Download Chapter Assessment f

Chapter 11 continued
27. Energy As shown in Figure 11-15, a
36.0-kg child slides down a playground slide
that is 2.5 mhigh. At the bottom of the
slide, she is moving at 3.0 rn/s. How much
energy was lost as she slid down the slide?
Chapter Assessment
Concept Mapping
page 306
29. Complete the concept map using the fol
lowing terms: gravitational potential energy,
elastic potential energy kinetic energy
Mastering Concepts
s Figure
U-iS
E=mgh
=
=
=
=
(38.0 kg)(9.80 m/s
)(2.5 m)
2
8803
jmv2
2
1(36.0 kg)(3.0 mis)
=160J
Energy loss
=
=
8803—1603
7203
28. Critical Thinking A ball drops 20 m.
When it has fallen half the distance, or
10 m, half of its energy is potential and half
is kinetic. When the ball has fallen for half
the amount of time it takes to fall, will
more, less, or exactly half of its energy be
potential energy?
The ball falls more slowly during the
beginning part of its drop. Therefore, in
the first half of the time that it falls, it
will not have traveled half of the dis
tance that it will fall. Therefore, the ball
will have more potential energy than
kinetic energy.
page 306
Unless otherwise directed, assume that air resis
tance is negligible.
30. Explain how work and a change in energy
are related. (11.1)
The work done on an object causes a
change In the object’s energy. This is
the work-energy theorem.
31. What form of energy does a wound-up
watch spring have? What form of energy
does a functioning mechanical watch have?
When a watch runs down, what has hap
pened to the energy? (11.1)
The wound-up watch spring has elastic
potential energy. The functioning watch
has elastic potential energy and rota
tional kinetic energy. The watch runs
down when all of the energy has been
converted to heat by friction in the
gears and bearings.
32. Explain how energy change and force are
related. (11.1)
A force exerted over a distance does
work, which produces a change in energy.
33.. A ball is dropped from the top of a build
ing. You choose the top of the building to
be the reference level, while your friend
chooses the bottom. Explain whether the
254
Solutions Manual
Physics: Principles and Problems
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Chapter 11 continued
41. Describe the transformations from kinetic
energy to potential energy and vice versa for
a roller-coaster ride. (11.2)
On a roller-coaster ride, the car has
mostly potential energy at the tops of
the hills and mostly kinetic energy at
the bottoms of the hills.
42. Describe how the kinetic energy and elastic
potential energy are lost in a bouncing
rubber ball. Describe what happens to the
motion of the ball. (11,2)
On each bounce, some, but not all, of the
ball’s kinetic energy is stored as elastic
potential energy; the ball’s deformation
dissipates the rest of the energy as ther
mal energy and sound. After the bounce,
the stored elastic potential energy is
released as kinetic energy. Due to the
energy losses in the deformation, each
subsequent bounce begins with a smaller
amount of kinetic energy, and results in
the ball reaching a lower height.
Eventually, all of the ball’s energy is dis
sipated, and the ball comes to rest.
Applying Concepts
pages 306—307
43. The driver of a speeding car applies the
brakes and the car comes to a stop. The
system includes the car but not the road.
Apply the work-energy theorem to the fol
lowing situations.
a. The car’s wheels do not skid,
If the car wheels do not skid, the
brake surfaces rub against each
other and do work that stops the car.
The work that the brakes do is equal
to the change in kinetic energy of
the car. The brake surfaces heat up
because the kinetic energy is trans
formed to thermal energy.
b. The brakes lock and the car’s wheels skid.
If the brakes lock and the car wheels
skid, the wheels rubbing on the road
are doing the work that stops the car.
The tire surfaces heat up, not the
brakes. This is not an efficient way to
stop a car, and it ruins the tires.
256
Solutions Manual
44. A compact car and a trailer truck are both
traveling at the same velocity. Did the car
engine or the truck engine do more work in
accelerating its vehicle?
The trailer truck has more kinetic energy,
KE =
mv, because it has greater mass
than the compact car. Thus, according
to the work-energy theorem, the truck’s
engine must have done more work.
45, Catapults Medieval warriors used catapults
to assault castles. Some catapults worked
by using a tightly wound rope to turn the
catapult arm, What forms of energy are
involved in catapulting a rock to the castle
wall?
Elastic potential energy is stored in the
wound rope, which does work on the
rock. The rock has kinetic and potential
energy as it flies through the air. When
it hits the wall, the inelastic collision
causes most of the mechanical energy
to be converted to thermal and sound
energy and to do work breaking apart
the wall structure. Some of the mechan
ical energy appears in the fragments
thrown from the collision.
