solucionario fundamentos de física 9na edición capitulo 5

5Energy

CLICKER QUESTIONS

Question C1.01

Description: Eliciting prior conceptions about “work” and introducing the physics defi nition.

Question

In which of the following cases would you do the least amount of work?

1. You push a box a distance d across a fl oor. 2. You pick up the box, walk a distance d, and gently put the box down on the fl oor. 3. You pick up the box, walk a distance d, and drop the box. 4. Both 1 and 2 are the least. 5. Both 1 and 3 are the least. 6. Both 2 and 3 are the least. 7. The amount of work done is the same in all three cases. 8. The amount of work cannot be determined from information given.

Commentary

Purpose: To introduce the formal defi nition of “work.”

Discussion: In physics, we have a mathematical defi nition for “work” based on the forces exerted on an object and the object’s motion. The work done by a constant force F is W FdF = || , where F is the magnitude of the force F, and d

|| is the component of the object’s displacement parallel to F. (There are other forms of

the defi nition of work, but this is the most useful here.)

Picking up the box requires a certain amount of positive work to be done. The force is mg (up) and the displacement is h (up), so the work is mgh. Carrying the box requires zero work to be done, because the displacement is horizontal and therefore the component parallel to the force is zero. (To be precise, some positive work is done when you start moving and some negative work is done when you stop, for zero total work.) And lowering the box requires a certain amount of negative work to be done. The force is still mg (up), but now the displacement is h (down), so the component parallel to the force is negative, and the work done is – mgh.

Thus, the total work done in case (2) is zero. In case (3), the total work is positive, since you do positive work to lift the box, zero work while walking, and zero work while the box falls since you aren’t exerting any force on it at all. In case (1), you will do some work as you apply a force in the direction you push the box; how much work depends on how much friction there is between the box and fl oor. Since even a “negligible” amount of friction is nonzero, answer (2) is best using a physics defi nition of work.

Other defi nitions of work are valid as well. If you apply one of several possible everyday defi nitions, other answers are defensible.

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Key Points:

• The physics defi nition of “work” does not always agree with everyday usage of the term. Sometimes you can get very tired doing no “work” at all!

• For a force exerted on an object, positive work is done if the force has a positive component in the direction of motion, and negative work is done if the force has a negative component in that direction (i.e., points opposite the direction of motion).

• A force acting perpendicularly to the direction of motion does zero work.

For Instructors Only

This problem is intended for use when fi rst introducing the concept of “work,” and is intended to elicit a vigorous discussion of the term’s meaning. We suggest using it before any formal defi nition has been presented, and then, after a discussion that draws out students’ preconceptions and colloquial defi nitions, introducing the physics usage of the word. This will help students better appreciate the defi nition when it comes, and be more defensive about confusing the technical and commonplace meanings.

When using this question before formally defi ning of work, “right or wrong” is not a useful distinction. Each one of the answers is defensible depending upon one’s perspective. Rather, strive to have students become conscious of and articulate their various interpretations of “work.”

Actually analyzing the exertion of a human in terms of the physics defi nition of work is possible, but quite diffi cult since humans are not conservative systems; the details of muscle fi ber operation, with their continual contraction and relaxation and energy dissipation even when holding something stationary, are relevant.

Question C1.02a

Description: Integrating work and force ideas.

Question

Two blocks with mass m2 > m

1 are pulled as shown along horizontal surfaces, each at a constant speed v.

Which tension force does more work moving the block a distance D along the surface?

T1m1

T2m2

1. T1

2. T2

3. Neither; both do zero work. 4. Neither; both do the same amount of negative work. 5. Neither; both do the same amount of positive work. 6. Impossible to determine without knowing the masses 7. Impossible to determine without knowing the coeffi cients of friction 8. Impossible to determine without knowing the speed v 9. Impossible to determine for some other reason


Energy 187

Commentary

Purpose: To integrate force ideas and work ideas.

Discussion: In each case, the block is not accelerating, so the net force is zero. Thus, the tension force must be equal in magnitude to the kinetic friction force, and the normal force must be equal in magnitude to the weight of the block. The kinetic friction force is proportional to the normal force f Nk k=( )µ . The normal force is larger in case 2, since the mass (and hence the weight) is larger.

The work done is the tension multiplied by the distance D. The distances are the same, so the question becomes, “Which tension force is larger?” Since the accelerations are zero, this is equivalent to asking, “Which friction force is larger?” If we assume that the surfaces are the same and the blocks are made of similar material, then the coeffi cients of friction will be the same. With this assumption, case 2 will have a larger friction force, which means T

2 is larger and therefore does more work. So, answer (2) is defensible.

On the other hand, since we are not told anything about the coeffi cients of friction, answer (7) is defensible.

Students might think that no work is being done on the blocks, since the kinetic energy is not changing. However, the question asks about the work done by a particular force, not the net work.

Students might ignore friction. Since the blocks are moving at constant velocity, friction must be present to oppose the tension force.

(Note that if the tension were being applied at an upward angle, the problem would become considerably more complicated.)

Key Points:

• The work done by a particular force is different from the total work done on an object.

• The work done by a particular force depends on the displacement of the object and the component of the force along the direction of motion.

• Even though we are not given numerical values for the masses, speed, etc., we can still reason through to an answer under certain assumptions.

• There is more than one defensible answer to this question. Make sure your answer is consistent with your chosen assumptions, and make sure you are aware of the assumptions you are making.


This is the fi rst of three related questions designed to explore student reasoning about work, the normal force, and the kinetic friction force. Here we highlight the reasoning needed to understand work. Even though we are talking about work and energy ideas, we still need to be able to analyze situations using forces; we cannot forget all that useful stuff!

Note that it is common to say that an object moving at constant speed has “no acceleration.” It is more proper to say “zero acceleration,” since the object still has an acceleration; its value is zero.

Yes, this question is ambiguous: should students assume that the coeffi cients of friction are the same? Rather than a liability, this ambiguity is a strength of the question: it focuses students’ attention on the signifi cance of that quantity and the implications of assuming one way or another. We suggest not mentioning anything about the coeffi cients of friction before or while students answer the question; instead, let disagreement among students raise the topic naturally. Then, stress that neither assumption (and corresponding answer) is “right”; rather, the point is the implications of the choice and the cogency of subsequent reasoning.


188 Chapter 5

Question C1.02b

Description: Integrating work and force ideas, emphasizing what the defi nition of work depends upon.

Question

Two blocks with mass m2 > m

1 are pulled as shown along rough horizontal surfaces with the same coef-

fi cient of friction. Both are pulled at constant speed, but v2 < v

1. Which tension force does more work

moving the block a distance D along the surface?

T1 m1

T2 m2

1. T1

2. T2

3. Neither; both do zero work. 4. Neither; both do the same amount of negative work. 5. Neither; both do the same amount of positive work. 6. Impossible to determine without knowing the masses 7. Impossible to determine without knowing the speeds 8. Impossible to determine for some other reason

Commentary

Purpose: To integrate force ideas with work ideas, exploring the defi nition of work and the quantities it does and does not depend upon.

Discussion: There are four forces on each block: (1) gravitation, down; (2) normal, up; (3) tension, left; and (4) kinetic friction, right. Since the vertical component of acceleration is zero, the normal force balances gravitation and is therefore larger for the heavier block (case 2). Since the coeffi cients of friction are the same, and f Nk k= µ , the friction force is larger in case 2 as well. Since neither block is accelerating, the tension force must balance the kinetic friction force, so the tension force is larger in case 2. Both blocks move through the same distance, so, according to the defi nition of work, T

2 does more work.

How about the sign? Both forces do positive work, even though they are pointed to the left. The displace-ment is also to the left, so the component of tension along the direction of motion is positive. Also, think about what would happen without any friction: Both blocks would speed up, which means tension (the only force doing work) is doing positive work.

The result does not depend on the speed of the block, since work depends on force and displacement (not velocity or time). Block 2 takes more time to move a distance D, but that does not affect the work.

Key Points:

• Work is positive when the component of force along the direction of motion is positive, and negative when the component is negative. Negative work does not mean work in the negative (leftward) direction; it means work due to a force that opposes the object’s motion.

• Work depends on force and displacement, not velocity or time. Because forces are related to acceleration, work is also related to acceleration.


Energy 189

• Two objects moving at constant velocity have the same acceleration (zero), even when they have different speeds.


This is the second of three related questions using similar situations. In the fi rst, we considered the impact of different masses (and left the coeffi cients of friction ambiguous). Here, we have equal masses but vary the block speeds.

Students commonly think work is a vector, whose sign is used to indicate direction; such a misconception should be elicited, made explicit, and dispelled.

A forward reference to the concept of power may be of benefi t here.

Question C1.02c

Description: Integrating work and force ideas, emphasizing the role of a force’s components in the work it does.

Question

Two identical blocks are pulled as shown along rough horizontal surfaces with the same coeffi cient of friction. Both are pulled at constant speed v. Which tension force does more work moving the block a distance D along the surface?

T1m

T2m

1. T1

2. T2

3. Neither; both do zero work. 4. Neither; both do the same amount of negative work. 5. Neither; both do the same amount of positive work. 6. Impossible to determine without knowing the mass m 7. Impossible to determine without knowing the speed v 8. Impossible to determine without knowing the angle of T

2

9. Impossible to determine for 2 of the reasons above 10. Impossible to determine for some other reason

Commentary

Purpose: To integrate force ideas with work ideas, highlighting the role of a force’s components in deter-mining the work it does.

Discussion: There are four forces on each block: (1) gravitation, down; (2) normal, up; (3) tension, right /up; and (4) kinetic friction, left. The normal force and the vertical component of tension balance gravita-tion, so the normal force is larger in case 1. Since the coeffi cients of friction are the same, and f Nk k= µ , the


190 Chapter 5

friction force is larger in case 1 as well. Since neither block is accelerating, the horizontal component of the tension force must balance the kinetic friction force, so this component is larger in case 1. The displacement is horizontal, so the work done depends only on the horizontal component of tension. Therefore, T

1 does

more work.

T2 might be larger than T

1, depending on the angle of the rope, but this is not relevant to the question

asked. It is not the larger force that does the larger amount of work, but the force with the larger component along the direction of motion.

Key Points:

• The work done by a force depends on the component of that force along the direction of motion. A large force does a small amount of work when the component along the direction of motion is small. And when the component is zero, the work is zero.

• The normal force is not always equal to the weight of the object. The normal force is a “constraint” force, which means it will balance all forces with components perpendicular to the surface.


This is the last of three related questions using similar situations. In the fi rst, we considered the impact of different masses (and left the coeffi cients of friction ambiguous); in the second, we had equal masses but varied the block speeds. Here, masses and speeds are equal, and we instead vary the angle and perhaps magnitude of the tension force.

This question highlighs the fact that the normal force is a constraint force (rather than as a force with an empirical law). It also highlights the fact that only the component of a force along the direction of motion contributes to the work done.

Some students may not notice that the masses of the blocks are the same. (In the two previous cases, the lower block was more massive than the upper one.) Rather than heading this off, let it happen and then come to light during the discussion. This is of greater long-term benfi t, helping students learn to be more careful in the future.

Question C1.03a

Description: Exploring work, preparing for energy conservation.

Question

Two identical blocks fall a distance H. One falls directly down; the other slides down a frictionless incline. Which has the larger speed at the bottom?

H

θ

1. The one falling vertically 2. The one sliding down the incline


Energy 191

3. Both have the same speed. 4. Cannot be determined

Commentary

Purpose: To hone the concept of “work.”

Discussion: We will ignore the effects of air resistance.

The gravitational force acts vertically downward. The work done by a constant force is the magnitude of the force times the displacement in the direction of the force. In both cases, the block’s vertical displace-ment is the same. Since the force is the same and the displacement in the direction of the force is the same, the work done by gravitation is the same.

For the block falling vertically, no other forces act, so this is the total work done on it.

For the block on the incline, a normal force also acts. The normal force is perpendicular to the displace-ment, however, so it does zero work. And we are told that the incline is frictionless, which means the force of friction is small enough to ignore and also does zero work. So, the total work done on this block is the same as for the vertical block.

According to the work–kinetic energy theorem, if both blocks have the same amount of work done on them, they must have the same kinetic energy at the bottom, and thus the same speed.

Note that the block on the incline takes longer to reach the bottom. Why? Because although it has the same speed as the falling block at any given height, that speed is not directed entirely downward. In other words, its acceleration is smaller, so to reach the same fi nal speed, it must take longer to reach the bottom.

Key Points:

• The total work done on an object by all forces (internal and external) is equal to the change in the object’s kinetic energy (work–kinetic energy theorem). In this case, zero work is done by internal forces.

• The work done on an object by a force depends on the component of the object’s displacement along the direction the force is applied, not on the object’s total displacement.

• Work and energy ideas are powerful for relating forces, distances, and speeds. They are not so useful for determining times.


This question is good for exploring the concept of work, and leading up to the defi nition of potential energy and the principle of conservation of energy.

Students commonly believe that since the falling block lands fi rst, it must be moving faster than the block on the incline.

Another common error is for students to run a “mental simulation” and decide that the falling block has the higher speed, not realizing that their intuition includes the effects of friction.

Since both blocks have the same speed regardless of their mass, an informative and valuable follow-up is to ask the same question again, but with different masses for the two blocks.


192 Chapter 5

Note that internal forces can do work on a system, though none do in this case. Thus, it is important to describe the work–kinetic energy theorem in terms that will generalize properly later and to avoid using phrases such as “net work,” “work done by the net force,” and “work done by external forces.” In other words, we generally use a “particle” model in which there is necessarily no deformation, no internal degrees of freedom, no moving parts, and no effects of rotation. However, students generally do not appreciate the subtleties of the particle model, especially when we use it with obvious non-particles. We also do not usually explain the transition from a particle model to a non-particle model, so it is recommended to avoid certain oversimplifi cations and overgeneralizations. The proper statement of the work–kinetic energy theorem is that the total work done by all forces on a system, internal and external, is equal to the change in total kinetic energy of the constituents of that system. That is, internal forces can do work on a system, and now we can properly apply the work–kinetic energy theorem to lots of more interesting situations, such as accelerating cars, bouncing balls, masses being “exploded” by a spring, and spinning disks.

Question C1.03b

Description: Exploring work, preparing for energy conservation.

Question

Two masses, M > m, travel down the surfaces shown. Both surfaces are frictionless. Which mass has the larger speed at the bottom?

m M

H

q

1. m 2. M 3. Both have the same speed. 4. Cannot be determined

Commentary

Purpose: To hone the concept of “work.”

Discussion: The surfaces are frictionless and we will ignore air resistance, so two forces act on each block: gravitation, down; and normal, perpendicular to the surface.

We cannot use Newton’s laws to answer the question, because our mathematics is not sophisticated enough to fi nd the speed as a function of time when the force is not constant or not known as a function of time. However, since the normal force always acts perpendicularly to the motion of the block, it does zero work in each situation. This means energy ideas are useful.

