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4. Mosca vol I -Fisica-Tipler-5ta-Edicion-Vol-1.pdf
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4. Mosca vol I -Fisica-Tipler-5ta-Edicion-Vol-1.pdf
4. Mosca vol I -Fisica-Tipler-5ta-Edicion-Vol-1.pdf
4. Mosca vol I -Fisica-Tipler-5ta-Edicion-Vol-1.pdf
4. Mosca vol I -Fisica-Tipler-5ta-Edicion-Vol-1.pdf
4. Mosca vol I -Fisica-Tipler-5ta-Edicion-Vol-1.pdf
4. Mosca vol I -Fisica-Tipler-5ta-Edicion-Vol-1.pdf
4. Mosca vol I -Fisica-Tipler-5ta-Edicion-Vol-1.pdf
4. Mosca vol I -Fisica-Tipler-5ta-Edicion-Vol-1.pdf
4. Mosca vol I -Fisica-Tipler-5ta-Edicion-Vol-1.pdf
4. Mosca vol I -Fisica-Tipler-5ta-Edicion-Vol-1.pdf
4. Mosca vol I -Fisica-Tipler-5ta-Edicion-Vol-1.pdf
4. Mosca vol I -Fisica-Tipler-5ta-Edicion-Vol-1.pdf
4. Mosca vol I -Fisica-Tipler-5ta-Edicion-Vol-1.pdf
4. Mosca vol I -Fisica-Tipler-5ta-Edicion-Vol-1.pdf
4. Mosca vol I -Fisica-Tipler-5ta-Edicion-Vol-1.pdf
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4. Mosca vol I -Fisica-Tipler-5ta-Edicion-Vol-1.pdf
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4. Mosca vol I -Fisica-Tipler-5ta-Edicion-Vol-1.pdf
4. Mosca vol I -Fisica-Tipler-5ta-Edicion-Vol-1.pdf
4. Mosca vol I -Fisica-Tipler-5ta-Edicion-Vol-1.pdf
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4. Mosca vol I -Fisica-Tipler-5ta-Edicion-Vol-1.pdf
4. Mosca vol I -Fisica-Tipler-5ta-Edicion-Vol-1.pdf
4. Mosca vol I -Fisica-Tipler-5ta-Edicion-Vol-1.pdf
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4. Mosca vol I -Fisica-Tipler-5ta-Edicion-Vol-1.pdf
4. Mosca vol I -Fisica-Tipler-5ta-Edicion-Vol-1.pdf
4. Mosca vol I -Fisica-Tipler-5ta-Edicion-Vol-1.pdf
4. Mosca vol I -Fisica-Tipler-5ta-Edicion-Vol-1.pdf
4. Mosca vol I -Fisica-Tipler-5ta-Edicion-Vol-1.pdf
4. Mosca vol I -Fisica-Tipler-5ta-Edicion-Vol-1.pdf
4. Mosca vol I -Fisica-Tipler-5ta-Edicion-Vol-1.pdf
4. Mosca vol I -Fisica-Tipler-5ta-Edicion-Vol-1.pdf
4. Mosca vol I -Fisica-Tipler-5ta-Edicion-Vol-1.pdf
4. Mosca vol I -Fisica-Tipler-5ta-Edicion-Vol-1.pdf
4. Mosca vol I -Fisica-Tipler-5ta-Edicion-Vol-1.pdf
4. Mosca vol I -Fisica-Tipler-5ta-Edicion-Vol-1.pdf
4. Mosca vol I -Fisica-Tipler-5ta-Edicion-Vol-1.pdf
ANSWERS
Chapter 1
1. (c)
3. (c)
5. (n)
7. (e)
9. (n) True
(b) False
(c) True
11. 1.19 X 1057
13. (n) 10-8 m
(b) 20 atoms
15. (a) 3 X 1010 diapers
(b) 1.5 x 107 m3
(c) 0.6 mi2
17. $177 M
19. (n) 0.000040 W
(b) 0.000000004 s
(c) 3,000,000 W
(d) 25,000 m
21. (n) C1 is in m; C
2 is in 111/S
(b) CI is in m/s2
(c) C1 is in m/s2
(d) CI is in m; C
2
is ill S-I
(e) C1 is in m/s; C
2
is in S-l
23. (a) 4 x 107 m
(b) 6.37 X 106111
(c) 2.48 X 104 mi; 3.96 X 103 mi
25. 210 em
27. 1.28 km
29. (n) 36.0 km/h·s
(b) 10.0 m/s2
(e) 88.0 ft/s
(d) 26.8 m/s
31. 4050 m2
33. (n) m/s2
(b) s
(c) m
35. T-I
37. IIIv2/r
41. M/L3
43. (n) 30,000
45.
47.
49.
51.
53.
55.
57.
(b) 0.0062
(c) 0.000004
(d) 217,000
(a) 1.44 X 105
(b) 255 X 10-8
(c) 8.27 X 103
(d) 6.27 X 102
4 X 106
(a) 1.69 X 103
(b) 4.8
(c) 5.6
(d) 10
31.7 Y
2.0 x 1023
(n) 1.41 X 1017 kg/m3
(b) 216 m
(a) 4.85 X 10-6 parsec
(b) 3.08 X 106 m
(c) 9.47 x 1015 m
(d) 6.33 X 104 AU
(e) 3.25 c·y
59. The claim is conservative as the actual weight of water used is
closer to 55,000 tons.
61. (n) 11 = 3/2; C = 17.0 y/(Gm)3/2
(b) 0.510 Gm
63. 1.16 X 1019 lb
Chapter 2
--------------------
1. 0
3. It is safer to land against the wind.
5. (a) Negative
(b) During the last five steps, gradually slow the speed of
walking, until the wall is reached.
(c) 1,-----------,
O�--�-�--�
t (5)
A-l
A-2 Answers
7. (a) True
(b) True in one dimension
9. False
11. (a)
13. (b)
15. Yes. In any round-trip, A to B, and back to A, the average veloc­
ity is zero.
17. No. If the velocity is constant, a graph of position as a function
of time is linear with a constant slope equal to the velocity.
19. (b)
21. (a) False
(b) True
23. Vtop of fHght = 0; atop of fHght = -g
25. (b)
27. (e)
29. (e)
31. (c)
33. (a)
35. (d)
37. (d)
39. Velocity: (a) negative at 10 and 11; (b) positive at 13, 14, 16, and 17;
(e) zero at 12 and 15
Acceleration: (a) negative at 14; (b) positive at 12 and 16; (e) zero
at 1o, I" 13, 15, and 17
41. (a) Graphs (a), (f), and (i)
(b) Graphs (e) and (d)
(e) Graphs (a), (d), (e), ( f), (11), and (i)
(d) Graphs (b), (e), and (g)
(e) Graphs (a) and (i) are mutually consistent. Graphs (d) and
(h) are mutually consistent. Graphs (f) and (i) are also
mutually consistent.
43. (a) 54.2 m/s; (b) -123g
45.
47.
49.
51.
53.
55.
4.02 m/s2
14.2 ms
(a) 0.278 km/min
(b) -0.0833 km/min
(e) 0
(d) 0.128 km/min
(a) 2.25 h
(b) 4.99 h
(e) 880 km/h
(d) 611 km/h
(n) 4.33 y
(b) 4.33 x 106y; No
35.8 m
57. (a) 0
(b) 0.333 m/s
(e) -2.00 m/s
(d) 1.00 m/s
59. 122 km/h; 1.04v,v
61. (a) 400 ,-�������--��
350
300
250
:§: 200
(b) 15 s
(e) 300 m
(d) 100 m
63. 6 h
65. -2.00 m/s2
67. (a) 2 m
"
150
100
50
O ¥-��L+�--������
o 2 4 6 8 10 12 14 16 18 20
I (s)
(b) �x = (21 - 5)M + (Mj2, where x is in meters if I is in
seconds.
(e) v = 21 - 5, where v is in meters per second if I is in seconds.
69. (a) a,v,All = 3.33 m/s2; a,v,BC = 0; a,v,CE = -7.50 m/s2
(b) 75.0 m
(c) 90
80
70
60
:§: 50
" 40
30
20
10
0
0 1 2 3 4 5 6 7 8 9
I (s)
A B C 0
10
E
(d) At point 0, I = 8 s, the graph crosses the time axis and so
v = 0,
71. (a) 80.0 m/s
(b) 400 11l
(e) 40,0 m/s
73. 15.6 Ill/S2
75. (a) 4,68 s
(b) 20,4 11l
(e) 0.991 s and 3.09 s
77. (a)
(b) 7.27 m
(c) 1,73 s
S
><
(d) 11.9 m/s
8
7
6
5
4
3
2
0.2 0.4 0.6 0.8 1.0 1,2 1.4 1.6 1.8
t (s)
79. 43.6 m
81. 68.0 m/s
83. (n) 666 m
(b) 13.6 m/s
85. (n) 10.4 5
(b) 27.4 5
(C) 12.8 5
87. (n) 19.0 km
(b) 2 min 18 5
(c) 610 m/s
89. 40.0 cm/s; -6.88 cm/s2
91. (n) 4.76 m/s or 10.7 mi/h
(b) 0.595
93. 10.9 m
95. 27.6 m
97. 4.59 km
99. 2.40 m; 1.40 5
103. (n) -25.7 m/s2
(b) 2.33 5
105. (n) 1.03 coy/y2 = 1.03 ely
(b) = 2 d
107. 4.80 m/5 60
50
40
5 30
'"'
20
10
2 4 6 8 10
1 (5)
109. ! "
111. (n) 34.7 5
(b) 1.20 km
(c) 1400
1200
1000
S
800
'"' 600
400
200
0
0 5 10 15 20
1 (5)
113. (n) Ll = � L
(b) I = � If;n
12 14 16
25 30 35
Chapter 2
115. (n) A = 90 m 35
30
25
(j) 20
"
r
15
S
;::>
10
5
0
0 0.5 1 1.5 2 2.5
1 (5)
(b) x(t) = (3 m/s2)t2 + (3 111/5)1; nx = 90.0 m
117. x(t) = (� 111/S3)13 - (5 m/s)1
119. (n) 0.250 m/s per box
3
A-3
(b) v(l 5) = 0.930 m/s; v(2 5) = 3.20 m/s; v(3 5) = 6.20 m/s
121.
(c) x(3 s) = 7.00 m
123. (n) v(l) = (0.1 m/s3)12
(b) 2.23 m/s
125. 12.8 m/s2; 30.5%
7
6
5
4
3
2
o
o
V
/
v vs. I
/
/
V
/
2 3
1 (5)
4
127. (n) a = wvmaxcos(wl); Because a varies sinusoidally with time,
it is not constant.
VOla>:
(b) X = Xo + -
' [1 - cos(wI)]
w
A-4
129. (a)
(b)
Answers
(b = 1 s)
Because the numerical value of b, expressed in 51 units, is
one, the numerical values of a, v, and x are the same at
each instant in time.
Chapter 3
1 . The magnitude o f the displacement o f a particle is less than or
equal to the distance it travels along its path.
3. The displacement for any trip around the track is ZERO. Thus
we see that no matter how fast the race car travels, the average
velocity is always ZERO at the end of each complete circuit.
5. No. The magnitude of a component of a vector must be less
than or equal to the magnitude of the vector. If the angle ()
shown in the figure is equal to 0° or multiples of 90°, then the
magnitude of the vector and its component are equal.
7. No
9. (e)
11. (c)
13. (a) The velocity vector, as a consequence of always being in
the direction of motion, is tangent to the path.
(b) Ij
v
v
x
15. (a) A car moving along a straight road while braking
17.
(b) A car moving along a straight road while speeding up
(c) A particle moving around a circular track at constant
speed
(a)
(b)
(c)
t -
UV?
V2 1 -",
A� = ,,- v,
V1 t a = /:'vl/1t
� -v21t�= v2- v1
V1
1
V2 t a = Mil/1t
�
Vl
- - - ""V2 . V2
t;.V = V2- Vl ,
�
-V1
t a = /:,vl/1t
19.
21. True
23. (d)
25. (a) False
(b) True
27.
-VA
VB -VA
VB
29. (a)
Direction of velocity
Path vector
AB north
BC northeast
CD east
DE southeast
EF south
(b)
Direction of acceleration
Path vector
AB north
BC southeast
CD 0
DE southwest
EF north
(c) The magnitudes are approximately equal.
