Download Study Questions/Problems Week 5 Chapters 7 and 8 deal with

Study Questions/Problems Week 5
Chapters 7 and 8 deal with rather abstract concepts that
nonetheless provide extremely important tools for problem
solving. Potential energy, energy conservation, and the
distinction between conservative and non-conservative forces
is a real taste of 19th–century physics. The pace of the course
is picking up.
Chapter 7:
Conceptual Questions 4, 5, 9, 10, 11
Conceptual Exercises 1, 4, 6, 8, 10, 13, 15
Problems 3, 11, 17, 20, 26, 27, 30, 33, 35, 37, 38, 41,
42
Chapter 8:
Conceptual Questions 1, 2, 3, 6, 11, 12, 14
Conceptual Exercises 2, 5, 8, 10, 12, 16
Problems 1, 2, 6, 12, 16, 18, 20, 24, 28, 31, 41
Answers/solutions for even numbered CQs and CEs, and for
all the problems listed above are on the following pages—don’t
peek until you have done your best to solve a problem.
Chapter 7: Work and Kinetic Energy
Answers to Even-Numbered Conceptual Questions
2.
False. Any force acting on an object can do work. The work done by different forces may
add to produce a greater net work, or they may cancel to some extent. It follows that the net
work done on an object can be thought of in the following two equivalent ways: (i) The sum
of the works done by each individual force; or (ii) the work done by the net force.
4.
If the net work done on an object is zero, it follows that its change in kinetic energy is also
zero. Therefore, its speed remains the same.
6.
Frictional forces do negative work whenever they act in a direction that opposes the motion.
For example, friction does negative work when you push a box across the floor, or when
you stop your car.
8.
The fact that the ski boat’s velocity is constant means that its kinetic energy is also constant.
Therefore, the net work done on the boat is zero. It follows that the net force acting on the
boat does no work. (In fact, the net force acting on the boat is zero, since its velocity is
constant.)
10.
No. What we can conclude, however, is that the net force acting on the object is zero.
12.
No. Power depends both on the amount of work done by the engine, and the amount of
time during which the work is performed. For example, if engine 2 does its work in less
than half the time of engine 1, it can produce more power even if the amount of work the
engines do is the same.
Answers to Even-Numbered Conceptual Exercises
2.
(a) Gravity does positive work on the bob as it moves from point A to point B. (b) The
tension in the string does zero work on the bob, because it always acts at right angles to the
motion of the bob.
4.
(a) The pitcher does positive work on the ball by exerting a force in the same direction as
the motion of the ball. This results in an increase of kinetic energy, and an increase in speed
from 0 to 90 mph. (b) The catcher does negative work on the ball. This is because the force
exerted by the catcher is opposite in direction to the motion of the ball. Since the work done
on the ball is negative, its speed decreases.
6.
(a) A car with a speed of v0/2 has a kinetic energy that is 1/4 the kinetic energy it has when
its speed is v0. Therefore, the work required to accelerate this car from rest to v0/2 is W0 /4.
(b) Suppose the car has a kinetic energy K0 when its speed is v0. As we saw in part (a), its
kinetic energy when the speed is v0/2 is K0/4. It follows that the increase in kinetic energy
in going from a speed of v0/2 to a speed of v0 is 3K0/4. As a result, the work required for
this increase in speed is 3W0/4.
8.
The appropriate ranking is C < A = D < B. To obtain this ranking, we use the fact that
kinetic energy is proportional to mass to the first power, and proportional to speed to the
second power. Thus, for example, jogger B has (1/2)(3)2 = 9/2 times the kinetic energy of
jogger A.
10.
x produces an increase in kinetic energy
from 0 to K = 12 mv2, where m is the mass of the car. That is, Fx = K. To increase the speed
We are given that the force F times the distance
from v to 2v implies a change in kinetic energy from K to K' =
1
2
m(2v)2 = 4( 12 mv2) = 4K, for
an increase of 3K. It follows, then, that the required distance x' satisfies the relation Fx' =
3K = 3(Fx), and hence that x' = 3x.
