physics chapter 8 answers

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8 – 1

Chapter 8: Potential Energy and Conservation of Energy

Answers to Even-Numbered Conceptual Questions 2. As water vapor rises, there is an increase in the gravitational potential energy of the system. Part of this

potential energy is released as snow falls onto the mountain. If an avalanche occurs, the snow on the mountain accelerates down slope, converting more gravitational potential energy to kinetic energy.

4. The initial mechanical energy of the system is the gravitational potential energy of the mass-Earth system. As the mass moves downward, the gravitational potential energy of the system decreases. At the same time, the potential energy of the spring increases as it is compressed. Initially, the decrease in gravitational potential energy is greater than the increase in spring potential energy, which means that the mass gains kinetic energy. Eventually, the increase in spring energy equals the decrease in gravitational energy and the mass comes to rest.

6. The object’s kinetic energy is a maximum when it is released, and a minimum when it reaches its greatest height. The gravitational potential of the system is a minimum when the object is released, and a maximum when the object reaches its greatest height.

8. The jumper’s initial kinetic energy is largely converted to a compressional, spring-like potential energy as the pole bends. The pole straightens out, converting its potential energy into gravitational potential energy. As the jumper falls, the gravitational potential energy is converted into kinetic energy, and finally, the kinetic energy is converted to compressional potential energy as the cushioning pad on the ground is compressed.

10. When the toy frog is pressed downward, work is done to compress the spring. This work is stored in the spring as potential energy. Later, when the suction cup releases the spring, the stored potential energy is converted into enough kinetic energy to lift the frog into the air.

12. The total mechanical energy E decreases with time if air resistance is present.

Solutions to Problems and Conceptual Exercises

1. Picture the Problem: The work done by a conservative force is

indicated at right for a variety of different paths connecting the points A and B.

Strategy: The work done by a conservative force is independent of the path taken. From the middle path, we see that the work to go from A to B is 15 J. Similarly, the work to go from B to A must be −15 J. Use this principle to determine the works done along paths 1 and 2.

Solution: 1. (a) For the total work along this path to be 15 J, the work on Path 1 must be −3 J.

2. (b) For the total work along this path to be −15 J, the work done on Path 2 must be −4 J.

Insight: If there were a nonconservative force such as friction the works would be path-dependent and we would need more information in order to answer the question.

Chapter 8: Potential Energy and Conservation of Energy James S. Walker, Physics, 4th Edition


8 – 2

2. Picture the Problem: The three paths of the object are depicted at right.

Strategy: Find the work done by gravity W mgy= when the object is moved downward, W mgy= − when it is moved upward, and zero when it is moved horizontally. Sum the work done by gravity for each segment of each path.

Solution: 1. Calculate the work for path 1:

[ ]( ) ( ) ( )

( )( )( )

1 1 2 3

21

0 04.0 m 1.0 m + 1.0 m

3.2 kg 9.81 m/s 2.0 m 63 J

W mg y y ymg

W

= − + + + += − +⎡ ⎤⎣ ⎦= − = −

2. Calculate W for path 2: [ ] ( )( )[ ]22 40 0 3.2 kg 9.81 m/s 2.0 m 63 JW mg y= − + = − = −

3. Calculate W for path 3: [ ] ( )( ) ( ) ( )2

3 5 60 3.2 kg 9.81 m/s 1.0 m 3.0 m 63 JW mg y y= + − = − = −⎡ ⎤⎣ ⎦

Insight: The work is path-independent because gravity is a conservative force. 3. Picture the Problem: The three paths of the sliding box are depicted at right.

Strategy: The work done by friction is kW mgdμ= − , where d is the distance the box is pushed irregardless of direction, because the friction force always acts in a direction opposite the motion. Sum the work done by friction for each segment of each path.

Solution: 1. Calculate the work for path 1:

[ ][ ]( )( )[ ]

1 k 1 2 3 4 5

k2

1

4.0 4.0 1.0 1.0 1.0 m0.26 3.7 kg 9.81 m/s 11.0 m 100 J

W mg d d d d dmg

W

μμ

= − + + + += − + + + +

= − = −

2. Calculate W for path 2: [ ]( )( ) ( ) ( ) ( )

2 k 6 7 820.26 3.7 kg 9.81 m/s 2.0 m 2.0 m 1.0 m 47 J

W mg d d dμ= − + +

= − + + = −⎡ ⎤⎣ ⎦

3. Calculate the work for path 3: [ ]( )( ) ( ) ( ) ( )

3 k 9 10 1120.21 3.2 kg 9.81 m/s 1.0 m 3.0 m 3.0 m 66 J

W mg d d dμ= − + +

= − + + = −⎡ ⎤⎣ ⎦

Insight: The amount of work done depends upon the path because friction is a nonconservative force. 4. Picture the Problem: The physical situation is depicted at right.

Strategy: Use equation 212W kx= (equation 7-8) to find the work done

by the spring, but caution is in order: This work is positive when the force exerted by the spring is in the same direction that the block is traveling, but it is negative when they point in opposite directions. One way to keep track of that sign convention is to say that

( )2 21i f2W k x x= − . That way the work will always be negative if you

start out at i 0x = because the spring force will always be in the opposite direction from the stretch or compression.

Solution: 1. (a) Sum the work done by the spring for each segment of path 1:

( ) ( )( ) ( ) ( ) ( ){ }

( ) ( )

2 2 2 211 1 2 2 32

2 22 212

1

550 N/m 0 0.040 m 0.040 m 0.020 m

0.44 J 0.33 J 0.11 J

W k x x x x

W

⎡ ⎤= − + −⎣ ⎦

⎡ ⎤ ⎡ ⎤= − + −⎣ ⎦ ⎣ ⎦

= − + = −



8 – 3

2. Sum the work done by the spring for each segment of path 2:

( ) ( )( ) ( ) ( ) ( ){ }

( ) ( )

2 2 2 212 1 4 4 32

2 22 212

2

550 N/m 0 0.020 m 0.020 m 0.020 m

0.1 J 0 J 0.11 J

W k x x x x

W

⎡ ⎤= − + −⎣ ⎦

⎡ ⎤ ⎡ ⎤= − − + − −⎣ ⎦ ⎣ ⎦

= − + = −

3. (b) The work done by the spring will stay the same if you increase the mass because the results do not depend on the mass of the block.

Insight: The work done by the spring is negative whenever you displace the block away from x = 0, but it is positive when the displacement vector points toward x = 0.

5. Picture the Problem: The two paths of the object are shown at right.

Strategy: The work done by gravity is W mgy= when the object is moved downward, W mgy= − when it is moved upward, and zero when it is moved horizontally. Sum the work done by gravity for each segment of each path.

Solution: 1. (a) Calculate the work for path 1: ( )( )( )( )

1 12

05.2 kg 9.81 m/s 1.0 m 51 J

W mg y= +

= =

2. Calculate the work for path 2: ( )( )( )( )

2 12

05.2 kg 9.81 m/s 1.0 m 51 J

W mg y= +

= =

3. (b) If you increase the mass of the object the work done by gravity will increase because it depends linearly on m.

Insight: The work is path-independent because gravity is a conservative force. 6. Picture the Problem: The physical situation is depicted at right.

Strategy: Use ( )2 21sp i f2W k x x= − for the work done by the spring.

That way the work will always be negative if you start out at i 0x = because the spring force will always be in the opposite direction

from the stretch or compression. The work done by kinetic friction is fr kW mgdμ= − , where d is the distance the box is pushed irregardless

of direction, because the friction force always acts in a direction opposite the motion.

Solution: 1. (a) Sum the work done by the spring for each segment of path 2:

( ) ( )( ) ( ) ( ) ( ){ }

( ) ( )

2 2 2 21sp 1 4 4 32

2 22 212

sp

480 N/m 0 0.020 m 0.020 m 0.020 m

0.096 J 0 J 0.096 J

W k x x x x

W

⎡ ⎤= − + −⎣ ⎦

⎡ ⎤ ⎡ ⎤= − − + − −⎣ ⎦ ⎣ ⎦

= − + = −

2. Sum the work done by friction for each segment of path 2:

( )( )( )( )( )

fr k 1 2

20.16 2.7 kg 9.81 m/s 0.020 0.040 m 0.25 J

W mg d dμ= − +

= − + = −

3. (b) Sum the work done by the spring for the direct path from A to B:

( )( ) ( )

2 21sp A B2

2212 480 N/m 0 0.020 m 0.096 J

W k x x= −

⎡ ⎤= − = −⎣ ⎦

4. Sum the work done by friction for the direct path from A to B:

( )( )( )( )2fr k 0.16 2.7 kg 9.81 m/s 0.020 m 0.085 JW mgdμ= − = − = −

Insight: The work done by friction is always negative, and increases in magnitude with the distance traveled.



8 – 4

7. Picture the Problem: Two identical balls are released from the same height, ball 1 is thrown downward with an initial

velocity and ball 2 is dropped from rest.

Strategy: Apply the concept of gravitational potential energy to answer the conceptual question.

Solution: 1. (a) Changes in gravitational potential energy mgy depend only upon any changes in an object’s weight mg or changes in its altitude y. Gravitational potential energy does not depend upon speed. We conclude that in this case the change in gravitational potential energy of ball 1 is equal to the change in gravitational potential energy of ball 2.

2. (b) The best explanation is II. The gravitational potential energy depends only on the mass of the ball and the drop height. Statement I is partly true (ball 1 does have greater total energy than ball 2) but the amount of gravitational potential energy that is converted into kinetic energy is determined by m and Δy. Statement III is true but irrelevant.

Insight: If ball 1 were thrown upward it would convert some of its initial kinetic energy into gravitational potential energy. However, the net change in gravitational potential energy from the release point to the landing point would still remain the same as ball 2.

8. Picture the Problem: A mass is attached to the bottom of a vertical spring, causing the spring to stretch and the mass to

move downward.

Strategy: Apply the concepts of gravitational and spring potential energies to answer the conceptual question.

Solution: 1. (a) As the mass moves downward the spring is stretched and the spring potential energy increases.

2. (b) As the mass moves downward its elevation y decreases and the gravitational potential energy decreases.

Insight: From an energy standpoint the gravitational potential energy of the mass is converted to spring potential energy during this process.

9. Picture the Problem: The cliff diver plunges straight downward due to the force of gravity.

Strategy: Solve equation 8-3 for the weight of the diver. Let y = 0 correspond to the surface of the water.

Solution: Solve equation 8-3 for mg: 25,000 J 540 N 0.54 kN46 m

UU mgy mgy

= ⇒ = = = =

Insight: If you set U = 0 at the top of the cliff, then U = −25 kJ and 46 my = − when the diver enters the water. 10. Picture the Problem: The climber stands at the top of Mt. Everest.

Strategy: Find the gravitational potential energy by using equation 8-3.

Solution: Calculate U mgy= : ( )( )( )2 688 kg 9.81 m/s 8848 m 7.6 10 J 7.6 MJU mgy= = = × =

Insight: You are free to declare that the climber’s potential energy is zero at the top of Mt. Everest and −7.2 MJ at sea level!

11. Picture the Problem: The Jeopardy button is a spring that must be compressed a certain amount to activate the switch.

Strategy: Use equation 8-5 to find the work required to compress the spring an estimated 1.0 cm in order to activate the switch. While the work the spring force does is c ,W U= −Δ the work a contestant must do is .W U= Δ

Solution: Calculate the work it takes by finding UΔ : ( ) ( )( )221 1

c 2 20 130 N/m 0.010 m 0.0065 J 6.5 mJW U kx= Δ = − = = =

Insight: The contestant does positive work on the spring because the force she exerts is in the same direction her finger is pushing. The energy she expends is briefly stored as potential energy in the spring.



8 – 5

12 Picture the Problem: The Hawkmoth wing behaves as an ideal spring.

Strategy: Use Hooke’s Law (equation 6-4) to find the force constant of the wing, then use equation 8-5 to find the energy stored. Combine both equations to find the force required to store twice the energy.

Solution: 1. (a) Solve equation 6-4 for k: 0.0030 N 0.625 N/m 0.63 N/m0.0048 m

Fkx

= − = − = =−

2. (b) Use equation 8-5 to find U: ( )( )22 61 12 2 0.625 N/m 0.0048 m 7.2 10 J 7.2 JU kx μ−= = = × =

3. (c) Solve equation 8-5 for x, taking the negative root because the wing is depressed in the negative direction as in step 1:

22 2 U Ux xk k

= ⇒ = −

4. Use Hooke’s Law to find the force: ( )( )( )6

2 2

2 2 7.2 10 J 0.625 N/m 0.0042 N 4.2 mN

F kx k U k Uk

−

= − = − − =

= × × = =

Insight: We bent the rules for significant figures for k a bit in steps 2 and 4 in order to avoid rounding error. Notice that in step 4 the force required to double the stored energy is 2 bigger than the original force.

13. Picture the Problem: A mass is suspended from a vertical spring. As the spring is stretched it stores potential energy.

Strategy: Doubling the mass doubles the force exerted on the spring, and therefore doubles the stretch distance due to Hooke’s Law .F kx= − Use a ratio to find the increase in spring potential energy 21

2U kx= when the stretch distance is doubled.

Solution: 1. (a) If the mass attached to the spring is doubled the stretch distance will double. Doubling the extension of the spring will increase the potential energy by a factor of 4.

2. (b) Doubling the mass doubles the force and therefore doubles the stretch distance:

1

2 12 2x mg kx mg k x== =

3. Calculate the ratio 2 1U U : ( )

( )

221122 2

2 211 1 12

2 1

24

4 4 0.962 J 3.85 J

xkxUU kx x

U U

= = =

= = =

Insight: Note that the change in gravitational potential energy also quadruples as the new mass is hung on the spring. It doubles because there is twice as much mass and it doubles again because the spring stretches twice as far. If you calculate the stretch distances you’ll find the spring stretched 1 1 12 5.6 cmx U m g= = when the 3.5-kg mass was suspended and it stretched 11.2 cm when the 7.0-kg mass was suspended from it.

14. Picture the Problem: The spring in the soap dispenser is compressed by the applied force.

Strategy: Use equation 8-5 to find the spring constant using the given energy and compression distance data. Solve the same equation for x in order to answer part (b).

Solution: 1. (a) Solve equation 8-5 for k:

( )( )2 2

2 0.0025 J2 200 N/m 0.20 kN/m0.0050 m

Ukx

= = = =

2. (b) Solve equation 8-5 for x:

( )2 0.0084 J2 0.92 cm200 N/m

Uxk

= = =

Insight: To compress the spring of this dispenser 0.50 cm requires 1.0 N or about ¼ lb of force.