Because there is no friction, we can use the work–kinetic energy theorem. The only force that does nonzero work is gravitation, and that acts downwards, so only the vertical component of each block’s displacement is relevant. Thus, the blocks will have kinetic energies of mgH and MgH at the bottom. Since kinetic energy is 1

22mv and 1

22M v , v2 = 2gH and both blocks have the same speed at the bottom.


Energy 193

In other words, the weight is proportional to mass, but the kinetic energy is also proportional to mass, so the mass cancels out. Without friction, the speed of each block at the bottom depends only on the vertical distance it falls, and not on the angle or shape of the incline or the mass of the block. (With friction, the fi nal speeds would be different.)

This does not mean that both blocks reach the bottom at the same time. One block will take longer, but when it does reach the bottom, it will have the same fi nal speed as the fi rst.

Key Points:

• The work–kinetic energy theorem is useful for problems where we want to relate forces, distances, and speeds, and have a complicated path or nonconstant acceleration and no friction.

• The work done on an object by a force depends on the component of the object’s displacement along the direction the force is applied, not on the object’s total displacement.

• Often, the mass of an object does not affect its motion because it cancels out of calculations. This generally happens for forces that are proportional to the mass of the object acted upon.


Students who choose answer (2) may be thinking that the larger mass has the larger potential energy change, and therefore the larger speed at the bottom.

Students who choose answer (1) may be applying the intuitive idea that since m reaches the bottom fi rst, it must have the larger fi nal speed.

Students may have other reasons for choosing an incorrect answer, making it worth your while to spend time soliciting all the different lines of thinking for each choice.

Note that when there is friction, we cannot properly apply the work–kinetic energy theorem.

Question B2.08

Description: Introducing the spring force.

Question

Three identical springs are arranged as shown below.

30m

8m

A

B

d

d

d

L0

L0

L0

m

C


194 Chapter 5

Put these in order from largest to smallest spring force:

1. A > B > C 2. A > B = C 3. A = B > C 4. B > A > C 5. B = A > C 6. B > A = C 7. None of the above 8. Impossible to determine

Commentary

Purpose: To introduce the spring force and stress the point that it depends only on the compression or extension of the spring from its resting length.

Introduction: The magnitude of the spring force is given by F kds = , where k is the spring constant (or elastic constant) measured in N/m, and d is how far the spring is compressed or stretched from its relaxed state. A spring that obeys this relationship is called “ideal.”

We sometimes write this relationship as F x��

s k= – , where x� is the displacement of the end of the spring

from its relaxed position. This is a vector equation, and the minus sign indicates that the direction of the force is opposite the displacement of the end of the spring. Another form of the same relationship is F

s = k |L – L

0|, where L is the length of the spring, L

0 is its relaxed length, and |x| is the absolute value of x.

In other words, d = |L – L0|.

An ideal spring has the same spring constant for both compression and extension. Also, the orientation of the spring does not matter.

The spring force does not depend on any factors other than its elastic constant k and how far it is com-pressed or stretched. In situation A, the spring is compressed a distance d, so the force is to the right with a magnitude of kd. In situation B, the spring is stretched an amount d, so the force is to the left with the same magnitude kd. In situation C, the spring is stretched an amount d, so the force is up with the same magnitude kd. Thus, the best answer is (7), “None of the above,” since “A = B = C” is the correct ordering.

It is easy to become distracted by the physical arrangement. In C, the spring force is certainly equal to the weight mg of the hanging block, but this is not relevant. In A and B, there is likely to be a maximum static friction force that is larger than the weight of the hanging block, but again, this is irrelevant. The actual static friction force must be kd in both cases.

Key Points:

• The force exerted by an “ideal” spring has a magnitude of kd, where k is a constant describing the strength of the spring and d is the distance the spring has been extended or compressed from its resting length. This is called Hooke’s law.

• The spring force does not depend on the orientation of the spring, whether the spring is stretched or compressed, or what the spring is attached to.

• The spring force is restoring, in that if the spring is stretched, the force pulls its ends together; if it is compressed, the force pushes its ends apart.


Students are likely to be distracted by irrelevant features, such as the masses of the blocks or the static friction forces in situations A and B.


Energy 195

Confusion about the static friction force can lead to diffi culties here. That is, if students think the static friction force is always equal to µs N , they might reason that the friction force on block A is larger than that on block B, so the spring force on A must also be larger to keep it stationary. To help resolve the confusion, you can point out that one could push block A to the left and have it stay in equilibrium, even though the spring force becomes larger. Likewise, block B could be pulled to the right and it would stay in equilibrium.

Answer (1) will be common if students approach this question via Fnet

= 0, since the maximum static friction force is likely to be larger in A than in B, and both are likely to be larger than the weight of block C.

Students sometimes don’t realize that k for compression is the same as k for extension.

Note that we really do use “None of the above” as the best or only correct answer sometimes. If we don’t, students quickly stop considering it as a serious option and then will assume they are incorrect whenever they don’t fi nd their preferred answer in the list. When any students choose “None of the above” for a question (whether appropriately or not), we advocate asking them what answer they would like to see on the list; this provides more information for the instructor and the rest of the class.

Question B2.09

Description: Reasoning about forces and accelerations with the spring force.

Question

A block is dropped onto a vertical spring. Which net force vs. distance graph best represents the net force on the block as a function of its distance traveled? Consider only the motion of the block from the time it is dropped until it fi rst comes to rest.

1.

d0

Fnet, y

2.

d

Fnet,y

0

3.

0 d

Fnet,y


196 Chapter 5

4.

0d

Fnet,y

5.

0d

Fnet,y

6.

0d

Fnet,y

7.

0d

Fnet,y

8.

0d

Fnet,y

9.

0d

Fnet,y

10. None of the above


Energy 197

Commentary

Purpose: To develop your understanding of forces and ability to reason with them, especially with the spring force.

Discussion: Before the block hits the spring, the only force exerted on it is gravitation, acting downward. Taking “up” to be positive, the net force in the vertical direction is – mg.

When the block hits the spring, there are two forces exerted on the block: gravitation, down; and the spring force, acting upward. The gravitational force is the same as before, – mg. According to Hooke’s law, the spring force is proportional to the compression x of the spring, kx. It is positive, because it is pointed up.

Drawing the two forces separately and then adding them up to fi nd the net force can be helpful. The gravi-tational force is constant and negative, as shown by the horizontal line. The spring force is zero initially, and then increases linearly with compression once the block reaches the spring, as shown. The sum of these two forces gives answer (4) above.

0Spring

Gravitation (weight)

d

Fnet,y

Note that the block does not stop when the net force is equal to zero, i.e., at the equilibrium point where kx = mg. This is when it starts to slow down. When the block stops, the net force is large and positive, causing the block to reverse direction and eventually rebound to its initial height above the spring.

Key Points:

• The gravitational force is exerted at all times; do not ignore it once another force is exerted.

• An object does not stop when the net force on it is zero. This is when the acceleration, not the velocity, is zero.

• The minus sign in Hooke’s law indicates the spring force is in the opposite direction of the spring’s compression or extension, which may or may not be the negative direction in your coordinate system.


Students often don’t realize that the block continues to speed up after fi rst touching the spring, only slowing down once it has passed the equilibrium point. In other words, students often think that an object starts to slow down the instant it fi rst touches the spring. (And for a very stiff spring, this is not far from what really happens.)

Students might object to answer (4) because the maximum net force (when the block stops) is so much larger than the weight of the block. However, this is a general result for ideal springs: If the block is dropped from any height above the spring, even a very small height, the maximum spring force will be more than twice as large as the weight. The maximum net force will therefore be larger than the weight.


198 Chapter 5

Students who choose a graph having a negative slope for the spring force might not understand what is meant by the minus sign in Hooke’s law. They should be encouraged to apply common sense, perhaps drawing a free-body diagram to help.

Answer (6) is the qualitative shape of a valid velocity vs. distance graph for the situation, and students who choose it might be confusing velocity with acceleration and/or force.

Note that the area under Fnet,y

vs. d is the total work done on the block. Since the block begins and ends at v = 0, its change in kinetic energy is zero, so the total area “under” the graph should be zero, that is, the total area below the d-axis should be exactly equal to the total area above the d-axis.

Question C2.01a

Description: Developing strategic problem-solving skills.

Question

Two identical steel balls are released from rest from the same height and travel along tracks as shown and labeled below.

Track A

Track B

Which ball reaches the end of its track fi rst?

1. The ball on track A 2. The ball on track B 3. Neither; it’s a tie. 4. Not enough information

Commentary

Purpose: To develop strategic problem-solving skills and learn to choose the best principle for answering a question.

Discussion: If the balls roll without slipping, energy losses due to friction are negligible, and both balls end up at the same speed at the end of the tracks. But this does not mean they get there at the same time.

It turns out that ball B, on the lower track, wins the race every time. Discussion of why this is the case is included with the commentary for the second question in this set.


This is the fi rst of two related questions. The discussion has been kept minimal so that students will con-sider the next arrangement before being shown the detailed solution.

Many students expect the two balls to tie, intuitively reasoning that ball B travels farther but faster and these two aspects should “about cancel out.” Careless reasoning with conservation of energy may support this way of thinking. In other words, they recognize that the initial speeds are the same and that the fi nal speeds are the same, and they incorrectly reason that the times must be the same too.


Energy 199

Most students ignore what is going on while ball B is going down into the “fl at valley” and again while it is going back up again. They often reason, even after seeing a demonstration, that the fl at part of the track is long enough so that the effect of the sloped regions is overcome. In other words, students expect ball B to fall behind on both slopes. The long straight section is key, in their minds, to the outcome.

Some people choose ball A because they think that ball B will get stuck in the valley due to frictional losses. If the balls roll without slipping, then both balls reach the end of the tracks. If desired with your own demo, you can roll ball B along its track (but not ball A!) to demonstrate that it does reach the end without getting stuck.

A popular answer is “Not enough information.” Students often think that if the horizontal section is very small, ball A wins; if the horizontal section is very long, ball B wins; and if the horizontal section is just right, the balls tie. Therefore, since students are not sure which category it is, the best answer must be “Not enough information.”

Other students choose “Not enough information” simply because ball A travels the shorter distance but ball B travels faster.

Note that this is really a kinematics demo, even though it is best done after conservation of energy has been covered. This is to help students sort through valid approaches and pick the one that fi ts the situation best.

If you can arrange it, this is an excellent question to do in conjunction with a demonstration. First pose the question, collect answers, and allow students to discuss and justify their responses while you maintain a “poker face” and don’t divulge the real result. Then, do the demonstration, and follow with more discussion and explanation from students. Do not offer any explanations of your own, and do not correct students’ explanations, even if many seem invalid or inadequate; note them for after the next question, when there will be a much stronger motivation to understand what is going on. In other words, use the next question to fully explore the reasoning needed to show that ball B wins the race every time, even if the horizontal section is taken away.

Question C2.01b

Description: Developing strategic problem-solving skills.

Question

Two identical steel balls are released from rest from the same height and travel along tracks as shown and labeled below.

Track A

Track B

Which ball reaches the end of its track fi rst?

1. The ball on track A wins by a large margin. 2. The ball on track A wins by a small margin. 3. It’s a tie. 4. The ball on track B wins by a small margin. 5. The ball on track B wins by a large margin. 6. There is not enough information.


200 Chapter 5

Commentary

Purpose: To check your understanding of the physics underlying the previous question.

Discussion: The only difference between this situation and the previous one is that here, track B has no fl at section between the two slopes. This does not matter: Even though the fl at section has been removed, the result is the same, with ball B winning the race.

It turns out that this is a kinematics demonstration. Force and energy ideas are not as useful as motion ideas for understanding the outcome. In particular, by focusing on the horizontal motion of the balls, we can prove that ball B wins the race every time.

Imagine a graph of each ball’s horizontal velocity component vx vs. time. Just before ball B starts down

into the “V,” the two balls are moving with the same vx and are side by side. As ball B rolls down into the

valley, it is speeding up, which means its acceleration vector points down the slope. This vector has an x component to the right (i.e., positive) and a y component down (negative). Meanwhile a

x for ball A is zero.

Since ax is positive for ball B, its v

x is increasing above that of ball A, reaching a maximum value at the

bottom of the V. Then ball B immediately starts to go up the other side of the V. Even though it is slowing down, it is still traveling faster horizontally than ball A, because at the top of the V, ball B has the same v

x

as ball A.

Thus, vx for ball B is equal to or larger than that for ball A during its entire motion. Since the horizontal

displacements are the same and vx is on average larger for ball B, ball B must win the race. In fact, ball B

will win by a large margin.

Note that many of these arguments are false if ball B slips while rolling or if it loses contact with the track. In other words, we are assuming that ball B rolls without slipping during its entire motion.

Key Points:

• To analyze this situation, x-velocity and x-acceleration, not speed and energy, are the useful quantities to consider.

• Don’t confuse speed, velocity, and the individual components of velocity.

• Try to recognize the difference between “kind of ” understanding the answer to a question and “really getting it.”


This is the second of two related questions. These questions can be answered using only kinematics and an intuitive understanding that ball B will speed up while rolling downhill. We have placed them later in the course, however, so they can serve to help students learn strategic problem-solving skills about selecting from among the various principles available to them. The questions are more meaningful when students can consider what they might (and might not) learn through energy arguments.

After discussing the previous question, students often believe that ball B somehow makes up for the fact that it travels a longer distance by going faster on the straightaway between the two slopes. In essence, they either ignore the sloped parts of the previous tracks, or they decide that the effect is negligible. This ques-tion forces students to confront what is happening on the slopes.

Some students try to second guess the instructor here, perhaps reasoning that the outcome must be different than the previous demo, otherwise, why would the instructor show it to them.


Energy 201

You might need to point out that the ball rolls without slipping during its entire motion. It turns out that this is a fairly restrictive constraint on the shape of track B.

Students often disregard any outcome with ball B winning, because they are using invalid reasoning. They take the extreme case of an infi nitely deep V, recognizing that ball B loses, and conclude that ball B loses in every case. The mistake is that since the ball rolls without slipping, it is inappropriate to compare a deep V with this situation. Ball B would literally fl y off the track and hit the other side of the V, completely nullify-ing any comparison.

Even those students who accept the idea that vx is increasing along the downhill sometimes have trouble

with the idea that ball B continues to pull ahead of ball A along the uphill part of the V. They simply can-not understand how something can be “slowing down” yet still “pull ahead.” Instead, they think that ball B pulls ahead of ball A during the downhill, and then, perhaps because the “motion is reversed,” ball A catches up as ball B travels on the uphill, with the balls ending up in a tie.

Some students might choose that ball B wins by a small margin. For most, it is a very small margin.

It is likely that no students will select choice (5). They simply cannot believe that ball B can win by a large margin.

As with the previous item, a demonstration is recommended. The outcome is truly sensational, with ball B winning by a large margin.