12.
The work done in compressing a spring a distance
x is
1
2
k(x)2, where k is the force
constant of the spring. If the force constant is increased by a factor of four, the compression
distance must be halved to produce the same amount of work; that is,
1
2
1
2
(4k)(x/2)2 =
k(x)2. Thus, the compression of the spring is x/2.
14.
(a) The work required to stretch a spring depends on the square of the amount of
stretch. Therefore, to stretch a spring by the amount x requires only 1/4 the work required
to stretch it by the amount 2x. In this case, the work required is W0/4. (b) To stretch this
spring by 3 cm from equilibrium requires 32 = 9 times the work to stretch it 1 cm.
Therefore, stretching to 3 cm requires the work 9W0/4. Subtracting W0, the work
required to stretch to 2 cm, we find that an additional work of 5W0/4 is required to stretch
from 2 cm to 3 cm. Thus, it takes 5 times as much work to produce a 1-cm stretch from 2
cm to 3 cm as it does to produce a 1 cm stretch from 0 to 1 cm.
See following pages for Problem solutions
Solutions to Ch. 7 Problems
3. Picture the Problem: The pumpkin is lifted vertically then carried
horizontally.
Strategy: Multiply the force by the distance because during the lift the two
point along the same direction.
Solution: 1. (a) Apply equation 7-1
directly:
W = mgd
)(
(
)(
)
= 3.2 kg 9.81 m/s2 1.2 m = 38 J
2. (b) The force is perpendicular to the displacement so W = 0 .
Insight: You can still get tired carrying a pumpkin horizontally even though you’re doing no work!
11. Picture the Problem: The plane and glider must be at
different altitudes. Since the altitudes are constant, both
are moving horizontally.
airplane
F
glider
d
Strategy: Use equation 7-3, solving for the angle between
the force and the direction of motion.
Solution: 1. Solve equation 7-3 for W = Fd cos or cos = W
Fd
the angle:
2. Calculate the angle:
2.00 105 J
W 1
=
cos
= cos1 Fd 2560 N 145 m
(
)(
)
= 57.4o
Insight: Only the component of the force along the direction of the motion does any work. The
vertical component of the force helps to lift the glider a little.
17. Picture the Problem: The bullet moves at high speed in a straight line.
Strategy: Calculate the kinetic energy using equation 7-6, then use ratios to find the new kinetic
energies in (b) and (c).
Solution: 1. (a) Apply equation
7-6 directly:
K = 12 mv 2 =
2. (b) Use a ratio to predict the
new kinetic energy:
K new
K old
=
1
2
(0.00950 kg ) (1300 m/s)2 = 8030 J
2
1
mvnew
2
2
1
mvold
2
1
v
2 old
2
vold
)
2
=
1
so that
4
(8.03 kJ ) = 2.01 kJ
2
2
1
K new 2 mvnew ( 2vold )
=
=
= 4 so that
K new = 14 K old =
3. (c) Use a ratio to predict the
new kinetic energy:
(
=
= 8.03 kJ
K old
1
2
2
mvold
1
4
(
2
vold
)
K new = 4K old = 4 8030 J = 32.1 kJ
Insight: The kinetic energy is proportional to the velocity squared, a useful thing to remember when
calculating ratios.
20. Picture the Problem: The object falls straight down under the influence of gravity.
Strategy: Use the dependence of kinetic energy upon mass and speed to answer parts (a) and (b). The work done by
gravity can be found from the change in the kinetic energy.
Solution: 1. (a) Apply Eq. 7-6 directly:
K = 12 mv 2 =
2. (b) Solve equation 7-6 for speed:
v=
3. (c) Calculate W = K :
2K
=
m
(0.40 kg ) (6.0 m/s)2 = 7.2 J
2 ( 25 J )
= 11 m/s
1
2
0.40 kg
W = K = K f K i = 25 J 7.2 J = 18 J
Insight: As an object falls, the work done by gravity increases the kinetic energy of the object.
