Download Problem Set 7 - Cabrillo College

Problem Set 7
Chapter 10: Energy & Work
Questions: 2, 14, 15, 19, 21
Exercises & Problems: 5, 10, 21, 28, 45, 49, 58, 60, 68, 74
**Midterm Exam Alert, Thursday, October 27th**
Q10.2: When you pound a nail with a hammer, the nail gets quite warm. Describe the
energy transformations that lead to the addition of thermal energy in the nail.
Q10.2. Reason: When you hit a nail with a hammer to pound it into some object, many processes are at work.
For example, some small amount of energy goes into temporarily increasing the nail’s kinetic energy as it moves
into the object. Part of the energy goes into permanently deforming the object to accept the nail. An appreciable
portion of the initial kinetic energy of the hammer is converted to thermal energy through, for example, friction
between the nail and the object as the nail moves into the object. Some gets directly converted to kinetic energy
of the molecules that make up the nail (see section 11.4 for an atomic view of thermal energy and temperature)
from the collision between the nail and the hammer.
Assess: If you ever try hammering nails, the thermal energy generated is appreciable. Note that energy can be
transformed directly into kinetic energy of atoms or molecules that make up an object. As a simpler example,
banging a hammer on a solid object directly will increase the temperature of the both the hammer and the object.
Q10.14: The meaning of the word "work" is quite different in physics from its everyday
usage. Give an example of an action a person could do that "feels like work" but that
does not involve any work as we've defined it in this chapter.
Q10.14. Reason: If you hold a heavy object over your head, you do no work on it because its displacement is
zero. Or, if you walk while carrying a heavy suitcase you do no work because the force you apply to the handle,
directed upward, is perpendicular to the forward-directed displacement.
Assess: Although you do no work in the physics sense of the word, your muscles quickly tire because they must
consume chemical energy to remain contracted under a heavy load. It is in this sense you feel that you are
“working hard” to hold the weight or carry the suitcase.
Q10.15: To change a tire, you need to use a jack to raise one corner of your car. While
doing so, you happen to notice that pushing the jack handle down 20 cm raises the car
only 0.2 cm. Use energy concepts to explain why the handle must be moved so far to
raise the car by such a small amount.
Q10.15. Reason: Neglecting frictional losses, the work you do on the jack is converted into gravitational
potential energy of the car as it is raised. The work you do is Fd , where F is the force you apply to the jack
handle and d is the 20 cm distance you move the handle. This work goes into increasing the potential energy by
an amount mgh  wh, where w is the car’s weight and h  0.2 cm is the change in the car’s height. So Fd  wh
so that F / w  h / d .
Assess: Because the force F you can apply is so much less than the weight w of the car, h must
be much less than d.
Q10.19: Sandy and Chris stand on the edge of a cliff and thrown identical mass rocks at
the same speed. Sandy throws her rock horizontally while Chris throws his upward at an
angle of 45° to the horizontal. Are the rocks moving at the same speed when they hit the
ground, or is one moving faster than the other? If one is moving faster, which one?
Explain.
Q10.19. Reason: Because both rocks are thrown from the same height, they have the same potential energy.
And since they are thrown with the same speed, they have the same kinetic energy. Thus both rocks have the
same total energy. When they reach the ground, they will have this same total energy. Because they’re both at the
same height at ground level, their potential energy there is the same. Thus they must have the same kinetic
energy, and hence the same speed.
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Assess: Although Chris’s rock was thrown angled upward so that it slows as it first rises, it then speeds up as it
begins to fall, attaining the same speed as Sandy’s as it passes the initial height. Sandy’s rock will hit the ground
first, but its speed will be no greater than Chris’s.
Q10.21: You are much more likely to be injured if you fall and your head strikes the
ground than if your head strikes a gymnastics pad. Use energy and work concepts to
explain why this is so.
Q10.21. Reason: As you land, the force of the ground or pad does negative work on your body, transferring
out the kinetic energy you have just before impact. This work is  Fd , where d is the distance over which your
body stops. With the short stopping distance involved upon hitting the ground, the force F will be much greater
than it is with the long stopping distance upon hitting the pad.
Assess: For a given amount of work, the force is large when the displacement is small.
P10.5:
a. At the airport, you ride a "moving sidewalk" that carries you horizontally for 25 m at
0.70 m/s. Assuming that you were moving at 0.70 m/s before stepping onto the
moving sidewalk and continue at 0.70 m/s afterward, how much work does the
moving sidewalk do on you? Your mass is 60 kg.
b. An escalator carries you from one level to the next in the airport terminal. The upper
level is 4.5 m above the lower level, and the length of the escalator is 7.0 m. How
much work does the up escalator do on you when you ride it from the lower level to
the upper level?
c. How much work does the down escalator do on you when you ride it from the upper
level to the lower level?