46. Two cars collide and come to a complete
stop. Where did all of their energy go?
The energy went into bending sheet
metal on the cars. Energy also was lost
due to frictional forces between the
cars and the tires, and in the form of
thermal energy and sound.
47. During a process, positive work is done on a
system, and the potential energy decreases.
Can you determine anything about the change
in kinetic energy of the system? Explain.
The work equals the change in the total
mechanical energy, W = A(KE + PE). If
W is positive and APE is negative, then
AKE must be positive and greater than W.
48. During a process, positive work is done on a
system, and the potential energy increases. Can
you tell whether the kinetic energy increased,
decreased, or remained the same? Explain.
Physics: Principles and Problems
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Chapter ‘11 continued
Mastering Problems
Unless otherwise directed, assume that air resistance
is negligible.
11.1 The Many Forms of Energy
pages 307—308
Level 1
54. A 1600-kg car travels at a speed of 12.5 rn/s.
What is its kinetic energy?
KE = Imv2
=
=
55. A racing car has a mass of 1525 kg. What is
its kinetic energy if it has a speed of
108 km/h?
=
5
6.86X10
3600s/h
0
=
=
1
‘d 2
m(
j
1
(10.0 mln)(60 s/mm)
kaI°-)
a. Find Tony’s kinetic energy.
=
I
2
my
=
2
I( kg)(1 0,0 mis)
3J
2.3X10
b. Tony’s speed changes to 5.0 m/s. Now
what is his kinetic energy?
I
KE
=
258
2
my
—
4
—
58. Katia and Angela each have a mass of 45 kg,
and they are moving together with a speed
of 10,0 rn/s.
a. What is their combined kinetic energy?
KE
mv
=
(m + mA)v
2
=
(45 kg + 45 kg)(1 0.0 mis)
2
J
3
=4.5X10
17¼+fl¼
=
(45 kg)( 5.0 rn/s)
2
—
—
45 kg + 45kg
45 kg
2
1
c. What is the ratio of their combined
kinetic energy to Katia’s kinetic energy?
Explain.
KEK
=
=
57. Tony has a mass of 45 kg and is moving
with a speed of 10.0 rn/s.
=
(1O.O)
(5,0)2
b. What is the ratio of their combined
mass to Katia’s mass?
203 J
KE
—
Twice the velocity gives four times
the kinetic energy. The kinetic energy
is proportional to the square of the
velocity.
)
56. Shawn and his bike have a combined mass
of 45.0 kg. Shawn rides his bike 1.80 km in
10.0 mm at a constant velocity. What is
Shawn’s kinetic energy?
mv2
—
2
V
=
=
1(1525
KE =
2
V
2
1(mv
2
my
=
=
1(mvi2)
2
160O kg)(12.5 rn/s)
1 .3x1
KE =
What is the ratio of the kinetic energies
in parts a and b? Explain.
KEc
KEK
j
m(v
(45 kg)(10.0 rn/s)
2
3J
2.3X10
(mK + mA)v
2
—
mKv
2
—
=
=
—
—
rnK + mA
mK
-
2
The ratio of their combined kinetic
energy to Katia’s kinetic energy is
the same as the ratio of their corn
bined mass to Katia’s mass. Kinetic
energy is proportional to mass.
2J
5.6X10
Solutions Manual
Physics: Principles and Problems
Chapter 11 continued
59. Train In the 1950s, an experimental train,
4 kg, was
which had a mass of 2.50X i0
powered across a level track by a jet engine
that produced a thrust of 5.OOX io N for a
distance of 509 m,
a. Find the work done on the train.
W = Fd
=
Now
m =
So d
=
F
9
g
imv2
2
F
2g
5 N)(509 m)
(5.00X10
F
—
=
8J
2.55X10
1
=
b. Find the change in kinetic energy.
J
8
KE= W=2.55X10
C.. Find the final kinetic energy of the train
if it started from rest.
1
KE = KE
1
so KE
—
KE
=
AKE +
=
8 J + 0.00 J
2.55X10
=
.ssxio j
8
2
9.80 flVS
7100 N
=66m
61. A 1 5.0-kg cart is moving with a velocity of
7.50 rn/s down a level hallway. A constant
force of 10,0 N acts on the cart, and its
velocity becomes 3.20 rn/s.
a. What is the change in kinetic energy of
the cart?
2)
2
m(v
= KE
—
d. Find the final speed of the train if there
had been no friction.
—
2
(15.0 kg)((3.20 mis)
—
2
1
KE= mv
)
2
(7.50 m/s)
=—345J
b. How much work was done on the cart?
W= KE= —345J
c.. How far did the cart move while the
force acted?
