Principles and practice of physics global 1st edition mazur solutions manual by alexandra.mitchell765

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WORK

Questions and Problems 9.1. Not necessarily. For work to be done, the object must be displaced along the direction of the force. So there could be force applied but work is still zero if the displacement is zero, or the angle between 𝐹⃗ and 𝛥𝑥⃗ is 90 degrees.

9.2. Hitting either surface reduces the kinetic energy of your fist to zero, which means work is done on the fist to change its energy. For a given fist speed, the amount of work (force times distance) needed to reduce the speed to zero is the same regardless of which surface you hit. This work is done over a very short distance when you hit the door, and the force exerted on your fist must be large and thus painful. This work is done over a long distance when your fist sinks into the sofa cushion, and so a relatively small force is exerted and hurts less.

9.3. When you drop the brick from the greater height, the force of gravity is exerted on the brick over a greater distance, thus doing more work on the brick. When more work is done on the brick, it has more kinetic energy when it hits your foot and thus hurts more.

9.4. The total force on an object moving at constant speed, so the work done in this case is zero. The total force on the freely falling object is a constant equal to mg. Work is a positive nonzero quantity, so (b) is more.

9.5. Initially, your hand exerts an upward force on the ball. The force displacement is nonzero because the point of application of the force is at the ball and therefore moves. Hence this force does work on the ball. Once you release the ball, the only force exerted on it is the gravitational force. The point of application is again at the ball, which is moving; thus the force displacement is nonzero, and the gravitational force does work on the ball as it rises. At the top of its path the ball reverses its motion and the gravitational force again does work on it because again the force displacement is nonzero. Finally, the laundry exerts an upward force on the ball as the ball moves downward; this force does work on the ball because again the force displacement is nonzero.

9.6. We consider only work resulting from forces acting along the x axis. (a) For the work to be positive, the force must be parallel to the displacement. This means the object must speed up. This is the case from b to c and from d to e. (b) If the work is negative, the force is applied opposite the displacement, and the speed will decrease. This is the case from a to b, from c to d, and from e to f. (c) If the work is zero, there cannot be an increase in speed. This is the case from f to g.

9.7. Work is positive when the force and displacement are in the same direction. (a) Force of gravity points down while the displacement is up. Work is negative. (b) Force balancing gravity by your hand is up and displacement is up. Work is positive. (c) Total force is zero for motion at constant velocity. Work is zero. © Copyright 2015 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

9-1

9-2

Chapter 9

9.8. As you move down the hill, Earth exerts a downward gravitational force that has a component in your direction of motion and so does positive work on you. This work would continue increasing your kinetic energy until it became dangerous, except that you use your legs to stop yourself from tumbling down the hill. On level ground, you don’t have a downward displacement over which the gravitational force could be exerted to speed you up. In walking down the hill, you must expend chemical energy to counteract the positive work done by gravity.

9.9. Both observers are right in their own reference frames. 9.10. It is concave upward. If a displacement requires that positive work be done on the system, but does not increase the kinetic energy of the system, then it must increase the potential energy of the system. Because this occurs for displacement in either direction, the potential energy curve must increase in either direction.

9.11. This could happen when dissipative forces are involved. Work can be done, but all lost to dissipative processes.

9.12. For the system of the jet plane alone, the drag force and the thrust are external forces that do work. The work done in each case is the product of force and force displacement, and the numbers given in the problem set a relative scale for the work by each force: force times force displacement d (the length of runway used). There is no potential energy to consider, so given the values of work we can compute the change in kinetic energy. However, there is a large amount of source energy converted (fuel being burned), and we have no way to determine the percentage dissipated as thermal energy, nor how much of that thermal energy is dissipated within the system and how much without. We suspect most of the dissipation is outside the system in the surrounding air, leaving an imbalance in the energy diagram (which is appropriate for a system that is not closed). But how much thermal energy is dissipated for a given amount of source energy? We must assume some energy efficiency for the jet engine. A quick check of the web suggests that roughly 30% of the source energy will go towards moving the plane, so we will assume that 70% of the source energy is dissipated. How much of that is dissipated in the plane and now much outside the system (in the air) is impossible to say, but we need that information to generate an energy diagram. To do so, we arbitrarily assume that 90% of the dissipation occurs outside the system. This is a blind assumption, with not much justification (it cannot be looked up on the web), so we add a question mark to the thermal energy change column in the energy diagram

Air drag at a boundary is very much like kinetic friction at a boundary. This is why the plane alone is not a good choice of system! Perhaps a better choice of system would be the plane plus the surrounding air. This system is approximately closed as the plane taxis toward takeoff, so the energy changes should balance. Note that both forces are now internal, so that there is zero work done on this system, and that all of the thermal energy dissipation now takes place within the system. This time the diagram has no questionable bars as long as we believe that our assumption about the efficiency of the jet is reasonable.

We could add other objects to this system, such as Earth, but would gain no useful information by doing so.

© Copyright 2015 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Work

9-3

9.13. (a)

(b)

(c) The block alone should not be used as the system because friction occurs at the block-incline interface and it is not possible to know how much thermal energy produced by friction goes into the block (i.e., into the system) and how much goes into the incline (i.e., out of the system).

9.14. The only forces exerted on the block are exerted by objects in the system. Therefore no work is done by any external force, and the W bar should be empty. Then the number of units in the energy bars must be revised to show that the energy of the system remains constant. (Because no quantitative values are given, you can either remove one unit from the U bar or add one unit to the  Eth bar.)

9.15. You push a block up an incline, with the block and incline part of a system that also includes Earth (but not

you). The block increases in speed as you push (positive K and positive U ), and some energy is lost to friction (positive Eth ). All these changes constitute work you do on the system (positive W).