QUICK QUIZZES

1. (c). The work done by the force is W F x= ( )∆ cosθ , where θ is the angle between the direction of the force and the direction of the displacement (positive x-direction). Thus, the work has its largest positive value in (c) where θ = °0 , the work done in (a) is zero since θ = °90 , the work done in (d) is negative since 90 180° < < °θ , and the work done is most negative in (b) where θ = °180 .

2. (d). All three balls have the same speed the moment they hit the ground because all start with the same kinetic energy and undergo the same total change in gravitational potential energy.

3. (c). They both start from rest, so the initial kinetic energy is zero for each of them. They have the same mass and start from the same height, so they have the same initial potential energy. Since neither spends energy overcoming friction, all of their original potential energy will be converted into kinetic energy as they move downward. Thus, they will have equal kinetic energies when they reach the ground.

4. (c). The decrease in mechanical energy of the system is f xk ∆( ). This has a smaller value on the

tilted surface for two reasons: (1) the force of kinetic friction fk is smaller because the normal

force is smaller, and (2) the displacement ∆ x is smaller because a component of the gravitational

force is pulling on the book in the direction opposite to its velocity.


202 Chapter 5

ANSWERS TO MULTIPLE CHOICE QUESTIONS

1. The net work done on the wheelbarrow is

W W W F x fnet appliedforce

friction= + = °( ) +cos co0 ∆ ss

. .

180

50 0 43 5 0 35

°( )= −( ) = −( )( ) = +

∆

∆

x

F f x N N m J

so choice (c) is the correct answer.

2. In the absence of any air resistance, the work done by nonconservative forces is zero. The work–energy theorem then states that KE PE KE PEf f i i+ = + , which becomes

1

2

1

22 2m m gy m m gyf f i iv v+ = + or v vf i i fg y y= + −( )2 2

Choosing the initial point to be where the skier leaves end of the jump and the fi nal point where he reaches maximum height, this yields

v f = ( ) + ( ) −( ) =15 0 2 9 80 4 50 11 7

2. . . . m s m s m m s2

making (a) the correct answer.

3. The mass of the crate is

m w g= = ( ) ( ) =40 0 9 80 4 08. . .N m s kg2

and we may write the work–energy theorem as

W W W KE KEf inet fric grav= + = −

Since the crate starts from rest, KE mi i= =1

22 0v , and we are left with

KE W W f x w xf k= + = ( )° + ( )°fric grav cos cos .180 60 0∆ ∆

so

KE f = −( )( ) + ( ) °6 00 6 00 40 0 60 0 6 00. . . cos . .N m N mm J 120 J J( ) = − =36 0 84 0. .+

and

v ffKE

m= =

( ) =2 2 84 0

6 42.

. J

4.08 kg m s

making choice (d) the correct response.

m

60.0°

30.0°w � 40.0 N

∆x � 6.00 m

fk � 6.00 N

yi

m

60.0°

30.0°w � 40.0 N

∆x � 6.00 m

fk � 6.00 N

yi


Energy 203

4. We assume the climber has negligible speed at both the beginning and the end of the climb. Then KE KEf i= ≈ 0, and the work done by the muscles is

W PE PE mg y ync f i f i= + −( ) = −( ) = ( )0 70 0 9 80. .kg m s2(( )( ) = ×325 2 23 10m J5.

The average power delivered is

P = = ×( ) =W

tnc

∆2 23 10

6039 1

..

5 J

95.0 min s 1 min WW

and the correct answer is choice (a).

5. The net work needed to accelerate the object from v = 0 to v is

W KE KE m m mf i112

2 12

2 12

20= − = − ( ) =v v

The work required to accelerate the object from speed v to speed 2v is

W KE KE m m m mf i212

2 12

2 12

2 2 122 4 3= − = ( ) − = −( ) =v v v v v22

13( ) = W

Thus, the correct choice is (c).

6. Assuming that the cabinet has negligible speed during the operation, all of the work Alex does is used increasing the gravitational potential energy of the cabinet. However, in addition to increas-ing the gravitational potential energy of the cabinet by the same amount as Alex did, John must do work overcoming the friction between the cabinet and ramp. This means that the total work done by John is greater than that done by Alex and the correct answer is (c).

7. Since the rollers on the ramp used by David were frictionless, he did not do any work overcoming nonconservative forces as he slid the block up the ramp. Neglecting any change in kinetic energy of the block (either because the speed was constant or was essentially zero during the lifting proc-ess), the work done by either Mark and David equals the increase in the gravitational potential energy of the block as it is lifted from the ground to the truck bed. Because they lift identical blocks through the same vertical distance, they do equal amounts of work and the correct choice is (b).

8. Once the athlete leaves the surface of the trampoline, only a conservative force (her weight) acts on her. Therefore, her total mechanical energy is constant during her fl ight, or KE PE KE PEf f i i+ = + . Taking the y = 0 at the surface of the trampoline, PE mgyi i= = 0. Also, her speed when she reaches maximum height is zero, or KE f = 0. This leaves us with PE KEf i= , or m gy m imax = 1

22v , which gives the maximum height as

ygi

max

.

..= =

( )( ) =v2 2

2

8 5

2 9 803 7

m s

m s m

2

making (c) the correct choice.

9. KE m m mcar car truck truck= = ( ) = ( )12

2 12

2 12

12

22v v v == KEtruck 2, so (b) is the correct answer.

10. The kinetic energy is proportional to the square of the speed of the particle. Thus, doubling the speed will increase the kinetic energy by a factor of 4. This is seen from

KE m m m KEf f i i i= = ( ) = ⎛⎝⎜

⎞⎠⎟ =1

2

1

22 4

1

242 2 2v v v

and (a) is the correct response here.


204 Chapter 5

11. The work–energy theorem states that W KE KEf inet = − . Thus, if Wnet = 0, then KE KEf i= or

12

2 12

2m mf iv v= , which leads to the conclusion that the speed is unchanged v vf i=( ). The velocity of the particle involves both magnitude (speed) and direction. The work–energy theorem shows

that the magnitude or speed is unchanged when Wnet = 0, but makes no statement about the direc-tion of the velocity. Therefore, choice (d) is correct but choice (c) is not necessarily true.

12. As the block falls freely, only the conservative gravitational force acts on it. Therefore, mechan-ical energy is conserved, or KE PE KE PEf f i i+ = + . Assuming that the block is released from rest

( )KEi = 0 , and taking y = 0 at ground level ( )PE f = 0 , we have that

KE PEf i= or 1

22m mgyf iv = and y

gif=

v2

2

Thus, to double the fi nal speed, it is necessary to increase the initial height by a factor of four, or the correct choice for this question is (e).

13. If the car is to have uniform acceleration, a constant net force F must act on it. Since the instan-taneous power delivered to the car is P = Fv, we see that maximum power is required just as the car reaches its maximum speed. The correct answer is (b).

ANSWERS TO EVEN-NUMBERED CONCEPTUAL QUESTIONS

2. (a) The chicken does positive work on the ground. (b) No work is done. (c) The crane does posi-tive work on the bucket. (d) The force of gravity does negative work on the bucket. (e) The leg muscles do negative work on the individual.

4. (a) Kinetic energy is always positive. Mass and speed squared are both positive. (b) Gravitational potential energy can be negative when the object is lower than the chosen reference level.

6. The total energy of the bowling ball is conserved. Because the ball initially has gravitational potential energy mgh and no kinetic energy, it will again have zero kinetic energy when it returns to its original position. Air resistance and friction at the support will cause the ball to come back to a point slightly below its initial position. On the other hand, if anyone gives a forward push to the ball anywhere along its path, the demonstrator will have to duck.

8. (a) The effects are the same except for such features as having to overcome air resistance outside. (b) The person must lift his body slightly with each step on the tilted treadmill. Thus, the effect is that of running uphill.

10. The kinetic energy is converted to internal energy within the brake pads of the car, the roadway, and the tires.


Energy 205

12. Work is actually performed by the thigh bone (the femur) on the hips as the torso moves upwards a distance h. The force on the torso F

��torso is

approximately the same as the normal force (since the legs are relatively light and are not moving much), and the work done by F

��torso minus the work

done by gravity is equal to the change in kinetic energy of the torso.

At full extension the torso would continue upwards, leaving the legs behind on the ground (!), except that the torso now does work on the legs, increasing their speed (and decreasing the torso speed) so that both move upwards together.

Note: An alternative way to think about problems that involve internal motions of an object is to note that the net work done on an object is equal to the net force times the displacement of the center of mass. Using this idea, the effect of throwing the arms upwards during the extension phase is accounted for by noting that the position of the center of mass is higher on the body with the arms extended, so that total displacement of the center of mass is greater.

14. If a crate is located on the bed of a truck, and the truck accelerates, the friction force exerted on the crate causes it to undergo the same acceleration as the truck, assuming that the crate doesn’t slip. Another example is a car that accelerates because of the frictional forces between the road surface and its tires. This force is in the direction of the motion of the car and produces an increase in the car’s kinetic energy.

PROBLEM SOLUTIONS

5.1 If the weights are to move at constant velocity, the net force on them must be zero. Thus, the force exerted on the weights is upward, parallel to the displacement, with magnitude 350 N. The work done by this force is

W F s= ( ) = ( ) °⎡⎣ ⎤⎦ ( ) =cos cos .θ 350 0 2 00 N m 700 J

5.2 (a) We assume the object moved upward with constant speed, so the kinetic energy did not change. Then, the work–energy theorem gives the work done on the object by the lifter as

W KE PE mgy mgy mg ync f i= + = + − = ( )∆ ∆ ∆0 ( ) , or

Wnc = ( )( ) ( )(281 5 9 80 17 1 1. . .kg m s cm m 10 cm2 2 ))⎡⎣ ⎤⎦ = 472 J

(b) If the object moved upward at constant speed, the net force acting on it was zero. There-fore, the magnitude of the upward force applied by the lifter must have been equal to the weight of the object

F mg= = ( )( ) = × =281 5 9 80 2 76 10 2 763. . . .kg m s N k2 NN

h

Ftorso→

Ftorso→

v→

h

Ftorso→

Ftorso→

v→


206 Chapter 5

5.3 Assuming the mass is lifted at constant velocity, the total upward force exerted by the two men equals the weight of the mass: F mgtotal

2 kg m s N= = ( )( ) = ×653 2 9 80 6 40 103. . . . They exert this upward force through a total upward displacement of 96 inches or 8 feet (4 inches per lift for each of 24 lifts). Thus, the total work done during the upward movements of the 24 lifts is

W F x F x= ( ) = °( ) = ×( )cos cos . cosθ ∆ ∆total N0 6 40 105 00 8 2 104°( ) ( )⎛

⎝⎜⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

= × ft1 m

3.281 ft J

5.4 (a) The 35 N force applied by the shopper makes a 25° angle with the displacement of the cart (horizontal). The work done on the cart by the shopper is then

W F xshopper N m= ( ) = ( ) °( )( ) =cos cos .θ ∆ 35 25 50 0 1..6 103× J

(b) Since the speed of the cart is constant, KE KEf i= and W KEnet = =∆ 0 .

(c) Since the cart continues to move at constant speed, the net work done on the cart in the second aisle is again zero. With both the net work and the work done by

friction unchanged, the work done by the shopper W W Wshopper net fric= −( ) is also

unchanged . However, the shopper now pushes horizontally on the cart, making

′ = ⋅ °F W x W xshopper shopper= smaller th( cos )∆ ∆0 aan before when the force was

F W x= ⋅ °( )shopper ∆ cos 35 .

5.5 (a) The gravitational force acting on the object is

w mg= = ( )( ) =5 00 9 80 49 0. . .kg m s N2

and the work done by this force is

W PE mg y y w y yg g f i i f= − = − −( ) = + −( )∆

or

W w Lg = °( ) = ( )( ) ° =sin . . . sin .30 0 49 0 2 50 30 0 N m 61..3 J

(b) The normal force exerted on the block by the incline is n mg= °cos .30 0 , so the friction force is

f nk k= = ( )( ) ° =µ 0 436 49 0 30 0 18 5. . cos . . N N

This force is directed opposite to the displacement (that is q = 180°), and the work it does is

W f Lf k= ( ) = ( ) °⎡⎣ ⎤⎦ ( ) = −cos . cos .θ 18 5 180 2 50 N m 466.3 J

(c) Since the normal force is perpendicular to the displacement, the work done by the normal force is W n Ln = °( ) =cos .90 0 0 .

(d) If a shorter ramp is used to increase the angle of inclination while maintaining the same vertical displacement y yf i− , the work done by gravity will not change , the

work done by the friction force will decreasse (because the normal force, and hence the

friction force, will decrease and also because the ramp length L decreases), and the

work done by the normal force remains zero (because the normal force remains perpen-dicular to the displacement).


Energy 207

5.6 (a) W F xF = ( ) = ( )( ) °=∆ cos . cosθ 150 6 00 0 N m 900 J

(b) Since the crate moves at constant velocity, a ax y= = 0. Thus,

ΣF f Fx k= ⇒ = =0 150 N

Also,

ΣF n mgy = ⇒ = = ( )( ) =0 40 0 392kg 9.80 m s N2.

so

µkkf

n= = =150

0 383N

392 N.

5.7 (a) ΣF F n mgy = + − =sinθ 0

n mg F= − sinθ

ΣF F nx k= − =cosθ µ 0

n

F

k

= cosθµ

∴ − =mg FF

k

sincosθ θµ

F

mgk

k

= =kg m s2µ

µ θ θsin cos

. . .

+( )( )(0 500 18 0 9 80 ))( ) ° + °0 500 20 0 20 0

79 4. sin . cos .

.= N

(b) W F sF = ( ) = ( )[ ⎤⎦ ( ) =cos . cos . . .θ 79 4 20 0 20 0 1 4N º m 99 10 1 493× =J kJ.

(c) f Fk = =cos .θ 74 6 N

W f sf k= = N º mcos . cos .θ( ) ( )[ ⎤⎦ ( ) =74 6 180 20 0 −− × = −1 49 10 1 493. .J kJ

mg→

∆x→

F→

n→

fk→

m

mg→

∆x→

F→

n→

fk→

m

F→

n→

q�20.0°18.0 kg

s �20.0 m

→mg

Fk�mknF→

n→

q�20.0°18.0 kg

s �20.0 m

→mg

Fk�mkn


208 Chapter 5

5.8 (a) W F sF = ( ) = ( )⎡⎣ ⎤⎦ ( )cos . cos . .θ 16 0 25 0 2 20 N ° m

WF = 31 9. J

(b) W n sn = °( ) =cos 90 0

(c) W mg sg = °( ) =cos 90 0

(d) W W W WF n gnet J 31.9 J= + + = + + =31 9 0 0.

5.9 (a) The work–energy theorem, W KE KEf inet = − , gives

5 0001

22 50 10 03 2 J kg= ×( ) −. v , or v = 2 00. m s

(b) W F s F= ( ) = °( )( ) =cos cos .θ 0 25 0 5000 m J, so F = 200 N

5.10 Requiring that KE KEping pong bowling= with KE m= 12

2v , we have

1

22 45 10

1

27 00 3 003 2 2

. . .×( ) = ( )( )− kg kg m sv

giving v = 160 m s .