-j——
Sov
=
2
1
m
2
J
8
2.55X10
=
1
So v
=
(2.50X1O kg)
/s
2
V2.04X10 m
=
60.. Car Brakes A 14,700-N car is traveling at
25 rn/s. The brakes are applied suddenly,
and the car slides to a stop, as shown in
Figure 11-17. The average braking force
between the tires and the road is 7100
How far will the car slide once the brakes
are applied?
After
Before
(final)
(initial)
w=
143 mIs
so d
62.
34.5 m
How much potential energy does DeAnna
with a mass of 60.0 kg, gain when she
climbs a gymnasium rope a distance of
35 m?
=
h
)(3.5 m)
2
= (60.0 kg)(9.80 m/s
=
v=25m/s
=
J
3
2.1X1O
v=0,OmIs
2.1 m into a storage rack. Calculate the
increase in the ball’s potential energy.
m= 14,IOON
PE= mgh
S Figure 11-17
W = Fd
2
tin’
Physics: Principles and Problems
=
(2.1 m)
mIs
)
(6.4 kg)(9.80 2
=
I .3X10
2J
Solutions Manual
259
Chapter 11 continued
59.
Train In the I 950s, au experimental train,
which had a mass of 2.5() i0 kg, was
powered across a level track by a jet engine
fi) N for a
hat produced a thrust of 5.00 X I
5fio)
distance of
a.
Fg
g
Nowm=
1
Sod=
mu.
2
my
F
92
1 (F
Find the work done on the train.
2\gj
5 N)(509 m)
W= Fd= (5.00X10
F
—
8J
255X10
=
b.
1(14,700 N
2
‘1(25.0 mis)
!
2
2\9.80 mis
Find the change in kinetic energy.
c.
Find the fijia! kinetic energy of the train
if it started from lest.
KE
KE
=
so KE
d.
-
=
61.
KE
=
1KE + KE
=
2.55X 108
j +
=
8
2.55X10
j
0.00 J
moVing with a velocity of
level hallway. A (onsta mit
down
a
mn/s
()
7.
ftrce of 1 fit) N acts on the cart, and its
velocity he(omnes .2() mn/s.
a.
What is the change in kinetic energy of
(lie cart?
=
KE
friction
iu
66 m
A I 5.() kg cart is
Iinl the fiuial speed of the tiain if there
had been
7100N
=
J
8
KE= W=2.55X10
KE
=
—
1
KE
=
m(v?
—
2
(15.0 kg)((3.20 rn/s)
v)
—
mvf
=
)
2
(7.50 mIs)
2
So v
KE
=
=
b.
—345 J
I lo’v much work was done on the cart?
W= KE= —345J
2,55X 108 j
I low far did the cart move while the
(2.50X 10 kg)
farce acted?
So v
=
1s
2
\ 2,04x io m
=
143 mIs
60. Car Brakes A 14,700-N cam is traveling at
s. 1 he hiakes are applied suddenly,
1
25 in
and the am slides to a 5101), as sllo\vn in
Figure 11-17. I he average braking force
between the tiles and the road is 7100 N.
I low far will the car slide once the brakes
are applied?
W
v=25 rn/s
Fd
sod=
62.
=
=34.5m
1 kyu,v much l)otential enem-gv does l)ei\mina
wi ili a mass f 60.() kg, gai ii when she
climbs a gymnnasitmmn mope a distance of
3.’ m?
PE= mgh
After
(final)
Before
(initial)
=
=
(3.5 m)
mIs
)
(60.0 kg)(9.80 2
=
J
3
2.1X1O
v=0.0 rn/s
63.
Bowling A (.4-kg bowling ball is lifted
2.1 m into a storage rack. (alculate the
increase in the ball’s potential energy.
m
=
14,700 N
PE
=
mgh
=
(2.1 m)
mIs
)
(6.4 kg)(9.80 2
=
2 J
1.3X10
• Figure 11-17
W=Fd=
P/)1sj(s: l’rui
mv2
iJ)I(’s ifl(I I’roh/ellN
SO/Il I 10115
5 Ia ii no!
259
___
Chapter 11 continued
64. Mary weighs 505 N. She walks down a
flight of stairs to a level 5.50 m below her
starting point. What is the change in Mary’s
potential energy?
M’E = mgh = Fh
= (505 N)(—5.50 m)
=
65
78X10
3
—2.
J
Weightlifting A weightlifter raises a 180-kg
barbell to a height of 1.95 m. What is the
increase in the potential energy of the
Level 2
69. Tennis It is not uncommon during the serve
of a professional tennis player for the racket
to exert an average force of 150.0 N on the
ball. If the ball has a mass of 0.060 kg and is
in contact with the strings of the racket, as
shown in Figure 11-18, for 0.030 s, what is
the kinetic energy of the ball as it leaves the
racket? Assume that the ball starts from rest.
barbell?