31. The droplet leaving the bottle has the same horizontal velocity
as the ship. During the time the droplet is in the air, it is also
moving horizontally with the same velocity as the rest of the
ship. Because of this, it falls into the vessel, whjch has the same
horizontal velocity. Because you have the same horizontal
velocity as the ship, you see the same thing as if the ship were
standing still.
33. True
35.
37.
39.
The principal reason is aerodynamic drag; when moving
through a fluid such as the atmosphere, the ball's acceleration
will depend strongly on its velocity.
14.8 m/s
R = 22.2 m; Cl' = 22.5°
N
Chapter 3 A-S
41. (11) Y 5l. D = (3 m){ + (3 m)] + (3 m)k; 0 = 5.20 m
53. v,v = (14.1 km/h)£ + (-4.1 km/h)]
�
55. (b)
A
57. (a) v". = (33.3 m/s)i + (26.7 m/s)]
A + B (b) a,v = (-3.00 m/s2)£ + (-1.77 m/s2)J
59. v = (30 m/s)£ + [40 m/s - (10 m/s2)11J; a = (-10 m/s2)]
x
61. (a) vov = (20 m/s)(- [ + J)
(b) y (b) a", = (-2 m/s2)i
(c) M = (600 m)(-£ + J)
63. (a) 13.1° west of north
(b) 300 km/h
A - B 65. 8.47°; 2.57 h
67. You should fly your plane acrOS5 the wind.
x 69. (a) rAil (6 5) = (120 m)i + (4 m)J
(b) VAB (6 5) = (-20 m/5)£ - (12 m/5)J
(c) y (c) aAB = (-2 m/s2)J
71. 1.52 X 10-6 m/52; 1.55 X 1O-7g
73. 3.44 X 1O-3g; 6.07 X 1O-4g
75. 33.4 min-1
77. 11 =
(va sin Ba)2
x 2g
(d) Y 79. 33.8 m/s
81. 20.3 m/s; 36.2°
x 83. 69.3°
85. (a) 18.0 111/5
B - A
(b) 14.0°
87. (11) 8.14 m/s
(b) 23.2 m/5
89. -63.4°
91. 209 m
(e) y 93. (a) 0.452 5
(b) 22.6 m
x
95. (a) 485 km
2B
(b) 1.70 km/5
101. L =
2v� tan B
g cos B
103. 10.8 m/s; v = (6.50 m/s)l + (-21.6 m/5)J
105. 40.5 m/5; 0.994 5
43. (b) 107. 7.41 111/S; 0.756 5; 15.9 m/5; 17.5 m/5; 25.0°
45. 109. 0.785 m
A () Ax Ay
lll. 4.91 m/s2; 8.50 m/52
(a) 10 m 30° 8.66 m 5 m 113. (a) y (m)
(b) 5 m 45° 3.54 m 3.54 m 25
(c) 7 km 60° 3.50 km 6.06 km
(d) 5 km 90° 0 5 km 20
(e) 15 km/s 150° - 13.0 km/5 7.50 km/s 15
(j) 10 m/5 240° -5.00 m/5 -8.66 m/5
(g) 8 m/s2 270° 0 -8.00 m/52 10
5
47. (a) 5.83; 31.0°
(b) 122; -35.0° 0 5 10 15
(c) 5.39; B = 42.1°; <p = 236° x (m)
49. (a) v = (5 m/s)i + (8.66 m/5)J (b) v = (5 m/5)[ + (10 m/5)J; 11.2 m/s
(b) A = (-3.54 m)[ + (-3.54 m)J
(c) r = (14 m)l - (6 m)J
A-6
115.
117.
119.
Answers
31.3°; S.06 m
Fourth step
x I g
(a) Vmin =
cos 0 j 2(x tan 0 - /1)
(b) vmin > 26.0 m/s = 5S.0 mi/h
(e) hmax < x tan 0
Chapter 4
1. Ifan object with no net force acting on it is at rest oris moving
with a constant speed in a straight line (i.e., with constant ve­
locity) relative to the reference frame, then the reference frame
is an inertial reference frame.
3. No. If the net force acting on an object is zero, its acceleration is
zero. The only conclusion one can draw is that the net force act­
ing on the object is zero.
5. No. Correctly predicting the direction of the subsequent
motion requires knowledge of the initial velocity as well as the
acceleration.
7. The mass of an object is an intrinsic property of the object
whereas the weight of an object depends directly on the local
gravitational field. Therefore, the mass of the object would not
change and Wgrav = I11g]oca]· Note that if the gravitational field is
zero then the gravitational force is also zero.
9. Your apparent weight would be greater than your true weight
when observed from a reference frame that is accelerating up­
ward. That is, when the surface on which you are standing has
an acceleration a such that ay is positive.
11. (a) Fn21 = 1111g
(b) F,112 = 1111g
(e) FnT2 = (1111 + 1112) g
(d) Fn2T = (1111 + 1112) g
13. (b)
15. (e)
17. (a)
y'
1
�
T1
w
1;'
19. (a) True
(b) False
(e) False
(d) False
21. (d)
(b)
Tf
F
T'
1
23. The velocity of the elevator has no effect on the person's
apparent weight.
25. (a) 7S2 N; 62.6 N
(b) Because there is no acceleration, the forces are the same
going up and going down the incline.
27. (a) 6.00 m/S2
(b) 1 /3
(e) 2.25 m/S2
29.
31.
33.
-3.75 kN
(a) 4.24 m/S2 @ 45.0° from each force
(b) S.40 m/s2 @ 14.6° from iFo
12.0 kg
35. (a) 4.00 m/S2
(b) 2.40 m/S2
37. (a) a = (1.50 m/s2)i + (-3.50 m/s2)j
(b) v = (4.50 m/s)i + (-10.5 m/s)j
(e) r = (6.75 m)i + (-15.S m)j
39. (a) 530 N
(b) 119 1b
41. (a) 60.0 N
(b) 57.7 N
43.
45.
47.
49.
51.
53.
T2 > T1
(a) 36.9°
(b) 4.0S N
(e) 3.43 N; 2.40 N; 3.43 N
(a) a = (0.500 m/ s2)i + (2.60 m/ S2)j
(b) 1 = (-5.00 N)i + (-26.0 N)j
(a)
(b)
(a)
(b)
(e)
(a)
(b)
T = _
,
_
v
_ . 0 = 90°· T -7 T as 0 -7 0°
2 sin Of I ITIi1X
19.6 N
11.S kN
9.S1 kN
7.S1 kN
3.S2 kN
4.30 kN
55. 56.0 N
57. (d)
59. (a) 50S N; 50S N
(b) 111g; 0
61. 552 N
63. (a)
65. (e)
67. (a) 19.6 N
(b) 19.6 N
(e) 25.6 N
(d) 14.6 N
69. (a) 1.31 m/S2
71.
75.
77.
79.
S1.
S3.
(b) 16.7 N; 21.3 N
(a)
(b)
(a)
(a)
(b)
(a)
(b)
(a)
(b)
(a)
(b)
a = _
_
F
_
_
. F
Fm1
1111 + 1112
' 2,1 111] + 111
2
0.400 m/s2; O.SOO N
g(m
2
- 1111 sin 0) gI111111
2
(1 + sin 0)
a = ; T = -
-'-
-=---
-
-
-
-
1111 + 111
2 1111 + 111
2
2.45 m/S2; 36.S N
1.37 m/s2; 61.4 N
1.19; The answer is the ratio of two quantities with the
same units and so has no units.
39S N
36S N
5.00 cm
aSkg = 4.91 m/s2; a20kg = 2.45 m/s2; T = 24.5 N
85.
87.
91.
93.
95.
97.
1.36 kg or 1.06 kg
4ml1Ti2g
F = 2T = ---
Ii'll + 111
2
(n) -100 m/52
(b) 6.13 cm
(c) 35.0 ms
305 N; 1.55 kN
(n)
(b)
(c)
(d)
(a)
(b)
(c)
(d)
1.50 m/s
1.50 m
0.500 m/s
12.0 N
F
a = ---
1111 + 1'/'12
Fm2
F = ---
net
IH 1 + m
2
Frll1
T = ---
m'1 + 1112
, �
�F
Yes . . . correct answers appear above.
99. (a) 55.0 g
(b) 2.45 m/s2; 2.03 N
101. (a) !(F
2
+ 2F1)
(b)
3To
4C
Chapter 5
1. The force of friction between the object and the floor of the
truck must be the force that causes the object to accelerate.
3. (d)
5. (b)
7. As thespring is extended, the force exerted by the spring on the
block increases. Once that force is greater than the maximum
value of the force ofstatic friction on theblock, the block will be­
gin to move. However, as it accelerates, it will shorten the length
of the spring, decreasing the force that the spring exerts on the
block. As this happens, the force of kinetic friction can then slow
the block to a stop, which starts the cycle over again. One inter­
esting application of this to the real world is the bowing of a vio­
lin string: the string under tension acts like the spring, while the
bow acts as the block, so as the bow is dragged across the string,
the string periodically sticks and frees itself from the bow.
9. (e)
11. Block 1 will hit the pulley before block 2 hits the wall.
13. (d)
15. (d)
17. For a rock, which has a relatively small surface area compared
to its mass, the terminal speed will be relatively high; for a light­
weight, spread-out object like a feather, the opposite is true.
Another issue is that the higher the terminal velocity is, the
longer it takes for a falling object to reach terminal velocity:
from this, the feather will reach its terminal velocity quickly,
19.
21.
23.
25.
27.
29.
31.
33.
35.
37.
39.
41.
43.
45.
47.
Chapter 5 A-7
and fall at an almost constant speed very soon after being
dropped; a rock, if not dropped from a great height, will have
almost the same acceleration as if it were in free-faU for the dura­
tion of its fall, and thus be continually speeding up as it falls.
(a) M/T; kg/s
(b) M/L; kg/m
(c) ML/T2
(d) 56.9 m/s
(e) 86.9 m/s
(b)
(a) 15.0 N
(b) 12.0 N
500 N
(a) -5.89 m/s2
(b) 76.4 m
(a) 49.1 N
(b) 123 N
4.57°
(a) 0.667
(b) 2.16 m/s2; 1.36 5
(a)
2.36 m/s2; 37.2 N
(n) 0.599
(b) 9.25 m
(c) 4.73 m/s
(a) 2.75 m/s2
(b) 10.1 s
(a) 0.965 m/s2directed up the incline
(b) 0.184 N
(a) 25.0°
(b) 0.118 N
(a)
(b)
The static-frictional force opposes the motion of the object,
and the maximum value of the static-frictional force is pro­
portional to the normal force FN• The normal force is equal
to the weight minus the vertical camponent Fv of the force
F. Keeping the magnitude F constant while increasing 0
from 0 results ill a decrease in F" and thus a corresponding
decrease in the maximum static-frictional force Im.x' The
object will begin to move if the horizontal component FI-I
of the force F exceeds Imax' An increase ill 0 results in a
decrease in FI-I' As 0 increases from 0, the decrease in FN is
larger than the decrease in FH, so the object is more
and more likely to slip. However, as 0 approaches 90°,
FH approaches 0, and no movement will be initiated. If F is
large enough and if 0 increases from 0, then at some value
of 0, the block will start to move.
240 .----,----,----,-----------,-----,-,
235 .
.. .........�...............:. . "......+.. ....... .+ ......... . : ........ .�

.. . +....... ......:.......... c · · · ,................:.. I
i
230 ' 
L
� 225
� :: "'
"...... ..
--!-
.
...
.... .... .f......... .....:............J',f!
210+. . ...... + ...
.'
...,.. ..........,..... .. ..j
/
/
..........:..........
.....
f"--�!
205 +-
--
--
�
--
-+
--
--
�
--
--
�
--
�
�
--
�
o 10 20 30 40 50 60
o (degrees)
A-8
49.
51.
53.
55.
57.
59.
61.
63.
65.
67.