9.16. (a) Yes. Positive work is done by the contact force you exert on the ball, and an equal amount of negative work is done by the gravitational force that Earth exerts on the ball. The algebraic sum of the positive and negative work is zero. (b) There is no change in the potential energy of the system because the ball alone is the system. In order for potential energy to have meaning, you must have two interacting objects. (c) Yes, positive work is done by the contact force that you hand exerts on the ball. (d) The potential energy of the system increases.

9.17. The downward trip takes longer because the ball’s initial kinetic energy is converted to other forms, making the downward speed lower than the upward speed. System 1––ball: gravity does negative work as the ball rises but does an equal amount of positive work as the ball falls. The air does negative work on the ball both as it rises and as it falls, decreasing the system’s energy. System 2––ball, Earth: as the ball rises, kinetic energy is converted to gravitational potential energy, and as the ball falls that potential energy is converted to kinetic energy. The air does negative work on the ball, making the system’s final energy lower than its initial energy System 3––ball, Earth, air: the same kinetic-potential-kinetic interconversion as in system 2, plus collisions between the ball and molecules in the air cause the molecules and ball to heated up, converting some mechanical energy to thermal energy.

9.18. The following verbal description is summarized in the table below. (a) The spring potential energy is converted to kinetic energy in the tomato. This kinetic energy is converted to gravitational potential energy as the tomato rises, and the © Copyright 2015 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Chapter 9

potential energy is converted back to kinetic energy as the tomato falls. As the tomato moves, collisions with molecules in the air cause the molecules and tomato to heat up, converting some mechanical energy to thermal energy. As the tomato lands, its kinetic energy is converted to thermal energy in it and in the pavement. (b) The spring potential energy is converted to kinetic energy in the tomato. Earth’s gravitational force does negative work on the system as the tomato rises and does positive work on the system as the tomato falls. As the tomato moves, collisions with molecules in the air cause the molecules and tomato to heat up, converting some mechanical energy to thermal energy. As the tomato lands, its kinetic energy is converted to thermal energy in it and in the pavement. (c) The spring force does positive work on the system as the tomato rises through the barrel, giving the tomato kinetic energy as it is launched. Earth’s gravitational force does negative work on the system as the tomato raises and positive work on the system as the tomato falls. As the tomato moves, collisions with molecules in the air cause the molecules and tomato to heat up, converting some mechanical energy to thermal energy. As the tomato lands, its kinetic energy is converted to thermal energy in it and in the pavement. (d) The spring force does positive work on the system as the tomato rises through the barrel, giving the tomato kinetic energy as it is launched. Earth’s gravitational force does negative work on the system as the tomato rises and does positive work on the system as the tomato falls. As the tomato moves, molecules in the air do negative work on it. When the tomato lands, the pavement exerts an upward force and does negative work on it.

(a) tomato, cannon, air, Earth, pavement during launch Uspring → Ktomato tomato rises

tomato falls

tomato lands

system (b) tomato, cannon, air, (c) tomato, air, pavement pavement Uspring → Ktomato +Wby spring on sys

(d) tomato +Wby spring on sys

Ktomato → Utomato

–Wby Earth on sys

Ktomato → Eth, air/tomato

Ktomato → –Wby air on sys and Eth, tomato

Utomato → Ktomato

+Wby Earth on sys

Ktomato → Eth, air/tomato

Ktomato → –Wby air on sys and Eth, tomato

Ktomato → Eth, pavement/tomato

–Wby pavement on sys

Ktomato → Eth, pavement/tomato Ktomato → Eth, pavement/tomato

Ktomato → Eth, tomato

9.19. No, the work magnitudes are the same, but the work done is positive for one ball and negative for the other. 9.20. The velocity at the max point is 0. So max height can be obtained from: 02 = 19.82 – 2  9.8  hmax. hmax = 19.82/(2  9.8) = 19.8 m. W = (mg)  19.8 = 0.1  9.8  19.8 J = 19.4 J.

9.21. Because the only force acting on the package in the horizontal direction is constant, the acceleration in the horizontal direction will also be constant. We assume that the initial speed of the package does not change. We can v 2 − vx2,i use a kinematic equation ax = x ,f to see that if the stopping distance doubles, the acceleration is halved. 2 x vx ,f − vx ,i , we see that halving the acceleration will double the time required to stop the package. Finally, using t = ax Thus the time interval doubles.

9.22. Because the only force acting on the package in the horizontal direction is constant, the acceleration in the horizontal direction will also be constant. We can use a kinematic equation vx2,i = vx2,f − 2xax = −2xax to see that if the stopping distance doubles, it means the initial x component of the velocity increased by a factor of

2. Finally,

Work

using t =

vx ,f − vx ,i ax

− vx ,i ax

, we see that increasing the initial x component of the velocity by

stopping time by a factor of

2. Thus the time interval increases by a factor of

9-5

2 increases the

9.23. (a) Treating the acrobat and Earth as the system, the work done by the acrobat’s legs on his center of mass is equal to his change in gravitational potential energy. Call the vertical distance the acrobat moves while in contact with the ground xcontact , and call the total vertical distance that the acrobat rises until he reaches his maximum c c height  xmotion . Thus Flegs cm x ,av  xcontact = mg  xmotion  Flegs cm x,av =

mg xmotion (55 kg)(9.8 m/s 2 )((1.20 m) − (0.400 m)) = xcontact ((0.900 m) − (0.400 m))

= 8.6  102 N. (b) The maximum speed will occur just before his feet leave the ground. At this moment, we know 1 the amount of work that has been done. So we can write W = K + U = mvx2,f + mg xcontact  vx,f = 2

2 2 (W − mg xcontact ) = ((55 kg)(9.8 m/s 2 )(0.8 m) − (55 kg)(9.8 m/s 2 )(0.500 m)) = 2.4 m/s. m (55 kg)