PE= mgh
=
=
150.0 N
5 m)
(1.9s
2
ml
(180 kg)(9.80 )
3J
3.4X10
66. A 10.0-kg test rocket is fired vertically from
Cape Canaveral. Its fuel gives it a kinetic
energy of 1960 1 by the time the rocket
engine burns all of the fuel, What additional
height will the rocket rise?
S
Ft= mv= mvf
=
PE= mgh= KE
1960J
KE
)
2
(10.0 kg)(9.80 m/s
mg
=
Ft
m
Figure 11-18
—
mv
and vi
=
0
0 kg
2
6..0X1
75 mIs
—
KE=
—
—
=20.Om
=
67.. Antwan raised a 12.0-N physics book from
a table 75 cm above the floor to a shelf
2.15 m above the floor. What was the
change in the potential energy of the
system?
h Fg(hf h)
9
PE = mgh = F
—
=
(12.0 N)(2,15 m
=
17 J
—
0.75 m)
=
70.
mv2
2
10 kg)(75 rn/s)
2
(6.0X
10
2
1..7X
J
Pam, wearing a rocket pack, stands on fric
tionless ice. She has a mass of 45 kg. The
rocket supplies a constant force for 22.0 m,
and Pam acquires a speed of 62.0 m/s.
a.. What is Pam’s final kinetic energy?
= Imvt2
1
KE
68. A hallway display of energy is constructed
in which several people pull on a rope that
lifts a block 1.00 m. The display indicates
that 100 J of work is done. What is the
mass of the block?
W= PE= mgh
—
-
=
w
gh
—
-
i.ooJ
(9.80 m/s2)(tOO m)
0.102 kg
2
kg)(62.0 rn/s)
=
=
4J
8.6X10
b. What is the magnitude of the force?
Work done on Pam equals her
change in kinetic energy.
KE
W= Fd= KE= K
-
KE=OJ
S 0, F—
KE
—
—
=
260
Solutions Manual
10
4
8.6X
J
22.Om
0N
3
3..9X1
Physics: Principles and Problems
Chapter 11 continued
71. Collision A 2.OOX 10
-kg car has a speed of
3
12.0 rn/s. The car then hits a tree. The tree
doesn’t move, and the car comes to rest, as
shown in Figure 11-19.
Before
(initial)
1 = 12.0 rn/s
V
Afterd
(final)
Vf = 0.0 rn/s
(2.00X
2
kg)((0.0 rn/s)
=
(98.0 N)(50.0 m)
=
3J
4,90X10
b. What is the increase in potential energy
of the sack of grain at this height?
APE= W=4.90X10
J
3
—
J
5
—1.44X10
c. The rope being used to lift the sack of
grain breaks just as the sack reaches the
storage room. What kinetic energy does
the sack have just before it strikes the
ground floor?
KE= APE= 4,90x10
J
3
b. Find the amount of work done as the
front of the car crashes into the tree.
J
5
W=AKE= —1.44X10
C.
Level 1
73. A 98.0-N sack of grain is hoisted to a stor
age room 50.0 m above the ground floor of
a grain elevator.
W=APE= mgAh= FAh
(12.0 mIs)
)
2
=
pages 308—309
a. How much work was done?
s Figure 11-19
a. Find the change in kinetic energy of
the car.
= m(vf2_2)
1
AKE=KEf-KE
=
11.2 Conservation of Energy
74. A 20-kg rock is on the edge of a 100-rn cliff,
as shown in Figure 11-20.
Find the size of the force that pushed in
the front of the car by 50.0 cm.
W=Fd
soF=
—
—
=
1C Drn
—1.44X10 J
0.500 rn
s Figure 11-20
5N
—2,88X10
72. A constant net force of 410 N is applied
upward to a stone that weighs 32 N. The
upward force is applied through a distance
of 2.0 m, and the stone is then released. To
what height, from the point of release, will
the stone rise?
W = Fd = (410 N)(2..0 m) = 8.2X 102 J
But W = APE = mgAh, so
W
mg
8.2X10
J
2
32N
26m
a. What potential energy does the rock
possess relative to the base of the cliff?
PE
(20 kg)(9.80 mIs
)(100 rn)
2
mgh
J
4
=2X10
b. The rock falls from the cliff. What is its
kinetic energy just before it strikes the
ground?
KE= APE= 2X10
J
4
c. What speed does the rock have as it
strikes the ground?