Answers
(b) 1400
1 200
'1000
� 800
..... 600
400
200
a 10
(a) 0.238
(b) 1.40 In/s2
(n) 17.7 N
(b) 1.47 In/s2; 5.88 N
(e) 1.96 In/s2; 7.87 m/s2
(n) 0.163 In/s2
(b) 0.0381 In
(e) -0.254 In/S2
-8.41 glapp/plipp2; 0.191
(n) -1.57 N; 83.8 N
(b) 6.49 N; 37.5 N
(n) -2.60 In/s2; 19.2 m
(b) -2.11 m/s2; 23.7 m
(n) 0.297
(b) 2.82 In/s
(e)
(n) 1.41 Ill/S
(b) 8.50 N
69. (n) 8.33 In/s2; upward
(b) 542 N; upward
(e) 1.18 kN; upward
15 20 25 30 35 40 45
o (degrees)
71. T2 = [m2(LJ + L2)le;tTJ = [m2(LJ + L2) + 1I"1JLJ1(2
;)2
73. 53.3°; 410 N
75. (n) 0.395 N
(b) 0.644
77. 3.44 x 1O-3g; 6.07 x 1O-4g
79.
81.
83.
85.
(n)
V6( 1 )2
nc=� l +(IL;Vo}
(b) a, = -ILknc
(e) n = nch + IL�
12.8 Ill/S
(n) 7.25 m/s
(b) 0.536
87. 21.7°
89. (n) 7.832 kN
(b) -766 kN
91. vm;n = 20.1 kln/h; v
Olax = 56.1 kln/h
93. 2.79 x 10-4 kg/In
95. 88.2 kln/h
50
97.
99.
101.
103.
105.
107.
109.
111.
113.
115.
117.
3.31 s; 100
y(3.5 s) = 60.4 Ill; Ymax = 60.6 m @ t = 3.3 s; tlUghl = 7 s; The ball
spends a little longer coming down than it does going up.
0.511
(n) 0.289
(b) 600 N
1.49 kN
n= g(sin 01 - tan 00 cos 01)
(n) 49.4 m/s2
(b) 4.49 s
(n) 193 N
(b) 51.8 N
(e) The sled does not move.
(d) ILk is undetermined.
(e) 536 N
0.433
23.6 rev/min
(n) Toward the earth's axis.
(b) A stone dropped from a hand at a location on the earth.
The effective weight of the stone is equal to 11last, surl'
where aSI, surl is the acceleration of the falling stone
(neglecting air resistance) relative to the local surface of
the earth. The gravitational force on the stone is equal to
rnast, ;ner'
where aSI, ;ner is the acceleration of the stone rela­
tive to an inertial reference frame. These accelerations are
related by as!, surf + asurf, iner = as!, iner' where asurf, iner is the
acceleration of the local surface of the earth relative to the
inertial frame (the acceleration of the surface due to the ro­
tation of the earth). Multiplying through this equation by J1l.
and rearranging gives 1nasI, surf = 111asl. iner - tnasurf, iller'
which relates the apparent weight to the acceleration due
to gravity and the acceleration due to the earth's rotation.
A vector addition diagram can be used to show that the
magnitude of mast, surl is slightly less than that of maSI, ;ner'
(e) 983 cm/s2
Chapter 6
1. (n) False
(b) True
(e) True
3.
5.
7.
9.
11.
13.
15.
False
No. TI::.
e work done on any o!?Ject by any force F is defined as
dW == F ' df. The direction of Fnel is toward the center of the cir­
cle in which the object is traveling, and df is taJ�
&ent to the cir­
cle. No work is done by the net force because Fnel and df are
perpendicular, so the dot product is zero.
Because W IX x2, doubling the distance the spring is stretched
will require four times as much work.
(d)
(n) False
(b) False
(e) True
(n) False
(b) False
(n) 0.245 In
(b) 120 J
17. "" 1%
19. 20.8 kJ
21. (n) 147 J
(b) 266 J
23. 10.6 kJ
25. (n) 6.00 J
(b) 12.0 ]
(e) 3.46 m/s
27. W = -�kx12 - laxf
29. (a) m(y) = 40 kg - (1 kg/m)y
(b) 5.89 kJ
31 . (a) 4.17 N
35.
37.
39.
(b) T, Fg, and Fn; Because all of these forces act perpendicu­
larly to the direction of motion of the object, none of them
do any work.
1800
(a) -24
(b) -10
(e) 0
(n) 1.00 J
(b) 0.213 N
43. No. Let A = i, B = 3i + 4J and C = 3i - 4] and form A . B
and A · C.
45. (b) The results of (a) and (b) tell us that a is perpendicular to v
and parallel (or antiparallel) to r.
47. (a) 98.1 W
(b) 392 J
49. (a) v = (� m/s2)1
(b) P = 3.13I W/s
(e) 9.38 W
51. 445 W
53. v = Vvl + 2gH
55. 4.71 kJ
57. (a) 392 J
59.
61.
63.
(b) 2.45 m; 4.91 m/s
(e)
(d)
(n)
(b)
(n)
(b)
(e)
(a)
(b)
(e)
(d)
24.1 J; 368 J
392 J; 19.8 m/s
0.100 m
0.141 m
U(O) = (m
2
C
2
- 1111C1)g sin 0
U is a minimum at 0 = - 7T/2 and a maximum at 0 = 7T/2
U = 0 independently of 0
(
Fx = -
2
X
F" is positive for x '* 0 and therefore F is directed away
from the origin.
U(x) decreases with increasing x.
F" is negative for x '* 0 and therefore F is directed toward
the origin. U(x) increases with increasing x.
a
65. U(x) = - + Uo
x
67. (a) F" = 4x(x + 2)(x - 2)
(b) -2 m, 0, 2 m
(e) Unstable equilibrium at x = -2 m; stable equilibrium at
x = 0; unstable equilibrium at x = 2 111
Chapter 6
69. (a) 0 and 2 m; neutral equilibrium for x > 3 m
(b) 4.a .------,_--,----,'"'"'---,
3.5
3.0
2.5
....-., 2.0
:::; 1.5
1.0
0.5
0.0 +----".;L--i--;-.---I
-1 a 1 2 3
x (m)
A-9
(e) Stable equilibrium at x = 0; unstable equilibrium at x = 2 m
(d) 2.00 m/s
71. (a) U(y) = -/JIgy - 2Mg(L - V
y2 + d2)
! 1112
(b) y = d f 4M2 _
1112
(e) Stable equilibrium
73. (a) 706 MJ
(b) 11.8 MW
75. 0.500 m
77. (a) 34.4 N
(b)
T
_ 4 nun
/ 3 f0. _ -
1.6
..:.-- - y
79.
81.
1.68 N
(e) 3.38 mJ
,?-- -
(a) F(x) = rn(2x
(b) W = 1 111(2X12
In the following, if I is in seconds and 111 is in kilograms, then
v is in m/s, a is in m/s2, P is in W, and W is in J.
(n) v = (612 - 81); a = (121 - 8)
(b) P = 8rlll(9t2 - 181 + 8)
(e) W = 21111� (311 - 4)2
83. 5.74 km
85. (a) x
(m)
-4
-3
-2
-1
a
2
3
4
(b)
W
(J)
6
4
2
0.5
0
0.5
1.5
2.5
3
�
::J
a
-1
-2
-3
-4
-5
-3 -2 -1 0 1 2 3 4
x (m)
A- l 0 Answers
87. (b) W = (107T m)Fo if the rotation is clockwise; -(107T m)Fo if
the rotation is counterclockwise. Because W '" 0 for a com­
plete circuit, F is not conservative.
89. (a)
"6-1 2" Potential
0.120 ,..--,------,-----,---..,----,----.,---,------,----,
(b)
(c)
0.100
0.080
0.060
0.040
0.020
0.000 1-=1;k;�:±=±�rJ[1=J
-0.020
0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75
r (nm)
The minimum value is about -0.0107 eV, occurring at a
separation of approximately 0.380 nm. Because the func­
tion is concave upward at this separation, this separation
is one of stable equilibrium, although very shallow.
-6.69 X 10-12 N; 7.49 X 10-11 N
Ch a pter 7
1. (a)
3. (a) False
(b) False
5. As she starts pedaling, chemical energy inside her body is con­
verted into kinetic energy as the bikepicks up speed.As she rides it
up the hill, chemical energy is converted into gravitational poten­
tial energy. While freewheeling down the hill, potential energy is
converted to kinetic energy, and while braking to a stop, kinetic
energy is converted into thermal energy (a more random form of
kineticenergy) by thefrictionalforcesactingon the bike.
7. (d)
9. No. From the work-kinetic energy theorem, no total work is
being done on the rock, as its kinetic energy is constant. How­
ever, the rod must exert a tangential force on the rock to keep
the speed constant. The effect of this force is to cancel the com­
ponent of the force of gravity that is tangential to the trajectory
of the rock.
11. 33.6 s
13. 3.04 X 1019 J/y; = 6%
15. 1.10 x 106 Lis
17. (c)
19. 3.89 m
21. 5.05 m
23. 25.6°
25.
[111g(Sin 8 + f.L cos 8)]2
U =
5
27.
29.
31.
33.
6mg
(c)
6mg
(a) 31.0 m
(b) -31.7 J
(c) 33.7 mls
2k
35.
37.
39.
41.
43.
45.
47.
(a) 151 m
(b) 45.3 mls
(a) � mgL
(b) 6mg
(a) 20.2°
(b) 6.39 mls
V = L)2t(1 - COS 8) + �(V¥ - 3 COS 8 - 1r
(a) 94.2 kJ
(b) The energy required to do this work comes from chemical
energy stored in the body.
(c) 471 kJ
(a) 104 J
(b) 70.2 J
(c) 33.8 J
(d) 2.91 m/s
(a) 7.67 m/s
(b) 58.9 J
(c) 0.333
49. (a)
(b)
(c)
Wf = (13.7 N)y
Emeeh = -(13.7 N)y
1.98 m/s
51. 0.875 m; 2.49 mls
53. (a) 9.00 x 1013 J
(b) $2.5 X 106
(c) 28,400 y
55. 1.88 X 10-28 kg
57. 3.56 X 1014 reactions
59. 0.782 MeV
61. (a) 3.16 kg
(b) 8.04 X 109 kg
63. (b)
65. 57.6 MJ
67. (a) 0.208
(b) 3.45 MJ
69. (a) From the FBD we can see that the forces acting on the box
are the normal force exerted by the inclined plane, kinetic
friction force, and the gravitational force (the weight of
the box) exerted by the earth.
(b) 0.451 m
(c) l.33 J
(d) 2.52 mls
71. 11.3 kW; -6.77 kW
73. (a) 1.60 kJ
(b) 619 J
(c) 23.4 m/s
75. (a) 147 J
(b) The energy is transferred to the girder from its surround­
ings, which are warmer than the girder. As the tempera­
ture of the girder rises, the atoms in the girder vibrate
with a greater average kinetic energy, leading to a larger
average separation, which causes the girder's expansion.
77.
79.
81.
83.
87.
89.
91.
93.
(n) 0.8
0.6
0.4
::::0 0.2
::::;
0
-0.2
-0.4
0
Yeq
(b) F = -ky + Ing
(c)
21ng
Yrnnx = T
(d)
mg
Yeq =
T
(e)
11l2g2
Wf = --
2k
(n) 17.3 m
(b) 4.91 kN
(c) 4.91 m/s2
(d) 13.4 kN, upward
(e) 5.46 kN; 63.9°
(j) 1.44 kN
(a) 491 N; 981 N
(b) 9.82 kW; 29.4 kW
(c) 8.85°
(d) 6.36 kmlL
(n) 17.4 MJ
(b) 1 .39 x 1010
J
(c) 9.73 x 109 J
(d) 1.59 MW
(n)
2111gY
v = M + In
(b) v =
0 = 2VHL(1 - cos 0)
(n)
(b)
(c)
(n)
(b)
11l2g2
K",ax = II/gh +
U
mg
x = - +
IllilX
k
Ing
x = T
+
246
245
244
g
243
::::; 242
241
240
239
238
5.39 kJ
0
11l2g2 2111gh
-- + -­
F k
Potential Energy
50 100
s (m)
150 200
Chapter 8 A- l l
Chapter 8
1. A doughnut.
3. (b)
5. No. Consider a I-kg block with a speed of 1 mls and a 2-kg
block with a speed of 0.707 m/s. The blocks have equal
kinetic energies but momenta of magnitude 1 kg·m/s and
1.414 kg-m/s, respectively.
7. Precoil = Prine = - Pbullet or Prine + Pbullet = 0
9. Conservation of momentum requires only that the net external
force acting on the system be zero. It does not require the pres­
ence of a medium such as air.
11. Think of someone pushing a box across a floor. Her push on the
box is equal but opposite to the push of the box on her, but the
action and reaction forces act on dif
ferent objects. You can only
add forces when they act on the same object.