9.24. (a) Call vertically upward the + x direction. The muscles do work on the beetle, giving it kinetic energy (and a small amount of gravitational potential energy), but as the beetle rises to its final height, all energy is converted to gravitational potential energy. Thus we can equate the work done on the beetle to that final gravitational potential c energy: W = Fmb x xcontact = mg xmotion . Here xcontact is the vertical distance the beetle moves while in contact with the ground, and xmotion is the total vertical distance that the beetle rises until it reaches its maximum height. Thus c Fmb x = mg

xmotion (0.300 m) = (4.0  10−6 kg)(9.8 m/s 2 ) = 16 mN. xcontact (7.5  10−4 m)

c G (b)  Fbx = Fmb x + FEbx = mab,x  ab,x =

c Fmb (0.0157 N) − (4.0  10 −6 kg)(9.8 m/s 2 ) x − mg = = 3.9  103 m/s 2 . So the m (4.0  10−6 kg)

beetle’s acceleration is 3.9 103 m/s 2 upward.

9.25. The system is the bicycle, the rider, Earth, the hill, and surrounding air.

9.26. The work done by friction reduces the kinetic energy to zero, mvf2 /2 = 0, and so Fd = mvi2 /2, d = mvi2 /2F. Thus for a constant force magnitude, increasing the initial speed from 10 m/s to 30 m/s increases the kinetic energy and thus the braking distance d by a factor of 9, making the new stopping distance 63 m.

9.27. (a) Yes (b) Gravitational force, tensile force exerted by rope, resistive force exerted by water, normal force exerted by bay floor. (c) We know that the change in gravitational potential energy is equal to mgx = (4.0 kg)(9.8 m/s2 )(−11.4 m) = −4.5 102 J. There is no change in kinetic energy, until the trap actually strikes the bottom of the bay. This means that the lost gravitational potential energy lost during the descent through air was lost because the tension in the rope did negative work on the system, and the gravitational potential energy lost during the descent through water was completely converted to thermal energy of the water and the trap. Thus

9-6

Chapter 9

G Wrope on trap = U air = mg xair = (4.0 kg)(9.8 m/s 2 )( −1.4 m) = 55 J,

and

G Eth,trap/water = −U water = − mg xwater

= −(4.0 kg)(9.8 m/s2 )(−10 m) = 3.9  102 J. The energy diagram should approximately reflect these values.

9.28. (a) Yes (b) Gravitational force, tensile force exerted by rope, resistive force exerted by water. (c) Clearly the total change in gravitational potential energy is mgx = (5.1 kg)(9.8 m/s 2 )(11.4 m) = 5.7 102 J. We might reasonably assume that the change in thermal energy of the trap and water will be similar to the change in thermal energy as the trap was lowered (from problem 9.27). In reality, resistive forces such as these would depend on the speed of the trap; but to a fair approximation we can say Eth,trap/water = 3.9  102 J. We know all this energy came from work done by the tension in the rope, so Wrope on trap = U G + Eth,trap/water = (5.7  102 J) + (3.9  10 2 J) = 9.6  10 2 J. The energy diagram should approximately reflect these values.

9.29. Let system contain the child only. No external forces act, so 𝛥(𝐾 + 𝑈) = 0. 1

(a) 𝐾1 + 𝑈1 = 𝐾2 + 𝑈2 → 0.0 + 25.0 × 9.8 × 2.0 = 2 × 25.0 × 𝑣 2 + 25.0 × 9.8 × 0.9 m . s (b) Same rule leads to 𝑣 = √2 × 9.8 × (2.0) = 6.26 m/s. → 𝑣 = √2 × 9.8 × (2.0 − 0.9) = 4.64

9.30. (a) The forces exerted on the block are the downward gravitational force and three upward tensile forces exerted by the ropes. Let the positive y direction be vertically upward. Because the block is not accelerating,  Fy = −mg + 3T = 0 and T = mg/3. (b) They have done the same amount of work, because the change in the block’s energy is the same in both cases. In terms of forces, call the magnitude of the force exerted by the worker using the straight rope Fs and call the distance over which that force is exerted d. The magnitude of the force exerted by the worker using the block-and-tackle is only Fs/3, but because of the way the three ropes are strung, that force must be exerted over a distance 3d.

9.31. Call vertically upward the + y direction throughout the problem. (a) The pulling rope makes contact with the lower pulley in two places. We momentarily treat the lower pulley and the hanging block as a single system. The sum of mah,y + mg all forces on that system in the y direction is  Fhy = 2Tp + FEhG y = 2Tp − mg = mah,y , such that Tp = = 2 (50 kg)((2.5 m/s 2 ) + (9.8 m/s 2 )) = 3.1102 N. (b) Now we look at the hanging block only, and write the sum of all 2 forces in the y direction:  Fby = Th + FEhG y = Th − mg = mab,y such that Th = mab,y + mg = (50 kg)((2.5 m/s 2 ) +

(9.8 m/s2 )) = 6.2 102 N. (c) The pulling rope makes contact with the top pulley in three places. So the sum of all forces in the y direction on the top pulley is

F

top y

c c = 3Tp + Fmount top y = 0. So Fmount top y = 3Tp = 3(308 N) =

Work

9-7

9.2 102 N. So the force exerted by the top mount on the top pulley is 9.2 102 N upward. (d) Note that when the person pulls the rope down by an amount , the block only rises a distance /2. So, the work done by the person is

equal to W = Fpcrope y ypulling rope = Tp (2yhanging block ) = (308 N)(2)(0.25 m) = 1.5  102 J. (e) The work done by the person in part (d) increases both the gravitational potential energy and the kinetic energy of the box. Thus W = K + U or K = W − mgy = (154 J) − (50 kg)(9.8 m/s 2 )(0.25 m) = 31 J. (f) This is the same as the work done by the person: 1.5 102 J. (g) The contact point between the ceiling and the system never moves. So the ceiling does no work. (h) WE on block = FEbG y y = − mg y = −(50 kg)(9.8 m/s 2 )(0.25 m) = −1.2  10 2 J. (i)

