WorkEnergy Theorem Reviewed - Physics and Astronomy at TAMU · child can push off and bounce....

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Chapter 6Due: 11:59pm on Sunday, October 9, 2016

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WorkEnergy Theorem ReviewedDescription: Simple questions about the workenergy theorem to test understanding, then one easy analytic questionapplying the theorem.

Learning Goal:

Review the workenergy theorem and apply it to a simple problem.

If you push a particle of mass in the direction in which it is already moving, you expect the particle's speed to increase. Ifyou push with a constant force , then the particle will accelerate with acceleration (from Newton's 2nd law).

Part A

Enter a one or twoword answer that correctly completes the following statement.

If the constant force is applied for a fixed interval of time , then the _____ of the particle will increase by an amount .

Hint 1. Kinematic equations recalled

Recall the kinematic equations

.

ANSWER:

Part B

Enter a one or twoword answer that correctly completes the following statement.

If the constant force is applied over a given distance , along the path of the particle, then the _____ of the particle willincrease by .

ANSWER:

The work done on the particle by the force over the distance is .

Chapter 6 [ Edit ]

Overview Summary View Diagnostics View Print View with Answers

MF a = F/M

t at

v = u + at ;s = ut+ a ;1

2 t2

= + 2asv2 u2

speed

Also accepted: velocity, v

DFD

kinetic energy

Also accepted: energy, KE, K

W F D FD

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Part C

If the initial kinetic energy of the particle is , and its final kinetic energy is , express in terms of and the work done on the particle.

ANSWER:

This is the workenergy theorem, often written . It is, essentially, a statement of energy conservationthat does not include potential energy explicitly. All forceseven conservative forces like gravitycontribute to thework.

Part D

In general, the work done by a force is written as

.Now, consider whether the following statements are true or false:

The dot product assures that the integrand is always nonnegative.The dot product indicates that only the component of the force perpendicular to the path contributes to theintegral.The dot product indicates that only the component of the force parallel to the path contributes to the integral.

Enter t for true or f for false for each statement. Separate your responses with commas (e.g., t,f,t).

ANSWER:

Part E

Assume that the particle has initial speed . Find its final kinetic energy in terms of , , , and .

Hint 1. Find the initial kinetic energy

Express the initial kinetic energy in terms of the particle's initial velocity and its mass .

ANSWER:

ANSWER:

Ki Kf Kf KiW

= Kf

W = −Kf Ki

F

W = ( )⋅d∫ fi F r r

f,f,t

vi Kf vi M F D

Ki vi M

= Ki

= Kf



Part F

What is the final speed of the particle?

Express your answer in terms of and .

ANSWER:

Work on a Sliding BlockDescription: A box is pushed up a frictionless incline. Find the work done by gravity, the pushing force, and the normalforce.

A block of weight sits on a frictionless inclined plane, which makes an angle with respect to the horizontal, as shown. Aforce of magnitude , applied parallel to the incline, pulls the blockup the plane at constant speed.

Part A

The block moves a distance up the incline. The block does not stop after moving this distance but continues to movewith constant speed. What is the total work done on the block by all forces? (Include only the work done after theblock has started moving, not the work needed to start the block moving from rest.)

Express your answer in terms of given quantities.

Hint 1. What physical principle to use

To find the total work done on the block, use the workenergy theorem:.

Hint 2. Find the change in kinetic energy

What is the change in the kinetic energy of the block, from the moment it starts moving until it has been pulled adistance ? Remember that the block is pulled at constant speed.

Hint 1. Consider kinetic energy

Kf M

= vf

w θF

LWtot

= −Wtot Kf Ki

L



If the block's speed does not change, its kinetic energy cannot change.

ANSWER:

ANSWER:

Part B

What is , the work done on the block by the force of gravity as the block moves a distance up the incline?

Express the work done by gravity in terms of the weight and any other quantities given in the problemintroduction.

Hint 1. Force diagram

Hint 2. Force of gravity component

What is the component of the force of gravity in the direction of the block's displacement (along the inclinedplane)?

Express your answer in terms of and .