P10.5. Prepare: We will use the definition of work, Equation 10.6 to calculate the work done. The sidewalk
and escalators exert a normal force on you, and may exert a force to propel you forward. We will assume that the
escalator propels you at constant velocity, as the sidewalk does.
Solve: (a) Since you get on the sidewalk moving at 0.70 m/s, and you continue at 0.70 m/s afterwards, there is
no acceleration and therefore no force on you in the horizontal direction. See the following diagram.
The work is then W  Fd cos  (F )(d )cos(90)  ( F )(d )(0)  0.0 J. The work done by the sidewalk is exactly
zero Joules.
(b) The escalator moves you across some distance like the sidewalk, but it also moves you upwards. See the
following diagram.
The force exerted on you by the escalator is the normal force, which is equal to your weight.
n  w  mg  (60 kg)(9.80 m/s2 )  588 N
which should be reported as 590 N to two significant figures.
Unlike the sidewalk case, there is a component of the displacement parallel to the normal force. The angle
between the force and displacement is cos1 (4.5/7)  50, so d cos( )  4.5 m. Then
W  Fd cos  (588 N)(4.5 m)  2.6 kJ
(c) Refer to the figure.
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Here, the displacement is in the opposite direction compared to part (b), so the angle between the force and the
displacement is now 180  50  130. So
W  Fd cos  (588 N)(7.0 m)cos(130)  2.6 kJ
Assess: In part (a), since the force has no component in the direction of your displacement, the force does no
work. In part (b), there is a component of the force along and in the same direction as the displacement, so the
force does positive work. In part (c), the component of the force along the displacement is in the opposite
direction to the displacement, so the force does negative work.
P10.10: Sam's job at the amusement part is to slow down and bring to a stop the boats
in the log ride. If a boat and its riders have a mass of 1200 kg and the boat drifts in at 1.2
m/s, how much work does Sam do to stop it?
P10.10. Prepare: We will assume that all the work Sam does goes into stopping the boat. We can use
conservation of energy as expressed in Equation 10.7 to calculate the work done from the change in kinetic
energy.
Solve: Refer to the before and after representation of Sam stopping a boat. Equation 10.7 becomes
1
1
mvf 2  mvi 2  W
2
2
Since the boat is at rest at the end of the process, vf  0 m/s. Therefore, the final kinetic energy is zero. The
work done on the boat is then
1
1
W   mvi 2   (1200 kg)(1.2 m/s)2  0.86 kJ
2
2
Assess: Note that the work done by Sam on the boat is negative. This is because the force Sam exerts on the
boat must be opposite to the direction of motion of the boat to slow it down.
P10.21: The elastic energy stored in your tendons can contribute up to 35% of your
energy needs when running. Sports scientists have studied the change in length of the
knee extensor tendon in sprinters and nonathletes. The find (on average) that the
sprinters' tendons stretch 41 mm, while nonathletes' stretch only 33 mm. The spring
constant for the tendon is the same for both groups, 33 N/mm. What is the difference in
maximum stored energy between the sprinters and the nonathletes?
P10.21. Prepare: We will assume the knee extensor tendon behaves according to Hooke’s Law and stretches
in a straight line. The elastic energy stored in a spring is given by equation 10.15, Us  12 kx 2 .
Solve: For athletes,
U s,athlete 
1 2 1
kx  (33,000 N/m)(0.041 m)2  27.7 J
2
2
For non-athletes,
1 2 1
kx  (33,000 N/m)(0.033 m)2  18.0 J
2
2
The difference in energy stored between athletes and non-athletes is therefore 9.7 J.
Assess: Notice the energy stored by athletes is over 1.5 times the energy stored by non-athletes.
U s,non-athlete 
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P10.28: What minimum speed does a 100 g puck need to make it to the top of a
frictionless ramp that is 3.0 m long and inclined at 20°?
P10.28. Prepare: Since the ramp is frictionless, the sum of the puck’s kinetic and gravitational potential
energy does not change during its sliding motion. Use Equation 10.4 for the conservation of energy.