9.32. (a) Wby hand on gelatin = Fhgc x x = (3.0 N)(0.050 m) = 0.15 J (b) Kcm = Fhccm x xcm = (3.0 N)(0.030 m) = 0.090 J 9.33. (a) W = FUc car x ,av xcm = K cm so FUc car x,av = −

1 1 2 mvcm (1000 kg)(5.0 m/s) 2 = −2.5  104 N. x ,i = − 2xcm 2(0.50 m)

So the magnitude of the average force exerted on the car is 2.5  104 N. (b) The work done on the car’s center of mass is W = FUc car x ,av xcm = (−2.5  104 N)(0.50 m) = −1.3  10 4 J. (c) The change in kinetic energy is equal to the work done on the car, in this case. So K = −1.3  104 J.

1 1 1 mvx2,f = (60 kg)(3.0 m/s) 2 = 5.4  102 N. 2 2xcm 2(0.50 m) (b) The wall itself does no work on the skater, because the contact point between the wall and the skater does not move. The skater’s muscles do work on his center of mass by pushing against the wall. This quantity of work is given by W = Fwsc x ,av xcm = (5.4  102 N)(0.50 m) = 2.7  102 J. (c) Here the change in kinetic energy is exactly equal to

9.34. (a) W = Fwsc x ,av xcm = Kcm = mvx2,f so Fwsc x,av =

the work done in part (b). Hence K = 2.7  102 J.

9.35. (a) W = Fhcred x xred = (2.0 N)(0.15 m) = 0.30 J. (b) xcm =

(mred xred + mgreen xgreen ) mred + mgreen

(0.50 kg)(0.15 m) = (1.0 kg)

0.075 m or 75 mm. (c) The increase in kinetic energy is equal to the work done by you: 0.15 J.

9.36. (a) The work done on the system is just the sum of the work done on each cart separately: W = Won 1 + Won 2 = Fh1c x x1 + Fh2c x x2 = (3.0 N)(1.0 m) + (−3.0 N)(−1.0 m) = 6.0 J. (b) Here, all the work done on the system went into increasing its kinetic energy, so K = 6.0 J. (c) Both carts are moving at the same speed, but in opposite directions. So the kinetic energy of the center of mass of the system is zero. Another way of looking at this is that the vector sum of forces acting on the system of two carts as a whole was zero.

9.37. They require the same amount of work. Both procedures end up placing the same amount of inertia at the same heights, meaning each process corresponds to the same change in energy.

9-8

Chapter 9

9.38. (a) W = Fhc1.0 kg cart x x1.0 kg cart = (2.0 N)(0.15 m) = 0.30 J. (b) xcm =

(m1.0 kg cart x1.0 kg cart + m0.5 kg cart x0.5 kg cart ) m1.0 kg cart + m0.5 kg cart

(1.0 kg)(0.15 m) = 0.10 m. (c) The change in kinetic energy of the center of mass of the system is (1.5 kg)

K = Fhccm x xcm = (2.0 N)(0.10 m) = 0.20 J.

9.39. (a) W = Fhgc x xg = (2.0 N)(1.0 m) = 2.0 J. (b) Here all work went to increasing the kinetic energy of the system, so K = 2.0 J. (c) The center of mass of the system only moved

xcm =

(mg xg + mt xt ) mg + mt

(1.0 kg)(1.0 m) = 0.50 m during the application of the force. Thus Kcm = Fhccm x xcm = (2.0 N)(0.50 m) = 1.0 J. (2.0 kg) (d) This is exactly the calculation done to answer part (c). This quantity is 1.0 J. (e) When the two blocks collide and stick together, they move as one object with a velocity equal to the velocity of the center of mass. Thus, the energy of the system is then the kinetic energy of the center of mass, which is 1.0 J. This can be confirmed through conservation of momentum.

9.40. (a) Call right the + x direction. The work done on the system is W = FhLc x xL = (5.0 N)(0.40 m) = 2.0 J. (b) Since the left cart moves 0.40 m to the right, and the spring is compressed 0.20 m, the cart on the right moved only (m x + mR xR ) (0.50 kg)(0.40 m) + (0.50 kg)(0.20 m) 0.20 m. So xcm = L L = = 0.30 m. (c) K = FhLc x xcm = mL + mR (1.0 kg) (5.0 N)(0.30 m) = 1.5 J.

9.41. As you throw a snowball, you might be able to exert your maximum 100 N of force over a distance of 1.0 m. 2Fhsc x xs 2(100 N)(1.0 m) = = 14.1 m/s. ms (1.0 kg) Suppose we throw N such snowballs at that speed, and we include the snowballs and the cooler in our system. The Nms vs,x + mcooler vcooler, x x component of the velocity of the center of mass is vcm,x = . Once the snowballs collide Nms + mcooler inelastically with the cooler, this will also be the x component of the velocity of the cooler. Therefore we require Nmsvs,x + mcooler vcooler,x m ((v ) − vcooler, x )  (vcooler ) min = 3.0 m/s. Solving for N  cooler cooler min yields N Nms + mcooler ms (vs,x − (vcooler )min ) This would result in each snowball having a speed given by vs,x =

(10 kg)((3.0 m/s) − 0) = 2.7. Thus 3 snowballs will be sufficient. (1.0 kg)((14.1 m/s) − (3.0 m/s))