26. Picture the Problem: The compressed spring pushes
the block from rest horizontally on a frictionless
surface. The block slides to the left as indicated in the
figure.
Strategy: The work to stretch or compress a spring a
distance x is 12 kx 2 . The work done by the spring equals
the kinetic energy gained by the block.
Solution: 1. Apply equations 7-7
and 7-8:
W = 12 kx 2 = K = 12 mvf2 0
2. Now solve for vf :
1.0 104 N/m k
vf = 0.15 m = 14 m/s
x=
1.2 kg
m
(
)
Insight: The work done on the spring in order to compress it becomes stored potential energy. That
stored energy becomes the kinetic energy of the block as the spring accelerates it.
27. Picture the Problem: The block slides toward the right
and into the spring. After compressing the spring the
block comes to rest.
Strategy: The work to stretch or compress a spring a
distance x is 12 kx 2 . The work done on the block by the
spring equals the kinetic energy lost by the block. The
work done on the block is negative because the force on
the block is toward the left while the motion is toward
the right.
Solution: 1. Apply equations 7-7
and 7-8:
2. Now solve for k:
Won block = K block = 0 12 mvi2 = 12 kx 2
2
2.2 m/s )
(
k = m 2 = (1.8 kg )
= 91 N/m
2
x
0.31
m
(
)
vi2
Insight: The kinetic energy of the block is transformed into the energy stored in the spring as it is
compressed.
30. Picture the Problem: The springs are stretched horizontally in
each case.
Strategy: The works and stretch distances can be used to find the
spring constants by applying equation 7-8.
Solution: 1. Solve equation 7-8
for k1 :
k1 =
2W1
x12
=
(
)
(0.20 m )2
2 150 J
= 7500 N/m
2. Solve equation 7-8 for k2 :
k2 =
2W2
x22
=
(
) = 4700 N/m
(0.30 m )2
2 210 J
3. Stiffer springs have larger spring constants so we conclude spring 1 is stiffer.
Insight: The work done on the spring in order to stretch it becomes stored potential energy. We will
discuss potential energy more in Chapter 8.
37. Picture the Problem: The microwave oven delivers energy to the ice cube via electromagnetic waves.
Strategy: The power required is the energy delivered divided by the time.
Solution: Solve equation 7-10 for t:
t=
W 32200 J
=
= 307 s = 5.11 min
P
105 W
Insight: Power can be regarded as the rate of energy transfer because work is essentially transferred
energy. We’ll learn more about melting ice cubes in Chapter 17 and electromagnetic waves in Chapter
25.
38. Picture the Problem: The bucket is lifted vertically upward.
Strategy: The power required is the force times the velocity, where the force is just the weight of the
bucket.
P
P
108 W
Solution: Solve equation 7-13 for v: v = =
=
= 2.20 m/s
F mg
5.00 kg 9.81 m/s2
(
)(
)
Insight: Lifting faster than this would require more power. If the rope’s mass were not ignored it
would require additional power since its center of mass is also lifted. If the force exceeded the weight,
the bucket would accelerate.
41. Picture the Problem: The aircraft flies horizontally at constant speed.
Strategy: The power required is the work required to fly the plane (against friction) divided by the
time.
Solution: 1. (a) Solve
equation 7-10 for W:
(
)(
W = Pt = 0.30 hp 746 W/hp 2 h 3600 s/h + 49 m 60 s/m
)
W = 2.3 10 J = 2.3 MJ
6
2. (b) Convert W into units of 2.27 106 J 1 Cal 1 Snickers bar = 1.9 bars or about 2 Snickers
4186 J
280 Cal
Snickers bars:
Insight: The energy required of the pilot is much higher than this because muscles aren’t 100%
efficient at converting food energy into mechanical energy, and the body requires additional energy to
stay warm, keep the heart pumping, etc.