Hint 1. Relative direction of the force and the motion

Remember that the force of gravity acts down the plane, whereas the block's displacement is directed upthe plane.

ANSWER:

= 0−Kf Ki

= 0Wtot

Wg L

w

w θ

= Fg||



ANSWER:

Part C

What is , the work done on the block by the applied force as the block moves a distance up the incline?

Express your answer in terms of and other given quantities.

Hint 1. How to find the work done by a constant force

Remember that the work done on an object by a particular force is the integral of the dot product of the force andthe instantaneous displacement of the object, over the path followed by the object. In this case, since the force isconstant and the path is a straight segment of length up the inclined plane, the dot product becomes simplemultiplication.

ANSWER:

Part D

What is , the work done on the block by the normal force as the block moves a distance up the inclined plane?

Express your answer in terms of given quantities.

Hint 1. First step in computing the work

The work done by the normal force is equal to the dot product of the force vector and the block's displacementvector. The normal force and the block's displacement vector are perpendicular. Therefore, what is their dotproduct?

ANSWER:

ANSWER:

± Baby Bounce with a HookeDescription: ± Includes Math Remediation. Application of Hooke's Law

=

Also accepted:

Wg

WF F L

F

L

= WF

Wnormal L

= 0⋅N L

= 0Wnormal



One of the pioneers of modern science, Robert Hooke (16351703), studied the elastic properties of springs and formulatedthe law that bears his name. Hooke found the relationship among the force a spring exerts, , the distance from equilibriumthe end of the spring is displaced, , and a number called the spring constant (or, sometimes, the force constant of thespring). According to Hooke, the force of the spring is directly proportional to its displacement from equilibrium, or

.

In its scalar form, this equation is simply

.

Where is the force exerted on the spring and is the extension of the spring from equilibrium caused by the force . Thevalue of depends on the geometry and the material of the spring; it can be easily determined experimentally using thisscalar equation.

Toy makers have always been interested in springs for the entertainment value of the motion they produce. One wellknownapplication shown in is a baby bouncer, which consists of a harnessseat for a toddler, attached to a spring. The entire contraption hooksonto the top of a doorway. The idea is for the baby to hang in theseat with his or her feet just touching the ground so that a goodpush up will get the baby bouncing, providing potentially hours ofentertainment.

Part A

The following chart and accompanying graph shown in depictan experiment to determine the spring constant for a babybouncer.

Displacement fromequilibrium,

( )

Force exerted onthe spring,

( )

0 0

0.005 2.5

0.010 5.0

0.015 7.5

0.020 10

What is the spring constant of the spring being tested for thebaby bouncer?

Express your answer to two significant figures in newtons per meter.

Hint 1. How to approach the problem

Look at the pattern in the data to determine what number must multiply the distance to achieve the force exertedon the spring. Look at both the table and the graph.

F x k

= −kF x

F = kx

F x Fk

xm

FN

k



Hint 2. Find the spring constant from the graph

The relationship between the displacement and force is linear. This set of data follows the form of ,where is the slope of the line and is the y intercept. For all springs, the force is 0 when the displacement is 0so . This leaves the slope of the line to determine the relationship between displacement and force. What isthe slope that you get from the graph?

Express your answer as a fraction in unsimplified form.

Hint 1. Slope equation

Slope is given by the change in divided by the change in . In this case, .

ANSWER:

All you need to do now is to convert the fraction to its decimal value.

ANSWER:

Part B

One of the greatest difficulties with setting up the baby bouncer is determining the right height above the floor so that thechild can push off and bounce. Knowledge of physics can be really helpful here.If the spring constant , the baby has a mass , and the baby's legs reach a distance

from the bouncer, what should be the height of the "empty" bouncer above the floor?

Express your answer in meters to two significant figures.

Hint 1. How to approach the problem

Use Hooke's law to determine the displacement of the spring from equilibrium given the force the spring mustexert to hold up the baby. The displacement must lower the baby toward the floor until the baby's feet can touch.

y = mx + bm b

b = 0

y x m = ΔF/Δx

= 500 k N/m

= 500 k N/m

k = 5.0 × N/m102 m = 11 kgd = 0.15 m h

x



Hint 2. Which force to use

The force the spring exerts is equal in magnitude but opposite in direction to the force exerted on it by the weight of the baby.