9.42. (a) We first look at conservation of momentum to determine the initial speed of the block. Call the direction in which the bullet is fired the + x direction. Then mv = mv/3 + 4mvblock x,f  vblock x,f =

1 v. So the initial kinetic energy 6

1 2 2 4mvblock x ,f = mv /18. When the block comes to rest, it means that sufficient work has been done to 2 f f 2 reduce this energy to zero. In that case W = Fsurface block x d = − Ki , or Fsurface block x = −mv /(18d ). So the magnitude of the

is K block,i =

frictional force between the block and the surface is mv2 /(18d ). (b) We first find the difference between the energies 2

before and after the collision. K collision = K bullet,f + K block,f − Kbullet,i =

1 v 1 7 v 1 m + (4m)   − mv 2 = − mv 2 . And 2  3  2 18 6 2

the change in kinetic energy as the block slides across the surface is −

1 mv 2 . So K collision / K friction = 7. 18

Work

9-9

9.43. We first look at conservation of momentum to determine the initial speed of the block after the bullet has

imbedded itself. Call the direction in which the bullet is fired the + x direction. Then mv = 1 1 2 2 Some (m + 4m)vblock x ,f  vblock x,f = v. So the initial kinetic energy is K block,i = 5mvblock x ,f = mv /10. 2 5 of this kinetic energy will turn into spring potential energy, but the rest is removed from the system through work 1 f done by friction between the block and the surface. Thus W = K block,i − U fspring or Fsurface ( K block,i − U fspring ) = block x = − d 1 1 1  −  mv 2 − kd 2  . So the magnitude of the frictional force between the block and the surface is d  10 2 

 1  mv 2 − kd 2  .  2d  5 

9.44. (a) A given force F can compress a spring x = F/k. The energy stored in a spring is

1 k x 2 , which we can 2

1 F2 . So if each spring is compressed using the same force, the spring with the smaller spring constant 2 k will store the most energy. A is less stiff, and therefore will store more potential energy. (b) Since the spring potential 1 energy can be written as k x 2 , it is clear that if both springs have the same compression, the spring with the 2 greater spring constant will store the most energy. Since spring B is stiffer, spring B would store more energy in this case. also write as

9.45. They require the same amount of work. The magnitude of a spring’s restoring force does not depend on the direction of the displacement; it is the same whether you stretch or compress the spring.

9.46. By counting the number of squares ( each square is equivalent to 0.5  0.5 = 0.25 Nm) we get 16 squares 4.0

under the curve between 1 m and 3 m: 𝑊 = 16 × 0.25 = 4 𝐽 = 𝐹𝛥𝑥 → 𝐹 = 2.0 = 2 Newtons.

9.47. Let the relaxed length be L. 1 𝐿 2 24 𝐽 = 𝑘 ( ) … … … . . (1) 2 2 1 3𝐿 2 𝐸 𝐽 = 𝑘 ( ) … … … … … (2) 2 4

Divide (2) by (1) to get: 𝐸 9 9 = ∙ 4 = → 𝐸 = 2.25 × 24 𝐽 = 54 𝐽. 24 16 4

9.48. No. In order to integrate the force over some distance, you must be able to write the time interval needed for the particle to travel that distance (and therefore the force) as a function of distance. You cannot do this because you do not know the particle’s initial velocity. Nor do you know the particle’s acceleration because you do not know its inertia.

9.49. (a) From the information given, we can determine the spring constant inside the dart gun: k = F/ x =

(6.0 N) = 50 N/m. So as the dart is fired, spring potential energy is converted to kinetic energy of the (0.12 m)

dart, such that we can write

1 1 k xi2 = mvx2,f  vx ,f = 2 2

k xi = m

(50 N/m) (0.12 m) = 5.5 m/s. (b) Yes, if the dart (0.024 kg)

is fired vertically, some of the spring potential energy is converted to gravitational potential energy as the dart moves upward. Hence, not all the initial energy is converted to kinetic energy, making the vertical launch speed slightly slower than the horizontal launch speed. © Copyright 2015 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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Chapter 9

9.50. Since (in one dimension) W =  F ( x)dx, we can also write F ( x) =

dW ( x) d = (ax + bx3 ) = a + 3bx 2 . So dx dx

F (x) = a + 3b(x)2 .

9.51. (a) We are always free to measure our gravitational potential energy relative to any height we like. As a result, answers involving gravitational potential energies may vary. Here, we choose to measure relative to the uncompressed bed. At each step we determine the gravitational potential energy using U g (x) = mg x, where x measures the change in vertical position relative to the height of the uncompressed mattress. The spring potential 1 energy is determined using U sp (x) = k x 2 . Finally, because energy is conserved, we can determine the kinetic 2 energy at each position by requiring Usp (x) + Ug (x) + K (x) = Usp (x = 0) + U g (x = 0) + K (x = 0) = K (x = 0) or equivalently K (x) = (0.59 J) − Usp (x) − Ug (x), where we have determined the kinetic energy of the bowling ball after falling the initial 10 mm to reach the mattress. Thus we can determine all the values required, as shown in the following table. x K (J) UG (J) U (J)

0 0.59 0 0

0.050 m 2.9 –2.9 0.63

0.10 m 4.0 –5.9 2.5

0.15 m 3.8 –8.8 5.6

0.20 m 2.3 –12 10

(b) The bowling ball will reach its maximum compression when it stops, and therefore momentarily has zero kinetic energy. We use K (xmax ) = (0.59 J) − Usp (xmax ) − Ug (xmax ) = 0 to find xmax . This yields a quadratic equation

1 k x 2 + mg xmax − (0.59 J) = 0, 2

which we solve to find x =

− mg  2kKi + m2 g 2 k

−(6.0 kg)(9.8 m/s 2 )  2(500 N/m)(0.59 J) + (6.0 kg)2 (9.8 m/s 2 )2 . The two solutions are x = −0.24 m and (500 N/m) x = 0.096 m. The second (positive) solution is non-physical because it refers to a positive height above the mattress. At those positions, there was no compression of the mattress and so our equations are not applicable there. The physical solution is x = −0.24 m, or a compression magnitude of 0.24 m.