Solve: The quantity K + Ug is the same at the top of the ramp as it was at the bottom. The energy conservation
equation Kf  U gf  Ki  U gi is
1 2
1
mvf  mgyf  mvi2  mgyi  vi2  vf2  2 g ( yf  yi )
2
2
 vi2  (0 m/s) 2  2(9.80 m/s 2 )(1.03 m  0 m)  20.1 m 2/s 2  vi  4.5 m/s
Assess: An initial push with a speed of 4.5 m/s  10 mph to cover a distance of 3.0 m up a 20 ramp seems
reasonable. Note that a ramp of any angle to the same final height would lead to the same final velocity for the
puck. Note that the mass cancels out in the equation since both kinetic energy and gravitational potential energy
are proportional to mass.
P10.45: In the winter sport of curling, players give a 20 kg stone a push across a sheet
of ice. A curler accelerates a stone to a speed of 3.0 m/s over a time of 2.0 s.
a. How much force does the curler exert on the stone?
b. What average power does the curler use to bring the stone up to speed?
P10.45. Prepare: Neglecting friction, the only horizontal force on the stone is the force exerted by the curler.
The work done by this force will be transferred entirely into the stone’s kinetic energy. We can use the
conservation of energy equation to calculate this work. We will assume the ice is frictionless, that the
acceleration of the stone is constant, and that the stone starts from rest.
Solve: (a) Refer to the diagram. We can find the force exerted on the stone by the curler from the acceleration
of the stone. Since vi  0 m/s,
a
vf  vi 3.0 m/s

 1.5 m/s 2
t
2.0 s
The force on the stone is given by
F  ma  (20 kg)(1.5 m/s)  30 N
(b) Since Eth  Echem  U g  U s  0 J, in this case, the law of conservation of energy (10.3) reads
W  K f  Ki 
1
1
mvf 2  mvi 2
2
2
Since vi  0 m/s
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W
1
1
mvf 2    (20 kg)(3.0 m/s) 2  90 J
2
2
The average power is given by Equation 10.22,
P
W 90 J

 45 W
t 2.0 s
Assess: Note that the amount of muscle power needed to quickly accelerate this relatively heavy
stone would not even fully light a 60 W light bulb.
P10.49: A 55 kg skateboarder wants to just make it to the
upper edge of a "half-pipe" with a radius of 3.0 m, as shown
in the figure. What speed vi does he need at the bottom if he
is to coast all the way up?
a. First do the calculation treating the skateboarder and
board as a point particle, with the entire mass nearly in
contact with the half-pipe.
b. More realistically, the mass of the skateboarder in a deep crouch might be thought of
as concentrated 0.75 m from the half-pipe. Assuming he remains in that position all
the way up, what vi is needed to reach the upper edge?
P10.49. Prepare: Assuming that the track offers no rolling friction, the sum of the skateboarder’s kinetic
and gravitational potential energy does not change during his rolling motion.
The vertical displacement of the skateboarder is equal to the radius of the track.
Solve: (a) The quantity K + Ug is the same at the upper edge of the quarter-pipe track as it was at the bottom.
The energy conservation equation Kf  U gf  Ki  U gi is
1 2
1
mvf  mgyf  mvi2  mgyi  vi2  vf2  2 g ( yf  yi )
2
2
vi2  (0 m/s) 2  2(9.80 m/s 2 )(3.0 m  0 m)  58.8 m/s  vi  7.7 m/s
(b) If the skateboarder is in a low crouch, his height above ground at the beginning of the trip changes to 0.75 m.
His height above ground at the top of the pipe remains the same since he is horizontal at that point. Following the
same procedure as for part (a),
1 2
1
mvf  mgyf  mvi2  mgyi  vi2  vf2  2 g ( yf  yi )
2
2
vi2  (0 m/s) 2  2(9.80 m/s 2 )(3.0 m  0.75 m)  44.1 m/s  vi  6.6 m/s
Assess: Note that we did not need to know the skateboarder’s mass, as is the case with free-fall motion. Note
that the shape of the track is irrelevant.
P10.58: The maximum energy a bone can absorb without breaking is surprisingly small.
For a healthy human of mass 60 kg, experimental data show that the leg bones can
absorb about 200 J.
a. From what maximum height could a person jump and land rigidly upright on both feet
without breaking his legs?
b. People jump from much greater heights than this; explain how this is possible.
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Hint: Think about how people land when they jump from greater heights.
P10.58. Prepare: We will take the system to be the person plus the earth. When a person drops from a
certain height, the initial potential energy is transformed to kinetic energy. When the person hits the ground, if
they land rigidly upright, we assume that all of this energy is transformed into elastic potential energy of the
compressed leg bones. The maximum energy that can be absorbed by the leg bones is 200 J; this limits the
maximum height.