5.11 (a) KE mi i= 12

2v where v v vi x y2

02

02 2 2

6 00 2 00 40 0= + = ( ) + −( ) =. . . m s m s m22 2s ,

giving KEi = ( )( ) =12 5 75 40 0 115. . kg m s J2 2 .

(b) KE mf f= 12

2v where v v vf x y2 2 2 2 2

8 50 5 00 97 3= + = ( ) + ( ) =. . . m s m s m s2 2,

so KE f = ( )( ) =12 5 75 97 3 280. . kg m s J2 2 , and the change in kinetic energy has been

KE KEf i− = − =280 115 165J J J .

5.12 (a) Since the applied force is horizontal, it is in the direction of the displacement giving θ = °0 . The work done by this force is then

W F x F x F xF0 0 0 00= ( ) = °( ) = ( )cos cosθ ∆ ∆ ∆

and

FW

xF

00 350

29 2= = =∆

J

12.0 mN.

(b) Since the crate originally had zero acceleration, the original applied force was just enough to offset the retarding friction force. Therefore, when the applied force is increased, it has a magnitude greater than the friction force. This gives the crate a resultant force (and hence an acceleration) in the direction of motion, meaning the speed of the crate will increase with time .

(c) If the applied force is made smaller than F0 , the magnitude of the friction force will be greater than that of the applied force. This means the crate has a resultant force, and accel-eration, in the direction of the friction force (opposite to the direction of motion).

The crate will now slow down and come to resst .

n→

25.0°

2.50 kg

s�2.20 mF�16.0 N

→mg

n→

25.0°

2.50 kg

s�2.20 mF�16.0 N

→mg


Energy 209

5.13 (a) We use the work–energy theorem to fi nd the work.

W KE m mf i= = − = − ( )( ) = −∆ 1

2

1

20

1

270 4 0 52 2 2

v v kg m s. ..6 102× J

(b) W F s f s n s mg sk k k= ( ) = °( ) = −( ) = −( )cos cosθ µ µ180 ,

so

sW

mgk

= − = −− ×( )

( )( )µ5 6 10

70 9 80

2.

.

J

0.70 kg m s2(( ) = 1 2. m

5.14 At the top of the arc, vy = 0, and v v vx x= = ° =0 0 30 0 34 6cos . . m s.

Therefore, v v v2 2 2 234 6= + = ( )x y . m s , and

KE m= = ( )( ) =1

2

1

20 150 34 6 90 02 2

v . . . kg m s J

5.15 (a) As the bullet penetrates the tree trunk, the only force doing work on it is the force of resist-ance exerted by the trunk. This force is directed opposite to the displacement, so the work done is W f x KE KEf inet = °( ) = −cos180 ∆ , and the magnitude of the force is

f

KE KE

xf i=

−( ) °

=− ×( )−

∆ cos

.

180

01

27 80 10 5753 kg mm s

m N

( )− ×( ) = ×−

2

24

5 50 102 34 10

..

(b) If the friction force is constant, the bullet will have a constant acceleration and its average velocity while stopping is v v v= +( )f i 2. The time required to stop is then

∆ ∆ ∆t

x x

f i

= =( )+

=×( )

+=

−

v v v2 2 5 50 10

0 5751

2..

m

m s991 10 4× − s

5.16 (a) KE mA A= = ( )( ) =1

2

1

20 60 2 0 1 22 2

v . . . kg m s J

(b) KE mB B= 12

2v , so

vBBKE

m=

( )=

( ) =2 2 7 5

0 605 0

.

..

J

kg m s

(c) W KE KE KEB Anet J J= = − = −( ) =∆ 7 5 1 2 6 3. . .

5.17 W F s F snet road resist N= ( ) + ( ) = ( )cos cosθ θ1 2 1 000 ccos cos0 950 180°⎡⎣ ⎤⎦ + ( ) °⎡⎣ ⎤⎦s s N

Wnet N N m J= −( )( ) = ×1 000 950 20 1 0 103.

Also, W KE KE mf inet = − = −1

202v , so

v = =×( )

=2 2 1 0 10

20001 0

3W

mnet

J

kg m s

..


210 Chapter 5

5.18 The initial kinetic energy of the sled is

KE mi i= = ( )( ) =1

2

1

210 2 0 202 2

v kg m s J.

and the friction force is f n mgk k k= = = ( )( ) =µ µ 0 10 98 9 8. .N N.

W f s KE KEk f inet = °( ) = −cos180 , so sKE

fi

k

= −°

= −−

=0

180

202 0

cos.

J

9.8 Nm

5.19 With only a conservative force acting on the falling ball,

KE PE KE PEg i g f+( ) = +( ) or

1

2

1

22 2m mgy m mgyi i f hv v+ = +

Applying this to the motion of the ball gives 0 012

2+ = +mgy mi fv

or ygif= =

( )( ) =

v2 2

2

9 0

2 9 804 1

.

..

m s

m s m

2

5.20 (a) The force stretching the spring is the weight of the suspended object. Therefore, the force constant of the spring is

k

F

x

mg

xg= = =

( )( )×

=−∆ ∆2 50 9 80

2 76 10 2

. .

.

kg m s

m

2

88 88 102. × N s

(b) If a 1.25-kg block replaces the original 2.50-kg suspended object, the force applied to the spring (weight of the suspended object) will be one-half the original stretching force. Since, for a spring obeying Hooke’s law, the elongation is directly proportional to the stretching force, the amount the spring stretches now is

∆ ∆x x( ) = ( ) = ( ) =

2 1

1

2

1

22 76 1 38. . cm cm

(c) The work an external agent must do on the initially unstretched spring to produce an elongation x f is equal to the potential energy stored in the spring at this elongation

W PE PE kxon s f s i fdone

spring= ( ) − ( ) = − =1

20

1

28 82 . 88 10 8 00 10 2 842 2 2

×( ) ×( ) =−N m m J. .

5.21 The magnitude of the force a spring must exert on the 3.65-g object to give that object an acceleration of a g= =0 500 4 90. . m s2 is given by Newton’s second law as

F ma= = ×( )( ) = ×− −3 65 10 4 90 1 79 103 2. . .kg m s N2

Then, by Newton’s third law, this object exerts an oppositely directed force of equal magnitude in the on the spring. If this reaction force is to stretch the spring 0.350 cm, the required force constant of the spring is

k

F

x= = × = ×

×

− −

∆1 79 10

0 350

1 79 102 2.

.

. N

cm

N

3.50 100 m N m-3 = 5 11.


Energy 211

5.22 (a) While the athlete is in the air, the interacting objects are the athlete and Earth . They interact through the gravitational force that one exerts on the other.

(b) If the athlete leaves the trampoline (at the y = 0 level) with an initial speed of vi = 9 0. m s, her initial kinetic energy is

KE mi i= = ( )( ) = ×1

2

1

260 0 9 0 2 4 102 2 3v . . .kg m s J

and her gravitational potential energy is PE mgy mgg i i( ) = = ( ) =0 0 J .

(c) When the athlete is at maximum height, she is momentarily at rest and KE f = 0 J . Because the only force acting on the athlete during her fl ight is the conservative gravitation force, her total energy (kinetic plus potential) remains constant. Thus, the decrease in her kinetic energy as she goes from the launch point where PEg =( )0 to maximum height is matched by an equal size increase in the gravitational potential energy.

∆ ∆PE KE PE PE KE KEg f i f i= − ⇒ − = − −( ) oor PE PE KE KEf i i f= + −

and

PE f = + × − = ×0 2 4 10 0 2 4 103 3. .J J

(d) The statement that the athlete’s total energy is conserved is summarized by the equation ∆ ∆KE PE+ = 0 or KE PE KE PE2 2 1 1+ = + . In terms of mass, speed, and height, this

becomes 12 2

22

12 1

21m mgy m mgyv v+ = + . Solving for the fi nal height gives

ymgy m m

mg2

1 12

221

2

1

2=+ −v v

or y yg2 1

12

22

2= +

−( )v v

The given numeric values for this case are y1 10 9 0= =, . m sv (at the trampoline level) and v2 0= ( ) at maximum height . The maximum height attained is then

y y

g2 112

22 2

20

9 0 0

2 9 80= +

−( )= +

( ) −

( ) =v v .

.

m s

m s244 1. m

(e) Solving the energy conservation equation given in part (d) for the fi nal speed gives

v v22

12

1 2

2 1

2= + −⎛

⎝⎜⎞⎠⎟m

m mgy mgy or v v2 12

1 22= ± + −( )g y y

With y1 0= , v1 9 0= . m s, and y y2 2 4 1 2= = ( )max . m , the speed at half the maximum height is given as

v2

29 0 2 9 80 0

4 1

26= ± ( ) + ( ) −⎛

⎝⎜⎞⎠⎟ = ±. .

.. m s m s

m2 44 m s


212 Chapter 5

5.23 The work the beam does on the pile driver is given by

W F x Fnc = °( ) = − ( )cos .180 0 180∆ m

Here, the factor cos180° is included because the force F exerted on the driver by the beam is directed upward, but the ∆x = =18 0 0 180. . cm m displacement undergone by the driver while in contact with the beam is directed downward.

From the work–energy theorem, this work can also be expressed as

W KE KE PE PE m mg y ync f i f i f i f= −( ) + −( ) = −( ) + −1

22 2v v ii( )

Choosing y = 0 at the level where the pile driver fi rst contacts the top of the beam, the driver starts from rest vi =( )0 at yi = +7 50. m and comes to rest again v f =( )0 at yf = − 0 180. m. Therefore, we have

− ( ) = −( ) + ( )( ) −F m0 180

1

20 0 2 300 9 80 0. . . m kg m s2 1180 7 50 m m−( ).

yielding F = − ×−

= ×1 73 10

0 1809 62 10

55.

..

J

m N directed uupward

5.24 When the child swings at the end of the ropes, she follows a circular path of radius r = 1 75. m as shown at the right. If we choose y = 0 at the level of the child’s lowest position on this path, then her y-coordinate when the rope is at angle θ with the vertical is y r r r= − = −( )cos cosθ θ1 . Thus, her gravitational potential energy at this position is

PE mgy mgr wrg = = −( ) = −( )1 1cos cosθ θ

(a) When the ropes are horizontal, θ = °90 0. , and

PEg = ×( )( ) −( )° = +3 50 10 1 75 1 90 0 6132. . cos . N m J

(b) When θ = °30 0. , PEg = ×( )( ) −( )° = +3 50 10 1 75 1 30 0 82 12. . cos . . N m J .

(c) At the bottom of the arc, y = 0, cos = cos 0°=1,θ and PEg = 0 .

5.25 While the motorcycle is in the air, only the conservative gravitational force acts on cycle and rider. Thus, 1

22 1

22m mgy m mgyf f i iv v+ = + , which gives

∆y y ygf i

i f= − =−

=( ) − ( )v v2 2 2 2

2

35 0 33 0

2 9

. .

.

m s m s

8806 94

m s m

2( ) = .

Note that this answer gives the maximum height of the cycle above the end of the ramp, which is an unknown distance above the ground.


Energy 213

5.26 (a) When equilibrium is reached, the total spring force supporting the load equals the weight of the load, or F F F ws s s, , ,total leaf helper load= + = . Let k kh� and represent the spring constants of the leaf spring and the helper spring, respectively. Then, if x� is the distance the leaf spring is compressed, the condition for equilibrium becomes

k x k x y wh� � �+ −( ) =0 load

or

xw k y

k kh

h�

�

= ++

=× + ×( )load

N N m0

5 55 00 10 3 60 10. . 00 500

5 25 10 3 60 100 7685 5

.

. ..

m

N m N m m

( )× + ×

=

(b) The work done compressing the springs equals the total elastic potential energy at equilibrium. Thus, W k x k xh= + −( )1

22 1

22

0 500� � � . m

or

W = ×( )( ) + ×( )1

25 25 10 0 768 3 60 105 2 1

25. . .N m m N m 00 268 1 68 102 5. .m J( ) = ×

5.27 The total work done by the two bicep muscles as they contract is

W F xbiceps av N m J= = ( )( ) = ×2 2 800 0 075 1 2 102∆ . .

The total work done on the body as it is lifted 40 cm during a chin-up is

W mghchin-up kg m s m= = ( )( )( ) = ×75 9 80 0 40 2 92. . . 1102 J

Since W Wchin-up biceps> , it is clear that addition muscles must be involved .

5.28 Applying W KE PE KE PEnc f i= +( ) − +( ) to the jump of the “original” fl ea gives

F d mgym f= +( ) − +( )0 0 0 or yF d

mgfm=

where Fm is the force exerted by the muscle and d is the length of contraction.

If we scale the fl ea by a factor f, the muscle force increases by f 2 and the length of contrac-tion increases by f. The mass, being proportional to the volume which increases by f 3, will also increase by f 3 . Putting these factors into our expression for yf gives

yf F fd

f m g

F d

mgyf

m mf( ) =

( )( )( ) = = ≈super

flea

2

30 5. m

so the “super fl ea” cannot jump any higher!

This analysis is used to argue that most animals should be able to jump to approximately the same height (~0.5 m). Data on mammals from elephants to shrews tend to support this.


214 Chapter 5

5.29 (a) Taking y = 0, and hence PE mgyg = = 0, at ground level, the initial total mechanical energy of the projectile is

E KE PE m mgyi i i i itotal

kg

( ) = + = +

= ( )

1

21

250 0 1

2v

. .220 10 50 0 9 80 142 42 2×( ) + ( )( )( ) = m s kg m s m2. . .330 105× J

(b) The work done on the projectile is equal to the change in its total mechanical energy.

W KE PE KE PE mnc f f i i f i( ) = +( ) − +( ) = −( ) +rise

1

22 2v v mmg y yf i−( )

= ( ) ( ) − ( )⎡1

250 0 85 0 120

2 2. . kg m s m s⎣⎣

⎤⎦ + ( )( ) −( )

= −

50 0 9 80 427 142

3 97

. .

.

kg m s m m2

×× 104 J

(c) If, during the descent from the maximum height to the ground, air resistance does one and a half times as much work on the projectile as it did while the projectile was rising to the top of the arc, the total work done for the entire trip will be

W W W Wnc nc nc nc( ) = ( ) + ( ) = ( )total rise descent rise++ ( )

= − ×( ) = − ×

1 50

2 50 3 97 10 9 93 104 4

.

. . .

Wnc rise

J J

Then, applying the work–energy theorem to the entire fl ight of the projectile gives

W KE PEnc( ) = +( )

totaljust beforehitting ground

−− +( ) = + − −KE PE m mgy m mgyf f i iat launch

1

2

1

22 2v v

and the speed of the projectile just before hitting the ground is

v vfnc

i i f

W

mg y y=

( )+ + −( )

=− ×(

22

2 9 93 10

2

4

total

J. ))+ ( ) + ( ) −( ) =

50.0 kg m s m s m2120 2 9 80 142 0 11

2. 55 m s

5.30 (a) The work done by the gravitational force equals the decrease in the gravitational potential energy, or

W PE PE PE PE mg y y mghg f i i f i f= − −( ) = − = −( ) =

(b) The change in kinetic energy is equal to the net work done on the projectile, which in the absence of air resistance is just that done by the gravitational force. Thus,

W W KE KE mghgnet = = ⇒ =∆ ∆

(c) ∆KE KE KE mghf i= − = so KE KE mgh m mghf i= + = +1

2 02v

(d) No . None of the calculations in Parts (a), (b), and (c) involve the initial angle.