9.52. (a) Call the distance through the haystack d , the initial speed of the ball v, and the frictional force between the hay and the ball F . The work done by the friction will be equal to the change in kinetic energy of the ball, such 2

that we can write W = − Fd = K =

1 v 1 2 m − mv . Solving for the speed yields v = 8dF/3m . (b) We use 2  2  2

v = 8dF/3m = 8(1.2 m)(6.0 N)/3(0.50 kg) = 6.2 m/s.

9.53. Work done by the external agent moving the proton is equal to minus the work done by the force from the first proton: 9𝑅

9𝑅

𝑊 = −∫

𝐹(𝑟)𝑑𝑟 = − ∫

𝑅

9𝑅 1 𝑒2 𝑒2 𝑒2 1 1 8𝑒 2 ∫ 𝑟 −2 𝑑𝑟 = + 𝑑𝑟 = − [ − ]= − 2 4𝜋𝜖0 𝑟 4𝜋𝜖0 𝑅 4𝜋𝜖0 9𝑅 𝑅 36𝜋𝜖0 𝑅

9.54. Accelerating from rest to a final speed of 20 m/s corresponds to an increase in kinetic energy of 1 2 1 mvf = (500 kg)(20 m/s) 2 = 1.0  105 J. We must be able to do this 50 times, meaning we must have a total 2 2 energy of 5.0 106 J at our disposal. This energy must initially be stored in spring potential energy. Thus we require 2

U isp =

8(5.0  106 J) 1   k   = 5.0  106 J or k = = 2.3  106 N/m. (4.2 m) 2 2 2

Work

9-11

9.55. (a)

(b) For this part, let us consider a system that includes the springs and the cart. If the cart is pushed to the left, the 1 2 1 2 initial kinetic energy will turn into spring potential energy in the left spring. Thus mvi = kl xmax  2 2

xmax =

m (0.15 kg) vi = (4.0 m/s) = 0.77 m. If the cart is initially pushed to the right, then the same principles kl (4.0 N/m)

apply, but it will be the right spring that is being compressed. Thus we have

xmax =

1 2 1 2 mvi = kr xmax  2 2

m (0.15 kg) vi = (4.0 m/s) = 0.63 m. (c) Here, we consider only the system of the cart, such that the kr (6.0 N/m)

springs exert external forces that do work. In either case, the work done by the cart is exactly enough to reduce the 1 1 cart’s kinetic energy to zero. Thus W = K = − mvi2 = − (0.15 kg)(4.0 m/s) 2 = −1.2 J. 2 2

9.56. Call the direction in which the ball is thrown the + x direction. The initial x component of the ball’s velocity is vb x ,i and its inertia is m. Then the conservation of momentum in the x direction reads mvb x,i = 4mvx,f , where

vx,f is the final x component of the velocity of the ball and mitt together. So the kinetic energy of the ball and mitt 2

v  1 1 1 immediately after the collision can be written K (x = 0) = (4m)vx2,f = (4m)  x ,i  = mvx2,i . When the ball and 2 2  4  8 mitt stop, all of this kinetic energy has been converted into spring potential energy. Thus K (x = 0) = U sp (xmax ) or k 1 2 1 2 . mvx ,i = k xmax . Solving for the initial speed yields vi = 2 xmax m 8 2

E Fajx x = = Fajdxvx = (25 N)(5.0 m/s) t t d

9.57. Call the direction of motion of the jogger the + x direction. Then P = = 1.3  102 W.

9.58. 𝑃=

𝛥𝑊 𝑚𝑔ℎ 2 × 9.8 × 0.5 = = = 1960 Watt = 1.96 kW. 𝛥𝑡 𝛥𝑡 5 × 10−3

9-12

Chapter 9

9.59. Because the zigzag path is spread out over a greater distance, the climber expends the same quantity of energy over a greater time interval, for a lower average power rating.

9.60. Work done by engine = 𝑃 × 𝛥𝑡 = 𝑃 × 20.0 × 60.0 = 1200 𝑃 𝐾=

1 × 1000 × 72 = 2.45 × 104 𝐽. 2

If the car just makes it to the top: 0.3 × 1200 × P + 0.3 × 2.45 × 104 𝐽 = 𝑚𝑔ℎ = 1000 × 9.8 × 20 Engine Power should be: 𝑃 = 0.875 Watts.

9.61. 1𝐹 2 𝑡 → 𝑑𝑊 = 𝐹𝑑𝑥 = 𝐹 𝑎𝑡 𝑑𝑡 2𝑚 𝑡 1 𝑊(𝑡) = ∫ 𝐹 𝑎𝑡 𝑑𝑡 = 𝐹𝑎 𝑡 2 2 0

𝐹 = 𝑚𝑎 → 𝑥(𝑡) =

Call the direction of motion the + x direction. Then, ignoring friction we have  (5.0 m/s) − (0)   v − vx ,i  2 Fdsc x = max = m  x ,f  = 3.3 10 N. (b) The work done by the dogs on the sled is  = (200 kg)   t (3.0 s)     1 1 the same as the change in the sled’s kinetic energy in this case. Thus W = Ks = mvf2 = (200 kg)(5.0 m/s) 2 = 2 2  E F  x = x = Fxvx = (333 N)(5.0 m/s) = 1.7  103 W. 2.5 103 J. (c) The instantaneous power output is given by P = t t  vx ,f − vx ,i   (5.0 m/s) − (0)  2 (d) If the acceleration is constant, it is given by ax =   = 1.67 m/s . Then the speed = (3.0 s)  t   