Hint 3. Find the force exerted by the baby

The weight of the baby is equal to the force exerted on the spring. What is the weight of the baby?

Express your answer in newtons to three significant figures.

Hint 1. Formula for weight

Recall that the weight of an object is given by,

where is the mass of the object and is the acceleration due to gravity.

ANSWER:

Hint 4. Find the displacement of the spring

Use Hooke's law to determine how far the spring would stretch downward once the baby is placed in the seat.How far does the bottom end of the spring move?

Express your answer in meters to two significant figures.

ANSWER:

To finish the problem, you must consider the length of the baby's legs.

ANSWER:

A displacement of for the spring holding up a baby may not seem very large but you must consider howsmall babies are. Also, once the baby begins jumping up and down, the extra energy allows the spring to stretchfurther than 0.22 and a resonant frequency may be achieved. At resonance the bouncing may become too violent,leading to a potentially dangerous situation for the little bouncer.

Exercise 6.10Description: An ##kg package in a mailsorting room slides ## m down a chute that is inclined at ## degree(s) belowthe horizontal. The coefficient of kinetic friction between the package and the chute's surface is ##. (a) Calculate the...

An 13.0 package in a mailsorting room slides 4.00 down a chute that is inclined at 55.0 below the horizontal. Thecoefficient of kinetic friction between the package and the chute's surface is 0.40.

w

w = mgm g

= 108 w N

= 0.22 x m

= 0.37 h m

−0.22 m

m

kg m ∘



Part A

Calculate the work done on the package by friction.

Express your answer with the appropriate units.

ANSWER:

Part B

Calculate the work done on the package by gravity.


ANSWER:

Part C

Calculate the work done on the package by the normal force.


ANSWER:

Part D

What is the net work done on the package?


ANSWER:

= = 120

Also accepted: = 117 , = 117 ,

= 120

Wfriction

= = 420

Also accepted: = 417 , = 418 , = 420

Wgravity

= 0WN

= = 300

Also accepted: = 301 ,

= 301 ,

= 300

Wnet



Exercise 6.20Description: A mkg watermelon is dropped from rest from the roof of a smtall building and feels no appreciable airresistance. (a) Calculate the work done by gravity on the watermelon during its displacement from the roof to the ground.(b) Just...

A 4.51 watermelon is dropped from rest from the roof of a 18.7 tall building and feels no appreciable air resistance.

Part A

Calculate the work done by gravity on the watermelon during its displacement from the roof to the ground.


ANSWER:

Part B

Just before it strikes the ground, what is the watermelon's kinetic energy?


ANSWER:

Part C

Just before it strikes the ground, what is the watermelon's speed?


ANSWER:

Part D

Would the answer in part A be different if there were appreciable air resistance?

ANSWER:

kg m

= = 827

Also accepted: = 827 , = 827

W

= = 827

Also accepted: = 827 , = 827

K

= = 19.1

Also accepted: = 19.2 , = 19.1

v



The work done by gravity would be the same.

Part E

Would the answer in part B be different if there were appreciable air resistance?

ANSWER:

Air resistance would do negative work and would be less than .

Part F

Would the answer in part C be different if there were appreciable air resistance?

ANSWER:

Air resistance would do negative work and would be less than .

Exercise 6.28Description: A soccer ball with mass m is initially moving with speed vi. A soccer player kicks the ball, exerting aconstant force of magnitude F in the same direction as the ball's motion. (a) Over what distance must the player's foot bein contact with the...

A soccer ball with mass 0.460 is initially moving with speed 2.30 . A soccer player kicks the ball, exerting a constantforce of magnitude 43.0 in the same direction as the ball's motion.

Part A

Over what distance must the player's foot be in contact with the ball to increase the ball's speed to 6.00 ?