Energy 215

5.31 (a) The system will consist of the mass, the spring, and Earth . The parts of this system interact

via the spring force, the gravitational force, aand a normal force .

(b) The points of interest are where the mass is released from rest at x = 6..00 cm( ) and

the equilibrium point, x = 0 .

(c) The energy stored in the spring is the elastic potential energy, PE kxs = 12

2.

At x = 6 00. cm ,

PEs = ( ) ×( ) =−1

2850 6 00 10 1 532 2

N m m J. .

and at the equilibrium position x =( )0 ,

PE ks = ( ) =1

20 02

(d) The only force doing work on the mass is the conservative spring force (the normal force and the gravitational force are both perpendicular to the motion). Thus, the total mechanical energy of the mass will be constant. Because we may choose y = 0, and hence PEg = 0, at the level of the horizontal surface, the energy conservation equation becomes

KE PE KE PEf s f i s i+ ( ) = + ( ) or

1

2

1

2

1

2

1

22 2 2 2m kx m kxf f i iv v+ = +

and solving for the fi nal speed gives

v vf i i f

k

mx x= + −( )2 2 2

If the fi nal position is the equilibrium position x f =( )0 and the object starts from rest vi =( )0 at xi = 6 00. cm, the fi nal speed is

v f = + ×( ) −⎡

⎣⎤⎦ =−0

850

1 006 00 10 0 3 062 2 N m

kg m

.. . m m s2 s2 1 75= .

(e) When the object is halfway between the release point and the equilibrium position, we have vi i fx x= = =0 6 00 3 00, . . cm, and cm, giving

v f = + ×( ) − ×− −0

850

1 006 00 10 3 00 102 2 2 N m

kg m

.. . mm m s( )⎡

⎣⎤⎦ =

21 51.

This is not half of the speed at equilibrium because the equation for fi nal speed is not a linear function of position .

5.32 Using conservation of mechanical energy, we have

1

2

1

202 2m mgy mf f iv v+ = +

or

ygf

i f=−

=( ) − ( )

( ) =v v2 2 2 2

2

10 1 0

2 9 80

m s m s

m s2

.

.55 1. m


216 Chapter 5

5.33 Since no nonconservative forces do work, we use conservation of mechanical energy, with the zero of potential energy selected at the level of the base of the hill. Then, 1

22 1

22m mgy m mgyf f i iv v+ = + with yf = 0 yields

y

gif i=

−=

( ) −

( ) =v v2 2 2

2

3 00 0

2 9 80

.

.

m s

m s0.459

2mm

Note that this result is independent of the mass of the child and sled.

5.34 (a) The distance the spring is stretched is x = − = − =� �0 41 5 35 0 6 5. . .cm cm cm. Since the object hangs in equilibrium on the end of the spring, the spring exerts an upward force of magnitude F mgs = on the suspended object. Then, Hooke’s law gives the spring constant as

k

F

x

mgs= =−

=( )( )

×=−� �0

2

7 50 9 80

6 5 10

. .

.

kg m s

m

2

11 1 10 1 13. .× =N m kN m

(b) Consider one person (say the one holding the left end of the spring) to simply hold that end of the spring in position while the person on the other end stretches it. The spring is then stretched a distance of x = − = −� � �0 0 350. m by a stretching force of magnitude Fs = 190 N. From Hooke’s law, x F ks= , we have

� − =×

=0 3 0190

0 17. .5 m N

1.1 10 N m m3 and � = =0 17 0 350 0 52. . . m + m m

5.35 (a) On a frictionless track, no external forces do work on the system consisting of the block and the spring as the spring is being compressed. Thus, the total mechanical energy of the

system is constant, or KE PE PE KE PE PEf g f s f i g i s i+ ( ) + ( ) = + ( ) + ( ) . Because the track

is horizontal, the gravitational potential energy when the mass comes to rest is the same as just before it made contact with the spring, or PE PEg f g i

( ) = ( ) . This gives

1

2

1

2

1

2

1

22 2 2 2m kx m kxf f i iv v+ = +

Since v f = 0 (the block comes to rest) and xi = 0 (the spring is initially undistorted),

x

m

kf i= = ( ) =v 1 500 250

4 600 350.

.

.. m s

kg

N m m

(b) If the track was not frictionless, some of the original kinetic energy would be spent overcoming friction between the mass and track. This would mean that less energy would be stored as elastic potential energy in the spring when the mass came to rest. Therefore,

the maximum compression of the spring would be less in this case.


Energy 217

5.36 (a) From conservation of mechanical energy,

1

2

1

22 2m mgy m mgyB B A Av v+ = +

or

v vB A A Bg y y= + −( )

= + ( )( ) =

2 2

0 2 9 80 1 80 5 94. . . m s m2 m s

Similarly,

v vC A A Cg y y g= + −( ) = + −( ) =2 2 0 2 5 00 2 00 7 67. . . m m m ss

(b) W PE PE mg y yg A C g A g C A C) = ( ) − ( ) = −( ) = ( )→

49 0 3 00. . N m J( ) = 147

5.37 (a) We choose the zero of potential energy at the level of the bottom of the arc. The initial height of Tarzan above this level is

yi = ( ) −( )° =30 0 1 37 0 6 04. cos . . m m

Then, using conservation of mechanical energy, we fi nd

1

20

1

22 2m m mgyf i iv v+ = +

or

v vf i igy= + = + ( )( ) =2 2 0 2 9 80 6 04 10 9. . . m s m m s2

(b) In this case, conservation of mechanical energy yields

v vf i igy= + = ( ) + ( )( ) =2 2

2 4 00 2 9 80 6 04. . . m s m s m2 111 6. m s

5.38 At maximum height, vy = 0 and v vx x= m s m s0 40 60 20= ( ) ° =cos .

Thus, v v vf x y= + =2 2 20 m s. Choosing PEg = 0 at the level of the launch point, conservation of mechanical energy gives PE KE KEf i f= − , and the maximum height reached is

ygf

i f=−

=( ) − ( )

( ) =v v2 2 2 2

2

40 20

2 9 806

m s m s

m s2.11 m

5.39 (a) Initially, all the energy is stored as elastic potential energy within the spring. When the gun is fi red, and as the projectile leaves the gun, most of the energy in the form of kinetic energy along with a small amount of gravitational potential energy. When the projectile comes to rest momentarily at its maximum height, all of the energy is in the form of gravitational potential energy.

A

B

C5.00 m3.20 m

2.00 m

A

B

C5.00 m3.20 m

2.00 m

yfyiq

��30.0 myf

yiq��30.0 m

continued on next page


218 Chapter 5

(b) Use conservation of mechanical energy from when the projectile is at rest within the gun vi i iy x= = = −( )0 0 0 120, , . and m until it reaches maximum height where v f = 0, y yf = =max .20 0 m, and x f = 0 (the spring is relaxed after the gun is fi red).

Then, KE PE PE KE PE PEg s f g s i+ +( ) = + +( ) becomes

0 0 0 0

1

22+ + = + +mgy kximax

or

kmgy

xi

= =×( )( )−

2 2 20 0 10 9 80 22

3

max

2 kg m s 0.0 . . mm

0.120 m N m

( )−( )

=2 544

(c) This time, we use conservation of mechanical energy from when the projectile is at rest within the gun vi i iy x= = = −( )0 0 0 120, , . and m until it reaches the equilibrium position

of the spring y xf f= + =( )0 120 0. ,m and . This gives

KE KE PE PE PE PEf g s i g s f= + +( ) − +( ) or

1

20 0

1

202 2m kx mgyf i fv = + +⎛

⎝⎜⎞⎠⎟ − +( )

v f i f

k

mx gy2 2

3

2

544

20 0 10

= ⎛⎝

⎞⎠ −

=×

⎛⎝⎜

⎞⎠⎟−

N m

kg.−−( ) − ( )( )0 120 2 9 80 0 120

2. . .m m s m2

yielding v f = 19 7. m s

5.40 (a) ΣF n F mgy = ⇒ + − =0 0sinθ , or n mg F= − sinθ .

The friction force is then

f n mg Fk k k= = −( )µ µ θsin

(b) The work done by the applied force is

W F FxF = =∆x

�cos cosθ θ

and the work done by the friction force is W ff kk= ∆x

�cosφ where φ is the angle between

the direction of f xk

�� and ∆ . Thus, W f x mg F xf k kk

= ° = − −( )cos sin180 µ θ .

(c) The forces that do no work are those perpendicular to the direction of the displacement ∆x�.

These are n g F� � ��, , and the vertical component of m .

(d) For part (a): n mg F= − = ( )( ) − ( )sin . . . sinθ 2 00 9 80 15 0 3 kg m s N2 77 0 10 6. .° = N

f nk k= = ( )( ) =µ 0 400 10 6 4 23. . .N N

For part (b): W FxF = = ( )( ) ° =cos . . cos . .θ 15 0 4 00 37 0 47 9 N m J

W f xf kk

= = ( )( ) ° = −cos . . cos .φ 4 23 4 00 180 16 9 N m J

n→

mg→

fk→

F→

qm

∆x→n→

mg→

fk→

F→

qm

∆x→


Energy 219

5.41 (a) When the child slides down a frictionless surface, the only nonconservative force acting on the child is the normal force. At each instant, this force is perpendicular to the motion and, hence, does no work. Thus, conservation of mechanical energy can be used in this case.

(b) The equation for conservation of mechanical energy, KE PE KE PEf i+( ) = +( ) , for this situation is 1

22 1

22m m gy m m gyf f i iv v+ = + . Notice that the mass of the child cancels out

of the equation, so the mass of the child is not a factor in the frictionless case.

(c) Observe that solving the energy conservation equation from above for the fi nal speed gives

v vf i i fg y y= + −( )2 2 . Since the child starts with the same initial speed (vi = 0) and has the same change in altitude in both cases, v f is the same in the two cases.

(d) Work done by a non conservative force must bbe accounted for when friction is present.

This is done by using the work–energy theorem rather than conservation of mechanical energy.

(e) From part (b), conservation of mechanical energy gives the fi nal speed as

v vf i i fg y y= + −( ) = + ( )( ) =2 2 0 2 9 80 12 0 15 3. . . m s m2 m s

5.42 (a) No . The change in the kinetic energy of the plane is equal to the net work done by all forces doing work on it. In this case, there are two such forces, the thrust due to the engine and a resistive force due to the air. Since the work done by the air resistance force is nega-tive, the net work done (and hence the change in kinetic energy) is less than the positive work done by the engine thrust. Also, because the thrust from the engine and the air resist-ance force are nonconservative forces, mechanical energy is not conserved in this ccase .

(b) Since the plane is in level fl ight, PE PEg f g i( ) = ( ) and the work–energy theorem reduces to

W W W KE KEnc f i= + = −thrust resistance , or

F s f s m mf icos cos0 1801

2

1

22 2( )° + ( )° = −v v

This gives

v vf i

F f s

m= +

−( ) = ( ) +−( ) ×⎡⎣2 2

42

602 7 5 4 0 10

m s N. . ⎤⎤⎦( )

×=

500

1 5 10774

m

kg m s

.

5.43 We shall take PEg = 0 at the lowest level reached by the diver under the water. The diver falls a total of 15 m, but the nonconservative force due to water resistance acts only during the last 5.0 m of fall. The work–energy theorem then gives

W KE PE KE PEnc g f g i= +( ) − +( )

or

Fav2 m 0 kg m scos . .180 5 0 0 0 0 7 9 80( )° ( ) = +( ) − + ( )(( )( )⎡⎣ ⎤⎦15 m

This gives the average resistance force as Fav N kN= × =2 1 10 2 13. . .


220 Chapter 5

5.44 (a) Choose PEg = 0 at the level of the bottom of the arc. The child’s initial vertical displacement from this level is

yi = ( ) −( )° =2 00 1 30 0 0 268. cos . . m m

In the absence of friction, we use conservation of mechanical energy as

KE PE KE PEg f g i+( ) = +( ) , or 1

22 0 0m mgyf iv + = + , which gives

v f igy= = ( )( ) =2 2 9 80 0 268 2 29. . . m s m m s2

(b) With a nonconservative force present, we use

W KE PE KE PE m mgnc g f g i f= +( ) − +( ) = +⎛⎝⎜

⎞⎠⎟ − +1

20 02v yyi( )

or

W m gyncf

i= −⎛

⎝⎜⎞

⎠⎟

= ( ) ( )−

v2

2

2

25 02 00

29.

.. kg

m s880 0 268 15 7 m s m J2( )( )

⎡

⎣⎢⎢

⎤

⎦⎥⎥

= −. .

Thus, 15.7 J of energy is spent overcoming friction.

5.45 Choose PEg = 0 at the level of the bottom of the driveway.

Then W KE PE KE PEnc g f g i= +( ) − +( ) becomes

f s m mg sfcos sin180

1

20 0 202° °( ) = +⎡

⎣⎢⎤⎦⎥

− + ( )[ ]v

Solving for the fi nal speed gives

v f gsf s

m= ( ) ° −2 20

2sin

or

v f = ( )( ) °−×( )

2 9 80 5 0 202 4 0 10 5 03

. . sin. .

m s m N

2 m

kg m s

( )×

=2 10 10

3 83..

n→

f→

20°

s

y→

mg

n→

f→

20°

s

y→

mg


Energy 221

5.46 (a) Yes . Two forces, a conservative gravitational force and a nonconservative normal force, act on the child as she goes down the slide. However, the normal force is perpendicular to the child’s motion at each point on the path and does no work. In the absence of work done by nonconservative forces, mechanical energy is conserved.

(b) We choose the level of the pool to be the y PEg= =( )0 0and hence, level. Then, when

the child is at rest at the top of the slide, PE mgh KEg = =and 0 . Note that this gives the

constant total mechanical energy of the child as E KE PE mghgtotal = + = . At the launching

point (where y h= 5), we have PE mgy mghg = = 5 and KE E PE mghg= − =total 4 5 .

At the pool level, PE KE mghg = =0 and .