KE= Imv2
v =
‘
=
Physics: Principles and Problems
m
=
V
20kg
40 m/s
Solutions Manual 261
Chapter 11 continued
75. Archery An archer puts a 0.30-kg arrow to
the bowstring. An average force of 201 N is
exerted to draw the string back 1.3 m.
a. Assuming that all the energy goes into
the arrow, with what speed does the
arrow leave the bow?
Work done on the string increases
the string’s elastic potential energy.
W= PE= Fd
All of the stored potential energy is
transformed to the arrow’s kinetic
energy
jmv2
KE=
2
v
=
v=
iPE= Fd
=
m
77.
A physics book of unknown mass is
dropped 4.50 m. What speed does the book
have just before it hits the ground?
KE= PE
Imv2
=
mgh
The mass of the book divides out, so
1v2
=
gh
v=
=
=
9.39 m/s
78. Railroad Car A railroad car with a mass of
kg collides with a stationary rail
5.0x
road car of equal mass. After the collision,
the two cars lock together and move off at
4.0 m/s, as shown in Figure 11-21.
/Li=
i
0.30kg
v m
=42m1s
b.. If the arrow is shot straight up, how
high does it rise?
The change in the arrow’s potential
energy equals the work done to pull
the string.
PE= mgih= Fd
Fd
mg
—
—
)
2
(0.30 kg)(9.80 nVs
=89m
m
KE=0
=
i/f
10 kg
5
5.0X
4.0 rn/s
s Figure 11-21
a. Before the collision, the first railroad car
was moving at 8.0 rn/s. What was its
momentum?
my
76. A 2.0-kg rock that is initially at rest loses
407 J of potential energy while falling to
the ground. Calculate the kinetic energy
that the rock gains while falling. What is
the rock’s speed just before it strikes the
ground?
P+K=PE+KE
50 rn)
(4.Is
2
m
\/(9.80 )
=
(5.0X i0 kg)(8.0 m/s)
10 kg’m/s
6
4.0X
b. What was the total momentum of the
two cars after the collision?
Because momentum is conserved, it
4.0X10 kgm/s
must be 6
c. What were the kinetic energies of the
two cars before and after the collision?
Before the collision:
=
So,
J
E
07
P
=4
E=
PE
1
K
—
KE
KE= mvj2
=
2
v
I
2KE
m
=
Imvi2
=
2
15.0X105 kg)(8.0 rn/s)
10 1J
7
1.6X
After the collision:
=
f2)(4o7 J)
(2.0 kg)
KE
=
1mvf2
‘.,/
=
262
10 rn/s
1
2.0X
Solutions Manual
Physics: Principles and Problems
Chapter 11 continued
b. If Kelli moves through the lowest point
at 2.0 rn/s, how much work was done
on the swing by friction?
5 kg)
15.0<b05 kg + 5.0X10
=
2
(4.0 m/s)
The work done by friction equals the
change in mechanical energy.
6j
8.0X10
d. Account for the loss of kinetic energy.
W= PE+ KE
While momentum was conserved
during the collision, kinetic energy
was not. The amount not conserved
was turned into thermal energy and
sound energy.
79.
[rorn what height would a compact car
have to he dropped to have the same
kinetic energy that it has when being driven
at l.OOX 102 km/h?
V
(420 N)(0.40 m
—
1.00 m) +
420N
1/
20
2 9.80 m/s2)(
2
mis)
J
2
—1.7X10
=
27.8 m/s
+PE
KKE
1
mv2
v2
=
=
—
i OOXlO2km y1±P
h A 1km A3600s
KE= PE
—
1mv2
mg(h
llakeem throws a 10.0-g ball straight down
from a height of 2.0 rn. The ball strikes the
floor at a speed of 7.5 rn/s. What was the
initial speed of the ball?
=
h
) +
1
h
=
=
81.
mv2+mgh
mv2=
mgh
The mass of the ball divides out, so
=
=
2
2gh,
—
v1
(27.8 rn/sI
2
2g
)
2
2(98O mis
39.4 m
=
=
\/v2
=
\/(7.5 m/s)
2
—
2gh
—
(2)(9.80 m/s
)(2.0 m)
2
4.1 m/s
Level 2
80. Kelli weighs 420 N, and she is sitting on a
playground swing that hangs 0.40 m above
the ground. Her mom pulls the swing back
and releases it when the seat is 1 .00 m
above the ground.
a. [low fast is Kelli moving when the swing
passes through its lowest position?
By conservation of mechanical energy:
+ KE
1
PE
=
PE+ KE
82. Slide Lorena’s mass is 28 kg. She climbs the
4.8-m ladder of a slide and reaches a velocity
of 3.2 rn/s at the bottom of the slide. How
much work was done by friction on Lorena?