ANSWER:

yes no

yes no

Wtot Wgrav

yes no

Wtot Wgrav

kg m/sN

m/s

= = 0.164 s m



Exercise 6.34Description: To stretch a spring l from its unstretched length, W of work must be done. (a) What is the force constant ofthis spring? (b) What magnitude force is needed to stretch the spring l from its unstretched length? (c) How much workmust be done to...

To stretch a spring 5.00 from its unstretched length, 13.0 of work must be done.

Part A

What is the force constant of this spring?

ANSWER:

If you need to use the value of the spring constant 'k' in subsequent parts, please use the unrounded full precisionvalue and not the one you submitted for this part rounded using three significant figures.

Part B

What magnitude force is needed to stretch the spring 5.00 from its unstretched length?

ANSWER:

Part C

How much work must be done to compress this spring 4.00 from its unstretched length?

ANSWER:

Part D

What force is needed to compress it this distance?

ANSWER:

Exercise 6.56

cm J

= = 1.04×104 k N/m

cm

= = 520 F N

cm

= = 8.32 W J

= = 416 F N



Description: When its engine of power P is generating full power, a small singleengine airplane with mass m gainsaltitude at a rate of v. (a) What fraction of the engine power is being used to make the airplane climb? (The remainder isused to overcome the...

When its engine of power 75 is generating full power, a small singleengine airplane with mass 710 gains altitude at arate of 2.6 .

Part A

What fraction of the engine power is being used to make the airplane climb? (The remainder is used to overcome theeffects of air resistance and of inefficiencies in the propeller and engine.)

Express your answer as a percentage using two significant figures.

ANSWER:

Alternative Exercise 6.113Description: The space shuttle Endeavour, with mass 86400 kg, is in a circular orbit of radius 6.66* 10^6 ( m) around theearth. It takes 90.1 min for the shuttle to complete each orbit. On a repair mission, the shuttle is cautiously moving xcloser to a disabled...

The space shuttle Endeavour, with mass 86400 , is in a circular orbit of radius around the earth. It takes 90.1 for the shuttle to complete each orbit. On a repair mission, the shuttle is cautiously moving 5.50 closer to a disabled

satellite every 16.5 .

Part A

Calculate the shuttle's kinetic energy relative to the earth.

ANSWER:

Part B

Calculate the shuttle's kinetic energy relative to the satellite.

ANSWER:

Problem 6.66Description: A mkg package slides s down a long ramp that is inclined at theta below the horizontal. The coefficient ofkinetic friction between the package and the ramp is mu_k=##. (a) Calculate the work done on the package by friction. (...

kW kgm/s

= 24 %

kg 6.66× m106

min ms

= 2.59×1012 K J

= = 4800 K J

kg ∘



A 5.00 package slides 2.80 down a long ramp that is inclined at 26.0 below the horizontal. The coefficient of kineticfriction between the package and the ramp is = 0.360.

Part A

Calculate the work done on the package by friction.


ANSWER:

Part B

Calculate the work done on the package by gravity.


ANSWER:

Part C

Calculate the work done on the package by the normal force.


ANSWER:

Part D

Calculate the total work done on the package.


ANSWER:

Part E

If the package has a speed of 2.50 at the top of the ramp, what is its speed after it has slid 2.80 down the ramp?

kg m ∘

μk

= = 44.4


Wf

= = 60.1


Wg

= 0Wn

= = 15.8


Wnet

m/s



If the package has a speed of 2.50 at the top of the ramp, what is its speed after it has slid 2.80 down the ramp?


ANSWER:

Problem 6.71Description: A small block with a mass of 0.0600 kg is attached to a cord passing through a hole in a frictionless,horizontal surface . The block is originally revolving at a distance of r_0 from the hole with a speed of v_0. The cord isthen...

A small block with a mass of 0.0600 is attached to a cord passing through a hole in a frictionless, horizontal surface . Theblock is originally revolving at a distance of 0.43 from the holewith a speed of 0.75 . The cord is then pulled from below,shortening the radius of the circle in which the block revolves to0.17 . At this new distance, the speed of the block is 1.90 .

Part A

What is the tension in the cord in the original situation when the block has speed = 0.75 ?