(c) At the launching point (i.e., where the child leaves the end of the slide),

KE mmgh= =1

2

4

502v

meaning that

v0

8

5= gh

(d) After the child leaves the slide and becomes a projectile, energy conservation gives

KE PE m mgy E mghg+ = + = =12

2v total where v v v2 2 2= +x y . Here, v vx x= 0 is constant, but

vy varies with time. At maximum height, y y y= =max and v 0, yielding

1

200

2m mgy mghxv +( ) + =max and y hgx

max = − v02

2

(e) If the child’s launch angle leaving the slide is θ , then v v0 0x = cosθ . Substituting this into the result from part (d) and making use of the result from part (c) gives

y hg

hg

ghmax cos cos= − = − ⎛

⎝⎜⎞⎠⎟

v02

2 2

2

1

2

8

5θ θ or y hmax cos= −⎛

⎝⎞⎠1

4

52 θ

(f ) No . If friction is present, mechanical energy would not be conserved, so her kinetic energy at all points after leaving the top of the waterslide would be reduced when compared with the frictionless case. Consequently, her launch speed, maximum height reached, and fi nal speed would be reduced as well.


222 Chapter 5

5.47 Choose PEg = 0 at the level of the base of the hill and let x represent the distance the skier moves along the horizontal portion before coming to rest. The normal force exerted on the skier by the snow while on the hill is n mg1 10 5= °cos . and, while on the horizontal portion, n mg2 = .

Consider the entire trip, starting from rest at the top of the hill until the skier comes to rest on the horizontal portion. The work done by friction forces is

W f fnc k k= ( )⎡⎣ ⎤⎦ ( ) + ( )⎡⎣ ⎤1 2

180 200 180cos cos° m °⎦⎦

= − ( )( ) − ( )

x

mg mg xk kµ µcos .10 5 200° m

Applying W KE PE KE PEnc g f g i= +( ) − +( ) to this complete trip gives

− ( )( ) − ( ) = +[ ] − +µ µk kmg mg x mgcos .10 5 200 0 0 0 20° m 00 10 5 m °( )⎡⎣ ⎤⎦sin .

or

xk

= ° −⎛⎝⎜

⎞⎠⎟

( )sin .cos .

10 510 5 200

µ° m . If µk = 0 0750. , then x = 289 m

5.48 The normal force exerted on the sled by the track is n mg= cosθ and the friction force is f n mgk k k= =µ µ θcos .

If s is the distance measured along the incline that the sled travels, applying

W KE PE KE PEnc g f g i= +( ) − +( ) to the entire trip gives

µ θ θk mg s mg s mcos cos sin( )⎡⎣ ⎤⎦ = + ( )⎡⎣ ⎤⎦ −180 0

1

2° vii

2 0+⎡⎣⎢

⎤⎦⎥

or

sg

i

k

=+( ) =

( )( )

v2 2

2

4 0

2 9 80sin cos

.

. siθ µ θ m s

m s2 nn . cos20 0 20 20° +( )° = 1.5 m

5.49 (a) Consider the entire trip and apply W KE PE KE PEnc g f g i= +( ) − +( ) to obtain

f d f d m f1 1 2 22180 180

1

20cos cos° °( ) + ( ) = +⎛

⎝⎜⎞⎠⎟ −v 00 +( )mgyi

or

v f ig yf d f d

m= − +⎛

⎝⎜⎞⎠⎟

= ( )(

2

2 9 80 1 000

1 1 2 2

. m s m2 )) −( )( ) + ( )( )⎛

⎝50 0 800 3 600 200. N m N m

80.0 kg⎜⎜⎞⎠⎟

which yields v f = 24 5. m s .



Energy 223

(b) Yes , this is too fast for safety.

(c) Again, apply W KE PE KE PEnc g f g i= +( ) − +( ) , now with d2 considered to be a variable,

d d1 21 000= − m , and v f = 5 00. m s. This gives

f d f d m f1 2 2 2

2180 1 000 1801

2cos cos° m °( ) −( ) + ( ) = v ++⎛

⎝⎜⎞⎠⎟ − +( )0 0 mgyi

which reduces to − ( ) + − = −1 000 1 1 2 2 212

2 m f f d f d m mgyf iv . Therefore,

dmg y f m

f f

i f

2

12

2 1

1 00012

784 1 00

=( ) − ( ) −

−

=( )

m

N

v

00 1 000 50 012

80 0 5 00m m N kg m s( ) − ( )( ) − ( )(. . . ))−

=

2

3 600 50 0N N206 m

.

(d) In reality, the air drag will depend on the skydiver’s speed. It will be larger than her 784 N weight only after the chute is opened. It will be nearly equal to 784 N before she opens the chute and again before she touches down, whenever she moves near terminal speed.

5.50 (a) W KE PEnc = +∆ ∆ , but ∆KE = 0 because the speed is constant. The skier rises a vertical

distance of ∆y = ( ) ° =60 m sin30 30 m. Thus,

Wnc = ( )( )( ) = × =70 9 80 30kg m s m 2.06 10 J 21 k2 4. JJ

(b) The time to travel 60 m at a constant speed of 2.0 m s is 30 s. Thus, the required power input is

P = = × = ( )⎛

⎝⎜⎞W

tnc

∆2 06 10

30686

14. J

s W

hp

746 W⎠⎠⎟ = 0.92 hp

5.51 As the piano is lifted at constant speed up to the apartment, the total work that must be done on it is

W KE PE mg y ync g f i= + = + −( ) = ×( )∆ ∆ 0 3 50 10 25 03. .N m(( ) = ×8 75 104. J

The three workmen (using a pulley system with an effi ciency of 0.750) do work on the piano at a rate of

P Pnet singleworker

W= ⎛⎝⎜

⎞⎠⎟

= (0 750 3 0 750 3 165. . ))[ ] = =371 371W J s

so the time required to do the necessary work on the piano is

∆tWnc= = × = = ( )Pnet

J

371 J s s s

8 75 10236 236

14. mmin

60 s min⎛

⎝⎜⎞⎠⎟ = 3 93.


224 Chapter 5

5.52 Let ∆ N be the number of steps taken in time ∆ t. We determine the number of steps per unit time by

Power = work done work per step per unit

∆t=

mmass mass steps( )( )( )#

∆t

or

70 0 60 60 W J step

kg kg=

⎛⎝⎜

⎞⎠⎟

( )⎛⎝⎜

⎞⎠⎟

.∆∆N

t, giving

∆∆N

t= 1.9 steps s

The running speed is then

vav distance traveled per st= =

⎛⎝⎜

⎞⎠⎟

∆∆

∆∆

x

t

N

teep

step

s

m

step m s( ) = ⎛

⎝⎜⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

=1 9 1 5 2 9. . .

5.53 Assuming a level track, PE PEf i= , and the work done on the train is

W KE PE KE PE

m

nc f i

f i

= +( ) − +( )

= −( ) =1

2

1

20 8752 2v v . kkg m s J( ) ( ) −⎡⎣ ⎤⎦ =0 620 0 0 168

2. .

The power delivered by the motor is then

P = =×

=−W

tnc

∆0 168

8 01.

. J

21.0 10 s W3

5.54 When the car moves at constant speed on a level roadway, the power used to overcome the total frictional force equals the power input from the engine, or P Poutput total input= =f v . This gives

ftotal

input hp

29 m s

W

1 hp= =

⎛⎝⎜

⎞⎠⎟

=P

v175 746

4..5 105× N

5.55 The work done on the older car is

W KE KE m mf inet old old( ) = −( ) = − =1

20

1

22 2v v

The work done on the newer car is

W KE KE m mf inet new new

( ) = −( ) = ( ) − = ⎛⎝

⎞1

22 0 4

1

22 2v v ⎠⎠ = ( )4 Wnet old

and the power input to this car is

P Pnewnet new net old

old=( )

=( )

=W

t

W

t∆ ∆4

4

or the power of the newer car is 4 times that oof the older car .


Energy 225

5.56 Neglecting any variation of gravity with altitude, the work required to lift a 3 20 107. × kg load at constant speed to an altitude of ∆y = 1 75. km is

W PE mg yg= = ( ) = ×( )( ) ×∆ ∆ 3 20 10 9 80 1 75 17. . .kg m s2 00 5 49 103 11m J( ) = ×.

The time required to do this work using a P = = ×2 70 2 70 103. . kW J s pump is

∆t

W= = ××

= × =P

5 49 10

2 70 102 03 10 2 0

11

38.

.. .

J

J s s 33 10

15 64 108×( ) ⎛

⎝⎜⎞⎠⎟

= × s h

3 600 s h4.

5.57 (a) The acceleration of the car is

at

= − = − =v v0 18 0 0

12 01 50

.

..

m s

s m s2

Thus, the constant

forward force due to the engine is found from ΣF F F ma= − =engine air as

F F maengine air

2N kg m s= + = + ×( )400 1 50 10 1 503. .(( ) = ×2 65 103. N

The average velocity of the car during this interval is v v vav m s= + =0 2 9 00/ . , so the average power input from the engine during this time is

Pav engine av N m s

hp

746= = ×( )( )F v 2 65 10 9 00

13. . W

32.0 hp⎛⎝⎜

⎞⎠⎟ =

(b) At t = 12 0. s, the instantaneous velocity of the car is v = 18 0. m s and the instantaneous power input from the engine is

P = = ×( )( )⎛

⎝Fengine N m s hp

746 Wv 2 65 10 18 0

13. . ⎜⎜⎞⎠⎟ = 63.9 hp

5.58 (a) The acceleration of the elevator during the fi rst 3.00 s is

a

t= − = − =v v0 1 0

30 583

.75 m s

.00 s m s2.

so F F mg manet motor= − = gives the force exerted by the motor as

F m a gmotor250 kg m s= +( ) = ( ) +( )⎡⎣ ⎤⎦ =6 0 583 9 80. . 66 75 103. × N

The average velocity of the elevator during this interval is v v vav m s= + =( ) .0 2 0 875 so the average power input from the motor during this time is

Pav motor av N .875 m s

hp

746= = ×( )( )F v 6 75 10 0

13. W

7.92 hp⎛⎝⎜

⎞⎠⎟ =

(b) When the elevator moves upward with a constant speed of v = 1 75. m s, the upward force exerted by the motor is F mgmotor = and the instantaneous power input from the motor is

P = ( ) = ( )( )( )mg v 650 9 80 1

1 kg m s .75 m s

hp

746 2.

WW14.9 hp⎛

⎝⎜⎞⎠⎟ =


226 Chapter 5

5.59 The work done on the particle by the force F as the particle moves from x xi= to x x f= is the area under the curve from x xi fto .

(a) For x = 0 to x = 8 00. m,

W ABC AC= = ×area of triangle altitude

1

2

W0 8

1

28 00 6 00 24 0→ = ( )( ) =. . .m N J

(b) For x = 8 00. m to x = 10 0. m,

W CDE CE8 10

1

2→ = = ×area of triangle altitude

= m N J1

22 00 3 00 3 00. . .( ) −( ) = −

(c) W W W0 10 0 8 8 10 24 0 3 00 21 0→ → →= + = + −( ) =. . .J J J

5.60 The work done by a force equals the area under the force versus displacement curve.

(a) For the region 0 5 00≤ ≤x . m,

W0 5

3 00 5 007 50to

N m

2J= ( )( )

=. ..

(b) For the region 5 00 10 0. .m m≤ ≤x ,

W5 3 00 5 00 15 0to 10 N m J= ( )( ) =. . .

(c) For the region 10 0 15 0. .m m≤ ≤x , W10

3 00 5 007 50to 15

N m

2J= ( )( )

=. ..

(d) KE KE W F xx x x xf f= =− = =0 0 to area under vs. currve from m tox x x f= =0 , or

1

2

1

22

02

0m m Wf x fv v= + to

giving

v vf xmW

f= + ⎛

⎝⎜⎞⎠⎟0

20

2 to

For x f = 5 00. m:

v vf mW= + ⎛

⎝⎜⎞⎠⎟ = ( ) +0

20 5

220 500

2

3 00 to m s kg

..

⎛⎛⎝⎜

⎞⎠⎟

( ) =7 50 2 29. . J m s

For x f = 15 0. m:

v v vf mW

mW= + ⎛

⎝⎜⎞⎠⎟ = + ⎛

⎝⎜⎞⎠⎟0

20 15 0

20 5

2 2 to to ++ +( )W W5 10 10 15 to to

or

v f = ( ) +⎛⎝⎜

⎞⎠⎟

+ +0 5002

3 007 50 15 0

2.

.. . m s

kg J J 77 50 4 50. . J m s( ) =

6

4

2

2 4

D

A

B

C E

Fx (N)

6 8 10 120

�2

�4

x (m)

6

4

2

2 4

D

A

B

C E

Fx (N)

6 8 10 120

�2

�4

x (m)

Fx (N)

x (m)

3

2

1

02 4 6 8 10 120 14 16

Fx (N)

x (m)

3

2

1

02 4 6 8 10 120 14 16


Energy 227

5.61 (a) F xx = −( )8 16 N See the graph at the right.

(b) The net work done is the total area under the graph from x x= =0 3 00to m. . This consists of two triangu-lar shapes, one below the axis (negative area) and one above the axis (positive). The net work is then

Wnet

m N m .00 N=

( ) −( )+

( )( )

= −

2 00 16 0

2

1 00 8

2

. . .

112 0. J

5.62 (b) Solving part (b) fi rst, we recognize that the egg has constant acceleration a gy = −( ) as it falls 32.0 m from rest before contacting the pad. Taking upward as positive, its velocity just before contacting the pad is given by v v2

02 2= + ( )a yy ∆ as

v1 0 2 9 80 32 0 25 0= − + −( ) −( ) = −. . . m s m m s2

The average acceleration as the egg comes to rest in 9.20 ms after contacting the pad is

a

tf

av 3

m s

9.20 10 s=

−=

− −( )×

= +−

v v1 0 25 0 25 0

9∆. .

.220103⎛

⎝⎜⎞⎠⎟ × m s2

and the average net force acting on the egg during this time is

F manet av av kg( ) = = ×( )⎛

⎝⎜⎞⎠⎟ ×−75 0 10

25 0

9 203.

.

.110 204 2043 m s N N upward2 = + =

(a) The egg has a downward displacement of magnitude ∆y as the upward force Fnet av( )

brings the egg to rest. The net work done on the egg in this process is

W F y KE KE mfnet net av°= ( )⎡⎣ ⎤⎦ = − = −cos180 0

1

21 12∆ v

so

∆y

m

F= −

( ) °=

×( )−v1

2 3

2 180

75 0 10 25 0

net av

kg

cos

. . m s

N m cm

( )( ) = =

2

2 2040 115 11 5. .

20(3, 8)

10

0

�10

�20

1 2 3 4

Fx (N)

x (m)

20(3, 8)

10

0

�10

�20

1 2 3 4

Fx (N)

x (m)


228 Chapter 5

5.63 The person’s mass is

mw

g= = =700

71 4N

9.80 m skg2 .

The net upward force acting on the body is Fnet N N N= ( ) − =2 355 700 10 0. . The fi nal upward velocity can then be calculated from the work–energy theorem as

W KE KE m mf i inet = − = −1

2

1

22 2v v

or

F snet N ° mcos . cos .θ( ) = ( )⎡⎣ ⎤⎦( ) =10 0 0 0 2501

271..4 02 kg( ) −v

which gives v = 0.265 m s upward .