The work done by friction on Lorena
equals the change in her mechanical
energy.
W= PE+ KE
=
mgh +
\/2g(h
—
=
(28 kg)(9.80 s2)(o.o m
mv
)(1 .00 m
2
\/(2)(9.80 m!s
=
3.4 m/s
Physics: Principles and Problems
—
2
1(28 kg)((3.2 rn/s)
I1)
=
1m(V2
mg(h
KE=0
1
mgh
) +
1
h
=
=
—
0.40 m)
—
—
2)
—
4.8 m) +
(0.0
)
2
mis)
J
3
—1.2x10
83. A person weighing 635 N climbs up a
ladder to a height of 5.0 rn. Use the person
and Earth as the system.
Solutions Manual
263
Chapter 11 continued
a. Draw energy bar graphs of the system
before the person starts to climb the
ladder and after the person stops at the
top. has the mechanical energy
changed? if so, by how much?
86. High Jump The world record for the men’s
high jump is about 2.45 m. To reach that
height, what is the minimum amount of
work that a 73.0-kg jumper must exert in
pushing off the ground?
W= E= mgh
3200 J
I
I
I
I
1
KE
I
I
PE
PEf
KE
Yes. The mechanical energy has
changed by an increase in potential
energy of (635 N)(5.0 m) = 3200 J.
b. \‘Vhere did this energy come from?
from the internal energy of the person
Mixed Review
pages 309—310
Level 1
84. Suppose a ch i iii I)flZCC swings through the
jungle on vines. If it swings from a tree on a
1 3-rn-long vine that starts at an angle of
45°, what is the chimp’s velocity when it
reaches the ground?
The chimpanzee’s initial height is
h
=
(13 m)(1
cos 45°)
—
=
3.8 m
Conservation of mechanical energy:
=
85.
1
V2gh
=
8.6 m/s
=
)(3.8 m)
2
\ 2(930m/s
=
d
=
mgh
=
=
1.75 kJ
87. A stuntwoman finds that she can safely
break her fall from a one-story building by
landing in a box filled to a i-rn depth with
foam peanuts. In her next movie, the script
calls for her to jump from a five-story build
ing. I low deep a box of foam peanuts
should she prepare?
Assume that the foam peanuts exert
a constant force to slow her down,
W = Fd = E = mgh. If the height is
increased five times, then the depth of
the foam peanuts also should be
increased five times to 5 m.
Level 2
88. Football A 110-kg football linebacker has a
head-on collision with a 1 50-kg defensive
end. After they collide, they come to a
complete stop. Before the collision, which
player had the greater momentum and
which player had the greater kinetic energy?
Planebacker
=
mlinebackervlinebacker
=
Pend
mend Vend. After the collision, each had
An 0.80-kg cart rolls down a frictionless hill
of height 0.32 m, At the bottom (>1 the hill,
the cart rolls on a fiat surface, which exerts
a frictional force of 2() N on the cart. I low
far does the cart roll on the flat surface
before it comes to a stop?
E
264
mvf
=
)(2.45 m)
2
(73.0 kg)(9.80 mIs
The momentum after the collision is
zero; therefore, the two players had
equal and opposite momenta before
the collision. That is,
PE + KE
mgh
=
W
=
)(0.32 m)
2
(0.80 kg)(9.80 m!s
F
2.ON
Solutions Manual
player was
2
my
=
)
v
2
(m
Because the momenta were equal but
milnebacker < mend the linebacker lost
more energy.
Fd
mgh
1,3 m
zero energy. The energy loss for each
89. A 2.0-kg lab cart and a 1.0-kg lab cart are
held together by a compressed spring. The
lab carts move at 2. 1 rn/s in one direction.
The spring suddenly becomes uncompressed
and pushes the two lab carts apart. The 2-kg
Physics: Principics and Problems
Chapter 1 1 continued
lab cart comes to a stop, and the 1.0-kg lab
cart moves ahead. How much energy did the
spring add to the lab carts?
E
Imv2
=
-(2.O kg + 1.0 kg2.1 mls$
=
6.6 J
(2.0 kg + 1.0 kg)(2.1 mis)
=
my
=
6.3 kgm/s
1
SO, V
=
=
=
1
p
1
(1.0 kg)v
=
6.3 mIs
_ 0 m)
2
g
2
/
1
(
55.0 kg + 21.0 kg)
6.28 m
=
91. An 0.80-kg cart rolls down a 30.0° hill from
a vertical height of 0.50 m as shown in
Figure 11-22. The distance that the cart
must roll to the bottom of the bill is
0.50 m/sin 30.0° = 1.0 m. The surface of the
hill exerts a frictional force of 5.0 N on the cart,
Does the cart roll to the bottom of the hill?
m
2
mv
=
2
(1 .0 kg)(6..3 mis)
=
tiE = 19.8 J
6.6 J
—
=
=
19.8
0.80 kg
F=
5.ON
13.2 J
=
13.2 J was added by the spring.
of
90. A 55.0-kg scientist roping through the top
a tree in the jungle sees a lion about to
attack a tiny antelope. She quickly swings
down from her 12.0-rn-high perth and grabs
the antelope (21.0 kg) as she swings. They
barely swing back up to a tree limb out of
reach of the lion. How high is this tree limb?