13. The problem is that the comic situations violate the conserva­
tion of momentum! To move forward requires pushing some­
thing backward, which Superman doesn't appear to be doing
when flying around. In a similar manner, if Superman picks up
a train and throws it at Lex Luthor, he (Superman) ought to be
tossed backward at a pretty high speed to satisfy the conserva­
tion of momentum.
15. The friction of the tire against the road causes the car to slow
down. This is rather subtle, as the tire is in contact with the
ground without slipping at all times, so as you push on the
brakes harder, the force of static friction of the road against
the tires must increase. Also, of course, the brakes heat up, and
not the tires.
17. Assume that the ball travels at 80 milh = 35 m/s. The ball
stops in a distance of about 1 cm, so the distance traveled is
about 2 cm at an average speed of about 18 m/s. The collision
0.02 m
time is -
-
1
- = 1 ms.
18 m s
19. (n) False
(b) True
(c) True
21. (a) The loss of kinetic energy is the same in both cases.
(b) The percentage loss is greatest for the case in which the
two objects have oppositely directed velocities of magni­
tude �v.
23. (b) is correct because all of l's kinetic energy is transferred to
2 when /1'12 = 1111,
25. The water is changing direction when it rounds the corner in
the nozzle. Therefore, the nozzle must exert a force on the
stream of water to change its direction, and, from Newton's 3rd
law, the water exerts an equal but opposite force on the nozzle.
27. No. FeXl,net = dpIdt defines the relationship between the net
force acting on a system and the rate at which its momentum
changes. The net external force acting on the pendulum bob is
the sum of the force of gravity and the tension in the string and
these forces do not add to zero.
29. Think of the stream of air molecules hitting the sail. Imagine
that they bounce off the sail elastically-their net change in mo­
mentulll is then roughly twice the change in momentum that
they experienced going through the fan. Another way of look­
ing at it: initially, the air is at rest, but after passing through the
fan and bouncing off the sail, it is moving backward; therefore,
the boat must exert a net force on the air pushiJlg it backward,
and there must be a force on the boat pushing it forward.
A- 1 2 Answers
31. (a) 2.33 s
(b) 6.74 m/s
33. (0.233 m, 0)
35. (2.00 m, 1.40 m)
37. (1.50 m, 1.36 m)
41. zem
= �R
43. vcm
= (3 m/s)f - (1.5 m/s)!
45. acm = (2.4 m/s2)1
47. (a) Fn = (rl'lp + 11Ib)g
(b) Fn = (l11p + 2l11b)g
(c) Fn = m�
49. (a) Fn = (11'lp + I11b)g
(b) F = 11'1 g + In g(l +
n
-p b
51. VlO = (4 m/s)f
53. v' = 2vf - v!
55. -Jii-f
57. (a) 43.5 J
(b) vcm
= (1.50 m/s)f
(c) vrcl = (3.50 m/s)f and V2•rel = (-3.50 m/s)f
(d) 36.8 J
(e) Kcm = 6.75 J = K - Krel
59. (a) 10.8 N·s
(b) 1.34 kN
61. 1.81 MN·s; 10.602 MN
63. 230 N
65. (a) 1 = (1.08 N·s)f (directed into wall)
67.
69.
71.
73.
(b) 360 N, into wall
(c)
(d)
(a)
(b)
(a)
(b)
(a)
(b)
(a)
(b)
(c)
0.480 N·s, away from wall
3.84 N, away from wall
20.0 m/s
20% of the initial kinetic energy is transformed into ther­
mal energy, sound, and the deformation of metal.
-2.00 m/s
The collision was inelastic.
von = (23.1 m/s)f
-254 m/s
5.00 m/s
0.250 m
Vlf = 0; V2f = 7.00 m/s
75. (a) 0.200vo
(b) 0.400vo
77. 450 m/s
81. h = - -
v2
(111,
)2
8g 1112
83. 0.0529
85. 1.50 X 106 m/s
87. (a) v, = (312 m/s)f + (66.6 m/s)!
(b) 5.61 km
(c) 35.8 kJ
89. 0.913
91. (a) 20% of its mechanical energy is lost.
(b) 0.894
93.
95.
97.
99.
101.
(a) 1.70 m/s
(b) 0.833
(a) 60°
(b) 2.50 m/s; 4.33 m/s
(a) 1.00 m/s; 1.73 m/s
(b) The collision was elastic.
v, = 8.66 mls; v2 = 5.00 mls
In an elastic collision
. = =
p,
2
[11l� + 6m,11I2 + In�
] =
p'f
[m�
.
+ 6111111
.
12 +. m�
]
K, Kr 2 2 ? ?
2 111,1112 + 11111112 2 rl'l,I1l2 + mimi
If p{ = +p" the particles do not collide.
103. (a) vcm = 0
(b)
(c)
(d)
(e)
105. (a)
(b)
(c)
107. (a)
(b)
(c)
u3 = (-5 m/s)f; Us = (3 m/s)f
u; = (5 m/s)f; u� = 0.75 m/s
v; = (5 m/s)f; v� = (-3 m/s)f
60.0J; 60.0 J
360 kN
120 s
1.72 km/s
ao
7 = 1 + -
o
g
( m 1
( 111
))
vr = gIsp ln� - - 1 - ----.!.
11Ir 70 mo
3
en 2 + ................: .
....
'-
E
C
"
O +---i---;--------;------+---I
o 2
(d) 28.1
109. 0.192 m/s; 31.3 mJ; 12.0 mJ
111. 0.462 m/s
4 6 8 10
113. (a) p = -(1.10 X 105 kg·km/h)f + (1.05 x 105 kg·km/h)!
(b) 43.4 km/h; 46.3° west of north
115. (a) 6.26 m/s
(b) 20.0 m
117. 3.72 m
119. (a) The velocity of the basketball will be equal in magnitude
but opposite in direction to the velocity of the baseball.
(b) Vlf = 0
(c) v2r = 2v
121. (a) 29.6 km/s
(b) 8.10; The energy comes from an immeasurably small
slowing of Saturn.
123. 3.00 x 105 m/s
125. (a) 0.600 m/s2
(b) 960 N
127. No. The driver was traveling at 23.3 km/h.
129. 8.85 kg
131. �r
133. (a) 0.716'1 Eo
(b) 55
135. (a)
(b)
(c)
v2
Ycm =
2L
12
v2
nCm = L
VI (v )
F = L
gt + 1 Mg
V
l f = 2 V
(m1 + I1lb) (/112 + I11b)
139. -0.960 m/s2
141. v = (1.70 mI/2/s) VL
Chapter 9
1. Because r is greater for the point on the rim, it moves the greater
distance. Both turn through the same angle. Because r is greater
for the point on the rim, it has the greater speed. .Both have the
same angular velocity. Both have zero tangential acceleration.
Both have zero angular acceleration. Because r is greater for the
point on the rim, it has the greater centripetal acceleration.
39.
41.
43.
45.
47.
49.
51.
55.
57.
(b) 124 rad/S2
(c) 620 rad/s
(a) g sin 8
Chapter 9 A- 1 3
(b) Because the line-of-action of the tension passes through
the pendulum'S pivot point, its lever arm is zero and it
causes no torque.
(c) g sin8
(a)
(b)
(c)
2i-<kMg
dT = --- r2dr
f
R2
Tf = �MRi-<kg
3Rw
M = --
4i-<kg
56.0 kgom2
(a) 28.0 kgom2
(b) 28.0 kg·m2
2.60 kg·m2
(b) [em = 12/11 (a2 + b2)
5.41 X 10- 47 kg·m2
1 = foMR2
(H2 R2)
J = 3M - + ­
x
5 20
3. � �. (a) 84.6 mJ
5. (d) (b) 347 rev/min
7. No. A net torque is required to change the rotational state of an 63. (a) 19.6 kN
object. A net torque may decrease the angular speed of an ob-
ject. AU we can say for sure is that a net torque will change the
angular speed of an object.
9. (b)
11. (b)
13. For a given applied force, this increases the torque about the
hinges of the door, which increases the door's angular accelera­
tion, leading to the door being opened more quickly. It is clear
that putting the knob far from the hinges means that the door
can be opened with less effort (force). However, it also means
that the hand on the knob must move through the greatest dis­
tance to open the door, so it may not be the quickest way to
open the door. Also, if the knob were at the center of the door,
you would have to walk aroLmd the door after opening it,
assuming the door is opening toward you.
15. (b)
17. (b)
19. (a)
21. True. If the sphere is slipping, then there is kinetic friction that
dissipates the mechanical energy of the sphere.
23. 10.3%
25.
27.
29.
31.
6.42
(a)
(b)
(c)
(d)
(a)
(b)
(a)
(b)
33. (d)
15.6 rad/s
46.8 rad
7.45 rev
73.0 m/s2
40.0 rad/s
0.960 m/s2; 192 m/s2
0.589 rad/s2
4.71 [ad
35. 1.04 radls; 9.92 rev/min
37. (a) 1 .87 Nom
(b) 5.89 kN-m
(c) 0.267 rad/s
(d) 1.57 kN
65. (a) 3.62 rad/s
(b) 3.62 rad/s
67. Unless M, the mass of the ladder, is zero, v, > vf. It is better to
let go and fall to the ground.
69. 3.11 m/s2; T1 = 12.5 N; T2 = 13.4 N
71. 8.23 m/s
73. (a)
g
a = ---
2M
1 + -
5rn
(b)
211lMg
T =
5111 + 2M
75. (a) 72.0 kg
(b) 1.37 rad/s2; TI = 294 N; T2 = 746 N
77. (a)
g sin 8
a = ---
1111
1 + -
2m2
(b) T =
�1111 g sin 8
1111
1 + -
21'/"12
(c) E = 1112gh
(d) Ebo!lom = 1112gh
(e) v = �
1111

1 + -
2m2
(j) For 8 = 0: a = T = 0
For 8 = 90°· a = -
-
g
-
- T = " 111 a and v =
. ' 1111 ' 2 l '
1 + -
2m2
For 1'1'11 = 0: a = g sin 8, T = 0, and v = V2ih
A- 1 4 Answers
79.
81.
83.
85.
87.
89.
9l.
93.
97.
99.
101.
103.
105.
107.
111.
113.
115.
117.
119.
121.
123.
125.
0.0864 m/s2; 3.14 m/s
0.192 m/s2; 0.962 N
1.13 kJ
45.9 m
19.5°
(a) a = �g sin 0
(b) I
s = �Illg sin 0
(c) Omax = tan-, (3,us)
v' = �v
223 J
(a)
2F
a = ; counterclockwise
R(M + 3111)
(b)
(c)
(a)
(b)
(a)
(b)
(c)
F
a = ---
C
M + 3m
2F
aCB = -
M + 3m
0.400 rad/S2; 0.200 rad/S2
4.00 N
12 v6 2 va
s, = - -- , tl = - --, and v
49 ,ukg 7 ,ukg
5/7
26.6 m; 3.88 s; 5.71 m/s
2rwo
v = --
7
(a) 360 kN
(b) 120 s
(c) 1.72 km/s
(a) v = 1.57 va
(b)
4 v
a
!It = --
7 ,ukg
v2
(c) �x = 0.735-
°
,ukg
1 = 2mR2
0.134 m
(a) 7.36 m/s2
(b) 14.7 m/S2
(c) 2.43 m/s
(a) 780 kJ
(b) 90.3 N·m; 150 N
(c)
(a)
(b)
(a)
(b)
(c)
(a)
(b)
1380 rev
15.0 m
15.4 rad/s
51
S3
S5
w - f4i
}3;
F = �Mg
,--
----:-
-
-
(a) v =
2MgD sin 0
I
M + -
(b) f =
Mg sin 0
s
R
1 + ­
r
r2
5
1 = "7vo
127. (a) 14.7 m/s2
(b) 66.7 cm
129. 41.7 J
131. The solid line on the graph shown below shows the position y
of the bucket when it is in free fall and the dashed line shows y
under the conditions modeled in this problem.
20 r---�---�--�--�
15
]: 10
;:,.,
5
0
0 0.5 1.0 1.5 2.0
t (s)
133. (a) 25.7 N
(b) 3.21 kg
(c) 1.10 m/s2
Chapter 1 0
1. (a) True
3.
5.
7.
9.
11.
13.
15.
17.
19.
21.
23.
25.
27.
(b) True
(c) True
90°
(a) Doubling p doubles L.