8 – 6

15. Picture the Problem: The spring is stretched by the applied force and stores potential energy.

Strategy: Use equation 6-4 F kx= − to find the spring constant, and then equation 8-5 212U kx= to find the stretch

distance. Solution: 1. (a) Solve equation 6-4 for k: 4.1 N 290 N/m

0.014 mFkx

−= − = − =

2. Solve equation 8-5 for x:

( )2 0.020 J2 1.2 cm290 N/m

Uxk

= = =

3. (b) Repeat step 2 with the new U:

( )2 0.080 J2 2.3 cm290 N/m

Uxk

= = =

Insight: Notice that in part (b) the stretch distance doubled but the stored potential energy quadrupled because U is proportional to x2.

16. Picture the Problem: A graph of the potential energy vs. stretch distance is depicted at right.

Strategy: The work that you must do to stretch a spring is equal to minus the work done by the spring because the force you exert is in the opposite direction from the force the spring exerts. Use equations 8-1 and 8-5 together to find the spring constant and the required work to stretch the spring the specified distance.

Solution: 1. (a) Because the stored potential energy in a spring is proportional to the stretch distance squared, the work required to stretch the spring from 5.00 cm to 6.00 cm will be greater than the work required to stretch it from 4.00 cm to 5.00 cm.

2. (b) Use equations 8-1 and 8-5 to find k:

( )( )

( )( ) ( )

req spring 5 42 2 2 21 1 15 4 5 42 2 2

req 42 2 2 25 4

2 2 30.5 J6.78 10 N/m

0.0500 m 0.0400 m

W W U U Ukx kx k x xW

kx x

= − = − −Δ = −

= − = −

= = = ×− −

3. Use k and equations 8-1 and 8-5 to find the new reqW : ( ) ( ) ( ) ( )2 22 2 41 1

req 2 12 2 6.78 10 N/m 0.0600 m 0.0500 m 37.3 JW k x x ⎡ ⎤= − = × − =⎣ ⎦

Insight: Using the same procedure we discover that it would take 44.1 J to stretch the spring from 6.00 cm to 7.00 cm.

17. Picture the Problem: The pendulum bob swings from point A to point B and loses altitude and thus gravitational potential energy. See the figure at right.

Strategy: Use the geometry of the problem to find the change in altitude yΔ of the pendulum bob, and then use equation 8-3 to find its change in gravitational potential energy.

Solution: 1. Find the height change yΔ of the pendulum bob: ( )cos cos 1y L L Lθ θΔ = − = −

2. Use yΔ to find UΔ : ( )( )( )( )( )2

cos 10.33 kg 9.81 m/s 1.2 m cos35 1

0.70 J

U mg y mgL

U

θΔ = Δ = −

= °−

Δ = − Insight: Note that yΔ is negative because the pendulum swings from A to B. Likewise, yΔ is positive and the

pendulum gains potential energy if it swings from B to A.



8 – 7

18. Picture the Problem: Two identical balls are released from the same height. You throw a ball upward and let it fall to

the ground, but your friend drops her ball straight down to the ground.

Strategy: Apply the concept of conservation of energy to answer the conceptual question.

Solution: 1. (a) For a ball in freefall the gravitational potential energy is converted into kinetic energy. If the ball is moving upward its kinetic energy is converted into gravitational potential energy. Therefore, any change in the kinetic energy of the ball must come at the expense of its gravitational potential energy (as long as there are no other forces on the ball). The gravitational potential energies of the two balls change by the same amount. We conclude that the change in kinetic energy of your ball is equal to the change in kinetic energy of your friend’s ball.

2. (b) The best explanation is III. The change in gravitational potential energy is the same for each ball, which means the change in kinetic energy must be the same also. Statement I is true but fails to compare the two balls, and statement II is true but the changes in kinetic and gravitational potential energies are independent of the time elapsed.

Insight: If you were to throw your ball downward it would begin and finish with more kinetic energy than your friend’s ball, but the net change in gravitational potential energy from the release point to the landing point would still remain the same for each ball.

19. Picture the Problem: A snowboarder on the fictional planet Epsilon coasts on a smooth track that rises from a lower

level to a higher level. The acceleration of gravity on Epsilon is less than it is on Earth. If the snowboarder’s initial speed on Earth is 4 m/s, the snowboarder just makes it to the upper level and comes to rest.


Solution: 1. (a) On Epsilon it will require less energy to move from the lower level to the higher level because the acceleration of gravity and the weight of the snowboarder are less than they are on Earth. The mass of the snowboarder remains the same, however, so the snowboarder’s initial kinetic energy is the same in each case. On Epsilon the snowboarder will therefore be able climb a higher hill with the same initial kinetic energy. We conclude that the height of a hill on Epsilon that causes a reduction in speed from 4 m/s to 0 is greater than the height of the corresponding hill on Earth.

2. (b) Conceptual Checkpoint 8-2 describes how on Earth the speed of the snowboarder at the top of the hill is 3 m/s if the initial speed at the bottom of the hill is 5 m/s. On Epsilon the hill described in part (a) requires the same increase in gravitational potential energy as the corresponding hill on Earth. Therefore the initial and final kinetic energies of the snowboarder on Epsilon are identical to the corresponding values for the snowboarder on Earth, and we conclude that the final speed at the top of the hill will be equal to 3 m/s.

Insight: If the change in gravitational potential energy mgΔy is the same on each planet, then on Epsilon a smaller g implies a larger Δy is required.

20. Picture the Problem: When a ball of mass m is dropped from rest from a height h, its kinetic energy just before landing

is K. A second ball of mass 4m is dropped from rest from a height h/4.


Solution: 1. (a) The change in gravitational potential energy for each ball is the same because ( ) ( )144 .mgh m g h= We

conclude that just before ball 2 lands its kinetic energy is also K.

2. (b) The best explanation is I. The two balls have the same initial energy. Statements II and III are each false because neither scenario would conserve mechanical energy.

Insight: If the second ball were dropped from the same height h as the first ball, it would have 4 times the initial potential energy and would land with a kinetic energy of 4K, but its speed would be the same as the first ball.



8 – 8

21. Picture the Problem: When a ball of mass m is dropped from rest from a height h, its speed just before landing is v. A

second ball of mass 4m is dropped from rest from a height h/4.


Solution: 1. (a) The change in gravitational potential energy for each ball is the same because ( ) ( )144 .mgh m g h= We

conclude that just before ball 2 lands its kinetic energy is the same as that of ball 1, or 212 .mv However, its speed will

be less because its mass is greater. Setting ( )2 21 122 2 4mv m v= and solving for v2 we can see that 2 2 .v v=

2. (b) The best explanation is II. The two balls land with the same kinetic energy; therefore, the ball of mass 4m has the speed v/2. Statements I and III are each false.

Insight: If the second ball were dropped from the same height h as the first ball, it would have 4 times the final kinetic energy, but its speed would be the same as the first ball. This is related to the principle that all objects, regardless of mass, are accelerated by gravity at the same rate.

22. Picture the Problem: For an object moving along the x axis, the potential

energy of the frictionless system is shown in the figure. The object is released from rest at point A.

Strategy: The principle of conservation of energy states that the sum of the potential and kinetic energies of an object remains constant if there are no nonconservative forces. Therefore, any decrease in potential energy is accompanied by an increase in the kinetic energy of the same magnitude. Use this principle to evaluate the speed of the object at the various other points in the figure.

Solution: Point G is at the same potential energy as point A, so the kinetic energy and the speed of the object will be zero at point G. For the other points, the lower the potential energy, the higher the kinetic energy and the speed. Using this reasoning we arrive at the ranking of speeds: A = G < B = D = F < E < C.

Insight: The object could not move to the right of point G unless it were given some additional mechanical energy. 23. Picture the Problem: For an object moving along the x axis, the potential

energy of the frictionless system is shown in the figure. The object is released from rest at a point halfway between the points F and G.

Strategy: The principle of conservation of energy states that the sum of the potential and kinetic energies of an object remains constant if there are no nonconservative forces. Therefore, any decrease in potential energy is accompanied by an increase in the kinetic energy of the same magnitude. Use this principle to evaluate the speed of the object at the various other points in the figure.

Solution: The object is released from a point below points A and G, so it will not be able to reach those locations. For the other points, the lower the potential energy, the higher the kinetic energy and the speed. Using this reasoning we arrive at the ranking of speeds: B = D = F < E < C.

Insight: The object could only reach points A and G if it were given some additional mechanical energy.



8 – 9

24. Picture the Problem: A swimmer descends through a vertical height of 2.31 m as she slides without friction.

Strategy: As the swimmer descends the slide her gravitational potential energy is converted into kinetic energy. Set the loss in gravitational potential energy equal to the gain in kinetic energy by setting her change in mechanical energy equal to zero, so that f i 0E E EΔ = − = or f iE E= . Let 0y = at the bottom of the slide, 0v = at the top.

Solution: Set bottom topE E= and solve for bottomv :

( )( )

bottom top

bottom bottom top top

21bottom top2

2bottom top

0 0

2 2 9.81 m/s 2.31 m 6.73 m/s

E E

K U K U

mv mgy

v gy

=

+ = +

+ = +

= = =

Insight: If she has a mass of 40 kg, the swimmer loses 906 J of potential energy and gains 906 J of kinetic energy. 25. Picture the Problem: A swimmer descends through a vertical height of 2.31 m as she slides without friction.

Strategy: As the swimmer descends the slide her gravitational potential energy is converted into kinetic energy. Set the loss in gravitational potential energy equal to the gain in kinetic energy by setting her change in mechanical energy equal to zero, so that f i 0E E EΔ = − = or f iE E= . Let 0y = at the bottom of the slide, 0.840 m/sv = at the top.

Solution: Set bottom topE E= and solve for bottomv :

( ) ( )( )

bottom top


2 21 1bottom top top2 2

2bottom top top

2 2

0

2

0.840 m/s 2 9.81 m/s 2.31 m

6.78 m/s

E E

K U K U

mv mv mgy

v v gy

=

+ = +

+ = +

= +

= +

=

Insight: Note that she is not going 0.840 m/s faster than the 6.73 m/s she would be traveling if she started from rest (see the previous problem). That’s because the 14.1 J of kinetic energy she has at the start (if she has a mass of 40 kg) is small compared with the 906 J of kinetic energy she gains on the way down.

26. Picture the Problem: As the ball flies through the air and gains altitude

some of its initial kinetic energy is converted into gravitational potential energy.

Strategy: Set the mechanical energy at the start of the throw equal to the mechanical energy at its highest point. Let the height be i 0y = at the start of the throw, and find fy at the highest point.

Solution: 1. (a) Set i fE E= and solve for fy :

( ) ( ) ( )( )

i f

i i f f2 21 1i f f2 2

2 22 2

f i f 2

0

8.30 m/s 7.10 m/s1 0.942 m2 2 9.81 m/s

E EK U K U

mv mv mgy

y v vg

=+ = +

+ = +

−= − = =

2. (b) The height change is independent of the mass, so doubling the ball’s mass would cause no change to (a).

Insight: A more massive ball would have more kinetic energy at the start, but would require more energy to change its height by 0.942 m, so the mass cancels out.



8 – 10

27. Picture the Problem: Three balls are thrown upward with the same

initial speed, but at different angles relative to the horizontal.

Strategy: Use the concept of the conservation of mechanical energy to evaluate the statements. Note that when each of the three balls increases its altitude by the same amount, they each convert the same amount of kinetic energy into gravitational potential energy.

Solution: 1. (a) “At the dashed level, ball 3 has the lowest speed” is incorrect because all three balls will have the same kinetic energy (if they have the same mass) and the same speed at the dashed level. Ball 1’s velocity is vertical and ball 3’s velocity is horizontal, but they have the same speed.

2. (b) “At the dashed level, ball 1 has the lowest speed” is incorrect because all three balls will have the same speed at the dashed level.

3. (c) “At the dashed level, all three balls have the same speed” is correct because all three balls will have the same kinetic energy (if they have the same mass), having had identical kinetic energies at the launch and having each converted the same amount of kinetic energy into gravitational potential energy.

4. (d) “At the dashed level, the speed of the balls depends on their mass” is incorrect because both gravitational potential energy and kinetic energy depend linearly upon mass. The above statements were evaluated while supposing the masses were the same, but even if they were different the speeds would remain the same. A more massive ball would begin with a larger kinetic energy and convert a larger amount into gravitational potential energy.

Insight: You can use a mathematical expression to support the claim in part (d). Set the mechanical energies at the launch and at the dashed level (at altitude h) equal to each other: 2 21 1

2 20 .i fmv mv mgh+ = + Now solve this expression

for fv to discover 2 2f iv v gh= + independent of the mass of the ball.

28. Picture the Problem: As the ball flies through the air and gains altitude some of its initial kinetic energy is converted into gravitational potential energy.

Strategy: Set the mechanical energy just after the bounce equal to the mechanical energy when it is caught. Let the height be i 0y = at the bounce, and find fy at the catch.

Solution: 1. (a) Set i fE E= and solve for fy :

( ) ( ) ( )( )

i f

i i f f2 21 1i f f2 2

2 22 2

f i f 2

0

16 m/s 12 m/s1 5.7 m2 2 9.81 m/s

E EK U K U

mv mv mgy

y v vg

=+ = +

+ = +

−= − = =

2. (b) The height change is independent of the mass, so it is not necessary to know the mass of the tennis ball.

Insight: A more massive ball would have more kinetic energy at the start, but would require more energy to change its height by 0.942 m, so the mass cancels out.

29. Picture the Problem: The apple falls straight down from rest, accelerating at a rate of 9.81 m/s2.

Strategy: As the apple falls its gravitational potential energy is converted into kinetic energy. The sum of the gravitational and kinetic energies equals the mechanical energy, which remains constant throughout the fall. Use equations 7-6 and 8-3 to find the kinetic and gravitational potential energies, respectively.

Solution: 1. (a) Find Ua when a 4.0 m:y = ( )( )( )2a a 0.21 kg 9.81 m/s 4.0 m 8.24 J 8.2 JU mgy= = = =

2. The apple falls from rest so a 0 m/s:v = a 0K =

3. The total energy is the sum of K and U: a a a 0 8.24 J 8.2 JE K U E= + = = + =



8 – 11

4. (b) Find Ub when b 3.0 m:y = ( )( )( )2b b 0.21 kg 9.81 m/s 3.0 m 6.18 J 6.2 JU mgy= = = =

5. The total energy remains 8.2 J always so find b :K b b 8.24 6.18 J 2.1 JK E U= − = − =

6. (c), (d), (e) Repeat steps 4 and 5:

y (m) 4.0 3.0 2.0 1.0 0

U (J) 8.2 6.2 4.1 2.1 0

K (J) 0 2.1 4.1 6.2 8.2

E (J) 8.2 8.2 8.2 8.2 8.2

Insight: We bent the rules a bit on significant figures in step 5 to avoid rounding errors. This is especially a problem when subtracting two large numbers in order to obtain a small number, as in step 5. Notice the progression as the mechanical energy begins as entirely gravitational potential energy and ends as entirely kinetic energy.