9.62. (a)

after only 1.5 s is vx ,f = vx ,i + ax t = (0) + (1.67 m/s 2 )(1.5 s) = 2.5 m/s. Now we can calculate the instantaneous power exactly as before: P =

E Fx x = = Fxvx = (333 N)(2.5 m/s) = 8.3 10 2 W. t t

9.63. (a) We cannot simply use the force and distance to calculate work because we do not immediately know the distance the child moves during this time interval. Instead we find the final speed of the child by using impulse: F t (5.0 N)(1.0 s) + 0 = 0.25 m/s. Now, we know the child’s final px = mvx,f − mvx,i = Fx t so vx ,f = x + vx,i = m (20 kg) kinetic energy came from the work done by the water on the child, so W = (b) The instantaneous power is given by P =

1 2 1 mvx ,f = (20 kg)(0.25 m/s) 2 = 0.63 J. 2 2

E Fx x = = Fx vx = (5.0 N)(0.25 m/s) = 1.3 W. t t

9.64. (a) The work done by the cog system equals the change in energy of the system of 25 cars. So 1 1  W = K + U G = (25) mvf2 + (25)mg y = (25)(579 kg)  (0.50 m/s 2 ) 2 + (9.8 m/s 2 )(100 m)  = 1.4  107 J. 2 2 

(b) P =

E (1.42  107 J) = = 2.4  105 W. (c) The only change in the energy requirement is that now the cars will t (60 s)

have to move faster at the beginning. But as long as there is no dissipation, this higher kinetic energy can still be converted to gravitational potential energy as the cars rise. Thus, as long as the final speed does not change, there is no change in the total energy provided by the cog system. It remains 1.4  107 J.

Work

9.65. (a) Let the positive x axis point to the right. The power from the child on the left is P =

9-13

FLbc x x = FLbc xvx = t

c FRb c x x = FRb x vx = (+2.0 N)(−3.0 m/s) = t −6.0 W. (b) The power being delivered to the block by both forces is PL + PR = (9.0 W) + ( −6.0 W) = 3.0 W. (c)

(−3.0 N)(−3.0 m/s) = 9.0 W. The power from the child on the right is P =

Yes. The box is accelerating because the vector sum of the forces exerted on it is nonzero. Because power is equal to the product of force and speed, if the box moves at a higher speed, then the power from the same forces will increase in magnitude.

9.66. (a) Choose a system that includes the elevator and Earth. First we note that the power is not constant. As the speed increases, the power needed to accelerate the elevator at a constant rate will increase. Also, as the speed increase, the gravitational potential energy will increase at an increasing rate. We will find the average power, and the peak power. The average power can be found by simply calculating the change in energy of the elevator and dividing by time. For this we need to know the height of the elevator as it reaches its cruising speed. We can use 1 1 kinematics to find y = (v y ,f + v y ,i )t = ((1.5 m/s) + (0))(2.0 s) = 1.5 m. To find the average power we write 2 2 1 2 1 2 2 E K + U G 2 mvcruising + mg y 2 (1400 kg)(1.5 m/s) + (1400 kg)(9.8 m/s )(1.5 m) Pav = = = = = 1.1 104 W. t t t (2.0 s) The maximum power comes just as the elevator reaches its cruising speed. We can calculate the instantaneous power by taking the time derivative of the energy at that moment t f . For this we will need the acceleration, which is given by a y =

2y 2(1.5 m) = = 0.75 m/s 2 . t 2 (2.0 s) 2 Pinst (tf ) =

dE (t ) d 1 2 dv dy  = mvcruising + mg y  = mvcruising + mg = mvcruising a y + mgvcruising dt tf dt  2 dt dt tf t f

= (1400 kg)(1.5 m/s)((0.75 m/s ) + (9.8 m/s 2 )) = 2.2  104 W 2

(b) Now the power goes entirely to increasing the gravitational potential energy of the elevator. Since the speed is U G mg y constant, so is the power: Pav = = = (1400 kg)(9.8 m/s 2 )(1.5 m/s) = 2.1 104 W 15 kW. t t

9.67. 𝑊𝑔 = (𝑚𝑔)𝛥𝑦 cos(0) = 10 × 9.81 × 10 × 1 = +981 𝐽. Since the object is being lowered at constant speed, the tension is balancing the weight! 𝑊𝑇 = (10 × 9.81) × 10 × (−1) = −981 𝐽.

9.68. Yes, if other forces are exerted on the object. For example, if a sled is being dragged across the snow at a constant speed by a rope, the direction of the tensile force exerted by the rope is the same as the sled’s direction of motion. There is no change in kinetic energy, however, because the amount of (negative) work done by friction is the same as the (positive) amount done by the rope.

9.69. (a) Call the direction of motion of the cart the + x direction. The center of mass moves 0.10 m as the force is applied to the grocery-cart system. We know the work changes the kinetic energy, so we can write mv 2 (50 kg)(2.0 m/s) 2 1 W = Fpcc x ,av x = K = − mvc2 x,i . Solving for the force yields Fpcc x ,av = − c x ,i = − = −1.0  103 N. 2 2 x 2(0.10 m) So the average force is 1.0 103 N toward the rear of the cart. (b) Zero, because the point of application does not move. (c) K = Fpcc x ,av x = (−1.0  103 N)(0.10 m) = −1.0  10 2 J. ⃗⃗⃗⃗⃗ = 𝛥𝐾 = 0 holds but only because the force is perpendicular to the displacement 9.70. In this case, using 𝑊 = 𝐹⃗ ∙ 𝛥𝑥 ⃗⃗⃗⃗⃗ = 𝐹 𝛥𝑥 cos(90) = 0, even though neither 𝐹 nor 𝛥𝑥 are zero. which implies 𝐹⃗ ∙ 𝛥𝑥

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Chapter 9

9.71. (a) Call vertically upward the + y direction. The work done by the man’s legs on his center of mass must be sufficient for his center of mass to reach the specified height. So W = Fl ccm y ,av ypushing = mg yto ladder , so

Fl ccm y ,av =

mg yto ladder (70 kg)(9.8 m/s 2 )(1.5 m) = = 1.9  103 N. So the minimum force is 1.9 103 N upward. ypushing (0.55 m)

(b) The fraction of a bandy bar needed is

1.2 10

−3

E jump Ecandy

(1.87  103 N)(0.55 m) = 1.2  10−3. So the jump requires (8.50  105 J)

candy bars.