(b) Doubling r doubles L.
False
(e)
It is easier to crawl radially outward. In fact, a radially inward
force is required just to prevent you from sliding outward.
The hardboiled egg is solid inside, so everything rotates with a
uniform velocity. By contrast, it is difficult to get the viscous
fluid inside a raw egg to start rotating; however, once it is rotat­
ing, stopping the shell will not stop the motion of the interior
fluid, and, for this reason, the egg may start rotating again after
momentarily stopping.
(b)
(b)
(a) The lifting of the nose of the plane rotates the angular mo­
mentum vector upward. It veers to the right in response to
the torque associated with the lifting of the nose.
(b) The angular momentum vector is rotated to the right
when the plane turns to the right. In turning to the right,
the torque points down. The nose will move downward.
(b)
The center of mass of the rod-and-putty system moves in a
straight line, and the system rotates about its center of mass.
4.17 rev/s
(a) 2.40 X 10-8 kg·m2/s
(b) 5.22 X 1052; 2.29 X 1026
(c) The quantization of angular momentum is not noticed in
macroscopic physics because no experiment can differen­
tiate between e = 2 X 1026 and e = 2 X 1026 + 1.
29. (a) 0.331
(b) Because experimentally C < 2/5 = 0.4, the mass density
must be greater near the center of the earth.
31. 10.1 rad/s
33. T = FRk
35. (n) 24k
(b) -24J
(c) -51(
39. B = 4J + 3k
45. (a) 54.0 kgom2/s
(b) w increases as the particle approaches the point and de­
creases as it recedes.
47. (a) 1.33 x 10-5 kgom2/s
(b) 1.33 x 10-5 kgom2/s
(c) 1.33 x 10-5 kgom2/s
(d) 8.83 x 10-5 kgom2/s; -6.17 X 10-5 kgom2/s
49. (n) 4.00 Nom
(b) (0.192 rad/s2)t
51. (a) Tnet = Rg(m
2 sin e -m,)
(b) L = VR(;2
+ 111, + 111
2
)
g(m
2
sin e -111,)
(c) a = =---
-=-
----'-
I
- + 111, + 111
2
R
2
55. (a) 5.00 rev/s
(b) 622 J
(c) The energy comes from your internal energy.
57.
59.
61.
63.
65.
67.
69.
71.
9.67 mm/s
(n)
(b)
(c)
54.7°
(n)
(b)
Lo = rolrlvo
Ko = � II/V2
0
v2 3
T = F = 111�; W = --mv2
e ,.o 2 0
3.46 X 10-47 kgom2
1.99 meV; 5.98 meV; 12.0 meV
82.5 m/s
mv mMvd
v = ---' w =
em M + 111 ' tzMU(M + 11'1) + Mmd2
(0.5 M + 0.8111)(�Md2 + 0.64md2)g
v =
0.32dm2
(n)
K
Vem =
M
(b)
4K
M
(c)
2K
--
M
(d) X = � e
73. 0.349
75. 12 rad/s; 10.8 J
77. (n) 18.1 Jos
(b) 0.414 rad/s
(c) 15.2 s
(d) 0.0791 Jos
79. (n) r =-(47. 7 kgom2/s)k
(b) T = (15.9 Nom)k
81. (a) 243 JoS
(b) 306 J
83. (a) No, L decreases.
Chapter R
(b) Its kinetic energy is constant.
(c) Vo (The kinetic energy remains constant.)
85. Yes.
87 v = ew , I(U -(2)
.
, 2L v
91. (n) 0.228 rad/s
(b) 0.192 rad/s
93. 4.47 x 1022 Nom
95. 12.5 rad/s
97. (a) 26.5 rad/s
(b) r = (0.303 kgom2/s)e(1.4'S-')1
Chapter R
1. The friend in the car.
A- 1 5
3. Yes. If two events occur at the same time and place in one
reference frame they occur at the same time and place in aU ref­
erence frames. (Any pair of events that occur at the same time
and at the same place in one reference frame are called a space­
time coincidence.)
5.
7.
9.
11.
13.
15.
17.
19.
21.
23.
25.
27.
29.
31.
33.
(a)
1 + (8.61 x 10-11)
6.00 ns
(n) 6.63 m
(b) 12.6 m
(a) 599 m
(b) 13.4 km
(a) 129 y
(b) 87.6 Y
(a) 0.600 m
(b) 2.50 ns
0.800e
(a) 4.50 x 10-10%
(b) 0.142 ms
25.0 min; 25.0 min
60.0 min
0.400c; event B can precede event A provided II > 0.400e
(a) 11.3 y
(b) 40.0 Y
(a) 1.005
(b) 1.155
(c) 1.667
(d) 7.089
(a) 0.155Eo
(b) 1.29Eo
(c) 6.09Eo
35. 2.97 GeV
39. (b) 0.866c
(c) 0.999c
A- 1 6 Answers
41. (a) 0.794%
(b) 68.7%
43. (a) 0.943
45. (a) 617 eV
(b) 79.6 eV
(e) 7.96 eV
47. (a) 0.745
(b) 5.00 ft
(e) No. In Keisha's rest frame, the back end of the ladder will
clear the doorbefore the front end hits the wall of the shed,
while in Ernie's rest frame, the front end will hit the wall of
the shed while the back end has yet to reach the door.
Chapter 1 1
1. (a) False
(b) True
3. (d)
5. (a)
7. The gravitational field is proportional to the mass within the
sphere of radius r and inversely proportional to the square of r,
i.e., proportional to ,.3/r2 = 1".
9. (d)
11. 1.08 X 1011 Ms
13. (a) 2.78 h
(b) 19.3 X 1042 kgom2/s; 7.85 X 1042 kgom2/s; 0.703%
(e) 4.80 X 10-4 rad/s
15. 84.0 Y
17. (a) 1.59 X 1011 m
(b) 2.71 X 1010 m; 2.91 X 1011 m
19. (a) 90°
(b) 0.731 AU
21. (a) 1.90 X 1027 kg
(b) 0.282 m/s2; 0.0356 m/s2
23. (a) 8.18 X 104 s
(b) 1.22 X 109 m
25. 1.99 X 1030 kg
27. lOw, where w is your weight on earth.
29. 2.27 X 104 m/s
31. 1.43
33. (a) 7.37 m
(b) 0.0319 mm
35. 0.605
37.
39.
41.
(a) 2.27kg
(b) It is the inertial mass of 111
2
,
109 m
GMmo
W = --
R
43. 6.94 km/s
45. (a)
(b)
--> GMmo ,
Foutside = --
r
-
Z - r
GMrl"lo GMmo
U(r) = ---; U(R) = ---
r R
GMmo
(d) U(r) = U(R) = --
R
-
47.
49.
51.
53.
55.
57.
59.
61.
63.
65.
(e)
2.38 km/s R
19.4 km/s
(a) 62.7 MJ
(b) 17.4 kWoh
(e) $139
(a) 7.31 h
(b) 1.04 GJ
(e) 8.72 X 1012 J-s
11.1 GJ
g = (4 N/kg)1
(a)
� Gm < Gm <
g
=
U I + U'
(b) v'2 Gm
g =
2
U
(a) g = (-1.67 X 10-11 N/kg)1
(b) g = (-8.34 X 1O-12 N/kg)1
(e)
(a)
(b)
(a)
(b)
(e)
2.48 m
M = 1 CU
� 2GM [ ( Xo
)
g = -- In -- -
U Xo - L
0
0
3.20 X 10-9 N/kg
GrnM,
(b) F =
3.61a2
(e) 0
Co :L)]I
71. (a)
mg
F = - r
g R
(b) FN = C�- mUlZ}-
(e) The change in mass between you and the center of the
earth as you move away from the center is more impor­
tant than the rotational effect.
73. g(x) = GC7T
�oR3
)[�-
8(x �1R)Z]
75. Ul
= �47T
;oG
77. 0.104 mm/s
79. (a)
--> GMm [ � l'
F = ----;r- 1 - {d2 +
�2y/2 i
-, GMm ,
(b) F(R) = -0.821 J?2i
81 . 249 Y
83. (n)
(b) W = IIlgR�(�
_ _
_
1
_)
.
RE RE + h
85. 8.96 x 107 m
87. 1.70 Mm
(GM
91. v = 1 .64 y ----;-
-
n
-
GM
93. For r < RI, g(r) =0; For r > R" g(r) = -
, ;
- r-
GM(r3 - Rt)
For R, < r < R2, g(r) =
r2(R� _
R�)
2GA
95. g = -
}-
.
0.12 -,--,--.,---,--.,----,--.,---,-----,
0.10
0.08
:s 0.06
bl)
0.04
002
o +--+-+--+-�-+-���
o 2 3 4 5 6 7 8
I I
Rl R2
GMmo
(Xo + L/2
)
97. (b) U = --- In
L Xo - L/2
99. 33.5 pN
101. (n) The gravitational force is greater on the lower robot, so if
it were not for the cable its acceleration would be greater
than that of the upper robot, and they would separate. In
opposing this separation the cable is stressed.
(b) 220 km
Chapter 1 2
----------------------
1 . (n) False
(b) True
(c) True
(d) False
3. No. The definition of the center of gravity does not require that
there be any material at its location.
5. Th.is technique works because the center of mass must be di­
rectly under the balance point. Hence the intersection of the
two lines must be at the center of mass.
7. (b)
9. (c)
11. The tensile strengths of stone and concrete are at least an order
of magnitude lower than their compressive strengths, so you
want to build compressive structures to match their properties.
13. (b) 200 N/m
15. 318 N
Chapter 1 2 A- 1 7
1 7. (b) Taking long strides requires a larger coefficient of static
friction because 0 is then large.
(c) If ILs is small, that is, there is ice on the surface, 0 must be
small to avoid slipping.
19.
. _
(�n2b - 7TnR2 + 7TR3 �
)
21. (xcg' Ycg) - ab _ 7TR2 ' 2
b
23. 692 N; 900; 2.54 kN; No block is required to prevent the mast
from moving.
25. 0.728 m
1 V3
27. F2
=
"2
W; FI = -
2
-
W
29. (n) 5.00 III
(b) 4.87 m
� MgV
�
h(
-
2-
R-
-
-
J
--
l) A A
31. FI =
h - R
i + Mgj
33. (n) F = (30.0 N)[ + (30.0 N)J
(b) F = (35.0 N)l + (45.0 N)J
35. (n)
N-
F = Mg - F - -
n
h
(b) FC•h = F
(c) _
)2R - h
Fe." - F
h
37. (n) 6.87 N
(b) 1.65 Nom
(c) -8.26 N; 15.1 N
39. 636 N; 21.50
41. (n) 70.7 N
(b) 1.77 m
(c) 3.54 m
(d) 497 N
43. 7
net = (69.3 N)b - (40.0 N)n
45. 0 = � ( V3b - n)
35.7 m - 30.4x
47.
Y
=
3.57 m - (294m-l)x
12.0
10.0
8.0
I 6.0
;0-,
4.0
2.0
0
0.0 0.2 0.4 0.6
x (111)
0.8 1 .0 1.2
A- 1 8 Answers
49.
51.
53.
55.
57.
61.
63.
65.
69.
71.
73.
75.
77.
79.
81.
83.
85.
h = f.LsL tan e sin e
21-1
f.Ls = -
L
-
t
-
an
-
e
-
s
-
il
-
l
-
e
59.00
(a) 41.6 N
(b) 0.136%
5.010
(a) 1.82 x 106 N/m2
(b) 6.62 mJ
0.686
It will not support the elevator.
FL = 117 N; FR = 333 N
WI = 1.50 N; w
2
= 7.00 N; W
3
= 3.50 N
0.148
f.Ls < 0.500
f.Ls = 1 (cot e - 1)
(a) 147 N
(b) 3.62 m
The block will tip before it slides.
f.Ls < 0.500
(a) The stick remains balanced as long as the center of mass is
between the two fingers. For a balanced stick the normal
force exerted by tbe finger nearest the center of mass is
greater than that exerted by the other finger. Consequently,
a larger static-frictional force can be exerted by the finger
closer to tbe center of mass, which means the slipping
occurs at tbe otber finger.
(b) The finger farthest from the center of mass will slide in­
ward until the normal force it exerts on the stick is suffi­
ciently large to produce a kinetic-frictional force exceed­
ing the maximum static-frictional force exerted by the
other finger. At that point the finger that was not sliding
begins to slide, the finger that was sliding stops sliding,
and the process is reversed. When one finger is slipping
the other is not.