30. Picture the Problem: A block slides on a frictionless, horizontal surface and encounters a horizontal spring. It

compresses the spring and briefly comes to rest.

Strategy: Set the mechanical energy when sliding freely equal to the mechanical energy when the spring is fully compressed and the block is at rest. Solve the resulting equation for the spring constant k, then repeat the procedure to find the initial speed required to compress the spring only 1.2 cm before coming to rest.

Solution: 1. (a) Set i fE E= where the initial state is when it is sliding freely and the final state is when it is at rest, having compressed the spring. ( ) ( )

( )

i i f f2 21 1i max2 2

22i

2 2max

0 0

2.9 kg 1.6 m/s3200 N/m 3.2 kN/m

0.048 m

K U K U

mv kx

mvkx

+ = +

+ = +

= = = =

2. (b) Solve the equation from step 1 for i :v

( )( )22max

i

3200 N/m 0.012 m0.40 m/s

2.9 kgkx

vm

= = =

Insight: The kinetic energy of the sliding block is stored as potential energy in the spring. Moments later the spring will have released all its potential energy, the block would have gained its kinetic energy again, and would then be sliding at the same speed but in the opposite direction.

31. Picture the Problem: The trajectory of the rock is depicted at right.

Strategy: The rock starts at height h, rises to maxy , comes briefly to rest, then falls down to the base of the cliff at 0.y = Set the mechanical energy at the point of release equal to the mechanical energy at the base of the cliff and at the maximum height maxy in order to find iv and maxy .

Solution: 1. (a) Set i fE E= and solve for iv :

( ) ( )( )

i i f f2 21 1i f2 2

2i f

2 2

i

0

2

29 m/s 2 9.81 m/s 32 m

15 m/s

K U K Umv mgh mv

v v gh

v

+ = ++ = +

= −

= −

=

2. (b) Now set ymax fE E= and solve for maxy : ( )

( )

ymax ymax f f22

2 f1max f max2 2

29 m/s0 0 43 m

2 2 9.81 m/s

K U K U

vmgy mv y

g

+ = +

+ = + ⇒ = = =

Insight: In part (a) the initial energy is a combination of potential and kinetic, but becomes all kinetic just before impact with the ground. In part (b) the rock at the peak of its flight has zero kinetic energy; all of its energy is potential energy.

maxy

h

iv



8 – 12

32. Picture the Problem: The block slides on a frictionless, horizontal surface, encounters a spring, compresses it, and

briefly comes to rest when the spring compression is 4.15 cm.

Strategy: As the block compresses the spring its kinetic energy is converted into spring potential energy. The sum of the spring potential and kinetic energies equals the mechanical energy, which remains constant throughout. Use equations 7-6 and 8-5 to find the kinetic and spring potential energies, respectively.

Solution: 1. (a) Find aK when a 0.950 m/s:v = ( )( )221 1a a2 2 1.40 kg 0.950 m/s 0.632 JK mv= = =

2. The spring is not compressed so a 0 cm:x = 21a a2 0U kx= =

3. The total energy is the sum of K and U: a a a 0.632 J 0 0.632 JE K U E= + = = + =

4. (b) Find bU when b 1.00 cm:x = ( )( )221 1b b2 2 734 N/m 0.0100 m 0.0367 JU kx= = =

5. The total energy remains 0.632 J always so find b :K b b 0.632 0.0367 J 0.595 JK E U= − = − =

6. (c), (d), (e) Repeat steps 4 and 5:

y (cm) 0.00 1.00 2.00 3.00 4.00

U (J) 0.00 0.037 0.147 0.330 0.587

K (J) 0.632 0.595 0.485 0.302 0.045

E (J) 0.632 0.632 0.632 0.632 0.632

Insight: The initial kinetic energy of the sliding block is stored as potential energy in the spring when it comes to rest. Moments later the spring will have released all its potential energy, the block would have gained its kinetic energy again, and would then be sliding at the same speed but in the opposite direction.

33. Picture the Problem: The rock falls straight down from rest, accelerating at a rate of 9.81 m/s2.

Strategy: The mechanical energy of the rock is initially stored as gravitational potential energy, but it is converted into kinetic energy as the rock falls. Set the mechanical energy equal to zero initially, so that i 0y = and i 0v = . The gravitational potential energy will decrease and have a negative value, but the kinetic energy will increase and have a positive value, so that the sum of the two will remain zero. The mechanical energy remains zero throughout.

Solution: 1. (a) The initial kinetic energy is zero because the rock falls from rest:

21ia ia2 0 JK mv= =

2. Set f iE E= after the rock falls to

fa 2.00 m:y = − ( )( )( )fa ia fa fa fa fa

2fa fa

0

5.76 kg 9.81 m/s 2.00 m 113 J

E E K U K mgy

K mgy

= = = + = +

= − = − − =

3. Calculate :KΔ fa ia 113 0 J 113 JK K KΔ = − = − =

4. (b) The initial kinetic energy is 98.1 J: ib 113 JK =

5. Set f iE E= after the rock falls to

fb 4.00 m:y = −

( )( )( ) ( )

ib ib fb fb

ib ib fb fb

fb ib ib fb

2113 J 5.76 kg 9.81 m/s 2.00 m 4.00 m

226 J

K U K UK mgy K mgy

K K mg y y

+ = ++ = +

= + −

= + − − −⎡ ⎤⎣ ⎦

=

6. Calculate :KΔ fb ib 226 113 J 113 JK K KΔ = − = − =

Insight: You can set the potential energy to any value you wish. For instance, we could claim that ia 12.00 m,y =

fa 10.00 m,y = and fb 8.00 my = and the results would be exactly the same. It is the change in the potential energy, not its particular value, that matters in physics!



8 – 13

34. Picture the Problem: The pendulum bob swings from point B to point A and gains

altitude and thus gravitational potential energy. See the figure at right.

Strategy: Use the geometry of the problem to find the change in altitude yΔ of the pendulum bob, and then use equation 8-3 to find its change in gravitational potential energy. Apply conservation of energy between points B and A to find the speed at A.

Solution: 1. (a) Find the height change yΔ of the pendulum bob:

( )cos 1 cosy L L Lθ θΔ = − = −

2. Use yΔ to find :UΔ ( )( )( )( )( )2

1 cos0.33 kg 9.81 m/s 1.2 m 1 cos35

0.70 J

U mg y mgL

U

θΔ = Δ = −

= − °

Δ = 3. (b) Set B AE E= and solve for A :v

( ) ( )

B B A A2 21 1B A2 2

22A B

2 0.70 J2 2.4 m/s 1.2 m/s0.33 kg

K U K U

mv mv U

Uv vm

+ = +

= + Δ

Δ= − = − =

4. (c) If the mass of the bob is increased the answer to part (a) will increase. The change in gravitational potential energy depends linearly on the mass.

5. (d) If the mass of the bob is increased the answer to part (b) will stay the same. Although the change in potential energy will increase, the change in kinetic energy will also increase.

Insight: Another way to look at the answer to (d) is that U m mg y m g yΔ = Δ = Δ independent of mass. That means the formula for Av in step 3 is independent of mass.

35. Picture the Problem: The pendulum bob swings from point B to point A and gains

altitude and thus gravitational potential energy. See the figure at right.

Strategy: Use equation 7-6 to find the kinetic energy of the bob at point B. Use the geometry of the problem to find the maximum change in altitude maxyΔ of the pendulum bob, and then use equation 8-3 to find its maximum change in gravitational potential energy. Apply conservation of energy between points B and the endpoint of its travel to find the maximum angle maxθ the string makes with the vertical.

Solution: 1. (a) Use equation 7-6 to find BK :

( )( )221 1B B2 2 0.33 kg 2.4 m/s 0.95 JK mv= = =

2. (b) Since there is no friction, mechanical energy is conserved. If we take the potential energy at point B to be zero, we can say that all of the bob’s kinetic energy will become potential energy when the bob reaches its maximum height and comes momentarily to rest. Therefore the change in potential energy between point B and the point where the bob comes to rest is 0.95 J.

3. (c) Find the height change maxyΔ of the pendulum bob:

( )max max maxcos 1 cosy L L Lθ θΔ = − = −

4. Use equation 8-3 and the result of part (b) to solve for maxθ :

( )

( )( ) ( )

max max

1 1max 2

1 cos

0.95 Jcos 1 cos 1 410.33 kg 9.81 m/s 1.2 m

U mg y mgL

UmgL

θ

θ − −

Δ = Δ = −⎡ ⎤⎛ ⎞Δ ⎢ ⎥= − = − = °⎜ ⎟⎢ ⎥⎝ ⎠ ⎣ ⎦

Insight: The pendulum bob cannot swing any farther than 41° because there is not enough energy available to raise the mass to a higher elevation.



8 – 14

36. Picture the Problem: The motions of the masses in the Atwood’s

machine are depicted in the figure at right:

Strategy: Mechanical energy is conserved because there is no friction. Set i fE E= and solve for fv . The speeds of each mass must always be the same because they are connected by a rope.

Solution: 1. (a) Set

i fE E= and solve for fv : ( ) ( )

( ) ( )

i i f f2 21 1

1 f 2 f 1 1 2 22 221

1 2 f 1 2221

2 1 1 2 f2

2 1f

1 2

0 00

2

K U K Um v m v m gy m gym m v m gh m g h

m m gh m m v

m mv gh

m m

+ = ++ = + + +

= + + + −− = +

⎛ ⎞−= ⎜ ⎟+⎝ ⎠

2. (b) Use the expression from part (a) to find fv :

( )( )2f

4.1 kg 3.7 kg2 9.81 m/s 1.2 m 1.1 m/s3.7 kg 4.1 kg

v⎛ ⎞−

= =⎜ ⎟+⎝ ⎠

Insight: The mass 2m loses more gravitational potential energy than 1m gains, so there is extra energy available to give the system kinetic energy. We bent the rules for significant figures a bit in step 2 because by the rules of subtraction, 4.1 − 3.7 kg = 0.4 kg, only one significant figure.

37. Picture the Problem: The motions of the masses in the Atwood’s machine are depicted in the figure at right:

Strategy: Mechanical energy is conserved because there is no friction. Set i fE E= and solve for h when f 0v = . The speeds of each mass must always be the same because they are connected by a rope.

Solution: 1. Set i fE E= and then set f 0v = :

( ) ( )

i i f f2 21 1

1 i 2 i 1 1 2 22 221

1 2 i 1 22

0 00

K U K Um v m v m gy m gy

m m v m gh m g h

+ = ++ + = + +

+ = + + −

2. Solve for h: ( ) ( )221 2

i 21 2

1 1 3.7 4.1 kg 0.20 m/s 0.04 m 4 cm2 4.1 3.7 kg2 9.81 m/s

m mh v

g m m⎛ ⎞ ⎛ ⎞+ +

= = = =⎜ ⎟ ⎜ ⎟− −⎝ ⎠⎝ ⎠

Insight: The kinetic energy of the two-mass system is converted to a difference in potential energy between mass 2 and mass 1. By the rules for subtraction of significant figures 4.1 − 3.7 kg = 0.4 kg, only one significant figure.

38. Picture the Problem: You coast up a hill on your bicycle with decreasing speed. Your friend pedals up the same hill with constant speed.

Strategy: Use the principle of the conservation of mechanical energy (if there are no nonconservative forces like friction) to answer the conceptual question. If there is a nonconservative force acting on the system, the mechanical energy of that system can change.

Solution: 1. (a) In the absence of friction there are no nonconservative forces acting on the you-bike-Earth system, so your mechanical energy will stay the same. Some of your kinetic energy is being converted into gravitational potential energy as you coast up the hill.



8 – 15

2. (b) The force your friend exerts on the pedals is a nonconservative force. The mechanical energy of the friend-bike-Earth system will increase as a result of the pedaling. Your friend’s kinetic energy remains the same while her gravitational potential energy is increasing.

Insight: In this case the nonconservative work is positive as energy is flowing into the friend-bike-Earth system. Your friend is converting chemical (food) energy into mechanical energy.

39. Picture the Problem: As the space shuttle returns from orbit and enters the atmosphere its protective heat tiles become extremely hot.

Strategy: Note that there is the nonconservative force of friction acting on the shuttle-Earth system, decreasing the mechanical energy of that system.

Solution: 1. (a) The mechanical energy of the shuttle-Earth system when the shuttle lands is less than when it is in orbit due to the negative work done by friction.

2. (b) The best statement is III. A portion of the mechanical energy has been converted to heat energy. Statement I is false (the kinetic energy increases in the absence of friction, but the potential energy decreases) and Statement II is true, but irrelevant because the nonconservative friction force is the one responsible for heating the protective tiles.

Insight: In this case friction has converted the space shuttle’s mechanical energy into thermal energy. Both its gravitational potential energy and its kinetic energy have decreased considerably since being in orbit.

40. Picture the Problem: The surfer changes altitude and speed, losing mechanical energy along the way due to friction.

Strategy: The nonconservative work equals the difference in mechanical energy between the beginning and the end of the run.

Solution: Use equation 8-9 to find nc :W ( ) ( )

( ) ( )( ) ( ) ( ) ( )( ){ }

nc f i2 21 1f f i i2 2

2 21f i f i2

2 2 212

nc

77 kg 8.2 m/s 1.3 m/s 9.81 m/s 0 1.65 m

1300 J 1.3 kJ

W E E Emv mgy mv mgy

m v v g y y

W

= Δ = −= + − +

⎡ ⎤= − + −⎣ ⎦⎡ ⎤= − + −⎣ ⎦

= =

Insight: We usually expect the nonconservative work to be negative because friction steals mechanical energy and converts it into heat. But in this case the force from the wave propels the surfer and he ends up with more mechanical energy ( 21

2 2600 Jfmv = ) than he had at the start ( 212 1300 Ji img y mv+ = ).

41. Picture the Problem: The child slides down the slide, changing altitude and speed, and loses mechanical energy along the way due to friction.

Strategy: The nonconservative work equals the difference in mechanical energy between the top and the bottom of the slide. Use equation 8-9 to find the mechanical energy at the bottom of the slide and equations 7-6 and 8-3 to find the speed of the child. Set y = 0 at the bottom of the slide.

Solution: 1. Use equations 8-9, 7-6, and 8-3 to solve for fK : ( ) ( )

nc f i2 21 1f f i i2 2

2 21 1f nc f i i2 2


mv W mgy mv mgy

= Δ = −= + − +

= − + +

2. Multiply both sides by 2 m and take the square root to find fv :

( )( ) ( ) ( )( )

2f nc i f i

2 2

f

2 2

2 361 J 19 kg 2 9.81 m/s 2.3 0 m 0

2.7 m/s

v W m g y y v

v

= + − +

= − + − +

=

Insight: If the slide had been frictionless the child’s final speed would have been 6.7 m/s and her mechanical energy would have remained 429 J throughout the trip down the slide.



8 – 16

42. Picture the Problem: The athlete accelerates horizontally through the water from rest to 1.20 m/s while doing

nonconservative work against the drag from the water.