42. Picture the Problem: The weight slowly descends straight down.
Strategy: The power delivered is the force (the weight) times the speed.
Solution: 1. (a) Apply equation
7-13:
)(
)
0.720 m
P = Fv = mgv = 4.15 kg 9.81 m/s2 3.25 d 86400 s/d (
P = 1.04 104 W = 0.104 mW
2. (b) To increase the power delivered you must either increase the force or the velocity. In this case,
the time it takes for the mass to descend should be decreased so the velocity will increase and so will
the delivered power.
Insight: The weight delivers energy to the clock by doing work. The downward force it exerts on the
clock is parallel to its displacement, so it is doing positive work on the clock.
Chapter 8: Potential Energy and Energy Conservation
Answers to Even-Numbered Conceptual Questions
2.
As water vapor rises, there is an increase in the gravitational potential energy of the system.
Part of this potential energy is released as snow falls onto the mountain. If an avalanche
occurs, the snow on the mountain accelerates down slope, converting more gravitational
potential energy to kinetic energy.
4.
The initial mechanical energy of the system is the gravitational potential energy of the massEarth system. As the mass moves downward, the gravitational potential energy of the
system decreases. At the same time, the potential energy of the spring increases as it is
compressed. Initially, the decrease in gravitational potential energy is greater than the
increase in spring potential energy, which means that the mass gains kinetic energy.
Eventually, the increase in spring energy equals the decrease in gravitational energy and the
mass comes to rest.
6.
The object’s kinetic energy is a maximum when it is released, and a minimum when it
reaches its greatest height. The gravitational potential of the system is a minimum when the
object is released, and a maximum when the object reaches its greatest height.
8.
The jumper’s initial kinetic energy is largely converted to a compressional, spring-like
potential energy as the pole bends. The pole straightens out, converting its potential energy
into gravitational potential energy. As the jumper falls, the gravitational potential energy is
converted into kinetic energy, and finally, the kinetic energy is converted to compressional
potential energy as the cushioning pad on the ground is compressed.
10.
When the toy frog is pressed downward, work is done to compress the spring. This work is
stored in the spring as potential energy. Later, when the suction cup releases the spring, the
stored potential energy is converted into enough kinetic energy to lift the frog into the air.
12.
The total mechanical energy decreases with time if air resistance is present.
14.
The distance covered by the ball is the same on the way down as it is on the way up, and
hence the amount of time will be determined by the average speed of the ball on the two
portions of its trip. Note that air resistance does negative, nonconservative work
continuously on the ball as it moves. Therefore, its total mechanical energy is less on the
way down than it is on the way up, which means that its speed at any given elevation is less
on the way down. It follows that more time is required for the downward portion of the
trip.
Answers to Even-Numbered Conceptual Exercises
2.
(a) The value you assign to the jumper’s initial gravitational potential energy is less than
the value assigned to that quantity by your friend. (b) The change in gravitational potential
energy in your calculation is equal to that in your friend’s calculation.
4.
The two balls have the same change in gravitational potential energy. Thus, ignoring air
resistance, they will also have the same change in kinetic energy.
6.
(a) Your answers will disagree. (b) Your answers will agree. (c) Your answers will agree.
8.
Note that the information given for part (b) shows that the kinetic energy at point B is K = 0
J (at rest) and the potential energy is U = 25 J. It follows, then, that the total mechanical
energy of the system is E = U + K = 25 J + 0 J = 25 J. Since the system is conservative, it will
have the same total mechanical energy at each point in its motion. (a) At point A, U = E – K
= 25 J – 12 J = 13 J. (b) At point C, K = E – U = 25 J – 5 J = 20 J.
10.
In the original situation, all of the initial gravitational potential energy of ball 1 (U = mgh) is
converted to kinetic energy. Therefore, K = mgh. In the second case, the initial gravitational
energy of ball 2 is U = (2m)g(h/2) = mgh. It follows that the kinetic energy of ball 2 just
before it lands is also K.