87. (a) 23.0 m/s
(b) 29.1 m/s
89. (c) Cs = 1.142 m; Clo = 1.464 m; Cwo = 2.594 m
91.
93.
(d) Increasing N in the spreadsheet solution suggests that the
sum of the individual offsets continues to grow as N in­
creases without bound. The series is, in fact, divergent
and the stack of bricks has no maximum offset or length.
566 N
mg R - r
Fn = 2111g; F = -- ; Fw = mg---:
=
=
=
=
cos e YR(2r - R)
Chapter 1 3
1. (e)
3. (d)
5. Nothing. The fish is in neutral buoyancy, so the upward accel­
eration of the fish is balanced by the downward acceleration of
the displaced water.
7. (b)
9. It blows over the ball, reducing the pressure above the ball to
below atmospheric pressure.
11.
13.
15.
17.
19.
21.
23.
25.
27.
29.
31.
33.
35.
False
The buoyant force acting on the ice cubes equals the weight of
the water they displace (i.e., B = wr = PrVrg). When the ice
melts, the volume of water displaced by the ice cubes will oc­
cupy the space previously occupied by the submerged part of
the ice cubes. Therefore the water level remains constant.
Because the pressure increases with depth, the object will be
compressed and its density will increase. Thus it will sink to
the bottom.
The drawing shows the beaker and a strip within the water. As
is readily established by a simple demonstration, the surface of
the water is not level while the beaker is accelerated, showing
that there is a pressure gradient. That pressure gradient results
in a net force on the small element shown in the figure.
From Bernoulli's principle, the opening above which the air
flows faster will be at a lower pressure than the other one,
which will cause a circulation of air in tbe tWlllel from opening
1 toward opening 2. It bas been shown tbat enough air will cir­
culate inside the tunnel even with the slightest breeze outside.
0.673 kg
103 kg
29.8 inHg
230 N
198 atm
(a) 14.8 kN
(b) 0.339 kg
0.453 m
pga3
F = -
8
37. 4.36 N
39. (a) 11.1 x 103 kg/m3
(b) lead
41. 800 kg/m3; 1.11
43. 250 kg/m3
45. 3.89 kg
47. 2.46 x 107 kg
49. 491 kN
51. (a) 9.28 cm/s
(b) 0.331 cm
(c) 8.31 cm, in reasonable agreement witb everyday experience.
53. (a) 12.0 m/s
(b) 133 kPa
55.
(c) The volume flow rates are equal.
(a) 4.58 L/min
(b) 763 cm2
57. 144 kPa
59. (a) 21.2 kg/s
(b) 636 kg·m/s
(c) 899 kg'm/s; 899 N
61. (a) x = 2Yh(H - 1-1)
(b) h = 1H ::': �YH2 - x2
63. (b) Ptop = Pot'" - pgd
65. 1.43 mm
67. 93.4 mi/h; Since most major league pitchers can throw a fast­
ball in the low-to-mid-90s, this drag crisis may very well play a
role in the game.
69. 0.0137; 0.0115
71. The net force is zero. Neglecting the thickness of the table, the
atmospheric pressure is the same above and below the surface
of the table.
73.
75.
77.
79.
81.
83.
1061 kg/m3
65.7%
If you are floating, the density (or specific gravity) of the liquid
in which you are floating is immaterial as you are in transla­
tional equilibrium under the influence of your weight and the
buoyant force on your body. Thus the buoyant force on your
body is your weight in both (a) and (b).
rn
V = --
0.96pw
11.8 cm
1 m is a reasonable diameter for the pipeline.
85. hA = 12.6 m; hs = 9.78 m
87. (a) 64.6%
(b) 10.7 kN
(c) 17.9 m/s2
Chapter 1 3
89. 3.31 X 10-3 mmHg or 3.31 /LmHg
91. 1.37
93. (a) 70.0 m3
(b) 7.47 m/s2
95. (c) 0.126 km-1
97. (a) 33.9 kN
(b) 39.8 kN; 36.1 kN
99. (c) 11 = Vii - -
- Vig t
( A,
Y
2A1
(d) 1 h 46 min
A- 1 9
A-20 Answers
Chapter 1 4
1. 0; 47T2j2A
3. (a) False
(b) True
(e) True
5. (a)
7. False
9. Assume that the first cart is given an initial velocity v by the
blow. After the initial blow, there are no external forces acting on
the carts, so their center of mass moves at a constant velocity v12.
The two carts will oscillate about their center of mass in simple
harmonic motion where the amplitude of their velocity is v12.
Therefore, when one cart has velocity vl2 with respect to the
center of mass, the other will have velocity -v/2. So, the velocity
with respect to the laboratory frame of reference will be +v
and 0, respectively. Half a period later, the situation is reversed;
so, one will appear to move as the other stops, and vice-versa.
11. True
13. Examples of driven oscillators include the pendulum of a clock,
31. (a) 7.85 m/s; 24.7 m/s2
(b) -6.28 m/s; -14.8 m/s2
33. (a) 0.313 Hz
(b) 3.14 s
(e) x = (40 cm)cos[(2 s-1)1 + 8]
35. 22.5 J
37. (a) 0.368 J
(b) 3.84 cm
39. 1.38 kN/m
41. (a) 6.89 Hz
(b) 0.145 s
(e) 0.100 m
(d) 4.33 m/s
(e) 187 m/s2
(j) 36.3 ms; 0
43. (a) 682 N/m
(b) 0.417 s
(e) 1.51 m/s
(d) 22.7 m/s2
a bowed violin string, and the membrane of any loudspeaker. 45. (a) 3.08 kN/m
15. Because f' varies inversely with the square root of 1n, taking
into account the effective mass of the spring predicts that the
frequency will be reduced.
17. (d)
19. (b)
21. 87T
23. (a) 3.00 Hz
(b) 0.333 s
(e) 7.00 cm
(d) 0.0833 s; Because v < 0, the particle is moving in the nega­
tive direction at I = 0.0833 s.
25. (a) x = (25 cm)cos[(4.19 s-1)1]
(b) v = -(105 cm/s)sin[(4.19 s-1)1]
(e) a = -(439 cm/s2)cos[(4.19 s-1)1]
27. (a) x = (27.7 cm)cos[(4.19 s-1)1 - 0.445]
(b) v = -(116 cm/s)sin[(4.19 s-1)1 - 0.445]
(e) a = -(486 cm/s2)cos[(4.19 s-1)1 - 0.445]
29. (a) 10 ",",--,--�--,------,------,--,----�...,...,
(b) If
(s)
1
2
3
4
I;
6
2
-2
-6
-10 +-�����L-�-+-��
0 1 2 3 4 5 6 7 8
I (s)
LlX
(s) (cm)
0 12.93 1
1 17.07 1
2 17.07 1
3 12.93 /
47.
49.
51.
53.
55.
57.
59.
(b) 4.16 Hz
(e) 0.240 5
(a) 0.438 m/s
(b) 0.379 m/s; 120 m/s2
(e) 95.5 ms
0.262 s
10.1 kJ
(a) 0.997 Hz
(b) 0.502 s
(e) 0.294 N
(a) 46.66 cm
(b) 0.261 s
(e) 0.767 m/s
(a) 0.270 J
(b) -0.736 J
(e) 1.01 J
(d) 0.270 J
(a) 1.90 cm
(b) 0.0542 J
(e) ::':0.224 J
(d) 0.334 J
61. 12.2 5
63. 11.7 s
65. T = 27T I L
jg(l - sin 0)
67. 1.10 5
69. 0.504 kg·m2
71. (b) 3.17 s
73. 21.1 cm from the center of the meter stick
77. (a) 1.63572 m
(b) 14.5 mm, upward
79. 13S
81.
85.
87.
89
91.
93.
95.
97.
99.
3.14%
(a) 0.314
(b) -3.13 X 10-2 percent
(a) 1.57%
(c) 0.430£0
(a) 1.01 Hz
(b) 2.01 Hz
(c) 0.352 Hz
(a) 4.98 cm
(b) 14.1 rad/s
(c) 35.4 cm
(d) 1.00 rad/s
(a) 0
(b) 4.00 m/s
(a) 14.1 cm; 0.444 s
(b) 23.1 cm; 0.363 s
(c) (14.1 cm)sin[(14.1 S-l)t]; (23.1 cm)sin[(17.3 S-l)t]
(a) v = -(1.2 m/s)sin [(3 rad/s)t + �]
(b) -0.849 m/s
(c) 1.20 m/s
(d) 1.31 s
(a) The normal force is identical to the tension in a string of
length r that keeps the particle moving in a circular path
and a component of mg provides, for small displacements
00 or 52' the linear restoring force required for oscillatory
motion.
(b) The particles meet at the bottom. Because 51 and 52 are
both much smaller than r, the particles behave like the
bobs of simple pendulums of equal length and, therefore,
have the same periods.
101. 1.62 s
103. 3.86 X 1O-7 N'm/rad
105. g' is closer tog than is goo. Thus the error is greater if the clock is
elevated.
107. (a)
Ak
f-L = -,---
-
-
-
5 (11'11 + m)g
(b) A is unchanged. £ is unchanged since £ = !kN. w is re­
duced by increasing the total mass of the system and T is
increased.
109. (b) 2.04 cm/s2
113. (a) x = 0
(b) v = x �
5
. uj�
(c) Xf = Xo
11'1
P
m + 11'1
b P
115. (a)
10TT-��-�-�-�--
8
:; 6
4
0.5 1.0 1.5 2.0 2.5 3.0
x/a
(b) Xo = a or ao = 1
Chapter 1 5
(c) U(xo + c:) = Uo[l + f3 + (1 + f3)-1]
(d)
82
U(Xo + 8) = constant + Uo?
a-
119. 6.44 X 1013 rad/s
121. 7.78 .Jff
123. (a) 0.0478
(b) 0.00228
127. (a)
A-2 1
0.6 ,--
--
--
--
--
--
--
--
---,-
--
--
----,--
--
--
�
--
-
/
----"
.
.
..
.
.
...
� �::+ ................... .-j..............................
j .. ................... +...................... ,..·......···..·..
7
·....
.
.
..
..
·
.
·
.
..
..
..
.
.
+
/
..
..
·..
..
..
..
..
..
···
..
-;/'
......
::J 0.2 + ........................,..............................ii .. . ............ . . !......
�
..............:;7'
C;
7
........................ .+- ....................... . 1
0.1 .... �
0.0
0.0 0.5 1.0 1.5
(b) r = ro; k = 2f320
(c) w = 2f3IQ
/-;;;
Chapter 1 5
r (nm)
2.0 2.5 3.0
1. The speed of a transverse wave on a rope is given by v =
�where F is the tension in the rope and f-L is its linear den­
sity. The waves on the rope move faster as they move up be­
cause the tension increases due to the weight of the rope below.
3. True
5. The speed of the wave v on the bullwhip varies with the ten­
sion F in the whip and its linear density f-L according to v =
�. As the whip tapers, the wave speed in the tapered end
increases due to the decrease in the mass density, so the wave
travels faster.
7. No; Because the source and receiver are at rest relative to each
other, there is no relative motion of the source and receiver and
there will be no Doppler shift in frequency.
9. The light from the companion star will be shifted about its
mean frequency periodically due to the relative approach to
and recession from the earth of the companion star as it re­
volves about the black hole.
11. (a) True
(b) False
(c) False
13. There was only one explosion. Sound travels faster in water
than air. Abel heard the sound wave in the water first, then,
surfacing, heard the sound wave traveling through the air,
which took longer to reach him.
15.
vy
4 6 8
X (cm)
A-22 Answers
17. Path C. Because the wave speed is highest in the water, and
more of path C is underwater than A or B, the sound wave will
spend the least time on path C.
19. (a) 78.5 m
(b) 69.7 m
(c) 70.5 m . . . about 1% larger than our result in part (b) and
11% smaller than our first approximation in (a).
21. 270 m/s; 20.6%
23. 1.32 km/s
25. 19.6 g
27. (a) 265 m/s
(b) 15.0 g
29. (b) 40.0 N
33. The lightning struck 680 m from the ball park, 58.4° W (or E) of
north.