Strategy: The total nonconservative work done on the athlete changes his mechanical energy according to equation 8-9. This nonconservative work includes the positive work nc1W done by the athlete’s muscles and the negative work nc2W done by the water. Use this relationship and the known change in kinetic energy to find nc2W .

Solution: Set the nonconservative work equal to the change in mechanical energy and solve for nc2W . The initial mechanical energy is zero: ( )( ) ( )

nc nc1 nc2 f i f21

nc2 f nc1 f nc1221

2

0

72.0 kg 1.20 m/s 161 J 109 J

W W W E E E KW K W mv W

= + = Δ = − = −= − = −

= − = − Insight: The drag force from the water reduced the swimmer’s mechanical energy, but his muscles increased it by a

greater amount, resulting in a net gain in mechanical energy. 43. Picture the Problem: The airplane travels in a straight line, slowing down and coming to rest after landing at high

speed.

Strategy: The total nonconservative work done on the airplane changes its mechanical energy according to equation 8-9. The nonconservative work equals the change in mechanical energy, which is known from the initial and final speeds of the airplane (there is no change to its gravitational potential energy).

Solution: Apply equation 8-9 directly: ( )( )22 21 1 1nc f i f i2 2 2

7nc

0 17,000 kg 82 m/s5.7 10 J 57 MJ

W E E E mv mvW

= Δ = − = − = −

= − × = − Insight: The aircraft brakes and the recovery cables remove the airplane’s kinetic energy, converting it into heat and

sound. Heat management in large braking systems like this one is an important engineering issue. 44. Picture the Problem: The car travels in a straight line and slows down after applying the brakes.

Strategy: The total nonconservative work done on the car changes its mechanical energy according to equation 8-9. The nonconservative work equals the change in mechanical energy, which is known from the initial and final speeds of the car (there is no change to its gravitational potential energy). In this case .E KΔ = Δ

Solution: 1. (a) Apply equation 8-9 directly:

( ) ( ) ( )

2 21 1nc f i f i2 2

2212

5

1300 kg 11 m/s 17 m/s

1.1 10 J 0.11 MJ

W E E E mv mv K

K

= Δ = − = − = Δ

⎡ ⎤Δ = −⎣ ⎦= − × = −

2. (b) The kinetic energy of the car is converted into heat by friction in the brakes.

Insight: Lots of hard braking in a car can cause heat management problems, because the heat created by friction can melt or warp parts of the brake system.

45. Picture the Problem: You ride down a hill on your bicycle with constant speed.

Strategy: The presence of a nonconservative force (friction) acting on the you-bike-Earth system causes the mechanical energy of that system to change.

Solution: 1. (a) As you ride down the hill your elevation decreases. Because your gravitational potential energy increases linearly with elevation, we conclude that the gravitational potential energy of the you-bike-Earth system will decrease as you ride downhill.

2. (b) As you ride downhill at constant speed the kinetic energy of you and your bike will stay the same because your kinetic energy depends upon your speed and your speed does not change.

3. (c) As you ride downhill at constant speed the mechanical energy of the you-bike-Earth system will decrease because your kinetic energy remains the same but your gravitational potential energy decreases.

Insight: In order for your speed and kinetic energy to remain constant as you pedal down the hill, a nonconservative force must have done negative work on you and your bicycle. For example, you may have applied the brakes to control your speed, or the ground may be soft or muddy.



8 – 17

46. Picture the Problem: The initial and final states of the system

are depicted at right. The values given in the example are 1 2.40 kg,m = 2 1.80 kg,m = 0.500 m,d = and k 0.450.μ =

Strategy: The nonconservative work done by friction changes the mechanical energy of the system. Use equation 8-9 to find

EΔ and set it equal to the work done by friction. Solve the resulting expression for the final velocity of the system.

Solution: 1. Write equation 8-9 to find ncW :

( ) ( )( ) ( )( )( ) ( )

nc f i f f i i

2 21 11 2 f 1 2 2f 1 2 i 1 2 2i2 2

2 21nc 1 2 f i 2 2f 2i2

W E E K U K U

m m v m gh m gy m m v m gh m gy

W m m v v m g y y

= − = + − +

⎡ ⎤ ⎡ ⎤= + + + − + + +⎣ ⎦ ⎣ ⎦

= + − + −

2. The nonconservative work is done by friction: nc k k 1W f d m gdμ= − = −

3. Substitute the expression from step 2 into step 1 and solve for 2 2

f iv v− :

( )( ) ( )( ) ( )( )( )( )

( )( ) ( )( )( )

2 21k 1 1 2 f i 22

2 212 k 1 1 2 f i2

22 k 1 2 2

f i1 2

2 2

0

2 9.81 m/s 0.500 m 1.80 kg 0.450 2.40 kg22.40 1.80 kg

1.68 m / s

m gd m m v v m g d

gd m m m m v v

gd m mv v

m m

μ

μ

μ

− = + − + −

− = + −

−⎡ ⎤− ⎣ ⎦= − =+ +

=

4. Finally, solve for fv : ( )22 2 2 2 2f i 1.68 m /s 1.3 m/s 1.68 m /s 1.8 m/sv v= + = + =

Insight: If the blocks are not given the initial speed of 1.3 m/s, their final speed is 1.30 m/s, as in Example 8-10. 47. Picture the Problem: The seal slides down the ramp, changing altitude

and speed, and loses mechanical energy along the way due to friction.

Strategy: The nonconservative work done by friction changes the mechanical energy of the seal. Use equation 8-9 and the given information to find the nonconservative work. Then use Newton’s Second Law to find the normal force on the seal, and use the normal force to find the force of kinetic friction. Solve the resulting expression for kμ . Let y = 0 at the water’s surface.

Solution: 1. (a) Use equations 8-9, 7-6, and 8-3 to find ncW : ( ) ( )

( ) ( )( ) ( ) ( )( ){ }

nc f i2 21 1f f i i2 22 21f i f i2

2 2 212

nc

42.0 kg 4.40 m/s 0 9.81 m/s 0 1.75 m

314 J


m v v mg y y

W

= Δ = −= + − +

= − + −

⎡ ⎤= − + −⎣ ⎦= −

2. Use Newton’s Second Law to find the normal force: cos 0 cosyF N mg N mgθ θ= − = ⇒ =∑

3. Use the right triangle formed by the ramp to find the distance the seal slides: sin

sinh hdd

θθ

= ⇒ =

4. (b) Set ncW equal to the work done by friction and solve for kμ :

( ) ( )

( )( )

( )( )( )

nc k k k

nck 2

cos sin314 J tan 35.0sin

0.305cos 42.0 kg 9.81 m/s 1.75 m

W f d Nd mg hW

mg h

μ μ θ θθ

μθ

= − = − = −− °

= − = − =

Insight: Verify for yourself that if the slide were frictionless the seal would enter the water at 5.86 m/s.



8 – 18

48. Picture the Problem: The rock falls straight downward under the influence of both gravity and water resistance. Strategy: The nonconservative work done by water resistance equals minus the force times the distance. This work

reduces the mechanical energy of the rock. Use equation 8-3 to find the potential energy of the rock and then equation 8-9 to determine the kinetic energy of the rock. Let y = 0 at the bottom of the pond.

Solution: 1. (a) Calculate the work done by water resistance: ( )( )nc 4.6 N 0 m 0 JW Fd= − = − =

2. Find the potential energy of the rock at 1.8 0.0 my = − : ( )( )( )2a a 1.9 kg 9.81 m/s 1.8 m 34 JU mgy= = =

3. Use equation 8-9 to find the kinetic energy of the rock: ( ) ( )( )

nc f f i i

af ai fa i nc 0 0 0 0 J

W E K U K U

K K U U W

= Δ = + − +

= − − + = − + =

4. Sum U and K to find E: a a a 34 J 0 J 34 JE U K= + = + =

5. (b), (c) Repeat steps 1 through 4: Wnc (J) U (J) K (J) E (J) (a) 0 34 0 34 (b) −2.3 24 7.0 31 (c) −4.6 15 14 29

Insight: If there were no water resistance the rock would be in freefall and the mechanical energy would remain constant at 34 J. Note that the nonconservative work only removes kinetic energy; the potential energy remains .mgy

49. Picture the Problem: The car drives up the hill, changing its kinetic and gravitational potential energies, while both the

engine force and friction do nonconservative work on the car. Strategy: The total nonconservative work done on the car changes its mechanical energy according to equation 8-9.

This nonconservative work includes the positive work nc1W done by the engine and the negative work nc2W done by the friction. Use this relationship and the known change in potential energy to find .KΔ

Solution: Set the nonconservative work equal to the change in mechanical energy and solve for :KΔ

( ) ( ) ( )( )

( ) ( ) ( )( )( )

nc nc1 nc2 f i

nc1 nc2 f f i i f i

nc1 nc2 f i5 5 2

5

6.44 10 J 3.11 10 J 1250 kg 9.81 m/s 16.2 m

1.34 10 J 134 kJ

W W W E E EW W K U K U K mg y y

K W W mg y y

K

= + = Δ = −+ = + − + = Δ + −Δ = + − −

= × + − × −

Δ = × = Insight: The friction force reduces the car’s mechanical energy, but the engine increased it by a greater amount,

resulting in a net gain in both kinetic and potential energy. The car gained speed while traveling uphill. 50. Picture the Problem: The skater travels up a hill (we know this for reasons given below), changing his kinetic and

gravitational potential energies, while both his muscles and friction do nonconservative work on him. Strategy: The total nonconservative work done on the skater changes his mechanical energy according to equation 8-9.

This nonconservative work includes the positive work nc1W done by his muscles and the negative work nc2W done by the friction. Use this relationship and the known change in potential energy to find .yΔ

Solution: 1. (a) The skater has gone uphill because the work done by the skater is larger than that done by friction, so the skater has gained mechanical energy. However, the final speed of the skater is less than the initial speed, so he has lost kinetic energy. Therefore he must have gained potential energy, and has gone uphill.

2. (b) Set the nonconservative work equal to the change in mechanical energy and solve for :yΔ

( ) ( ) ( )( )

( ) ( ) ( ) ( ) ( ){ }( )( )

nc nc1 nc2 f i2 21

nc1 nc2 f f i i f i2

2 21nc1 nc2 f i2

2 212

2

3420 J 715 J 81.0 kg 1.22 m/s 2.50 m/s3.65 m

81.0 kg 9.81 m/s

W W W E E EW W K U K U m v v mg y

y W W m v v mg

= + = Δ = −+ = + − + = − + Δ

⎡ ⎤Δ = + − −⎣ ⎦⎡ ⎤+ − − −⎣ ⎦= =

Insight: Verify for yourself that if the skates had been frictionless but the skater’s muscles did the same amount of

work, the skater’s final speed would have been 4.37 m/s. He would have sped up if it weren’t for friction!



8 – 19

51. Picture the Problem: The initial and final states of the system are

depicted at right. The masses given in the example are 1 2.40 kg,m = 2 1.80 kg,m = but k 0.350μ = and d is unknown in

this problem.

Strategy: The nonconservative work done by friction changes the mechanical energy of the system. Use equation 8-9 to find EΔ and set it equal to the work done by friction. Solve the resulting expression for the travel distance d.

Solution: 1. (a) Write eq. 8-9 to obtain an expression for ncW :

( ) ( )( ) ( )( )( ) ( )

nc f i f f i i

2 21 11 2 f 1 2 2f 1 2 i 1 2 2i2 2

2 21nc 1 2 f i 2 2f 2i2

W E E K U K U


W m m v v m g y y

= − = + − +

⎡ ⎤ ⎡ ⎤= + + + − + + +⎣ ⎦ ⎣ ⎦

= + − + −


3. Substitute the expression from step 2 into step 1 and solve for d:

( )( ) ( )( ) ( )

( )( )

( )( )( ) ( )( )

2 21k 1 1 2 f 22

212 k 1 1 2 f2

221 2 f

22 k 1

0 0

2.40 1.80 kg 2.05 m/s2 2 9.81 m/s 1.80 kg 0.350 2.40 kg

0.937 m

m gd m m v m g d

gd m m m m v

m m vd

g m m

μ

μ

μ

− = + − + −

− = +

+ += =

− −⎡ ⎤⎣ ⎦

=

4. (b) The conservative work changes the potential energy: ( )( )( )2

c 2 1.80 kg 9.81 m/s 0.937 m 16.5 JW U m g y= −Δ = − Δ = − − =

5. (c) The nonconservative work is done by friction: ( )( )( )( )2

nc k 1 0.350 2.40 kg 9.81 m/s 0.937 m 7.72 JW m gdμ= − = − = −

6. (d) Verify total :W K= Δ ( ) ( )2 2 2 21 11 2 f 02 2

total c nc

2.40 1.80 kg (2.05 m/s) 0 8.83 J16.5 7.72 J 8.8 J verified

K m m v vW W W K

⎡ ⎤ ⎡ ⎤Δ = + − = + − =⎣ ⎦ ⎣ ⎦= + = − = = Δ

7. Verify c :W U= −Δ c 16.5 JW U= = −Δ was verified in step (4)

8. Verify nc :W E= Δ ( ) nc8.83 J 16.5 J 7.7 J verifiedE K U WΔ = Δ + Δ = + − = − =

Insight: The blocks travel almost twice as far in this problem when compared with Example 8-10, and are traveling almost twice as fast when 2m hits the floor, but the similarity of these comparisons is simply coincidental.

52. Picture the Problem: The truck travels down a hill, changing its kinetic and gravitational potential energies.

Strategy: Find the change in gravitational potential energy by using equation 8-3 and then the change in kinetic energy by using equation 7-6. Sum the two changes to find the change in mechanical energy.

Solution: 1. (a) Use equation 8-3 to find :UΔ ( )( )( )2f i

7

( ) 15,800 kg 9.81 m/s 1440 m 1630 m

2.94 10 J 29.4 MJ

U mg y yΔ = − = −

= − × = − 2. (b) Use equation 7-6 to find :KΔ ( ) ( ) ( ) ( )2 22 21 1

f i2 2

6

15,800 kg 29.0 m/s 12.0 m/s

5.51 10 J 5.51 MJ

K m v v ⎡ ⎤Δ = − = −⎣ ⎦= × =

3. (c) No. The total mechanical energy changes by 23.9 MJ.E K UΔ = Δ + Δ = −

Insight: The loss in mechanical energy means a nonconservative force has removed some kinetic energy. In this case the truck driver either applied the brakes or relied on frictional forces to slow the truck.



8 – 20

53. Picture the Problem: The block slides horizontally on a rough surface, encounters a spring, and compresses it a

distance of 11.0 cm before coming to rest.

Strategy: The nonconservative work done by friction changes the mechanical energy of the system. Use equation 8-9 to find EΔ and set it equal to the work done by friction. Solve the resulting expression for the spring constant k.