12.
(a) The potential energy of the system—which is gravitational potential energy—decreases
as you move down the hill. (b) Your kinetic energy remains the same, since your speed is
constant. (c) In order for your speed and kinetic energy to remain constant as you pedal
down the hill, a nonconservative force must have done negative work on you and your
bicycle. For example, you may have applied the brakes to control your speed, or the
ground may be soft or muddy. In any case, the mechanical energy of this system has
decreased.
14.
The mechanical energy of the shuttle when it lands is considerably less than when it is in
orbit. First, its speed on landing is low compared to its speed in orbit, giving a much smaller
kinetic energy. Second, the source of energy responsible for heating the tiles is the
gravitational potential energy that is released as the shuttle loses altitude. Therefore, both
the kinetic and potential energy of the shuttle have much smaller values when the shuttle
lands, leading to a smaller value for the total mechanical energy.
16.
Ignoring any type of frictional force, the speed of the object is the same at points A and G.
Similarly, the speed of the object at point B is the same as its speed at points D and F. In
addition, the lower a point on the curve, the greater its speed. Combining these
observations, we arrive at the following ranking: A = G < B = D = F < E < C.
See following pages for Problem solutions
Solutions to Ch. 8 Problems
1. Picture the Problem: The three paths of the object are depicted at
right.
Strategy: Find the work done by gravity W = mgy when the object is
moved downward, W = mgy when it is moved upward, and zero
when it is moved horizontally. Sum the work done by gravity for each
segment of each path.
Solution: 1.
Calculate the work
for path 1:
W1 = mg y1 + 0 + y2 + 0 + y3 = mg 4.0 m + 1.0 m + 1.0 m (
(
)(
) (
)(
) (
)
)
W1 = 3.2 kg 9.81 m/s2 2.0 m = 63 J
)(
(
)
2. Calculate W for
path 2:
W2 = mg 0 y4 + 0 = 3.2 kg 9.81 m/s2 2.0 m = 63 J
3. Calculate W for
path 3:
W3 = mg y5 + 0 y6 = 3.2 kg 9.81 m/s2 1.0 m 3.0 m = 63 J
)(
(
)(
) (
)
Insight: The work is path-independent because gravity is a conservative force.
2.
Picture the Problem: The three paths of the sliding box are depicted at right.
Strategy: The work done by friction is W = μk mgd , where d is the distance the box
is pushed irregardless of direction, because the friction force always acts in a
direction opposite the motion. Sum the work done by friction for each segment of
each path..
Solution: 1. Calculate the
work for path 1:
W1 = μk mg d1 + d2 + d3 + d4 + d5 = μk mg 4.0 + 4.0 + 1.0 + 1.0 + 1.0 m (
)(
)
W1 = 0.21 3.2 kg 9.81 m/s2 11.0 m = 73 J
2. Calculate W for path 2:
W2 = μk mg d6 + d7 + d8 (
)(
)(
) (
) (
)
)(
) (
) (
)
= 0.21 3.2 kg 9.81 m/s2 2.0 m + 2.0 m + 1.0 m = 33 J
3. Calculate the work for
path 3:
W3 = μk mg d9 + d10 + d11 (
)(
= 0.21 3.2 kg 9.81 m/s2 1.0 m + 3.0 m + 3.0 m = 46 J
Insight: The amount of work done depends upon the path because friction is a nonconservative force.
6.
Picture the Problem: The cliff diver plunges straight downward due to the force of gravity.
Strategy: Solve equation 8-3 for the weight of the diver. Let y = 0 correspond to the surface of the
water.
Solution: Solve equation 8-3 for
mg:
U = mgy mg =
U 25,000 J
=
= 540 N = 0.54 kN
y
46 m
Insight: If you set U = 0 at the top of the cliff, then U = 25 kJ and y = 46 m when the diver enters
the water.