39. (a) y(x,t) = A sin k(x - vt)
(b) y(x,t) = A sin 27TG- ft)
(c)
C
1
)
y(x,t) = A sin 27T i - yt
(d)
27i
y(x,t) = A sinA (x - vt)
(e) y(x,t) = A sin 27Tf (;- t)
41. 9.87 W
43. (a) The wave is traveling in the -x direction.; 5.00 m/s
(b) 10.0 cm; 50.0 Hz; 0.0200 s
(c) 0.314 m/s
45. (a) 6.82 J
(b) 44.0 W
47. (a) 79.0 mW
(b) Increasingfby a factor of 10 would increase Pa" by a factor
of 100. Increasing A by a factor of 10 would increase Pa" by
a factor of 100. Increasing F by a factor of 104 would in­
crease v by a factor of 100 and Pa" by a factor of 100.
(e) Depending on the adjustability of the power source, in­
creasingfor A would be the easiest.
49. (a) 0.750 Pa
(b) 4.00 m
(e) 85.0 Hz
(d) 340 m/s
51. (a)
(b)
53. (a)
(b)
55. (a)
(b)
(e)
57. (a)
(b)
3.68 X 10-5 m
8.27 X 10-2 Pa
The displacement s is zero.
3.68 fLm
138 Pa
21.7 W1m2
0.217 W
50.3 W
2.00 m
(e) 4.45 x 10-3 W1m2
59. (a) 20.0 dB
(b) 100 dB
61. 90.0 dB
65. (a) 100 m
(b) 0.126 W
67. (a) 100 dB
(b) 50.3 W
(e) 2.00 m
(d) 96.5 dB
69. (a) 81.1 dB
(b) 80.0 dB; Eliminating the two least intense sources does not
reduce the intensity level significantly.
71. 87.8 dB
73. 57.0 dB
75. (a) 260 m/s
(b) 1.30 m
(e) 262 Hz
77. (a) 1.70 m
(b) 247 Hz
79. 153 Hz
81. 1021 Hz or a fraction increase of 2.06%; Because this fractional
change in frequency is less than the 3% criterion for recognition
of a change in frequency, it would be impossible to use your
sense of pitch to estimate your running speed.
83. 349 mi/h
85. 7.78 kHz
87. 15.0 km west of P
89. (a) f' = (1 - u,/v)(l - usIV)-l fo
91. 1.33 m/s
93. (a) 824 Hz
(b) 849 Hz
95. 184 m
97. -2.07 x 10-5 nm; 99 2.25 x 108 m/s
99. 2.25 x 1018 m/s . . . where the upper arrow means the 8 is an
exponent.
101. 20.8 cm
103. 3.42 m/s
105. 529 Hz; 474 Hz
107. 7.99 m
109. (a) 55.1 NIm2
(b) 3.46 W1m2
(e) 0.109 W
111. 77.0 kN
113.
115.
117.
119.
204 m
24.0 cm
(b) Vo = Jf;
(e) As seen by an observer at rest, the pulse remains at the
same position because its speed along the chain is the
same as the speed of the chain. With respect to a fixed
point on the chain, the pulse travels through 360°.
(b) 2.21 s
Chapter 1 6
1.
t = 1 �I--------T--,..-----'If-------r-----!RI-------+------""'----'
1 = 2 1 �
1 = 3 r
l --
,-
-.
--
�
R
--
�
--
r-
-+
--
.-
-.
--
�
-.
3. (c)
5. (b)
7. (a)
9. since v C/. T, increasing the temperature increases resonant
frequencies.
11. No; the wavelength of a wave is related to its frequency and
speed of propagation (, = vlj). The frequency of the plucked
string will be the same as the wave it produces in air, but the
speeds of the waves depend on the media in which they are
propagating. Since the velocities of propagation differ, the
wavelengths will not be the same.
13. When the edges ofthe glass vibrate, sound waves areproduced
in the air in the glass. The resonance frequency of the air
columns depends on the length of the air column, which de­
pends on how much water is in the glass.
15. (b)
17. The pitch is determined mostly by the resonant cavity of the
mouth, and the frequency of sounds he makes is directly pro­
pOI·tional to their speed. Since vHe > va;r (see Equation 15-5), the
resonance frequency is higher if helium is the gas in the cavity.
19. Pianos are tuned by ringing the tuning fork and the piano note
simultaneously and tuning the piano string until the beats are
far apart (i.e., the time between beats is very long). If we assume
that 2 s is the maximum detectable period for the beats, then
one should be able to tune the piano string to at least 0.5 Hz.
21. 34.0 Hz; Because v C/. T, the frequency will be somewhat higher
in the summer.
23. 7.07 em
25. (a) 90.0°
(b) V
2A
27. (a) 0
(b) 210
(c) 410
29. (a) � ,
(b) � ,
31. (a) 60.0 cm
(b)
2'71"
-
5
(c) 24.0 mls
33. 4726 Hz; 9452 Hz
35. (b)
Y
(c) 0.500 mls
37. 1.81; 51.5"
39. (a) 0.279 m
(b) 1.22 kHz
(c) 111
3
8m
(rad)
0.432
4 0.592
5 0.772
6 0.992
7 1.354
8 undefined
(d) 0.0698 rad
41. 1.98 rad or 113°
43. (a) 70.5 Hz
Chapter 1 6 A-23
x (s)
-1 = 0.0
-I = 0.53
1 = 1.05
(b) The person on the street hears no beat frequency as the
sirens of both ambulances are Doppler shifted up by the
same amount (approximately 35 Hz).
45. (a) 2.00 m; 25.0 Hz
(b) Y3(x,l) = (4 mm)sin kx coswl, where k = '71"m-1 and w =
50'71"S-1
47. (a) 521 mls
(b) 2.80 m; 186 Hz
(c) 372 Hz; 558 Hz
49. 141 Hz
51. (a) 31.4 cm; 47.7 Hz
(b) 15.0 mls
(c) 62.8 cm
A-24 Answers
53. (a)
4 .... 4
2 2
E 0
E 0
"" ""
"., ".,
-2 -2
-4
0 0.5 1.0 1.5 2.0 2.5 2.0 2.5
x lm)
4
2
E 0
""
'"
-2
-4 .
0 0.5 1.0 1.5 2.0 2.5
x (m)
(b) 12.6 ms
(c) Since the string is moving either upward or downward
when y(x) = 0 for all x, the energy of the wave is entirely
kinetic energy.
55. (a) 70.8 Hz
(b) 4.89 Hz
(c) 35
57. 452 Hz; It would be better to have the pipe expand so that vlL,
where L is the length of the pipe, is independent of temperature.
59. (a) 80 cm
(b) 480 N
(c) You should place your finger 9.23 cm from the scroll
bridge.
61. (a) 75.0 Hz
(b) The harmonics are the 5th and 6th.
(c) 2.00 m
63. (a) 0.574 glm
(b) 1.29 g/m; 2.91 g/m; 6.55 glm
65. (a) The two sounds produce a beat because the th.irdharmonic
of the A string equals the second harmonjc of the E string,
and the original frequency of the E string is sligh.tly
greater than 660 Hz. If fE = (660 + Llf)Hz, a beat of 2Llf
will be heard.
(b) 661.5 Hz
(c) 79.6 N
69. 76.8 N; 19.2 N; 8.53 N
71. (a) Nlfa
(b) Llx/N
(c) 27TNILlX
(d) N is uncertain because the waveform dies out gradually
rather than stopping abruptly at some time; hence, where
the pulse starts and stops is not well defined.
73. (a) 3.40 kHz; 10.2 kHz; 17.0 kHz
(b) Frequencies near 3400 Hz will be most readily perceived.
75. ! A
77. 6.62 m
79. (a) 1.90 cm; 3.59 mls
(b) 0; 0
(c) 1.18 cm; 2.22 m/s
(d) 0; 0
81. (a) At resonance, standing waves are set up in the tube. At a
displacement antinode, the powder is moved about; at a
node the powder is stationary, and so it collects at the
nodes.
(b) 2fD
(c) If we let the length L of the tube be 1.2 m and assume that
Va;, = 344 mls (the speed of sound in air at 20°C), then the
10th harmonic corresponds to 0 = 25.3 cm and a driving
frequency of 680 Hz.
(d) Iff = 2 kHz and vHe = 1008 mls (the speed of sound in
helium at 20°), then 0 for the 10th harmonic in helium
would be 25.3 em, and 0 for the 10th harmonic in air
would be 8.60 cm. Hence, neglecting end effects at the
driven end, a tube whose length is the least common
multiple of 8.60 cm and 25.3 cm (218 cm) would work
well for the measurement of the speed of sound in either
air or helium.
83. (a) The pipe is closed at one end.
(b) 262 Hz
(c) 32.4 cm
85. (a) Y'(X,t) = (0.01 m)Sin[(�m-')x - (407TS-')t ];
)
/2
(x,t) = (0.01 m)Sin[(�m-I)x + (407T S-I)t];
(b) 2.00 m
(c) vy{l m,t) = -(2.51 m/s)sin(407T s-')t
(d) ay(l m,t) = -(316 m/s2)cos(407T s-')t
87. Y,es(x,t) = 0.1 sin(kx - wt)
89. (b) 203 Hz
91. (a) What you hear is the fundamental mode of the tube and
its overtones. A more physical explanation is that the echo
of the finger snap moves back and forth along the tube
with a characteristic time of 2L/c, leading to a series of
clicks from each echo. Since the clicks happen with a fre­
quency of e/2L, the ear interprets this as a musical note of
that frequency.
(b)
93. (a)
(b)
95. (a)
(b)
(c)
97. (a)
1 .5
38.6 cm
Since no conditions were placed on its derivation, this ex­
pression is valid for all harmonics.
1.54%
vy(x,t) = -w,AI sin wIt sin k,x - w2A2 sin w
2
t sin k2x
dK = �fL[w� Ai sin2 WIt sin2 k,x + 2w,w2A,A2 sin wIt
sin k,x sin w2t sin k
2x + wiAi sin2 W
2
t sin2 k2xJdx
K = ! l11wiAi sin2 WIt + ! l11w�A� sin2 wi
-1.5 L-----�----�----�----�----�----�----�
(b) f(27T) = 1 which is equivalent to the Liebnitz formula.
99. (b) 0.014 .,------r--------,,------,---------,
0.012
0.010 t ························
······,········r
.--
c............. ,
L--
...............................,....·················.···· ···· 1
/
0 008 · · · · ·· · · ··
1
<i 0.006 / ........; . .......................... ,................................;.
....................... . 1
0.00. /
/-
···
.
·
.
·
.
·
.
·
..............,...............................�................................;...........·················
·
··1
0.002 17
0.000 +------i----�c----+-----I
o 50 100
1/
150 200
(c) The frequency heard at any time is l/Llt", so because Llt"
increases over time, the frequency of the culvert whistler
decreases.; 7.65 kHz
Chapter 1 7
1. (n) False
(b) False
(c) True
(d) False
3. Mert's room was colder.
5. From the ideal-gas law we have P = nRTN. In the process de­
picted, both the temperature and the volume increase but the
temperature increases faster than does the volume. Hence the
pressure increases.
7. True
9. Kav increases by a factor of 2; Kav is reduced by a factor of � .
11. False
13. Since 107 » 273, it does not matter.
15. (b)
17. (d)
19. The ratio of the rms speeds is inversely proportional to the
square root of the ratio of the molecular masses. The kinetic
energies of the molecules are the same.
21. Because the temperature remains constant, the average speed
of the molecules remains constant. When the volume decreases,
the molecules travel less distance between collisions, so the
pressure increases because the frequency of collisions increases.
23. The average molecular speed of He gas at 300 K is about
1.4 km/s, so a significant fraction of He molecules have speeds
in excess of earth's escape velocity (11.2 km/s), and thus "leak"
away into space. Over time, the He content of the atmosphere
decreases to almost nothing.
25. (n) 3.61 X 103 K
(b) 225 K
(c) If Ven" > !ve or T 2:: 25T"m' H2 molecules escape. Therefore,
the more energetiC Hz molecules escape from the upper at­
mosphere.
(d) 164 K; 10.3 K; If we assume that the temperature on the
moon with an atmosphere would have been approximately
1000 K, then all O2 and H2 would have escaped during the
time since the formation of the moon to the present.
27. (a) 1.24 km/s
(b) 310 m/s
29.
31.
33.
35.
37.
39.
41.
Chapter 1 8
(c) 264 m/s
(d) O2, CO2, and H2 should be found on Jupiter.