Solution: 1. Write equation 8-9 to obtain an expression for ncW :

( ) ( )( ) ( )

nc f i f f i i

2 21 1i2 20 0

W E E K U K U

kx mv

= − = + − +

= + − +

2. The nonconservative work is done by friction as the block travels a distance x:

nc k kW f d mg xμ= − = −

3. Substitute the expression from step 2 into step 1 and solve for k:

( ) ( ) ( )( )( )( )

2 21 1k i2 2

22i ki k

2 2

2 2

2

22

1.80 kg 2.00 m/s 2 0.560 9.81 m/s 0.110 m

0.110 m

415 N/m

mg x kx mv

m v g xmv mg xk

x x

k

μ

μμ

− = −

⎡ ⎤−− ⎣ ⎦= =

⎡ ⎤−⎣ ⎦=

=

Insight: The force exerted by the spring at 0.110 mx = is 45.7 N.− Verify for yourself that if the coefficient of static friction with the rough surface is less than s 2.59μ = , then the spring is strong enough to accelerate the block again in the opposite direction and the block will not remain at rest.

54. Picture the Problem: The U vs. x plot for an object is depicted at right.

Strategy: Describe the motion of the object, keeping in mind that objects tend to move to the minimum potential energy, and when they do their kinetic energy is maximum. The turning points on the potential energy plot are points A and E.

Solution: At point A, the object is at rest. As the object travels from point A to point B, some of its potential energy is converted into kinetic energy and the object’s speed increases. As the object travels from point B to point C, some of its kinetic energy is converted back into potential energy and its speed decreases. From point C to point D, the speed increases again, and from point D to point E, the speed decreases.

Insight: The object momentarily comes to rest at point E, but then turns around and accelerates back toward D and retraces its path all the way to point A, at which time the cycle begins again.

55. Picture the Problem: The U vs. x plot for an object is depicted at right.

Strategy: The loss of potential energy from point A to point B becomes a gain in kinetic energy. Use the magnitude of UΔ to determine the velocity at points B, C, and D by using the conservation of mechanical energy (equation 8-9).

Solution: 1. (a) Set i fE E= and solve for fv :

( )

i i f f21f f2

ff

0 10.0 J

2 10.0 J

K U K U

mv U

Uv

m

+ = +

+ = +

−=

2. Use the expression in step 1 to find Bv :

( )B

2 10.0 J 2.0 J3.8 m/s

1.1 kgv

−= =



8 – 21

3. (b) Use the expression in step 1 to find Cv :

( )C

2 10.0 J 6.0 J2.7 m/s

1.1 kgv

−= =

4. (c) Use the expression in step 1 to find Dv :

( )D

2 10.0 J 5.0 J3.0 m/s

1.1 kgv

−= =

5. (d) If the object starts at rest at point A, the potential energy at point A is the maximum energy of the object. Therefore points A and E are the turning points of its motion.

Insight: Note that the lower the object is on the potential energy curve, the higher its speed. 56. Picture the Problem: The U vs. x plot for an object is depicted at right.

Strategy: The object will be at its turning point when its total mechanical energy equals its potential energy. Use the provided information to find the total mechanical energy of the particle, then draw a horizontal line at the corresponding value on the U axis to find the turning points.

Solution: 1. Determine E from the provided information: ( )( )

21C C2

212 1.34 kg 1.25 m/s 6.0 J

E 7.0 J

E mv U= +

= +=

2. Draw a horizontal line on the plot at U = 7.0 J and determine where the line crosses the potential energy curve. In this case it crosses at approximately 0.6 mx = and 4.6 mx = . These are the locations of the turning points.

Insight: In order for the turning points to be at A and E, the object needs a speed of 2.36 m/s at point C. 57. Picture the Problem: The geometry of a simple pendulum representing the

child on a swing is shown at right, together with the plot of U versus θ.

Strategy: Determine an expression for the potential energy of a pendulum as a function of θ. Then find the energy for θ = 90° and θ = −90° and construct the plot. Let y = 0 correspond to the lowest point of the child’s motion.

Solution: 1. Use the right triangle in the diagram to find the height y of the child:

( )cos 1 cosy L L Lθ θ= − = −

2. Use Eq. 8-3 to find ( )U θ : ( )1 cosU mgy mgL θ= = − 3. Find the maximum U: ( ) ( )

( )( )( )( )( )

( ) ( )

2

90 1 cos90

23 kg 9.81 m/s 2.5 m 1 0

90 560 J

90 1 cos 90 560 J

U mgL

U

U mgL

° = − °

= −

° =

− ° = − − ° =⎡ ⎤⎣ ⎦

4. The plot of U versus θ is shown at right.

Insight: The angle θ = 0° corresponds to the minimum of the potential energy and is the equilibrium position for the child on the swing. Once the child acquires mechanical energy she starts swinging back and forth between the turning points. If there is friction the mechanical energy slowly decreases and the turning points occur at smaller and smaller angles until the child comes to rest at θ = 0°.



8 – 22

58. Picture the Problem: The plot of U versus θ for the child is shown at right.

Strategy: From the previous problem we know the height of the child is given by ( )1 cosy L θ= − . Set the potential energy at the turning points equal to the kinetic energy when the ropes are vertical (and when U = 0) because that is the total mechanical energy of the child.

Solution: 1. Use equation 8-3 to find ( )U θ : ( )1 cosU mgy mgL θ= = − 2. Set maxU K= and solve for the angle maxθ : 21

max max22max

max

(1 cos )

cos 12

mv mgLv

gL

θ

θ

= −

= −

3. Determine the two values of maxθ :

( )( ) ( )

221 1max

max 2

0.89 m/scos 1 cos 1 10

2 2 9.81 m/s 2.5 mv

gLθ − −

⎡ ⎤⎡ ⎤⎢ ⎥= ± − = ± − = ± °⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

Insight: It is possible for a child to do nonconservative work on herself, “pumping” the swing to increase her energy.

59. Picture the Problem: The U vs. x plot is depicted at right.

Strategy: Use the provided information to find the total mechanical energy of the particle, then draw a horizontal line at the corresponding value on the U axis to find the turning points.

Solution: 1. (a) Determine E from the provided information. From the graph we see that ( )1.0 m 5.0 JU ≅ :

3.6 5.0 J 8.6 JE K U= + = + =

2. Draw a horizontal line on the plot at U = 8.6 J and determine where the line crosses the potential energy curve. In this case it crosses at approximately 0.2 mx = and 4.8 m.x = These are the locations of the turning points. Therefore, the smallest value of x the particle can reach is (b) 0.2 m and the largest value of x the particle can reach is (c) 4.8 m.

Insight: In order for the turning points to be at A and C the object needs a kinetic energy of 5.0 J at 1.0 m.x =

60. Picture the Problem: The plot of U versus x is shown at right.

Strategy: Use equation 8-5 to find an expression for the potential energy as a function of x. Use the provided information to find the total mechanical energy of the block. Set the potential energy at the turning points equal to the kinetic energy at x = 0 because that is the total mechanical energy of the block.

Solution: 1. (a) Determine an expression for ( )U x and draw the plot. The plot is shown at right.

( )( )

( ) ( )( )

2 21 12 2

2

2

775 N/m388 N/m

5.00 cm 388 N/m 0.0500 m0.969 J

U kx xx

U

= ==

==

2. (b) Set maxU K= and solve for maxx :

( )

2 21 1max max2 2

max max0.95 kg 1.3 m/s 0.046 m 4.6 cm775 N/m

kx mvmx vk

=

= ± = ± = ± = ±

Insight: In order to make the turning points equal to ±5.00 cm the block needs a speed of 1.43 m/s at x = 0.

U (J)

x (cm)−5.00 5.00

0.969



8 – 23

61. Picture the Problem: A ball is thrown straight upward.

Strategy: Use equation 8-3 to find an expression for the potential energy as a function of y. Use the mass and velocity to find the kinetic energy of the ball when it is thrown. The kinetic energy at that point is also its total mechanical energy of the block. Set the potential energy equal to the total mechanical energy to find the turning point.

Solution: 1. (a). Determine an expression for ( )U y and draw the plot. The plot is shown at right.

( )( )( )

20.75 kg 9.81 m/s7.4 N

U mgy yy

= =

=

2. (b) Set maxU K= and solve for maxy : ( )( )

21max max2

22

max 2

8.9 m/s4.0 m

2 2 9.81 m/s

mgy mv

vyg

=

= = =

Insight: The ball is thrown with 30 J of kinetic energy and rises 4.0 m until its gravitational potential energy is 30 J. 62. Picture the Problem: The two blocks are connected by a spring and oscillate on a frictionless horizontal surface. The

two blocks always move in opposite directions as they oscillate about a fixed position. Strategy: When the two blocks are not oscillating their separation is L. When they are pulled apart a distance x (so that

the block separation is L x+ ) the spring stores potential energy in the amount of 212 kx . This potential energy becomes

the total kinetic energy of both blocks when they return to a separation L. The blocks’ kinetic energy becomes potential energy again as the spring compresses a distance x and the block separation is L x− .

Solution: 1. Set max maxU K= and solve for maxx : ( )2 21 1

max max max max2 222 mkx m v x vk

= ⇒ =

2. Now write an expression for the block separation: max max

2Separation mL x L vk

= ± = ±

Insight: Since each block receives only half the available energy, the maximum speed and amplitude of the motion is less than if the same spring were anchored on one side and had one block attached on the other end, and the spring were stretched the same amount maxx .

63. Picture the Problem: You and your friend each solve the same physics problem involving the motion of a skier down a

slope, but choose different locations to correspond to y = 0.

Strategy: Note that only changes in the gravitational potential energy mgy are important for predicting speeds in physics problems like this one, not the absolute value of the gravitational potential energy.

Solution: 1. (a) The choice of the location of y = 0 affects the absolute value of the potential energy so that your answer and your friend’s answer will disagree on this quantity.

2. (b) The choice of the location of y = 0 does not affect the calculated change in potential energy so that your answer and your friend’s answer will agree on this quantity.

3. (c) The choice of the location of y = 0 does not affect the calculated change in potential energy. Since the kinetic energy depends only on change of potential energy, your answer and your friend’s answer will agree on this quantity.

Insight: From the standpoint of predicting the skier’s speed, it makes no difference whether the skier’s potential energy changes from 1075 J to 1000 J or from 575 J to 500 J. The skier gains 75 J of kinetic energy in either case.

U (J)

y (m) 4.0 5.0

3037



8 – 24

64. Picture the Problem: A particle moves under the influence of a conservative force. The particle’s potential and kinetic

energies are to be considered at points A, B, and C.

Strategy: Use the conservation of mechanical energy to answer the questions. At each point the kinetic and gravitational potential energies will sum to the same total energy.

Solution: 1. (a) At point B the particle is at rest and has no kinetic energy, but it has 25 J of potential energy. We conclude that the total energy of the particle is 25 J. That means the potential energy at point A is 25 J − 12 J = 13 J.

2. (b) At point C the total energy is 25 J and the potential energy is 5 J, so the kinetic energy is 25 J − 5 J = 20 J.

Insight: If there were a point D at which the particle had 40 J of total energy, we would conclude that 15 J of positive work was done on the particle between points C and D.

65. Picture the Problem: A leaf falls to the ground with constant speed.

Strategy: Consider the value of the mechanical energy of the leaf in order to answer the question.

Solution: Since the leaf’s speed does not change its kinetic energy remains constant. Its gravitational potential energy, however, decreases as the leaf descends to a lower elevation. We conclude that the value of Ki + Ui is greater than the value of Kf + Uf.

Insight: As the leaf falls air friction is changing its mechanical energy into thermal energy (it heats up the air slightly). 66. Picture the Problem: Two blocks are connected by a string and

move in a clockwise fashion as depicted at right. There is friction between block 1 and the table.

Strategy: The nonconservative work done by friction changes the mechanical energy of the system.

Solution: 1. (a) Friction is doing negative work on the system, converting some of the mechanical energy into thermal energy. The speed of block 2 just before it hits the ground will be less than it would be if it were in freefall, in which case its mechanical energy would be conserved We conclude that as block 2 descends through the distance d its mechanical energy will decrease.

2. (b) The tension in the rope acts in a direction that is opposite to the direction of motion of block 2. Therefore, the tension does a negative nonconservative work on block 2.

Insight: Another reason we expect the work done by the tension in the rope to be negative is the decrease in the mechanical energy of block 2 as explained in part (a). If the work were positive block 2’s energy would have increased.

67. Picture the Problem: A bungee jumper steps off a platform and falls until the cord brings her to rest. You analyze this

system by choosing y = 0 at the platform level, and your friend chooses y = 0 at ground level.

Strategy: Consider the value of the gravitational potential energy of the bungee jumper in order to answer the question.

Solution: 1. (a) Your value of the jumper’s initial gravitational potential energy is zero, a value that is less than the value assigned to that quantity by your friend.

2. (b) In both cases the jumper falls through the same distance, so your value for the change in the jumper’s gravitational potential energy is equal to the value assigned to that quantity by your friend.

Insight: From the standpoint of finding the bungee jumper’s speed, it makes no difference whether the potential energy changes from 50,000 J to 0 J or from 0 J to −50,000 J. The jumper gains 50,000 J of kinetic energy in either case.



8 – 25

68. Picture the Problem: The sled slides down the hill without friction.

Strategy: As the sled descends the hill its gravitational potential energy is converted into kinetic energy. Find the loss in gravitational potential energy by setting the change in mechanical energy equal to zero, so that f i 0E E EΔ = − = or

f iE E= . Let 0y = at the bottom of the hill and either 0v = or 1.50 m/sv = at the top.

Solution: 1. (a) When the sled reaches the bottom of the hill after starting with a speed of 1.50 m/s, it will arrive at the bottom of the hill with a speed less than 9.00 m/s. That’s because the kinetic energy at the top of the hill is small compared to the gain in kinetic energy as the sled descends the hill. Another way to say the same thing is to recognize that the energies add but the speeds do not.

2. (b) Set bottom topE E= and solve for topU , letting bottom top 0U K= = :

bottom,1 top,1

bottom,1 bottom top,1 top

21bottom,1 top2 0 0

E E

K U K U

mv U

=

+ = +

+ = +

3. Now find bottom,2v after the second run using the same approach:

( ) ( )

bottom,2 bottom top,2 top

2 2 2 21 1 1 1bottom,2 top,2 top top bottom,12 2 2 2

2 2bottom,2 top bottom,1

2 2

0

1.50 m/s 7.50 m/s 7.65 m/s

K U K U

mv mv mgy mv mv

v v v

+ = +

+ = + = +

= +

= + =

Insight: Note that the sled is not going 1.50 m/s faster than the 7.50 m/s it would be traveling if it started from rest. That’s because the 45 J of kinetic energy it has at the start (if the sled and passengers have a mass of 40 kg) is tiny compared with the 1125 J of kinetic energy it gains on the way down. The energies add but the speeds do not.