5.64 Taking y = 0 at ground level, and using conservation of energy from when the boy starts from rest ( )vi = 0 at the top of the slide ( )y Hi = to the instant he leaves the lower end ( )y hf = of the frictionless slide with a horizontal velocity ( , )v v v0 0 0x f y= = , yields

1

202m mgh mgHfv + = + or v f g H h2 2= −( ) [1]

Considering his fl ight as a projectile after leaving the end of the slide, ∆y t a ty y= +v012

2 gives the time to drop distance h to the ground as

− = + −( )h g t01

22 or t

h

g= 2

The horizontal distance traveled (at constant horizontal velocity) during this time is d, so

d th

gx f= =v v0

2 and v f d

g

h

gd

h= =

2 2

2

Substituting this result into Equation [1] above gives

gd

hg H h

2

22= −( ) or H h

d

h= +

2

4


Energy 229

5.65 (a) If y = 0 at point B, then yA m ° m= ( ) =35 0 50 0 26 8. sin . . and yB = 0. Thus,

PE mgyA A

2kg m s m= = ×( )( )( ) =1 50 10 9 80 26 8 33. . . .994 105× J

PE mgyB B= = 0 and ∆PE PE PEA B B A J J→ = − = − × = − ×0 3 94 10 3 94 105 5. .

(b) If y = 0 at point C, then yA m ° m= ( ) =50 0 50 0 38 3. sin . . and

yB m ° m= ( ) =15 0 50 0 11 5. sin . . . In this case,

PE mgyA A

2kg m s m= = ×( )( )( ) =1 50 10 9 80 38 3 53. . . .663 105× J

PE mgyB B

2kg m s m= = ×( )( )( ) =1 50 10 9 80 11 5 13. . . .669 105× J

and

∆PE PE PEA B B A J J→ = − = × − × = − ×1 69 10 5 63 10 3 945 5. . . 1105 J

5.66 The support string always lies along a radius line of the circular path followed by the bob. This means that the tension force in the string is always perpendicular to the motion of the bob and does no work. Thus, mechanical energy is conserved and (taking y = 0 at the point of support) this gives

12

2 12

2 0m m mg y y mg L Li i f iv v= + −( ) = + − − −( )⎡⎣ ⎤⎦cosθ

or

v = −( ) = ( )( ) −( )2 1 2 9 8 2 0 1 25gL icos . . cosθ m s m °2

v = 1 9. m s

5.67 (a) The equivalent spring constant of the bow is given by F kx= as

kF

xf

f

= = 230

0 400575

N

m= N m

.

(b) From the work–energy theorem applied to this situation,

W KE PE PE KE PE PE kxnc g s f g s i f= + +( ) − + +( ) = + +⎛ 0 01

22

⎝⎝⎞⎠ − + +( )0 0 0

The work done pulling the bow is then

W kxnc f= = ( )( ) =1

2

1

2575 0 4002 2N m m 46.0 J.

Lqi

vi �0

∆y

v→

Lqi

vi �0

∆y

v→


230 Chapter 5

5.68 Choose PEg = 0 at the level where the block comes to rest against the spring. Then, in the absence of work done by nonconservative forces, the conservation of mechanical energy gives

KE PE PE KE PE PEg s f g s i

+ +( ) = + +( ) or

0 01

20 02+ + = + +kx mg Lf sinθ

Thus,

x

mg L

kf = =( )( )( )2 2 12 0 9 80 3 00sin . . . siθ kg m s m2 nn .

..

35 0

3 00 100 1164

°×

=N m

m

5.69 (a) From v v202 2= + ( )a yy ∆ , we fi nd the speed just before touching the ground as

v = + ( )( ) =0 2 9 80 1 0 4 4. . . m s m m s2

(b) Choose PEg = 0 at the level where the feet come to rest. Then

W KE PE KE PEnc g f g i= +( ) − +( ) becomes

F s m mg sia °v cos180 0 0

1

22( ) = +( ) − +⎛

⎝⎞⎠v

or

Fm

smgi

av

kg m s

m= + =

( )( )×( )−

v2 2

32

75 4 4

2 5 0 10

.

.++ ( )( ) = ×75 9 80 1 5 105kg m s N2. .

5.70 From the work–energy theorem,

W KE PE PE KE PE PEnc g s f g s i= + +( ) − + +( )

we have

f s m kxk f icos1801

20 0 0 0

1

22 2°( ) = + +⎛

⎝⎞⎠ − + +⎛

⎝⎞⎠v

or

v fi kkx f s

m= − =

( ) ×( ) −−2 2 22 8 0 5 0 10 2 0 032. . .N m m NN m

kgm s

( )( )×

=−

0 15

5 3 101 43

.

..


Energy 231

5.71 (a) The two masses will pass when both are at yf = 2 00. m above the fl oor. From conservation

of energy, KE PE PE KE PE PEg s f g s i+ +( ) = + +( )

1

20 0 01 2

21 2 1 1m m m m gy m gyf f i+( ) + +( ) + = + +v

or

v fi

f

m gy

m mgy=

+−

=( )( )

22

2 5 00 9 80 4

1 1

1 2

. . . kg m s2 000

8 002 9 80 2 00

m

kg m s m 2

( )− ( )( )

.. .

This yields the passing speed as v f = 3 13. m s .

(b) When m1 5 00= . kg reaches the fl oor, m2 3 00= . kg is y f2 4 00= . m above the fl oor.

Thus, KE PE PE KE PE PEg s f g s i+ +( ) = + +( ) becomes

1

20 0 01 2

22 2 1 1m m m gy m gyf f i+( ) + + = + +v

or

v f

i fg m y m y

m m=

−( )+

2 1 1 2 2

1 2

Thus,

v f =( ) ( )( ) − ( )2 9 80 5 00 4 00 3 00 4. . . . . m s kg m kg2 000

8 004 43

m

kg m s

( )⎡⎣ ⎤⎦ =.

.

(c) When the 5.00-kg hits the fl oor, the string goes slack and the 3.00-kg becomes a projectile launched straight upward with initial speed v0 4 43y = . m s. At the top of its arc, v

y = 0

and v vy y ya y202 2= + ( )∆ gives

∆ya

y y

y

=−

=− ( )−( ) =

v v202 2

2

0 4 43

2 9 80

.

.

m s

m s1.0

200 m


232 Chapter 5

5.72 The normal force the incline exerts on block A is n m gA A= ( ) °cos37 , and the friction force is f n m gk k A k A= = °µ µ cos37 . The vertical distance block A rises is ∆yA = ( ) =20 37 12 m ° msin ,

while the vertical displacement of block B is ∆yB = −20 m.

We fi nd the common fi nal speed of the two blocks by use of

W KE PE KE PE KE PEnc g f g i g= +( ) − +( ) = +∆ ∆

This gives − ( ) = +( ) −⎡⎣⎢

⎤⎦⎥

+µk A A B f A Am g s m m m g ycos371

202° v ∆(( ) + ( )⎡⎣ ⎤⎦m g yB B∆ , or

v f

B B A A k A

A

g m y m y m s

m2

2 37=

− ( ) − ( ) − ( )⎡⎣ ⎤⎦+

∆ ∆ µ cos °

mmB

=( ) − ( ) −( ) − ( )2 9 80 100 20 50 12. m s kg m kg m2 (( ) − ( )( )⎡⎣ ⎤⎦0 25 50 20 37

150

. cos kg m °

kg

which yields v f2 2157= m s2 .

The change in the kinetic energy of block A is then

∆KE mA A f= − = ( )( ) = ×1

20

1

250 157 3 9 102 3v kg m s2 2 . JJ 3.9 kJ=

5.73 Since the bowl is smooth (that is, frictionless), mechanical energy is conserved or

KE PE KE PEf i+( ) = +( )

Also, if we choose y = 0 (and hence, PEg = 0)at the lowest point in the bowl, then y R y y RA B C= + = =, , and0 2 3.

(a) PE mgy mgRg A A( ) = = , or

PEg A( ) = ( )( )( ) =0 200 9 80 0 300 0 588. . . .kg m s m2 JJ

(b) KE KE PE PE mgy mgyB A A B A B= + − = + − = − =0 0 588 0 0 588. .J J

(c) KE mKE

mB B BB= ⇒ = =

( ) =12

2 2 2 0 588v v

J

0.200 kg

.22 42. m s

(d) PE mgyg C C( ) = = ( )( ) ( )0 200 9 80

2 0 300

3. .

.kg m s

m2 ⎡⎡⎣⎢

⎤⎦⎥

= 0 392. J

KE KE PE PEC B B C= + − = + − =0 588 0 0 392 0 196. . .J J J

A

R

B

C

2R/3

A

R

B

C

2R/3


Energy 233

5.74 (a) KE mB B= = ( )( ) =1

2

1

20 200 1 50 0 2252 2

v . . . kg m s J

(b) The change in the altitude of the particle as it goes from A to B is y y RBA − = , where R = 0 300. m is the radius of the bowl. Therefore, the work–energy theorem gives

W KE KE PE PE

KE mg y y KE

nc B A B A

B B A B

= −( ) + −( )= − + −( ) =0 ++ −( )mg R

or

Wnc = + ( )( ) −( ) =0 225 0 200 9 80 0 300. . . .J kg m s m2 −−0 363. J

The loss of mechanical energy as a result of friction is then 0 363. J .

(c) No. Because the normal force, and hence the friction force, vary with the position of the

particle on its path, it is not possible to use the result from part (b) to determine the coeffi cient of friction without using calculus.

5.75 (a) Consider the sketch at the right. When the mass m = 1 50. kg is in equilibrium, the upward spring force exerted on it by the lower spring (i.e., the tension in this spring) must equal the weight of the object, or F mgS2 = . Hooke’s law then gives the elongation of this spring as

x

F

k

mg

kS

22

2 2

= =

Now, consider point A where the two springs join. Because this point is in equilibrium, the upward spring force exerted on A by the upper spring must have the same magnitude as the downward spring force exerted on A by the lower spring (that is, the tensions in the two springs must be equal).

The elongation of the upper spring must be

xF

k

F

k

mg

kS S

11

1

2

1 1

= = =

and the total elongation of the spring system is

x x xmg

k

mg

kmg

k k= + = + = +

⎛⎝⎜

⎞⎠⎟

= ( )

1 21 2 1 2

1 1

1 50. kg 99 801

1 20 10

1

1 80 103 3.. .

m sN m N m

2( ) ×+

×⎛⎝⎜

⎞⎠⎟

= 22 04 10 2. × − m

(b) The spring system exerts an upward spring force of F mgS = on the suspended object and undergoes an elongation of x. The effective spring constant is then

kF

x

mg

xS

effective

2 kg m s= = =

( )( )×

1 50 9 80

2 04

. .

. 1107 20 102

2− = ×

m N m.

vA� 0

vB�1.50 m/s

R�30.0 cm

m

m

A

B

vA� 0

vB�1.50 m/s

R�30.0 cm

m

m

A

B

m

k1

k2

A

m

k1

k2

A


234 Chapter 5

5.76 Refer to the sketch given in the solution of Problem 5.75.

(a) Because the object of mass m is in equilibrium, the tension in the lower spring, FS2, must equal the weight of the object. Therefore, from Hooke’s law, the elongation of the lower spring is

xF

k

mg

kS

22

2 2

= =

From the fact that the point where the springs join (A) is in equilibrium, we conclude that the tensions in the two springs must be equal, F F mgS S1 2= = . The elongation of the upper spring is then

x

F

k

mg

kS

11

1 1

= =

and the total elongation of the spring system is

x x x mgk k

= + = +⎛⎝⎜

⎞⎠⎟1 2

1 2

1 1

(b) The two spring system undergoes a total elongation of x and exerts an upward spring force F mgS = on the suspended mass. The effective spring constant of the two springs in series is then

k

F

x

mg

x

mg

mgk k

kS

effective = = =+

⎛⎝⎜

⎞⎠⎟

= +1 1

1 1

1 2

1 kk2

1⎛⎝⎜

⎞⎠⎟

−

5.77 (a) The person walking uses Ew = ( )( ) = ×2 4 186 9 21 10520 kcal J 1 kcal J. of energy while going 3.00 miles. The quantity of gasoline which could furnish this much energy is

V1

539 21 10

7 08 10= ××

= × −..

J

1.30 10 J galgal8

This means that the walker’s fuel economy in equivalent miles per gallon is

fuel economy mi

7.08 10 gal mi gal3=

×=−

3 00423

.

(b) In 1 hour, the bicyclist travels 10.0 miles and uses

EB = ( )⎛⎝

⎞⎠ = ×400

4 186

11 67 106kcal

J

kcalJ.

which is equal to the energy available in

V2

621 67 10

1 29 10= ××

= × −..

J

1.30 10 J galgal8

of gasoline. Thus, the equivalent fuel economy for the bicyclist is

1

1 29 107762

0.0 mi

galmi gal

. ×=−


Energy 235

5.78 When 1 pound (454 grams) of fat is metabolized, the energy released is E = ( )( ) = ×454 9 00 4 09 103g kcal g kcal. . . Of this, 20.0% goes into mechanical energy (climbing stairs in this case). Thus, the mechanical energy generated by metabolizing 1 pound of fat is

Em = ( ) ×( ) =0 200 4 09 10 8173. . kcal kcal

Each time the student climbs the stairs, she raises her body a vertical distance of ∆y = ( )( ) =80 0 150 12 0steps m step m. . . The mechanical energy required to do this is

∆ ∆PE mg yg = ( ), or

∆PEg = ( )( )( ) = ×50 0 9 80 12 0 5 88 103. . . .kg m s m J2 (( )⎛

⎝⎞⎠ =1 kcal

4186 Jkcal1 40.

(a) The number of times the student must climb the stairs to metabolize 1 pound of fat is

N

E

PEm

g

= = =∆

817582

kcal

1.40 kcal triptrips

It would be more practical for her to reduce food intake.

(b) The useful work done each time the student climbs the stairs is W PEg= = ×∆ 5 88 103. J. Since this is accomplished in 65.0 s, the average power output is

Pav

J

65.0 s90.5 W W

hp= = × = = ( )W

t

5 88 1090 5

13..

7746 W0.121 hp⎛

⎝⎜⎞⎠⎟ =

5.79 (a) Use conservation of mechanical energy, KE PE KE PEg f g i+( ) = +( ) , from the start

to the end of the track to fi nd the speed of the skier as he leaves the track. This gives 1

22 0m mgy mgyf iv + = + , or

v = −( ) = ( )( ) =2 2 9 80 40 0 28 0g y yi f . . . m s m m s2

(b) At the top of the parabolic arc the skier follows after leaving the track, vy = 0 and

vx = ( ) =28 0 45 0 19 8. cos . . m s ° m s. Thus, v v vtop m s= + =x y2 2 19 8. . Applying

conservation of mechanical energy from the end of the track to the top of the arc gives 1

2

2 12

219 8 2 10 0m mg y m mg. . m s 8.0 m s mmax( ) + = ( ) + (( ), or

ymax 2m

m s 9.8 m s

2 9.80 m s= +

( ) − ( )(10 0

28 0 12 2

..