Solve: (a) The initial potential energy can be at most 200 J, so the height h of the jump is limited by
mgh  200 J For m  60 kg , this limits the height to
h  200 J mg  200 J 60 kg  9.8 m/s 2   0.34 m
(b) If some of the energy is transformed to other forms than elastic energy in the bones, the initial height can be
greater. If a person flexes her legs on landing, some energy is transformed to thermal energy. This allows for a
greater initial height.
Assess: There are other tissues in the body with elastic properties that will absorb energy as well, so this limit is
quite conservative.
P10.60: The 5.0-m-long rope in the figure hangs vertically from a
tree right at the edge of a ravine. A woman wants to use the rope to
swing to the other side of the ravine. She runs as fast as she can,
grabs the rope, and swings out over the ravine.
a. As she swings, what energy conversion is taking place?.
b. When she's directly over the far edge of the ravine, how much
higher is she than when she started?
c. Given your answers to parts a and b, how fast much she be
running when she grabs the rope in order to swing all the way
across the ravine?
P10.60. Prepare: Since there is no friction, total mechanical energy is conserved.
Solve: (a) As she swings, her height above the cliff increases since the rope doesn’t stretch. Her initial kinetic
energy is being converted to gravitational potential energy during the swing.
(b) Refer to the diagram. When she is directly above the opposite side of the ravine, she has moved 3.0 m
horizontally while the rope is also swinging her upwards. Using the Pythagorean theorem, we can find the
distance between the branch and her new height. l  (5.0 m)2  (3.0 m) 2  4.0 m. Therefore, she is 5.0 m – 4.0
m = 1.0 m above the cliff.
(c) To make it to the other side of the ravine, she must have enough kinetic energy to be converted to the
equivalent gravitational potential energy of her additional 1.0 m of height. The minimum initial velocity she will
need will be when she just makes it to the other side of the ravine with no kinetic energy (all her initial kinetic
energy being converted to potential energy of her height above the cliff). Using the cliff as the reference for
gravitational potential energy, the conservation of energy equation reads
 
K1  U g
1
 
 K2  U g
2
1
mv12  mgy2
2
v1  2 gy2  2(9.80 m/s 2 )(1.0 m)  4.4 m/s
Assess: We calculated for the case where all her initial kinetic energy is converted to gravitational potential
energy at the other side of the ravine. Note she could start with a greater initial kinetic energy, which would also
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get her to the other side of the ravine. In this case, when she’s above the other side of the ravine, she will have
some additional kinetic energy instead of just making it to the other side.
P10.68: A 70 kg human sprinter can accelerate from rest to 10 m/s in 3.0 s. During the
same interval, a 30 kg greyhound can accelerate from rest to 20 m/s. Compute (a) the
change in kinetic energy and (b) the average power output for each.
P10.68. Prepare: We can find the kinetic energies directly from the runner’s masses and final speeds. The
power is the rate at which each runner’s internal chemical energy is converted into kinetic energy, so the average
power is K / t.
1
1
1
1
Solve: (a) We have KS  mSvS2  (70 kg)(10 m/s) 2  3500 J, while KG  mG vG2  (30 kg)(20 m/s) 2
2
2
2
2
 6000 J (here, S stands for sprinter and G for greyhound).
(b) We have PS  KS / t  (3500 J)/(3.0 s)  1200 W, and PG  KG / t  (6000 J)/(3.0 s)  2000 W.
Assess: Although the greyhound has less than half the mass of the human, its final speed is twice as great. And
since kinetic energy depends on the square of the speed, the higher speed of the greyhound means that its kinetic
energy is greater than the human’s. Because 1 hp = 760 W, a human can output about 2 hp and a greyhound
about 3 hp in very short bursts.
P10.74: The human heart has to pump the average adult's 6.0 L of blood through the
body every minute. The heart must do work to overcome frictional forces that resist the
blood flow. The average blood pressure is 1.3 × 104 N/m2.
a. Compute the work done moving the 6.0 L of blood completely through the body,
assuming the blood pressure always takes it average value.
b. What power output must the heart have to do this task once a minute?
Hint: When the heart contracts, it applies force to the blood. Pressure is just
force/area, so we can write work = (pressure)(area)(distance). But (area)(distance) is
just the blood volume passing through the heart.
P10.74. Prepare: The heart provides the pressure to move blood through the body and therefore does work on
the blood. We assume all the work goes into pushing the blood through the body.
Solve: (a) Using the hint, W  PAd  PV  (1.3 104 N/m2 )(6.0 103 m3 )  78 J (in this equation P
represents pressure, not power).
W 78 J

 1.3 W.
(b) P 
t 60 s
Assess: This seems like a reasonable answer, as it is a small fraction of the power required for most human
activities.
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