69. Picture the Problem: The sled slides down the hill without friction.

Strategy: As the sled descends the hill its gravitational potential energy is converted into kinetic energy. Find the loss in gravitational potential energy by setting the change in mechanical energy equal to zero, so that f i 0E E EΔ = − = or

f iE E= . Let 0y = at the bottom of the hill and 0v = at the top.

Solution 1. Set bottom topE E= and solve for topU , letting bottom top 0U K= = :


21bottom top2 0 0

K U K U

mv U

+ = +

+ = +

2. Now use equation 8-3 to find the height of the hill:

( )( )

21bottom top2

22bottom

top 2

7.50 m/s2.87 m

2 2 9.81 m/s

mv mgy

vy

g

=

= = =

Insight: The gravitational potential energy of the sled at the top of the hill has been converted into kinetic energy at the bottom of the hill.



8 – 26

70. Picture the Problem: The physical situation is depicted at right.

Strategy: If there are no frictional forces then the mechanical energy at point A equals the mechanical energy at point B. Use that relation to find the speed at point B, then use Newton’s Second Law and the speed of the skier to find the normal force at B. Let A 1.75 my = and B 0y = .

Solution: 1. Set A B :E E= A A B B2 21 1A A B2 2 0

K U K U

mv mgy mv

+ = +

+ = +

2. Now solve for Bv : ( ) ( )( )22 2B A A2 8.0 m/s 2 9.81 m/s 1.75 m 9.9 m/sv v gy= + = + =

3. Write Newton’s Second Law in the vertical direction at point B and find the normal force: ( ) ( )

2

222 9.9 m/s

68 kg 9.81 m/s 1200 N 1.2 kN12 m

yF N mg mv r

vN m gr

= − =

⎡ ⎤⎛ ⎞= + = + = =⎢ ⎥⎜ ⎟

⎢ ⎥⎝ ⎠ ⎣ ⎦

∑

Insight: At point B the skier feels 84% heavier than her normal 670-N weight. Or put another way, she experiences an acceleration of 2 218 m/s 9.81 m/s 1.8 ’sg= . Her actual acceleration is 8.2 m/s2, but when speaking of g forces we include the 1.0g she experiences when standing still on level ground.

71. Picture the Problem: The weight of the runner vertically compresses the “springs” of the soles.

Strategy: Use Newton’s Second Law to set the weight of the person equal to the force exerted by the “springs” of both soles in order to find the compression distance. Use the compression distance together with equation 8-5 to find the energy stored in the soles.

Solution: 1. (a) Write Newton’s Second Law in the vertical direction to find the compression distance:

( )( )( )

sole

sole

2

5sole

2 0

2

62 kg 9.81 m/s0.0015 m 1.5 mm

2 2 2.0 10 N

yF F mg

k x mg

mgxk

= − =

=

= = = =×

∑

2. (b) Use equation 8-5 to find U: ( )( )22 51 12 2 2.0 10 N/m 0.0015 m 0.23 J/shoe 2 shoes 0.46 JU kx= = × = × =

Insight: The energy stored in the shoes is enough to accelerate the runner from rest to 0.12 m/s or 0.27 mi/h or 0.39 ft/s. Not very impressive, but she still literally has a “spring in her step”!

72. Picture the Problem: The nasal strip is compressed as a spring when it is applied to the nose.

Strategy: The nasal strip acts as an ideal spring, exerting a force according to Hooke’s Law (equation 6-4) and storing energy according to equation 8-5. Write the two equations for force and potential energy and substitute one into the other to solve for the spring constant k. Once k is known, the compression distance x can be found from Hooke’s Law.

Solution: 1. (a) Write equations 6-4 and 8-5 and substitute for x to solve for k: ( )

( )( )

22 21 12 2

22

2

0.25 N14 N/m

2 2 0.0022 J

F kx x F k

U kx k F k F k

FkU

= − ⇒ = −

= = − =

= = =

2. (b) Use the known k together with Hooke’s Law to find x :

0.25 N 0.018 m 1.8 cm14 N/m

Fxk

= = = =

Insight: The force exerted by the spring on the nose helps the person to breathe a bit easier. 0.25 N is about 0.90 ounce of force.



8 – 27

73. Picture the Problem: The pendulum begins at point A, swings through point B,

and is vertical at point C as shown in the diagram at right.

Strategy: Use equation 8-3 and the geometry of the situation to find the potential energy at point B. Then use equation 8-3 again to find the potential energy at point C. Let y = 0 correspond to position A.

Solution: 1. (a) Use equation 8-3 to find BU : ( )

( )( )( )

B A

2

B

0cos 0

0.13 kg 9.81 m/s 0.95 m cos 45

0.86 J

U U U mg ymg L

U

θ= + Δ = + Δ

= − −

= − °

= −

2. (b) The magnitude of the change in potential energy from 90° to 45° is greater than the magnitude of the change from 45° to 0° because 71% of the height change occurs between 90° and 45°.

3. (c) Use equation 8-3 to find CU : ( ) ( )( )( )

C A

2

0

0 0.13 kg 9.81 m/s 0.95 m 1.2 J

U U U mg y

mg L

= + Δ = + Δ

= − − = − = −

Insight: You can mathematically show the answer to (b) by writing ( )cos 45 1 cos 45mgL mgL− ° > − ° which is true

because 12cos 45° > .

74. Picture the Problem: The pendulum bob swings between points A and B, changing altitude

and thus gravitational potential energy. See the figure at right.

Strategy: Use the geometry of the problem to find the change in altitude yΔ of the pendulum bob, and then use equation 8-3 to find its change in gravitational potential energy. Let y = 0 correspond to point B:

Solution: 1. (a) Find the height change yΔ of the pendulum bob as it goes from A to B:

( ) ( )0 cos cos 1y L L Lθ θΔ = − − = −

2. Use yΔ to find UΔ as it goes from A to B:

( )( )( )( )( )2

cos 10.25 kg 9.81 m/s 1.2 m cos35 10.53 J

U mg y mgL

U

θΔ = Δ = −

= °−Δ = −

3. Now use equation 8-1 to find

the work done by gravity cW : ( )c, A to B 0.53 J 0.53 JW U= −Δ = − − =

4. (b) The work to go from B to A is the opposite of the work to go from A to B:

( )c, B to A c, A to B 0.53 J 0.53 JW W= − = − = −

5. (c), (d) The force exerted by the string is always perpendicular to the motion so the work done by the string is zero.

Insight: Gravity does positive work as the bob falls from A to B because the force and displacement are along the same direction, but it does negative work as the bob rises because the force is downward but the displacement is upward.

y = 0.0 m A

B

C

θ yΔ

y = −0.95 m



8 – 28

75. Picture the Problem: The airplane takes off, changing both its gravitational potential and kinetic energies.

Strategy: The airplane engines do nonconservative work on the airplane to change its mechanical energy, including its kinetic and gravitational potential energies. Let U = 0 on the runway and let the airplane start from rest. Use the given altitude and speed values to find the change in its mechanical energy. Then set its kinetic energy change equal to its gravitational potential energy change and solve for fv .

Solution: 1. (a) Use the given fy and fv values and equation 8-6 to find EΔ : ( ) ( )

( )( ) ( ) ( )( )

2 21f f2

22 12

7 6 7

0 0

1865 kg 9.81 m/s 2420 m 1865 kg 96.5 m/s

4.43 10 J 8.68 10 J 5.30 10 J 53.0 MJ

E U Kmg y m v

E

Δ = Δ + Δ= − + −

= +

Δ = × + × = × =

2. (b) Set K UΔ = Δ and solve for fv :

( )( )

21f f2

2f f2 2 9.81 m/s 2420 m 218 m/s

K mv U mgy

v gy

Δ = = Δ =

= = =

Insight: Note that in part (a) the change in potential energy (44.3 MJ) is 5.1 times larger than the change in kinetic energy (8.68 MJ). The primary struggle for airplanes is gaining altitude, not speed.

76. Picture the Problem: The child on a swing is a simple pendulum like the one

depicted in the figure at right. The child swings from point B to point A.

Strategy: Set the mechanical energy at point B equal to the mechanical energy at point A and solve for A .y Let y = 0 correspond to point B and let point A be the turning point ( A 0v = ).

Solution: 1. (a) Use equation 8-6 to find Ay :

( )( )

B B A A21B A2

22B

A 2

0 02.02 m/s

0.208 m2 2 9.81 m/s

K U K Umv mgy

vy

g

+ = ++ = +

= = =

2. (b) Use a ratio to find the relationship between the old and new maximum vertical heights:

( )( )

( )

22 1B, old2A, new B, new

2 2A, old B, old B, old

1 1A, new A, old4 4

2 142

0.208 m 0.0520 m

vy v gy v g v

y y

= = =

= = =

Insight: Cutting the initial speed in half reduced the maximum altitude by a factor of four. It would not cut the maximum deflection angle by a factor of four. Instead, the angle would decrease from 18.6° to 9.25° for a L = 4.00 m swing, roughly factor of two, because of the nonlinear relationship between y and θ : ( )1 cos .y L θ= −

77. Picture the Problem: The child slides from rest at point A and

lands at point B as indicated in the figure at right.

Strategy: Use the conservation of mechanical energy to find the horizontal speed of the child at the bottom of the slide in terms of h. Then use equation 4-9, the landing site of a projectile launched horizontally, to find the speed the child should have in order to land 2.50 m down range. Set the speeds equal to each other and solve for h.



8 – 29

Solution: 1. Use equation 8-3 and let A bottomE E= to find bottomv :

( ) ( )

A A bottom bottom2 21 1A A bottom bottom2 2

21bottom2

bottom

0 1.50 m 1.50 m

2

K U K U

mv mgy mv mgy

mg h mv mg

gh v

+ = +

+ = +

+ + = +

=

2. Use equation 4-9 to find the appropriate bottomv for the child to land 2.50 m down range:

bottom bottom bottom bottom2 2x v y g v x g y= ⇒ =

3. Set the two velocities equal to each other and solve for h:

( )( )

bottom

2bottom

22

bottom

2 2

2 2

2.50 m1.04 m

4 4 1.50 m

gh x g y

gh x g y

xhy

=

=

= = =

Insight: These are the sort of calculations an engineer might make to determine how high to build a slide so that a child will land in a certain place. However, an engineer should account for the nonzero friction when making his design.

78. Picture the Problem: The child slides from rest at point A and

lands at point B as indicated in the figure at right.

Strategy: Use the conservation of mechanical energy to find the horizontal speed of the child at the bottom of the slide. Then use equation 4-9 to find the landing site of the child because she is launched horizontally.

Solution: 1. Use equation 8-3 and let A bottomE E= to find bottomv :

( ) ( )

A A bottom bottom2 21 1A A bottom bottom2 2

2 21 1A bottom2 2

2A bottom

1.50 m 1.50 m

2

K U K U

mv mgy mv mgy

mv mg h mv mg

v gh v

+ = +

+ = +

+ + = +

+ =

2. Use equation 4-9 to find x: ( )( )

( ) ( )( ) ( ) ( )

2bottom bottom A bottom

2 2 2

2 2 2

0.54 m/s 2 9.81 m/s 3.2 m 2 1.50 m 9.81 m/s

4.4 m

x v y g v gh y g

x

= = +

⎡ ⎤ ⎡ ⎤= + ⎣ ⎦⎣ ⎦=

Insight: Her initial speed doesn’t make much difference in the landing spot; verify for yourself that she lands at 4.39 m with the initial speed as opposed to 4.38 m when starting from rest. This is quite a slide—4.4 m is 14 ft from the base of the slide, and the intrepid swimmers hit the water at 9.6 m/s (21 mi/h)!



8 – 30

79. Picture the Problem: The swing is essentially a simple pendulum as shown at

right. The person swings from point A to point B.

Strategy: Use the conservation of mechanical energy and the geometry of the problem to find the speed of the person at point B. Then use Newton’s Second Law in the vertical direction at point B to find the force the ropes must exert to support the weight of the person and produce their centripetal acceleration. Solve the resulting expression for the weight of the person.

Solution: 1. (a) Find the height of the person at point A if B 0y = : ( ) ( )A cos 1 cosy L L Lθ θ= − = −

2. Set B AE E= to find Bv :

( )

B B A A21B A2

B A

0 02 2 1 cos

K U K Umv mgy

v gy gL θ

+ = ++ = +

= = −

3. Write Newton’s Second Law in the vertical direction at B to find maxW : ( ) ( )

( )

( )

2max max B

max max max max

max max

max max

2

2 1 cos2 2 2 1 cos

1 2 1 cos 2

2 2355 N 634 N3 2cos 3 2cos 20.0

yF T W mv L

gLW T m T W

LW T

W T

θθ

θ

θ

= − =

−= − = − −

+ − =⎡ ⎤⎣ ⎦

= = =− − °

∑

4. (b) If the person is released at an angle greater than 20.0°, cosθ will decrease, the denominator 3 2cosθ− will in-crease, and maxW will decrease. This is because the person’s speed will be greater at B, requiring more centripetal force.

Insight: Note that maxW is 710 N if the swing is at rest and hanging vertically. The maximum weight in this problem is less than 710 N because the ropes must also provide the centripetal force to keep the person going in a circle.

80. Picture the Problem: The physical situation is depicted in the figure.

Strategy: Use the conservation of mechanical energy and the geometry of the problem to find the speed of the car at point B. Then write Newton’s Second Law in the vertical direction at point B and set the normal force N equal to zero. Solve the resulting expression to find the minimum 0v that will allow the car’s wheels to lose contact with the road. Let A 0y = and

B .y h=

Solution: 1. (a) Set B AE E= to find Bv : B B A A2 21 1B 02 2

2B 0

0

2

K U K Umv mgh mv

v v gh

+ = ++ = +

= −

2. Write Newton’s Second Law in the vertical direction at B to find Bv :

2B

2B

2B

0yF N mg mv r

g v r

gr v

= − = −

− = −

=

∑

3. Substitute the expression from step 1 into step 2 and solve for 0v :

( )20 02 2gr v gh v g r h= − ⇒ = +

4. (b) If the initial speed is greater than the value found in part (a), the car will become airborne at the top of the hill!

Insight: Note that this is a larger 0v than the 0v gr= we found in Chapter 6, problem 63. In that problem, the car’s engine kept its speed constant. In this problem, the car is coasting, so that some of its initial kinetic energy is needed to climb the hill of height h. That is why the initial speed needs to be greater in this case.



8 – 31

81. Picture the Problem: The physical situation is depicted in the figure.

Strategy: Use the conservation of mechanical energy and the geometry of the problem to find the speed of the skateboarder at point A. Let A 0y = and

B 0v = .

Solution: Set A BE E= to find Av :

( )( )

A A B B21A B2

2A B

0 0

2 2 9.81 m/s 2.64 m

7.20 m/s

K U K Umv mgy

v gy

+ = ++ = +

= =

=

Insight: The solution assumes there is no friction as the skateboarder travels around the half pipe. In real life the skater’s speed must exceed 7.20 m/s at point A because friction will convert some of his kinetic energy into heat.