E
mgh
=
The velocity of the botanist when she
reaches the ground is
2
v
3
m
=
=
mgh
/E
=
L2rnB9hV/
m
Momentum is conserved when the
botanist grabs the antelope.
8 + mA)vf
v = (in
8
m
v
8
m
1
50, V
/
=
=
mB
rn
rn
+
8
KEof both is
1
KE
=
=
mA)v1
1(mB + 2
--—(2gh)
8 + mA)(——
1(m
mB + inA)
S Figure 11-22
1
E
=
mgh
=
39 J
Level3
92. Object A, sliding on a frictionless surface at
3.2 rn/s, hits a 2.0-kg object, B, which is
motionless. The collision of A and B is corn
pletely elastic. After the collision, A and B
move away from each other at equal and
opposite speeds. What is the mass of object A?
1 + 0
pI = mftv
1
p
=
p
=
2
) + mB V
2
mA(— V
1 (conservation of momentum)
p
therefore, mAy
1
—
v
2
mjj
=
=
=
)+
2
mA(— v
1
mAy
1
mAy
=
KE
50 m)
(0.Is
2
m
(0.80 kg)(9.80 )
The work done by friction over 1.0 m
would be
W = Fd = (5.0 N)(1 .0 m) = 5,0 J.
The work done by friction is greater
than the energy of the cart. The cart
would not reach the bottom of the hill.
(mB
h
mB
g
2
=
PEwhen they swing to the limb,
h
m
g
2
8
8 + mA
m
(
,h
1
So
—
—
=
2 =
V
I
5 + m)ghf
(ni
E
8
in
v
V
A
B
m
m
=I
+
Ef
2
2
=
2l
mAv
mB+mA)
Physics: Principles and Problems
Solutions Manual
265
g
:&
-
mZk1
;
=
11
x
C
xx
II
0
—
we
ft
I
C
ft
•
M
o
—
ft
m
>
—.
oIs,
I’h
o 0
[!H
—
m
C
ft
+
I
if
M-’
ft
4
P%
+
if
‘I
m
P
Z
.
+
C
ft
S
+
SJ
+
+
I
c
I
+
—I
0
ft
ii
C
I
sa
aI
0
ft
MM
+
ft
H
c
0
j
S
S
C
MM
+
MM
I r
ft
I
ft
a
a
z
0
z
Ig
I
it
ew
z
S
C
w
+
14
c;
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ft
I
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I
w
W
+
1
+
II
+
L
a
+I
if
.
m
+
ç
s a
I>
+
+
ç
t4,
—
+
+
if
+
if
Uentoe/MCraw Hffl a dwfsfoo of Ibe MKeawfhli (omparnee IOL
+a
if
Copynght
&
+
if
if
i
I
+
o
0
I
I
a
i
1
+
a
c
uiZ
+
•
c
0
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a
—
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—
a
0
cr1
p
=
0 —
0
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0
I
I
Chapter 11 continued
1 toward
97. Analyze and Conclude A 25-g ball is fired with an initial speed of v
a I 25-g hail that is hanging motionless from a I .25-rn string. The halls have a
perfectly elastic collision. As a result, the I 25-g ball swings otLt until the string
1
makes an angle of 37.O° with the vertical. What is V
Object 1 is the incoming ball. Object 2 is the one attached to the string.
In the collision, momentum is conserved.
1 + P2f or
P-
Pu
1
m
1
V
1
m
=
+ m
2
2V
1f
V
In the collision, kinetic energy is conserved.
11
V
12
m
2
1v
m
2
m. v
=
1 v2
m
=
(mivii2)(_)
m
2
1
1
m
2
P-it
1
m
1
m
+
f
2
2V
m
(mivif2)(_) + (m2v2f2)()
m
v
2
1
1
m
=
=
=
+
v
2
m
+
+
m
v
2
2
m
2
P2f
2
m
2 + ()P2t2
P
2
1
P
so get rid of Pj using Pit
We don’t care about
2
1
P
2 +
P2f)
11
(P
=
=
11
1v
m
(
—
P2f
m
2
P2f
(i + rn)P2f
)(i +
=
1
P
1
H
2
P2f
2 +
ijP2f + P2f
2
P
—
=
=
=
)p
2
()(m
v=()(mn2
+
v
2
)
1
m
+1)v2f
Now consider the pendulum.