12. Picture the Problem: The spring is stretched by the applied force and stores potential energy.
Strategy: Use equation 6-4 F = kx to find the spring constant, and then equation 8-5 U = 12 kx 2 to find the stretch
distance.
Solution: 1. (a) Solve equation 6-4 for k:
k=
F
4.7 N
=
= 360 N/m
x
0.013 m
2. Solve equation 8-5 for x:
x=
2U
=
k
(
2 0.020 J
360 N/m
(
) = 1.05 cm = 1.1 cm
)
2 0.080 J
2U
=
= 2.1 cm
360 N/m
k
Insight: Notice that in part (b) the stretch distance doubled but the stored potential energy quadrupled because U is
proportional to x2.
x=
3. (b) Repeat step 2 with the new U:
16. Picture the Problem: The swimmer descends through a vertical height of 2.61 m as she slides without
friction.
Strategy: As the swimmer descends the slide her gravitational potential energy is converted into
kinetic energy. Set the loss in gravitational potential energy equal to the gain in kinetic energy by
setting her change in mechanical energy equal to zero, so that E = Ef Ei = 0 or Ef = Ei . Let
y = 0 at the bottom of the slide, v = 0.840 m/s at the top.
Solution: Set Ebottom = Etop and solve
for vbottom :
Ebottom = Etop
K bottom + U bottom = K top + U top
1
2
2
2
mvbottom
+ 0 = 12 mvtop
+ mgytop
2
vbottom = vtop
+ 2gytop
=
(0.840 m/s)2 + 2 (9.81 m/s2 ) ( 2.61 m )
= 7.21 m/s
Insight: Note that she is not going 0.840 m/s faster than the 7.16 m/s she would be traveling if she
started from rest. That’s because the 14.1 J of kinetic energy she has at the start (if she has a mass of
40 kg) is tiny compared with the 1020 J of kinetic energy she gains on the way down.
18. Picture the Problem: As the ball flies through the air and gains altitude some of its initial kinetic
energy is converted into gravitational potential energy.
Strategy: Set the mechanical energy just after the bounce equal to the mechanical energy when it is
caught. Let the height be yi = 0 at the bounce, and find yf at the catch.
Solution: 1. (a) Set Ei = Ef and solve
Ei = Ef
Ki + U i = Kf + U f
for yf :
1
2
mvi2 + 0 = 12 mvf2 + mgyf
(
)
(
)2 (
16 m/s 12 m/s
1 2
yf =
vi vf2 =
2g
2 9.81 m/s2
(
)
)2 = 5.7 m
2. (b) The height change is independent of the mass, so it is not necessary to know the mass of the
tennis ball.
Insight: A more massive ball would have more kinetic energy at the start, but would require more
energy to change its height by 0.942 m, so the mass cancels out.
20. Picture the Problem: The block slides on a frictionless, horizontal surface, encounters a spring,
compresses it, and briefly comes to rest.
Strategy: Set the mechanical energy when sliding freely equal to the mechanical energy when the
spring is fully compressed and the block is at rest. Solve the resulting equation for the spring constant
k, then repeat the procedure to find the initial speed required to compress the spring only 1.2 cm before
coming to rest.
Solution: 1. (a) Set Ei = Ef
Ki + U i = Kf + U f
where the initial state is when it is 1 mv 2 + 0 = 0 + 1 kx 2
max
2
sliding freely and the final state is 2 i
when it is at rest, having
2.7 kg 1.4 m/s
mv 2
k= 2i =
compressed the spring.
2
xmax
0.048 m
(
2. (b) Solve the equation from
step 1 for vi :
vi =
2
kxmax
m
=
(
)(
)
)2 = 2300 N/m = 2.3 kN/m
( 2300 N/m ) (0.012 m )2
2.7 kg
= 0.35 m/s
Insight: The kinetic energy of the sliding block is stored as potential energy in the spring. Moments
later the spring will have released all its potential energy, the block would have gained its kinetic
energy again, and would then be sliding at the same speed but in the opposite direction.