9.72. Estimates may vary. Here, we will assume the trash collector takes approximately 1 minute to lift a trash can, empty it and replace it, and move on to the next can, on average. Assume the average trash can has inertia of about 3  101 kg. Finally, assume that emptying a trash can requires lifting it to a height of 1 m, and that keeping the can aloft while trash falls out, and replacing the can adds an additional 50% to this energy. Then the total work is  60 min  1 min  1 2 5 W = N (1.5)( mg y ) = (8 hr)    (1.5)(3  10 kg)(9.8 m/s )(1 m) = 2 10 J.  1 hr  1 can  𝑑𝑊

9.73. Power = 𝑑𝑡 =

𝑑(𝐹∙𝑥) 𝑑𝑡

=𝐹∙

𝑑𝑥 𝑑𝑡

= 𝐹. 𝑣

km 1000 m m 750 = 750 = 208.3 hr 3600 sec s In this case, 4 × 30000 × 746 = 𝐹 × (208.3) → 𝐹 ≈ 4.3 × 105 Newtons.

9.74. (a) Gravitational force exerted by Earth and spring restoring force. The gravitational force is always directed downward. The spring force is directed upward until the object passes through its equilibrium position. At that instant, the force direction changes to downward until the object passes through its equilibrium position again. At that instant, the force direction changes to upward again. (b) 0

9.75. (a) With no dissipation, it doesn’t matter which way you roll the ball. (b) With energy dissipation, you want to roll the ball down hill A, so that it travels a shorter distance. The shorter the path, the less energy dissipated, so the ball has a better chance of having sufficient mechanical energy to get over hill B.

9.76. The dog can exert a maximum force of 2.5mg and can exert this force over a maximum of 0.20 m. From this we can find the maximum work that his legs can do on his center of mass, and equate this the dog’s maximum change in gravitational potential energy. Wmax = Fl ccm y ,max ymax = mg y jump,max , so y jump,max =

Fl ccm y ,max ypush,max

(2.5)mg ypush,max

= 2.5(0.20 m) = 0.50 m. Since the dog started at 0.10 m off the ground, this mg mg corresponds to a maximum height of 0.60 m.

9.77. We equate the work done by your legs to the required change in your gravitational potential energy: W = Fl ccm y ,av ypush = mg y jump . In terms of the center of mass height at the moment your jump begins, ymin , we have

(2.3)mg ( ypush,max − ymin ) = mg ( y jump,max − ymin ) or ymin =

y jump,max − (2.3) ypush,max (−1.3)

(2.0 m) − (2.3)(1.0 m) = 0.23 m. ( −1.3)

Your center of mass would have to be 0.23 m above the floor. This is not practical.

9.78. If we assume that the elevator is designed to move upward at that speed, then the only factor that is increasing the power output of the motor is the increased inertia. In that case, increasing the power output by 10% would correspond to increasing the inertia being moved by 10%. Here, we are 200 kg over the limit, where the limit is 3200 kg. This is only 6.3%. So, no, this should not be over the 10% tolerance.

Work

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9.79. In this problem we will ignore the inertia of the forearm itself, and focus on forces required to lift the cup, specifically. If we consider the system of the coffee cup and Earth, then we know it takes a certain amount of work to lift the cup a distance ycup , which is given by W = mgycup . This work is done on the cup by the hand beneath it, so Fhcc y ycup = mg ycup . As this motion occurs, you bend your elbow. The contact point where your bicep attaches to your forearm moves a much smaller distance upward as the cup is lifted. Specifically, the vertical distance that the d bicep contact point moves (ybicep ) is related to ycup through simple geometry, and one finds ybicep = ycup . But the force exerted by the bicep on the forearm is ultimately the source of the work being done on the cup; forearm mg ycup mg simply acts as a lever. So Fbfc y ybicep = Fhcc y ycup = mg ycup and we find Fbfc y = = . Coffee cups vary ybicep d greatly in inertia. For a typical inertia around 0.30 kg, we find Fbfc y =

(0.3 kg)(9.8 m/s 2 )(0.350 m) = 2  101 N. (0.050 m)

Answers in the range 20 N  10 N are reasonable.

9.80. Consider a system of you, the bungee cord, and Earth. During your fall, you will first convert gravitational potential energy into kinetic energy alone, for the distance of the unstretched cord, which we will call

. Then, you

will convert gravitational potential energy and kinetic energy into spring potential energy, as you fall the remaining distance. If we call the height of the bridge h, then this second distance must be equal to h − 0 . Then we have U iG = mgh = U fsp =

1 k (h − 0 ) 2 , so 2

=h−

and 80 kg. Using an inertia of 70 kg, we find use an inertia of 80 kg, we find

2mgh . Let us assume that your inertia is somewhere between 70 kg k 0

= (150 m) −

2(70 kg)(9.8 m/s 2 )(150 m) = 78 m. Similarly, if we (40 N/m)

= 73 m. Any answer between these two values is reasonable.