ANSWER:

m/s m

= = 3.54

Also accepted: = 3.54 ,

= 3.54

v

kgm

m/s

m m/s

v0 m/s

= = 7.8×10−2

Also accepted: = 7.85×10−2 , = 7.8×10−2

T



Part B

What is the tension in the cord in the final situation when the block has speed = 1.90 ?


ANSWER:

Part C

How much work was done by the person who pulled on the cord?


ANSWER:

Alternative Exercise 6.116Description: A block of ice with mass m is initially at rest on a frictionless, horizontal surface. A worker then applies ahorizontal force F_vec to it. As a result, the block moves along the xaxis such that its position as a function of time isgiven by...

A block of ice with mass 6.30 is initially at rest on a frictionless, horizontal surface. A worker then applies a horizontal force to it. As a result, the block moves along the xaxis such that its position as a function of time is given by ,

where = 0.200 and = 2.05×10−2 .

Part A

Calculate the velocity of the object at time = 4.40 .

Express your answer to three significant figures.

ANSWER:

v1 m/s

= = 1.3


T

= = 9.1×10−2

Also accepted: = 9.11×10−2 , = 9.1×10−2

W

kgF x(t) = α + βt2 t3

α m/s2 β m/s3

t s

= = 2.95 v m/s



Part B

Calculate the magnitude of at time = 4.40 .


ANSWER:

Part C

Calculate the work done by the force during the first time interval of 4.40 of the motion.


ANSWER:

Problem 6.76Description: The spring of a spring gun has force constant k = 400 N/m and negligible mass. The spring is compressed6.00 cm and a ball with mass 0.0300 kg is placed in the horizontal barrel against the compressed spring. The spring isthen released,...

The spring of a spring gun has force constant k = 400 and negligible mass. The spring is compressed 6.00 and aball with mass 0.0300 is placed in the horizontal barrel against the compressed spring. The spring is then released, andthe ball is propelled out the barrel of the gun. The barrel is 6.00 long, so the ball leaves the barrel at the same point that itloses contact with the spring. The gun is held so the barrel is horizontal.

Part A

Calculate the speed with which the ball leaves the barrel if you can ignore friction.

ANSWER:

Part B

Calculate the speed of the ball as it leaves the barrel if a constant resisting force of 6.00 acts on the ball as it movesalong the barrel.

ANSWER:

Part C

F t s

= = 5.93 F N

F s

= = 27.4 W J

N/m cmkg

cm

= 6.93 v m/s

N

= 4.90 v m/s



For the situation in part B, at what position along the barrel does the ball have the greatest speed? (In this case, themaximum speed does not occur at the end of the barrel.)

ANSWER:

Part D

What is that greatest speed?

ANSWER:

Problem 6.82Description: Consider the system shown in the figure. The rope and pulley have negligible mass, and the pulley isfrictionless. The coefficient of kinetic friction between the 8.00kg block and the tabletop is mu_k=0.250. The blocks arereleased from rest. (a)...

Consider the system shown in the figure. The rope and pulley have negligible mass, and the pulley is frictionless. Thecoefficient of kinetic friction between the 8.00kg block and the tabletop is . The blocks are released from rest.

Part A

Use energy methods to calculate the speed of the 6.00kg block after it has descended 1.50 .

ANSWER:

Problem 6.81

= 4.50×10−2 from start point x m

= 5.20 vmax m/s

= 0.250μk

m

= 2.90 v m/s



Description: Consider the system shown in the figure . The rope and pulley have negligible mass, and the pulley isfrictionless. Initially the 6.00kg block is moving downward and the 8.00kg block is moving to the right, both with a speedof v. The blocks come to ...

Consider the system shown in the figure . The rope and pulley havenegligible mass, and the pulley is frictionless. Initially the 6.00block is moving downward and the 8.00 block is moving to theright, both with a speed of 0.400 . The blocks come to restafter moving 7.00 .

Part A

Use the workenergy theorem to calculate the coefficient of kinetic friction between the 8.00 block and the tabletop.

ANSWER:

kgkg

m/sm

kg

= = 0.752μk

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WorkEnergy Theorem Reviewed - Physics and Astronomy at TAMU · child can push off and bounce....

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