)) = 30 0. m

(c) Using ∆y t a ty y= +v012

2 for the fl ight from the end of the track to the ground gives

− = ( )⎡⎣ ⎤⎦ + −(10 0 28 0 45 01

29 80. . sin . . m m s ° m s2t )) t 2

The positive solution of this equation gives the total time of fl ight as t = 4 49. s . During this time, the skier has a horizontal displacement of

∆x tx= = ( )⎡⎣ ⎤⎦( ) =v0 28 0 45 0 4 49. cos . . m s ° s 89.0 mm


236 Chapter 5

5.80 First, determine the magnitude of the applied force by considering a free-body diagram of the block. Since the block moves with constant velocity, Σ ΣF Fx y= = 0.

From ΣFx = 0, we see that n F= °cos30 .

Thus, f n Fk k k= = °µ µ cos30 , and ΣFy = 0 becomes

F mg Fksin cos30 30° = + °µ

or

F

mg

k

=−

=( )( )

sin cos

. .

sin30 30

5 0 9 80

30° °

kg m s2

µ °° ° N

− ( ) = ×0 30 30

2 0 102

. cos.

(a) The applied force makes a 60° angle with the displacement up the wall. Therefore,

W F sF = ( ) = ×( )⎡⎣ ⎤⎦ ( ) =cos . cos .60 2 0 10 60 3 02° N ° m 33 1 102. × J

(b) W mg sg = ( ) = ( ) −( )( ) = − ×cos . . .180 49 1 0 3 0 1 5 10° N m 22 J

(c) W n sn = ( )° =cos90 0

(d) PE mg yg = ( ) = ( )( ) = ×∆ 49 3 0 1 5 102N m J. .

5.81 We choose PEg = 0 at the level where the spring is relaxed (x = 0), or at the level of position B.

(a) At position A, KE = 0 and the total energy of the system is given by

E PE PE mgx k xg s A= + +( ) = +0

1

21 12 , or

E = ( )( ) −( ) + ×25 0 9 80 0 100

1

22 50 104. . . .kg m s m2 NN m m 101 J( ) −( ) =0 100 2.

(b) In position C, KE = 0 and the spring is uncompressed, so PEs = 0. Hence,

E PE mg xg C= + +( ) =0 0 2

or

xE

mg2

101

9 80= = ( )( ) =J

25.0 kg m s0.410 m

2.

(c) At position B, PE PEg s= = 0 and E KE mB B= + +( ) =0 0 12

2v

Therefore,

vB

E

m= = ( )

=2 2 101

25 02 84

J

kgm s

..

5 kgn

30°

F→

→

fk→

→mg

5 kgn

30°

F→

→

fk→

→mg



Energy 237

(d) Where the velocity (and hence the kinetic energy) is a maximum, the acceleration is a F my y= =( )Σ 0 (at this point, an upward force due to the spring exactly balances the

downward force of gravity). Thus, taking upward as positive, ΣF kx mgy = − − = 0, or

x

mg

k= − = −

×= − × = −−245

9 80 10 3kg

2.50 10 N mm 9.84 . 00 mm

(e) From the total energy, E KE PE PE m mgx kxg s= + + = + +12

2 12

2v , we fi nd

v = − −2

2 2E

mgx

k

mx

Where the speed, and hence kinetic energy is a maximum (that is, at x = − 9 80. mm ), this gives

vmax .

. .=( ) − ( ) − × −2 101

25 02 9 80 9 80 10 3 J

kg m s 2 mm

N m

kg m( ) −

×( )− ×( )−2 50 10

25 09 80 10

43 2.

..

or

vmax .= 2 85 m s

5.82 When the hummingbird is hovering, the magnitude of the average upward force exerted by the air on the wings (and hence the average downward force the wings exert on the air) must be F mgav = where mg is the weight of the bird. Thus, if the wings move downward distance d during a wing stroke, the work done each beat of the wings is

W F d mgdbeat av

2 kg m s= = = ×( )( ) ×−3 0 10 9 80 3 5 13. . . 00 1 0 102 3− −( ) = × m J.

In 1 minute, the number of beats of the wings that occur is

N = ( )( ) = ×80 60 4 8 103beats s s min beats min.

so the total work preformed in 1 minute is

W NWtotal beat min

beats

min= ( ) = ×⎛

⎝⎜⎞⎠⎟1 4 8 103. 11 0 10 1 4 93. .×⎛

⎝⎜⎞⎠⎟ ( ) =−

J

beat min J

5.83 Choose PEg = 0 at the level of the river. Then yi = 36 0. m, yf = 4 00. , the jumper falls 32.0 m, and the cord stretches 7.00 m. Between the balloon and the level where the diver stops momentarily,

KE PE PE KE PE PEg s f g s i+ +( ) = + +( ) gives

0 700 4 00

1

27 00 0 700 362+ ( )( ) + ( ) = + ( )N m m N. . .k 00 0m( ) +

or

k = 914 N m


238 Chapter 5

5.84 If a projectile is launched, in the absence of air resistance, with speed v0 at angle θ above the horizontal, the time required to return to the original level is found from ∆y t a ty y= +v0

12

2 as 0 20

2= ( ) −v sin ( )θ t g t , which gives t g= ( )sin2 0v θ . The range is the horizontal displacement occurring in this time.

Thus,

R tgx= = ( )⎛

⎝⎜⎞⎠⎟

=(

v vv v

0 00 0

22 2cos

sin sin cosθ θ θ θ)) =

( )g g

v02 2sin θ

Maximum range occurs when θ = °45 , giving R gmax = v02 or v0

2 = g Rmax. The minimum kinetic energy required to reach a given maximum range is then

KE m mg R= =1

2

1

202v max

(a) The minimum kinetic energy needed in the record throw of each object is

Javelin: KE = ( )( )( ) = ×1

20 80 9 80 98 3 8 102. . .kg m s m J2

Discus: KE = ( )( )( ) = ×1

22 9 80 74 7 3 102.0 kg m s m J2. .

Shot: KE = ( )( )( ) = ×1

27 9 80 2 8 1 102.2 kg m s 3 m J2. .

(b) The average force exerted on an object during launch, when it starts from rest and is given the kinetic energy found above, is computed from W F s KEnet av= = ∆ as F KE sav = −( )0 . Thus, the required force for each object is

Javelin: Fav

J

mN= × = ×3 8 10

2 001 9 10

22.

..

Discus: Fav

J

mN= × = ×7 3 10

2 003 6 10

22.

..

Shot: Fav

J

mN= × = ×8 1 10

2 004 1 10

22.

..

(c) Yes . If the muscles are capable of exerting 4 1 102. × N on an object and giving that object a kinetic energy of 8 1 102. × J, as in the case of the shot, those same muscles should

be able to give the javelin a launch speed of

v0

22 2 8 1 10

45= =×( )

=KE

m

. J

0.80 kg m s

with a resulting range of

Rgmax 2

m s

m s m= =

( )= ×v0

2 2

245

9 802 1 10

..

Since this far exceeds the record range for the javelin, one must conclude that air resistance plays a very signifi cant role in these events.


Energy 239

5.85 From the work–energy theorem, W KE KEf inet = − . Since the package moves with constant velocity, KE KEf i= giving Wnet = 0 .

Note that the above result can also be obtained by the following reasoning: Since the object has zero acceleration, the net (or resultant) force acting on it must be zero. The net work done is W F dnet net= = 0 .

The work done by the conservative gravitational force is

W PE mg y y mg dg f igrav = − = − −( ) = − ( )∆ sinθ

or

Wgrav2 kg m s m °= − ( )( )( ) = −50 9 80 340 7 0 2 0. sin . . ×× 104 J

The normal force is perpendicular to the displacement. The work it does is

W ndnormal °= =cos 90 0

Since the package moves up the incline at constant speed, the net force parallel to the incline is zero. Thus, ΣF f mg f mgs s� = ⇒ − = =0 0 or sin , sinθ θ .

The work done by the friction force in moving the package distance d up the incline is

W f d mg dkfriction2 kg m s= = ( ) = ( )( )sin . siθ 50 9 80 nn . .7 0 340 2 0 104° m J⎡⎣ ⎤⎦( ) = ×


240 Chapter 5

5.86 Each 5.00-m length of the cord will stretch 1.50 m when the tension in the cord equals the weight of the jumper (that is, when F w mgs = = ). Thus, the elongation in a cord of original length L when F ws = will be

xL

L= ⎛⎝

⎞⎠ ( ) =

5 001 50 0 300

.. .

mm

and the force constant for the cord of length L is

kF

x

w

Ls= =

0 300.

(a) In the bungee-jump from the balloon, the daredevil drops y yi f− = 55 0. m.

The stretch of the cord at the start of the jump is xi = 0, and that at the lowest point is x Lf = −55 0. m . Since KE KEi f= = 0 for the fall, conservation of mechanical energy gives

0 0

1

22+ ( ) + ( ) = + ( ) + ( ) ⇒ −PE PE PE PE k xg f s f g i s i f xx mg y yi i f

2( ) = −( ) giving

1

2 0 30055 0 55 02mg

LL mg

.. .

⎛

⎝⎜⎞

⎠⎟−( ) = ( )m m and 55 0 33 02. .m m−( ) = ( )L L

which reduces to

55 0 110 33 02 2. .m m m( ) − ( ) + = ( )L L L

or

L L2 2143 55 0 0− ( ) + ( ) =m m.

and has solutions of

L =− −( ) ± −( ) − ( )( )

( )143 143 4 1 55 0

2 1

2 2m m m.

This yields

L = ±143 91 4

2

m m. and L L= =117 25 8m or m.

Only the L = 25 8. m solution is physically acceptable!

(b) During the jump, ΣF ma kx mg mam g

Ly y y= ⇒ − =⎛⎝⎜

⎞⎠

, or 0 300. ⎟⎟ − =x m g m ay

Thus,

ax

Lgy = −⎛

⎝⎞⎠0 300

1.

which has maximum value at x x L= = − =max . .55 0 29 2m m.

a g gy( ) =( ) −

⎡⎣⎢

⎤⎦⎥ = =

max

..

29 21 2 77

m

0.300 25.8 m227 1. m s2


Energy 241

5.87 (a) While the car moves at constant speed, the tension in the cable is F mg= sinθ , and the power input is P = =F mgv v sinθ , or

P = ( )( )( ) = ×950 9 80 2 20 30 0 1 02 kg m s m s °2. . sin . . 110 10 24 W kW= .

(b) While the car is accelerating, the tension in the cable is

F mg ma m gta = + = +

⎛⎝⎜

⎞⎠⎟

= ( )

sin sin

.

θ θ ∆∆

v

950 9 80 kg m s ° m s

s2( ) + −⎡

⎣⎢⎤⎦⎥

= ×sin ..

..30 0

2 20 0

12 04 83 1103 N

Maximum power input occurs the last instant of the acceleration phase. Thus,

Pmax max . . .= = ×( )( ) =Fav 4 83 10 2 20 10 63 N m s kW

(c) The work done by the motor in moving the car up the frictionless track is

W KE PE KE PE KE PE m mnc f i f g f f= +( ) − +( ) = + ( ) − = +01

22v gg L sinθ( )

or

Wnc = ( ) ( ) + ( )( )9501

22 20 9 80 1 250

2 kg m s m s m2. . ssin . .30 0 5 82 106° J⎡

⎣⎢⎤⎦⎥

= ×

5.88 (a) Since the tension in the string is always perpendicular to the motion of the object, the string does no work on the object. Then, mechanical energy is conserved:

KE PE KE PEg f g i+( ) = +( )

Choosing PEg = 0 at the level where the string attaches to the cart, this gives

01

2 02+ −( ) = + −( )mg L m mg Lcosθ v

or

v0 2 1= −( )g L cosθ

(b) If L = 1.20 m and θ = °35 0. , the result of part (a) gives

v0 2 9 80 1 20 1 35 0 2 06= ( )( ) −( ) =. . cos . . m s m ° m s2


242 Chapter 5

5.90 (a) Realize that, with the specifi ed arrangement of springs, each spring supports one-fourth the weight of the load (shelf plus trays). Thus, adding the weight ( w mg= ) of one tray to the load increases the tension in each spring by ∆F mg= 4 . If this increase in tension causes an additional elongation in each spring equal to the thickness of a tray, the upper surface of the stack of trays stays at a fi xed level above the fl oor as trays are added to or removed from the stack.

(b) If the thickness of a single tray is t, the force constant each spring should have to allow the fi xed-level tray dispenser to work properly is

kF

x

mg

t

mg

t= = =∆

∆4

4

or

k =( )( )

×( ) =−

0 580 9 80

4 0 450 10316

2

. .

.

kg m s

mN

2

mm

The length and width of a tray are unneeded pieces of data.

5.91 When the cyclist travels at constant speed, the magnitude of the forward static friction force on the drive wheel equals that of the retarding air resistance force. Hence, the friction force is proportional to the square of the speed, and her power output may be written as

P = = ( ) =f k ksv v v v2 3

where k is a proportionality constant.

If the heart rate R is proportional to the power output, then R k k k k k= ′ = ′ ( ) = ′P v v3 3 where ′k is also a proportionality constant.

The ratio of the heart rate R2 at speed v2 to the rate R1 at speed v1 is then

R

R

k k

k k2

1

23

13

2

1

3

= ′′

=⎛⎝⎜

⎞⎠⎟

vv

vv

giving

v v2 12

1

1 3

=⎛⎝⎜

⎞⎠⎟

R

R

Thus, if , R = =90 0 22 0. . beats min at km hv , the speed at which the rate would be 136 beats min is

v = ( )⎛⎝⎜

⎞⎠⎟

22 0136

90 0.

. km h

beats min

beats min

11 3

= 25.2 km h

and the speed at which the rate would be 166 beats min

v = ( )⎛⎝⎜

⎞⎠⎟

22 0166

90 0.

. km h

beats min

beats min

11 3

= 27.0 km h


Energy 243

5.92 (a) The needle has maximum speed during the interval between when the spring returns to nor-mal length and the needle tip fi rst contacts the skin. During this interval, the kinetic energy of the needle equals the original elastic potential energy of the spring, or 1

22 1

22m kxivmax = .

This gives

vmax ..

= = ×( ) ×=−

−xk

mi 8 10 10375

5 60 1022

3 m N m

kg11 0. m s

(b) If F1 is the force the needle must overcome as it penetrates a thickness x1 of skin and soft tissue while F2 is the force overcame while penetrating thickness x2 of organ material, application of the work–energy theorem from the instant before skin contact until the instant before hitting the stop gives

W F x F x m mfnet = − − = −1 1 2 2

12

2 12

2v vmax

or

v vf

F x F x

m= −

+( )max2 1 1 2 22

v f = ( ) −( ) ×( ) +−

21 02 7 60 2 40 10 9 202

2

.. . .

m s N m N(( ) ×( )⎡⎣ ⎤⎦

×=

−

−

3 50 10

5 60 1016 1

2

3

.

..

m

kg m s


solucionario fundamentos de física 9na edición capitulo 5

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