82. Picture the Problem: The motions of the masses in the Atwood’s machine

are depicted in the figure at right: Strategy: Mechanical energy is conserved because there is no friction. Set

i fE E= and solve for fv . The speeds of each mass must be the same at first because they are connected by a rope. When mass 2m lands, mass 1m will still have kinetic energy because it will be traveling at speed fv . Convert this kinetic energy into potential energy to find how high 1m will rise after 2m lands.

Solution: 1. Set i fE E= and substitute for y1 and y2:

( ) ( )

i i f f2 21 1

1 f 2 f 1 1 2 22 221

1 2 f 1 22

0 00


+ = ++ = + + +

= + + + −

2. Now solve for fv : ( ) ( ) 2 2 2 11

2 1 1 2 f f21 2

2m m

m m gh m m v v ghm m

⎛ ⎞−− = + ⇒ = ⎜ ⎟+⎝ ⎠

3. Convert the K of 1m into U and solve for yΔ :

211 f 12

2 2 1 2 1f

1 2 1 2

1 1 22 2

m v m g y

m m m my v gh h

g g m m m m

= Δ

⎛ ⎞ ⎛ ⎞− −Δ = = =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠

4. Enter the given values to find a numerical result: ( ) 4.1 3.7 kg1.2 m 0.06 m 6 cm

3.7 4.1 kgy

⎛ ⎞−Δ = = =⎜ ⎟+⎝ ⎠

Insight: We end up with only one significant figure because by the rules of subtraction 4.1 − 3.7 kg = 0.4 kg. Note that yΔ is rather small because the energy from 2m falling the much larger distance h went into lifting 1m and giving kinetic

energy to both 1m and 2m . The kinetic energy of 2m is converted to heat as a result of the collision with the floor.



8 – 32


Strategy: Use equation 8-9 to set the mechanical energy loss equal to the work done by friction. Then solve for the distance d that the block slides along the surface of the inclined plane. Use Newton’s Second Law to find the normal force between the block and plane, and use the relation

siny d θ= to convert the distance along the incline into a gain of altitude. Let 0y = at the start of the slide and f 0v = at the end.

Solution: 1. Write out equation 8-9 to find ncW :

( ) ( )( ) ( )

nc f i f f i i

21i20 0

W E E K U K U

mgy mv

= − = + − +

= + − +

2. Use Newton’s Second Law to find N:

cos 0 cosyF N mg N mgθ θ= − = ⇒ =∑

3. Set ncW equal to the work done by friction, set sin ,y d θ= and solve for d:

( ) ( )( )

( )( )

( ) ( )

21nc k k i2

21k i2

21i k2

2i

k2

2

cos sinsin cos

2 sin cos1.56 m/s

0.121 m2 9.81 m/s sin 28.4 0.62 cos 28.4

W f d Nd mgy mvmg d mg d mv

mv mgdv

dg

μμ θ θ

θ μ θ

θ μ θ

= − = − = −− = −

= +

=+

= =°+ °⎡ ⎤⎣ ⎦

Insight: At first glance it may seem like the k 0.62μ = limits us to two significant figures, but by the rules of addition

( ) ( )sin 28.4 0.62 cos 28.4 0.476 0.55 1.02° + ° = + =⎡ ⎤⎣ ⎦ (if no rounding) which still has three significant figures.


Strategy: Use equation 8-9 to set the mechanical energy loss equal to the work done by friction. Then solve for the distance d that the block slides along the surface of the inclined plane. Use Newton’s Second Law to find the normal force between the block and plane, and use the relation

siny d θ= − to convert the distance along the incline into a loss of altitude. Let 0y = at the start of the slide and f 0v = at the end.

Solution: 1. Write out equation 8-9 to find ncW :

( ) ( )( ) ( )

nc f i f f i i

21i20 0

W E E K U K U

mgy mv

= − = + − +

= + − +

2. Use Newton’s Second Law to find N: cos 0

cosyF N mg

N mg

θ

θ

= − =

=∑

3. Set ncW equal to the work done by friction, set sin ,y d θ= and solve for d:

( ) ( )( )

( )( )

( ) ( )

21nc k i2

21k i2

21i k2

2i

k2

2

cos sincos sin

2 cos sin1.56 m/s

1.8 m2 9.81 m/s 0.62 cos 28.4 sin 28.4

W fd Nd mgy mvmg d mg d mv

mv mgdv

dg

μμ θ θ

μ θ θ

μ θ θ

= − = − = −− = − −

= −

=−

= =°− °⎡ ⎤⎣ ⎦

Insight: The work done by friction decreases the kinetic energy at a faster rate than the conversion of gravitational potential energy can increase it, and the block eventually comes to rest.

y

d

θ

N

kf

Wiv

y

d

θ

N

kfW cosmg θ

iv



8 – 33

85. Picture the Problem: The vine is essentially a simple pendulum as shown at right.

Jeff swings from point A to point B.

Strategy: Use the conservation of mechanical energy and the geometry of the problem to find Jeff’s speed at point B. Then use Newton’s Second Law in the vertical direction at point B to find the force the vine must exert to support Jeff’s weight and provide his centripetal force.

Solution: 1. Find Jeff’s height at point A if B 0y = :

( ) ( )A cos 1 cosy L L Lθ θ= − = −


( )

B B A A21B A2

B A

0 02 2 1 cos

K U K Umv mgy

v gy gL θ

+ = ++ = +

= = −

3. Write Newton’s Second Law in the vertical direction at B to find T: ( ) ( )

( )( )( )

2B

2

2 1 cos1 2 1 cos

78 kg 9.81 m/s 3 2cos37 1100 N 1.1 kN

yF T mg mv L

gLT m g mg

Lθ

θ

= − =

−⎡ ⎤= + = + −⎡ ⎤⎢ ⎥ ⎣ ⎦

⎣ ⎦

= − ° = =

∑

Insight: The tension in the vine would only be 765 N if Jeff were simply hanging at rest from the vertical vine. The additional 310 N is the centripetal force required to keep Jeff moving in a circle.

86. Picture the Problem: The block slides from rest at point A and is

launched horizontally at point B as indicated in the figure at right.

Strategy: Use the conservation of mechanical energy to find the horizontal speed of the block at the bottom of the ramp. Then use equation 4-9 to find the landing site of the block because it is launched horizontally.

Solution: 1. Set A BE E= and use equation 8-3 to find Bv :

( )

A A B B21

A B B2

A B B

0

2

K U K U

mgy mv mgy

g y y v

+ = +

+ = +

− =

2. Use equation 4-9 to find d: ( ) ( ) ( )( )( )

B B A B B B A B2 2 2 4

4 0.25 m 1.5 0.25 m

1.1 m

d v y g g y y y g y y y

d

= = − = −⎡ ⎤⎣ ⎦= −

=

Insight: If there were friction between the block and the ramp it would convert some of the kinetic energy of the block into heat. The block would then be launched with less speed and would land a shorter distance d from the ramp.



8 – 34

87. Picture the Problem: The block slides from rest at point A and is

launched horizontally at point B as indicated in the figure at right. Strategy: Use equation 8-9, ( ) ( )nc B B A AW K U K U= + − + , to find the

change in mechanical energy due to the presence of friction. Use the resulting expression to find the horizontal speed of the block at the bottom of the ramp. Then use equation 4-9 to find the landing site of the block because it is launched horizontally.

Solution: 1. Write equation 8-9 and solve for Bv :

( )

A A nc B B21

A nc B B2

A B nc B

0

2 2

K U W K U

mgy W mv mgy

g y y W m v

+ + = +

+ + = +

− + =

2. Use equation 4-9 to find d: ( ) ( ) ( )

( ) ( ) ( )( )( )

B B A B nc B B A B nc

2

2 2 2 2 4

9.7 J4 0.25 m 1.50 0.25 m

1.9 kg 9.81 m/s

0.85 m

d v y g g y y W m y g y y y W mg

d

= = − + = − +⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

⎡ ⎤−⎢ ⎥= − +⎢ ⎥⎣ ⎦

=

Insight: The block does not travel as far because friction has converted some of the kinetic energy into heat. 88. Picture the Problem: The weight of the runner vertically compresses the “spring” of the soft track surface.

Strategy: Use Newton’s Second Law to set the weight of the person equal to the force exerted by the “spring” of the track in order to find the compression distance. Use the compression distance together with equation 8-5 to find the energy stored in the track surface.

Solution: 1. (a) Write Newton’s Second Law in the vertical direction to find the compression distance:

foot

track foot

foot5

track

0

2700 N 0.011 m 1.1 cm2.5 10 N

yF N F

k x FF

xk

= − =

=

= = = =×

∑

2. (b) Use equation 8-5 to find U: ( )( )22 51 12 2 2.5 10 N/m 0.011 m 15 JU kx= = × =

Insight: As we discovered in problem 71, the soles of the runner’s shoes also act like springs and store energy. Note that the track surface stores a lot more energy than the sole of the shoe in problem 71 because the instantaneous force exerted by the foot is much greater than the weight of the runner.

89. Picture the Problem: The energy stored in the “springs” of the flea’s legs is converted into kinetic energy after the

jump, and then the kinetic energy becomes gravitational potential energy. Strategy: Let point A correspond to the flea at rest with its compressed “spring” legs and point B correspond to the flea

at rest at its maximum altitude. In this way the spring potential energy has been converted into gravitational potential energy. Use the conservation of mechanical energy to find the maximum altitude attained by the flea. Let A 0y = and

A B 0.v v= = Note that there are two legs, so the total energy stored in the flea’s “springs” is 2122 kx× .

Solution: Set A BE E= and solve for B :y ( ) ( )

( )( )( )( )

A A B B

21B2

242

B 6 2

0 2 0 0 0

26 N/m 1.0 10 m0.053 m 5.3 cm

0.50 10 kg 9.81 m/s

K U K U

kx mgy

kx ymg

−

−

+ = +

+ × + = + +

×= = = =

×

Insight: If the flea stands 1.0 mm tall then the flea can jump straight upward over 53 times its own height! That would be 318 feet for a 6-ft-tall human!



8 – 35

90. Picture the Problem: The trapeze is essentially a simple pendulum as shown at

right. The trapeze artist swings from point A to point B.

Strategy: Use the conservation of mechanical energy and the geometry of the problem to find the trapeze artist’s speed at point B. Then use Newton’s Second Law in the vertical direction at point B to find the force the rope must exert to support the artist’s weight and provide her centripetal force.

Solution: 1. (a) Find the artist’s height at point A if B 0y = : ( ) ( )A cos 1 cosy L L Lθ θ= − = −


( )

B B A A21B A2

B A

0 02 2 1 cos

K U K Umv mgy

v gy gL θ

+ = ++ = +

= = −

3. Write Newton’s Second Law in the vertical direction at B to find T:

( ) ( )

( )

2B

2 1 cos1 2 1 cos

3 2cos

yF T mg mv L

gLT m g mg

L

T mg

θθ

θ

= − =

−⎡ ⎤= + = + −⎡ ⎤⎢ ⎥ ⎣ ⎦

⎣ ⎦

= −

∑

4. (b) Increasing L increases the change in potential energy as the artist swings from A to B, but decreases the centripetal acceleration. The effect of L cancels out.

Insight: The tension in the rope would only be mg if the trapeze artist were simply hanging at rest from the vertical rope. The additional factor of ( )3 2cosθ− is due to the centripetal force required to keep her moving in a circle.

91. Picture the Problem: The ball on a string is a simple pendulum as shown at right.

The ball swings from point A to point B.

Strategy: Use the conservation of mechanical energy and the geometry of the problem to find the ball’s speed at point B. Then use Newton’s Second Law in the vertical direction at point B to find the force the string must exert to support the ball’s weight and provide its centripetal force.

Solution: 1. (a) Set B AE E= to find Bv . Let Ay l= and B 0y = :

B B A A21B2

B

0 02

K U K Umv mgl

v gl

+ = ++ = +

=

2. Write Newton’s Second Law in the vertical direction at B to find T:

( )( )

2cp B

2 3yF T mg ma mv l

T m g gl l mg

= − = =

= + =

∑

3. (b) Tension depends on cp ,a which is proportional to 2Bv and inversely proportional to the radius r. Therefore, l

cancels out because both 2Bv and r are proportional to l.

Insight: The tension in the string would only be mg if the ball were simply hanging at rest from the vertical string. The additional factor of 3 is due to the centripetal force required to keep the ball moving in a circle.



8 – 36

92. Picture the Problem: The ball on a string is a simple pendulum as shown at right.

The ball swings from point A to point C.

Strategy: Use the conservation of mechanical energy and the geometry of the problem to find the ball’s position at point C. Let Ay l= and A B 0.v v= =

Solution: 1. (a) Find the ball’s height at point C if B 0y = : ( ) ( )C cos 1 cosy L L Lθ θ= − = −

2. Set A CE E= to find θ: ( )

( )( )

A A C C

1

1

0 0 1 cos1 cos

cos 10.325 mcos 1 59.90.652 m

K U K Umgl mgL

l Ll L

θθ

θ −

−

+ = ++ = + −

= −= −

⎛ ⎞= − = °⎜ ⎟⎝ ⎠

3. (b) From steps 1 and 2 we can see that ( )C 1 cos 0.325 my L lθ= − = = . The mass rises to the same height from which it started because mechanical energy is conserved.

4. (c) As shown in step 2, ( )1cos 1 l Lθ −= − . Insight: When all of its kinetic energy has been converted to gravitational potential energy, the vertical height of the

ball will always be the same at point A no matter how complicated the arrangement of pegs in the apparatus. 93. Picture the Problem: The physical situation is depicted at right.

Strategy: Use the conservation of mechanical energy and the geometry of the problem to find the ice cube’s speed at point B. Then write Newton’s Second Law in the radial direction at point B, set the normal force to zero, and find the angle θ. Let the table height correspond to y = 0, so that

Ay r= and B cos .y r θ=

Solution: 1. (a) Set B AE E= to find Bv : ( )

( )( )

B B A A21B2

21B2

B

cos 01 cos

2 1 cos

K U K Umv mg r mgr

mv mgr

v gr

θθ

θ

+ = ++ = +

= −

= −

2. Write Newton’s Second Law in the radial direction at B and set N = 0. Then solve for θ :

( )

( )

2cp B

1

cos

0 cos 2 1 cos /

cos 2 2cos

2 3 cos cos 2 3 48.2

rF N mg ma mv r

mg m gr r

θ

θ θ

θ θ

θ θ −

= − = = −

− = − −⎡ ⎤⎣ ⎦− = − +

= ⇒ = = °

∑

Insight: Note that this angle is independent of the mass of the ice cube, the radius of the bowl, even the acceleration of gravity (it’d be the same on the Moon!). If there were friction, the speed of the cube would be smaller and the angle θ would be larger.