24. Picture the Problem: The pendulum bob swings from point B to point A
and gains altitude and thus gravitational potential energy. See the figure at
right.
Strategy:
Use the geometry of the problem to find the change in
altitude y of the pendulum bob, and then use equation 8-3 to find its
change in gravitational potential energy. Apply conservation of energy
between points B and A to find the speed at A.
Solution: 1. (a) Find the y = L L cos = L 1 cos height
change y of the
pendulum bob:
(
2. Use y to find U :
(
)
U = mgy = mgL 1 cos (
)(
)(
)
)(
= 0.33 kg 9.81 m/s2 1.2 m 1 cos 35°
)
U = 0.70 J
3. (b) Set EB = EA and
solve for vA :
KB + U B = KA + U A
1
2
mvB2 = 12 mvA2 + U
vA = vB2 2U
=
m
( 2.4 m/s)2 (
2 0.70 J
0.33 kg
) = 1.2 m/s
4. (c) If the mass of the bob is increased the answer to part (a) will increase. The change in
gravitational potential energy depends linearly on the mass.
5. (d) If the mass of the bob is increased the answer to part (b) will stay the same. Although the change
in potential energy will increase, the change in kinetic energy will also increase.
Insight: Another way to look at the answer to (d) is that U m = mgy m = gy independent of
mass. That means the formula for vA in step 3 is independent of mass.
28. Picture the Problem: The surfer changes altitude and speed, losing mechanical energy along the way
due to friction.
Strategy: The nonconservative work equals the difference in mechanical energy between the
beginning and the end of the run.
Solution: Use equation 8-9 to
find Wnc :
Wnc = E = Ef Ei
=
(
1
2
) (
mvf2 + mgyf (
)
1
2
mvi2 + mgyi
(
)
)
= m 12 vf2 vi2 + g yf yi 2
2
1 = 72 kg 2 8.2 m/s 1.3 m/s + 9.81 m/s2 0 1.75 m (
)
(
) (
)
(
)(
)
Wnc = 1100 J = 1.1 kJ
Insight: We usually expect the nonconservative work to be negative because friction steals mechanical
energy and converts it into heat. But in this case the force from the wave propels the surfer and he
ends up with more mechanical energy (1200 J) than he had at the start (61 J).
31. Picture the Problem: The airplane travels in a straight line, slowing down and coming to rest after
landing at high speed.
Strategy:
The total nonconservative work done on the airplane changes its mechanical energy
according to equation
8-9. The nonconservative work equals the change in mechanical energy, which is known from the
initial and final speeds of the airplane (there is no change to its gravitational potential energy).
Solution: Apply equation 8-9 directly: W = E = E E = 1 mv 2 1 mv 2
nc
f
i
f
i
2
2
= 0
1
2
(17,000 kg ) (82 m/s)2
Wnc = 5.7 107 J = 57 MJ
Insight: The aircraft brakes and the recovery cables remove the airplane’s kinetic energy, converting it
into heat and sound. Heat management in large braking systems like this one is an important
engineering issue.
41. Picture the Problem: The U vs. x plot is depicted at right.
Strategy: Describe the motion of the object, keeping in
mind that objects tend to move to the minimum potential
energy, and when they do their kinetic energy is maximum.
The turning points on the potential energy plot are points A
and E.
Solution: At point A, the object is at rest. As the object
travels from point A to point B, some of its potential energy
is converted into kinetic energy and the object’s speed
increases. As the object travels from point B to point C,
some of its kinetic energy is converted back into potential
energy and its speed decreases. From point C to point D, the
speed increases again, and from point D to point E, the
speed decreases.
Insight: The object momentarily comes to rest at point E, but then turns around and accelerates back
towards D and retraces its path all the way to point A, at which time the cycle begins again.