∆∆

=∆

=

=

We know that initial kinetic energy came from elastic potential energy, therefore:

T1 T2

e K

2

4 2

e

121 (5.5 10 kg)(8.4 m/s)20.019 J

E EE E

mv

E

−

==

=

= ×

=

The elastic potential energy stored was 0.019 J.


37. y1 = 16 m y2 = 9.0 m H = ? First we must solve for the launch speed if v1 = 0:

T1 T2

2 21 1 2 2

21 2 222 1 2

2 1 2

2

2

1 12 2

2 2

2 2

2 ( )

2(9.8 m/s )(16 m 9.0 m)11.71 m/s

E E

mv mgy mv mgy

gy v gy

v gy gy

v g y y

v

=

+ = +

= +

= −

= −

= −=

Resolve vector components:

As just clearing the wall, the horizontal component will be the same, and the vertical component must produce a 30º angle

to the vertical.


To calculate the vertical speed:

( )

2

2

cos 45tan 30

cos 45tan 30

11.71 m/s cos 45tan 30

14.35 m/s [down]

y

y

y

vv

vv

v

°° =

°=

°°

=°

=

Determine how far the skier has dropped from the launch point:

( )( )( )

2 2f i

2 2f i

22

2

2

2

( 14.35 m/s) 11.71 m/s sin 45

2( 9.8 m/s )7.0 m

v v a d

v vda

d

= + ∆

−∆ =

− − °=

−∆ = −

Therefore, the wall must be 9.0 – 7.0 = 2.0 m tall. 38. y1 = 2.5 m

v1 = 9.0 m/s y2 = 3.0 m v2 = ?

T1 T2

2 21 1 2 2

2 21 1 2 2

2 22 1 1 2

22 1 1 2

2 2

2

1 12 2

2 2

2 2

2 ( )

(9.0 m/s) 2(9.8 m/s )(2.5 m 3.0 m)8.4 m/s

E E

mv mgy mv mgy

v gy v gy

v v gy gy

v v g y y

v

=

+ = +

+ = +

= + −

= + −

= + −=

The speed of the ball when it “swishes” through the hoop is 8.4 m/s. 39. g = 2.0

v1 = 6.0 y2 = 5.0 S = ?

Solve for the launch speed at the top of the ramp where y1 = 0:

T1 T2

2 21 1 2 2

2 21 2 2

2 22 1 2

22 1 2

2

2

1 12 2

0 2

2

2

(6.0) 2(2.0)(5.0)4.0 units

E E

mv mgy mv mgy

v v gy

v v gy

v v gy

v

=

+ = +

+ = +

= −

= −

= −=


Analyze projectile motion to find the change in time:

2i

2

2

2

1 ( )2

15.0 (4.0sin 30 ) ( 2.0)( )2

5 2 ( )

( ) 2 5 0

y yd v t a t

t t

t t

t t

∆ = ∆ + ∆

− = ° ∆ + − ∆

− = ∆ − ∆

∆ − ∆ − =

Using the quadratic equation:

2( 2) ( 2) 4(1)( 5)2(1)

2 4.92

3.45 units or 1.45 units (dismiss negative answer)3.45 units

t

t

− − ± − − −∆ =

±=

= −∆ =

To calculate horizontal range (S):

(4.0cos30 )(3.45)12 units

xS v t

S

= ∆= °=

The shuttle lands a distance of 12 units from the ramp.

4.1 WORK DONE BY A CONSTANT FORCE · energy. 4.2 KINETIC ENERGY AND THE WORK–ENERGY THEOREM...

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Transcript of 4.1 WORK DONE BY A CONSTANT FORCE · energy. 4.2 KINETIC ENERGY AND THE WORK–ENERGY THEOREM...