8 – 37


Strategy: Since the system is frictionless, we can set the mechanical energy at configuration i equal to the mechanical energy at configuration f . Solve the resulting expression for the distance d. The work done by the rope on 2m is nonconservative and therefore changes the mechanical energy of 2m according to equation 8-9. Use this fact to find the work the rope does on 2m .

Solution: 1. (a) Set i fE E= and solve for d: ( )

( )

i i f f21

1 2 22

21 2

2

0 0

2

K U K U

m m v m gd

m m vd

m g

+ = +

+ + = +

+=

2. (b) The rope pulls upward on 2m , the same direction it is displaced, so we conclude the work done by the rope on

2m is positive.

3. (c) Set the work done by the rope equal to the change in mechanical energy for 2m . The mass 2m starts out with kinetic energy 21

22 m v and finishes with gravitational potential energy 2 :m gd

( )

( )

nc 2,f 2,i

212 22

21 2 21

nc 2 222

2 2 21 1 11 2 2 12 2 2

2

W E E

m gd m v

m m vW m g m v

m g

m m v m v m v

= −

= −

⎡ ⎤+= −⎢ ⎥

⎢ ⎥⎣ ⎦

= + − =

Insight: The nonconservative work done by the rope is positive, so 2m gains mechanical energy during this episode. However, the conservative work done by gravity is negative, because 2m gains potential energy (equation 8-1,

cW U= −Δ ). Therefore we conclude that the gain in potential energy for 2m is greater than its loss of kinetic energy.



8 – 38


Strategy: Use equation 8-9 to relate the nonconservative work of friction on the entire system equal to the energy difference between configuration f and configuration i . Solve the resulting expression for the initial speed v. The work done by the rope on 2m is nonconservative and therefore changes the mechanical energy of

2m according to equation 8-9. Use this fact to find the work the rope does on 2m .

Solution: 1. (a) Write equation 8-9 to obtain an expression for ncW :

( ) ( )( ) ( )

nc f i f f i i

212 1 220 0

W E E K U K U

m gd m m v

= − = + − +

⎡ ⎤= + − + +⎣ ⎦


3. Substitute the expression from step 2 into step 1 and solve for v:

( )( ) ( )

( )( )

( )( ) ( )( )( )

21k 1 2 1 22

212 k 1 1 22

2 k 1

1 2

2

2

2 9.81 m/s 0.065 m 1.1 kg 0.25 2.4 kg

2.4 1.1 kg

0.79 m/s

m gd m gd m m v

gd m m m m v

gd m mv

m m

v

μ

μ

μ

− = − +

− + = − +

+=

+

+⎡ ⎤⎣ ⎦=+

=

4. (b) The rope pulls upward on 2m , the same direction it is displaced, so we conclude the work done by the rope on

2m is positive.

5. (c) Set the work done by the rope equal to the change in mechanical energy for 2m . The mass 2m starts out with kinetic energy 21

22 m v and finishes with gravitational potential energy 2m gd :

( )( )( ) ( )( )

nc 2,f 2,i

212 22

22 1nc 21.1 kg 9.81 m/s 0.065 m 1.1 kg 0.79 m/s 0.36 J

W E E

m gd m v

W

= −

= −

= − =

Insight: The nonconservative work done by the rope is positive, so 2m gains mechanical energy 0.36 J during this episode. However, the conservative work done by gravity is negative, because 2m gains potential energy (equation 8-1,

cW U= −Δ ). Therefore we see that the 0.70 JUΔ = for 2m is greater than its loss of kinetic energy, 0.34 J.KΔ = − 96. Picture the Problem: The physical situation is depicted at right.

Strategy: Use the conservation of mechanical energy and the geometry of the problem to find the block’s speed at point B. Then write Newton’s Second Law in the vertical direction at point B, set the normal force to zero, and find the initial height h. Let the table height correspond to y = 0, so that Ay h= and B 2 .y r=

Solution: 1. (a) Set B AE E= to find Bv : ( )

( )( )

B B A A21B2

21B2

B

2 02

2 2

K U K Umv mg r mgh

mv mg h r

v g h r

+ = ++ = +

= −

= −



8 – 39

2. Write Newton’s Second Law in the radial direction at B and set N = 0. Substitute for

Bv and multiply both sides by r mg− and solve for h:

( )

2cp B

52

0 2 2 /

2 4

rF N mg ma mv r

mg m g h r r

r h r h r

= − − = = −

− = − −⎡ ⎤⎣ ⎦

= − ⇒ =

∑

3. (b) The release height is independent of the block’s mass because although a larger mass would have a larger gravitational force on it, it would also have a larger inertia, so the effect of mass cancels out.

Insight: Note that this angle is independent of the mass of the block and even the acceleration of gravity (it’d be the same on the Moon!). If there were friction, the speed of the block would be smaller and h would need to be larger.


Strategy: Use equation 8-9 to relate the nonconservative work of friction on the entire system equal to the energy difference between the initial and final states. Solve the resulting expression for the initial height h.


( ) ( )( ) ( )

nc f i f f i i

21f2 0 0

W E E K U K U

mv mgh

= − = + − +

= + − +

2. The nonconservative work is done by friction:

nc k kW f d mgdμ= − = −

3. Substitute the expression from step 2 into step 1 and solve for v: ( )

( )( ) ( )( )

21k f2

21f k2

2f k

2

2

2

3.50 m/s0.640 0.100 m 0.688 m

2 9.81 m/s

mgd mv mgh

mgh m v gd

h v g d

h

μ

μ

μ

− = −

= +

= +

= + =

Insight: Note that the required h is independent of mass. Friction has an appreciable effect; if there were no friction the initial height would only need to be 0.624 m for the final speed to be 3.50 m/s.


Strategy: Use equation 8-9 to relate the nonconservative work of friction on the entire system equal to the energy difference between the initial and final states. Solve the resulting expression for the spring compression distance d.


( ) ( )( ) ( )

nc f i f f i i

2 21 1f2 20 0

W E E K U K U

mv kd

= − = + − +

= + − +

2. The nonconservative work is done by friction: nc k kW f x mg xμ= − Δ = − Δ

3. Substitute the expression from step 2 into step 1 and solve for v: ( )

( )( )

( ) ( ) ( )( )( )

2 21 1k f2 2

2 21 1f k2 2

21f k2

2 212

2

2 1.2 kg2.3 m/s 0.44 9.81 m/s 0.050 m 0.097 m

730 N/m

mg x mv kd

kd m v g x

md v g xk

d

μ

μ

μ

− Δ = −

= + Δ

= + Δ

⎡ ⎤= + =⎣ ⎦

Insight: The compression distance d needs to be larger if you desire a larger fv or if the rough patch xΔ is longer. Friction has a small effect here; if there were no friction d would still need to be 9.3 cm for the final speed to be 2.3 m/s.



8 – 40

99. Picture the Problem: The motions of the masses in the Atwood’s machine

are depicted in the figure at right: Strategy: Mechanical energy is conserved because there is no friction. Set

i fE E= and solve for fv . The speeds of each mass must always be the same because they are connected by a rope. Reapply conservation of energy for the individual masses to find the change in mechanical energy for just mass 2. Use that result and W Fd= to find the tension in the rope.

Solution: 1. (a) The mass 2m gains kinetic energy but loses gravitational potential energy. The way to tell the sign of EΔ is to concentrate on the nonconservative work of the rope on 2m , which is negative because the rope pulls upward but the mass moves downward. Since ncW E= Δ we conclude that EΔ is negative.

2. (b) Set i fE E= and substitute for y1 and y2:

( ) ( )

i i f f2 21 1

1 f 2 f 1 1 2 22 221

1 2 f 1 22

0 00


+ = ++ = + + +

= + + + −

3. Now solve for fv : ( ) ( ) 2 2 2 11

2 1 1 2 f f21 2

2m mm m gh m m v v ghm m

⎛ ⎞−− = + ⇒ = ⎜ ⎟+⎝ ⎠

4. Now set i fE E= again and solve for 2EΔ :

i i f f

1i 2i 1i 2i 1f 2f 1f 2f

2f 2f 2i 2i 1i 1i 1f 1f21

2 1 1f 1 1f20 0

K U K UK K U U K K U U

K U K U K U K UE m v m gy

+ = ++ + + = + + ++ − − = + − −

Δ = + − −

5. Substitute fv from step 3 into the expression in step 4: ( ) ( )( )

2 112 1 12

1 2

22 11

1 2

2

4.1 3.7 kg1 3.7 kg 9.81 m/s 1.2 m 1 46 J3.7 4.1 kg

m mE m gh m gh

m mm m

m ghm m

⎡ ⎤⎛ ⎞−Δ = − −⎢ ⎥⎜ ⎟+⎢ ⎥⎝ ⎠⎣ ⎦

⎛ ⎞ ⎛ ⎞− −= − + = − + = −⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠

6. (c) Find the work done by the rope on 2m and solve for T:

2nc 2

46 J 38 N1.2 m

EW F d T h E Th

Δ −= − = − = Δ ⇒ = − = − =

Insight: The mass 2m loses more gravitational potential energy than the kinetic energy it gains, so its net change in mechanical energy is negative. Using Newton’s Second Law we find 20.50 m/sa = and ( )2 38 NT m g a= − = .

100. Picture the Problem: The model wing of a dragonfly behaves as an ideal

spring when a force is applied to its tip.

Strategy: The slope of a force vs. stretch distance graph for a spring is equal to the spring constant. In this case, the inverse of the slope of the deflection vs. force graph is the spring constant.

Solution: 1. Find the slope of the line labeled #1:

5.3 mm 0slope 10.6 mm/N0.5 N 0

−= =

−

2. Take the inverse to find k:

1.0 N 94 N/m10.6 mm 1 m /1000 mm

k = =×

Insight: The graph is presented in this fashion (stretch distance vs. force instead of force vs. stretch distance) because when the measurements were made, the independent variable was the magnitude of the force applied to the wing, and the dependent variable was the measured deflection.



8 – 41


spring when a force is applied at two-thirds the distance from the base of the wing to the tip.



1.7 mm 0slope 3.4 mm/N0.5 N 0

−= =

−


1.0 N 290 N/m3.4 mm 1 m /1000 mm

k = =×

Insight: The wing is almost three times stiffer when the force is applied to two-thirds the distance from the base of the wing to the tip as it is when the force is applied to the tip. This is due to the fact that the ends of the wings of the dragonfly are very thin and flexible.


spring.


Solution: The forewing is the stiffer wing because it takes a larger force to produce the same deflection as the hindwing.

Insight: Another way to approach the problem is to say that the stiffer the wing, the smaller the slope of the deflection vs. force graph, and the forewing clearly has the smaller slope.


spring.



4.0 mm 0slope 8.0 mm/N0.5 N 0

−= =

−


1.0 N 125 N/m8.0 mm 1 m/1000 mm

k = =×

3. Find the energy stored: ( )( )22 41 12 2 125 N/m 0.0035 m 7.66 10 J 0.766 mJU kx −= = = × =

Insight: Although the wing stores only a tiny amount of energy, it flaps many times a second, generating sufficient power to allow the dragonfly to maneuver with great agility.



8 – 42


Strategy: The system is frictionless so that the mechanical energy is conserved. In each case set the initial mechanical energy equal to the final mechanical energy and solve for fv .

Solution: 1. (a) Set i fE E= : i i f f

2 2 21 1 1i f f2 2 20

K U K U

kx mv kx

+ = +

+ = +

2. Now solve for fv : ( ) ( )

( ) ( ) ( )2 22 2f i f

955 N/m0.0500 m 0.0250 m 1.03 m/s

1.70 kgkv x xm

⎡ ⎤= − = − − − =⎣ ⎦

3. (b) Set i fE E= and solve for fv : ( ) ( )

( ) ( )22 2 2f i f

955 N/m0.0500 m 0 1.19 m/s

1.70 kgkv x xm

⎡ ⎤= − = − − =⎣ ⎦

Insight: In part (a) the transformation of spring potential energy into kinetic energy is incomplete, but you can see that although the spring is halfway compressed, the block has more than half of its final kinetic energy and is already at 87% of its final speed.


Strategy: The system is frictionless so that the mechanical energy is conserved. In each case set the initial mechanical energy equal to the final mechanical energy and solve for either iv or m.

Solution: 1. (a) Set i fE E= : i i f f2 21 1i f2 20 0

K U K U

mv kx

+ = +

+ = +

2. Now solve for i :v ( )

( ) ( )22i f

955 N/m0.0400 m 0.948 m/s

1.70 kgkv xm

= = − =

3. (c) Solve the expression from step 1 for m: ( ) ( )

( )

22f2 2i

0.0400 m955 N/m 1.29 kg

1.09 m/sx

m kv

−= = =

Insight: In part (a) the initial speed is less than the 1.09 m/s required to compress the spring a distance 4.60 cm as in Example 8-8. In part (b) the mass is less than the 1.70 kg required to compress the spring a distance 4.60 cm as in Example 8-8.



8 – 43

106. Picture the Problem: The initial and final states of the system are

depicted at right. The values given in the example are 1 2.40 kg,m =

2 1.80 kg,m = and d = 0.500 m.

Strategy: The nonconservative work done by friction changes the mechanical energy of the system. Use equation 8-9 to find EΔ and set it equal to the work done by friction. Solve the resulting expression for the coefficient of kinetic friction kμ .

Solution: 1. (a) The coefficient of kinetic friction should be decreased in order for the system to keep more kinetic energy and have a larger final speed than the 1.30 m/s found in Example 8-10.

2. (b) Write equation 8-9 to obtain an expression for ncW :

( ) ( )( ) ( )( )( ) ( )

nc f i f f i i

2 21 11 2 f 1 2 2f 1 2 i 1 2 2i2 2

2 21nc 1 2 f i 2 2f 2i2

W E E K U K U


W m m v v m g y y

= − = + − +

⎡ ⎤ ⎡ ⎤= + + + − + + +⎣ ⎦ ⎣ ⎦

= + − + −


4. Substitute the expression from step 2 into step 1 and solve for kμ :

( )( ) ( )( ) ( )

( )( )( ) ( )( )( )( )( )

2 21k 1 1 2 f 22

2 21 11 2 f 2 2 1 2 f2 2

k1 1

22 12

k 2

0 0

1.80 kg 9.81 m/s 0.500 m 4.20 kg 1.50 m/s0.349

2.40 kg 9.81 m/s 0.500 m

m gd m m v m g d

m m v m gd m gd m m vm gd m gd

μ

μ

μ

− = + − + −

+ − − += − =

−= =

Insight: The coefficient of kinetic friction is smaller than the k 0.450μ = from Example 8-10 as expected.



8 – 44

physics chapter 8 answers

Documents

gravitational energy

gravitational potential

spring energy

spring potential energy

compressional potential

stored potential energy

objects kinetic energy

initial mechanical energy