2
v
2
m
or
2
V
=
where h
Thus,
f
2
V
=
=
268
Solutions I\4an no!
=
h
m
g
2
V2gh
=
2
V
=
L(1
—
cos 0)
\/2gL(1
—
cos 0)
(1 .25 m)(1
m/s
)
\(2)(9.80 2
—
cos
37.00)
2.22 m/s
JThvsj
S:
Prjm iJ )JOS 311(1 Problems
Chapter 11 continued
vii
=
=
1 (1259 + 1)(2.22 mis)
2\25g
6.7 mis
Writing in Physics
page 310
98. All ci ‘EV comes from the Sun. In what forms
has this solar energy come to us to allow us to
I ive and to operate our Society? Research the
wa’s that the Sun’s energy is turned into a
form that we can use. i\Eer we use the Sun’s
((gy, ‘here does it go? Explain.
The Sun produces energy through
nuclear fusion and releases that energy
in the form of electromagnetic radia
tion, which is transferred through the
vacuum of space to Earth. Earth
absorbs that electromagnetic radiation
in its atmosphere, land, and oceans in
the form of thermal energy or heat. Part
of the visible radiation also is converted
by plants into chemical energy through
photosynthesis. There are several other
chemical reactions mediated by sun
light, such as ozone production. The
energy then is transferred into various
forms, some of which are the chemical
processes that allow us to digest food
and turn it into chemical energy to build
tissues, to move, and to think. In the
end, after we have used the energy, the
remainder is dispersed as electromag
netic radiation back into the universe.
99.
All fun us of energy can be classified as
eitheu kinetic or potential energy. I low
would on des nbc nu leai electric, chemi
al. biological, solai and light energy, and
vhv Eur each of these types of energy,
research what ‘biects are moving and how
energy is stored in those objects.
Potential energy is stored in the binding
of the protons and neutrons in the nucle
us. The energy is released when a heavy
nucleus is broken into smaller pieces (fis
sion) or when very small nuclei are com
bined to make bigger nuclei (fusion). In
the same way, chemical potential energy
is stored when atoms are combined to
i1’ifl If)!( 111(1 P,ol)!cIllc
make molecules and released when the
molecules are broken up or rearranged.
Separation of electric charges produces
electric potential energy, as in a battery.
Electric potential energy is converted to
kinetic energy in the motion of electric
charges in an electric current when a con
ductive path, or circuit, is provided.
Biological processes are all chemical, and
thus, biological energy is just a form of
chemical energy. Solar energy is fusion
energy converted to electromagnetic radi
ation. (See the answer to the previous
question.) Light is a wave form of electro
magnetic energy whose frequency is in a
range detectible by the human eye.
Cumulative Review
page 310
100. A satellite is placed in a circular orbit with
7 m and a period of
a radius oil OX i0
9.9 . 10 s. Calculate the mass of Earth.
/ lint: Gravity is the net fiiie on sue/i a sate/
li Ic. Scu’i ii isis ha i’e actual/i’ nieasu reti the
mass (4 Earth this u’aE (Chapter 7)
Fnet
=
mv
2
r
because v
=
r
2
/ m v4ir
Gmsme
2-r
Gmsme
)
r
2
T
me—
4’r
3
r
2
2
GT
r2
3
(1.Ox iü rn)
2
42
kg 1 O s)
(9.9X
/
Nm
)
(6.67x 1 0 11 2
=
24 kg
6.0X10
101. A .00-g bullet is fired with a velocity of
lOOt)
m/s toward
a
block resting
(Chapter 9)
solid
1000-kg stationary
on a frictionless surface.
a. What is the change in momentum
the bullet if it is embedded in the
block?
mbvbl
=
2
2 + mv
mbv
=
2
(mb + m)v
Solutions tlanua/
of
269
I
ft
ø
C
x
0
H
I
w
M
M
I
Mj
a
g
Sn9
H
U
.
H
0
—
11
if
0
c
0
a
S
0
z
c
I
%*k
ft
ft
91
dl fl
—
ft
o
-‘
—
S
0
—
a
ft
fl
ft
39g
z
p
O
c
H
I
7$
(Il
ft
0
I
<0
0
—
0
b
a:
4
a
ft
$x
Ia,
E
=
,
11
H
L
—
0
+
L
U
0q
0
00
0
ft
I
L
II
ft
0
x
a
H
H