1063°C
(n) 8.40 cm
(b) 107°C
-319°F
(n) 54.9 torr
(b) 3704 K
-40°C = -40°F
-183°C; -297°F
(n) B = 3.94 X 103 K; Ro = 3.97 X 10-3 fl
(b) 1.31 kO
(c) -389 fl/K; -433 fl/K
A-2S
(d) The thermistor is more sensitive (i.e., has greater sensitiv­
ity, at lower temperatures).
43. 1.79 mol; 1.08 X 1024 molecules
45. -83.2glips
47. (n) 3.66 X 103 mol
(b) 60.0 mol
49. 10.0 atm
51. 1.19 kg/m3
53. 2.56 N
55. (n) 276 m/s
(b) 872 m/s
57. 499 km/s; 2.07 X 10-16 J
61. K/LlU = 7.95 X 104
65. (n) 0.142 s
(b) 0.143 s
67. (n) 122 K
(b) 244 K
(c) 1.43 atm
69. 111 mol; 55.5 mol
71. 711'1H
73. 400.49 K
75. (n) 4.10 X 10-26 m
(b) 4.28 nm; The mean free path is larger by approximately a
factor of 1000.
77. (a) 48.9%
(b) 70.6%
Chapter 1 8
1. LlTB = 4LlT"
3. (c)
5. Yes, if the heat adsorbed by the system is equal to the work
done by the system.
7. Won + Qm = LlE;nt; For an ideal gas, LlE;nt is a hmction of T only.
Since W = 0 and Q = 0 in a free expansion, LlE;nt = 0 and T is
constant. For a real gas, LlE;nt depends on the density of the gas
because the molecules exert weak attractive forces on each other.
In a free expansion, these forces reduce the average kinetic
energy of the molecules and, consequently, the temperature.
9. The temperature of the gas increases. The average kinetic en­
ergy increases with increasing volume due to the repulsive in­
teraction between the ions.
A-26
11.
13.
15.
(a)
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(d)
Answers
False
False
True
True
True
True
True
17. If V decreases, the temperature decreases.
19. Theheatcapacity of a substance is proportional to thenumber of
degrees of freedom per molecule associated with the molecule.
Since there are 6 degrees of freedom per molecule in a solid, and
only 3 per molecule (translational) for a monatomic liquid, you
would expect the solid to have the higher heat capacity.
21. 1.63 min, an elapsed time that seems to be consistent with
experience.
23. ep = (1.01%)ew.,.,
25. (a) 10.5 MJ
(b) 121 W
27. 7.48 kcal
29. 48.8 mg
31. 365°C
33. 20.8°C
35. 453 kg
37. (a) O°C
(b) 125 g
39. (a) 4.94°C
(b) No ice is left.
41. (a) 2.99°C
(b) 199.8 g
(c) The answer would be the same.
43. 618°C
45. 2.21 kJ
47. 176°C
49. 53.7 J
51.
53.
55.
57.
59.
61.
63.
(a) 6.13 W
(b) 19.0 min
(a) 405 J
(b) 861 J
(a) 507 J
(b) 963 J
�PoVo
(a) 555 J
(b) 555 J
(a) 55.7 g/mol
(b) Fe
(a) 0; 6.24 kJ; 6.24 kJ
(b) 8.73 kJ; 6.24 kJ; 2.49 kJ
(c) 2.49 kJ
65. 59.6 L
67. �Cp = -!iNk
69. Cv,w.'er = 5Nk
71. (a) 465 K
(b) 387 K
73. (a) 300 K; 7.80 L; 1.14 kJ; 1.14 kJ
(b) 208 K; 5.41 L; 574 J; 0
75. (a) 263 K
(b) 10.8 L
(c) -1.48 kJ
(d) 1.48 kJ
79. -142 J
81. QO-;A = 8.98 kJ; QA-;B = 13.2 kJ; QB-;C = -8.98 kJ;
Qc-;o = -6.56 kJ; Wcycle = 6.62 kJ
83. (a) p
,

,
P2 - -,
"
P4 "
- - -' 4'
P3 - - -,- ,- - -
, , ,
Vj V4 V2
85. 180 kJ
87. (a) 65.2 K; 81.2 K
(b) 1.62 kJ
(c) 2.22 kJ
89. (a) 65.2 K; 81.2 K
(b) 2.65 kJ
(c) 3.25 kJ
91. (a) 9.20 X 10-2 J/kg'K
(b) 0.0584 J/kg
93. 47.6 kPa; 51.5 K; 71.2 K; 148 kPa
95. (a) 2.49 kJ
(b) 3.20 kJ
97. 171 K
99. (a) W = 0; Q = 3.74 kJ
T,;
T'
c
, 3
V3
V
(b) �U = 3.74 kJ; Q = 6.24 kJ; W = 2.50 kJ
101.
110 - - - - -
-10 "
�
__
__
__
__
__
__
__
__
__
__
__
L--
103. 4RT
105. 396 K
107. (a) 1Po
20353.5 771.5
(b) diatomic
3028.5 3048.5
t (s)
(c) In the isothermal process, T is constant, and the transla­
tional kinetic energy is unchanged. In the adiabatic process,
T3 = 1.32To, and the translational kinetic energy increases
by a factor of 1.32.
109. (a) 93.5 kPa
(b) 6266 K; 1.30 MPa
(c) 56.7 kPa
111. (b) flU = 4621 J, a result in good agreement with the result of
Problem 106.
Chapter 1 9
1. Friction reduces the efficiency of the engine.
3. Increasing the temperature of the steam increases the Carnot effi­
ciency, and generally increases the efficiency of any heat engine.
5. (c)
7. (d)
9. Note that A--7B is an adiabatic expansion. B--7C is a constant
volume process in which the entropy decreases; therefore heat
is released. C--7D is an adiabatic compression. D--7A is a con­
stant volume process that returns the gas to its original state.
The cycle is that of the Otto engine (see Figure 19-3).
11. 5
�
_
_
---, 3
1
V
13. p
C
"-----�A
�---------V
15. 56.5%
17. (a) 1.66 x 1017 W
(b) 5.66 x 1014 JIK·s
(c) 3.09 x 1013 J/K·s
19. 29.8 kJ/K
21. (a) 500 J
(b) 400 J
23. (a) 40.0%
(b) 80.0 W
25. (a)
2
tJ
1 4
10 20 30 40 50
V (L)
27.
Chapter 20
W1....2 = 0; Ql....2 = 3.74 kJ
W2....3 = 4.99 kJ; Q2....2 = 12.5 kJ
W3....4 = 0; Q3....4 = -7.48 kJ
W4....1 = 2.49 kJ; Q4....1 = -6.24 kJ
(b) 15.4%
2 

1.5 
]' - - �- �
�
,
1 ,
P.., 4, 2
0.5
- - - - , - - - .... ...._400 K
A-27
, 3.... -
- -
, - -300 K
,
0
0 10 20 30 40 50 60
V (L)
13.1%
29. (a) 600 K; 1800 K; 600 K
(b) 15.4%
31. (a) 5.16%; The fact that this efficiency is considerably less
than the actual efficiency of a human body does not con­
tradict the Second Law of Thermodynamics. The applica­
tion of the second law to chemical reactions such as the
ones that supply the body with energy have not been
discussed in the text.
(b)
35. (a)
(b)
(c)
(d)
Most warm-blooded animals survive under roughly the
same conditions as humans. To make a heat engine work
with appreciable efficiency, internal body temperatures
would have to be maintained at an unreasonably high level.
33.3%
33.3 J
66.7 J
2.00
37. Let the first engine be run as a refrigerator. Then it will remove
140 J from the cold reservoir, deliver 200 J to the hot reservoir,
and require 60 J of energy to operate. Now take the second
engine and run it between the same reservoirs, and let it eject
140 J into the cold reservoir, thus replacing the heat removed
by the refrigerator. If 82, the efficiency of this engine, is greater
than 30%, then Qh2' the heat removed from the hot reservoir by
this engine, is 140 J1(1 - 82) > 200 J, and the work done by this
engine is W = 82Qh2 > 200 J. The end result of all this is that the
second engine can run the refrigerator, replacing the heat taken
from the cold reservoir, and do additional mechanical work.
The two systems working together then convert heat into me­
chanical energy without rejecting any heat to a cold reservoir,
in violation of the second law.
39. (a) 33.3%
(b) If COP > 2, then 50 J of work will remove more than 100 J
of heat from the cold reservoir and put more than 150 J of
heat into the hot reservoir. So running engine (a) to oper­
ate the refrigerator with a COP > 2 will result in the trans­
fer of heat from the cold to the hot reservoir without doing
any net mechanical work in violation of the second law.
41. (a) 100°C
(b) Ql....2 = 3.12 kJ; Q2....3 = 0; Q3....1 = -2.91 kJ
(c) 6.73%
(d) 35.5%
43. (a) 5.26
(b) 3.l9 kW
(c) 4.81 kW
A-28 Answers
45. (a) 173 kJ
(b) 121 kJ
47. t.Su = 2.40 JIK
49. (a) 11.5 JIK
(b) Since the process is not quasi-static, it isnonreversible and
the entropy of the universe must increase.
51. 1.22 kJIK
53.
55.
57.
59.
61.
63.
65.
67.
(a) 0
(b) 267 K
(a) 244 kJ/K
(b) -244 kJ/K
(c) t.Su > 0
(a) - 117 J/K
(b) 137 J/K
(c) 20.3 J/K
1.97 kJ/K
(a) 0.417 J/K
(b) 125 J
(a) 20.0 J
(b) 66.7 J; 46.7 J
(a) 51.0%
(b) 102 kJ
(c) 98.0 kJ
113 W/K
69. (a) Process (1) is more wasteful of mechanical energy. Process
(2) is more wasteful of total energy.
(b) 1.67 JIK; 0.833 JIK
71. 313 K
73. 10.0 W
75. (a) 253 kPa
(b) 462 K
(c) 6.96 kJ; 25.9%
77. (a) 253 kPa
(b) 416 K
(c) 6.58 kJ; 34.8%
79. 180 J
83. =10478
-
Chapter 2 0
1. The glass bulb warms and expands first, before the mercury
warms and expands.
3. (c)
5. (a) With increasing altitude P decreases; from curve OF, T of
the liquid-gas interface diminishes, so the boiling temper­
ature decreases. Likewise, from curve OH, the melting
temperature increases with increasing altitude.
(b) Boiling at a lower temperature means that the cooking
time will have to be increased.
7. The thermal conductivity of metal and marble is much greater
than that of wood; consequently, heat transfer from the hand is
more rapid.
9. (c)
11. In the absence of matter to support conduction and convection,
radiation is the only mechanism.
13. (a)
15. The temperature of an object is inversely proportional to the
maximum wavelength at which the object radiates (Wein's dis­
placement law). Since blue light has a shorterwavelength than
red light, an object for which the wavelength of the peak of
thermal emission is blue is hotter than one which is red.
17. 18.1 mWl(m·K)
19. 2.90 nm
21. (a)
t.AIA
y = �
(b) y = 2at.T
23. 217°C
25. 15.4 x 10-6 K-l
27. 5.24 m
29. 0.255 mm
31. (a) The clock runs slow.
(b) 8.21 s
33. 3.68 x 10-12 N1m2
35. (a) 90°C
(b) 82°C
(c) 170 kPa
37. (b) (pr+ �2}3Vr - 1) = 8Tr
39. 2.07 kBtu/h
41. (a) leu = 962 W; fAl = 569 W
(b) 1.53 kW
(c) 0.0523 K/W
43. (a) Conservation of energy requires that the thermal current
through each shell be the same.
21TkL
f =
I ( I )
(T2 - TI)
n 1'1 1'2
45.
47. 9.35 X 10-3 m2
49. 1598°C
51. 2.10 km
53. 5767 K
55.
57.
59.
1.18 cm
f3exp - 13th �
(b) < �
f3t11
1.26 X 1010 kW; <0.002%
61. 132 W ignoring the cylindrical insulation; 142 W taking the in­
sulation into account.
63. L2 = LI; W2 = (1 - 2at.T)wI; E2 = EI(l - 2at.T)
65. (a) 0.698 cm/h
(b) 11.9 d
67. (b) 40.5 min
600 ,-------------------------.
�o -- - - -- - - - - - - - ---- - - - - - -
� 400
�
h 300
200
100
O +----,-----,----,----.--�
o 500 1000 1500
t (h)
2000 2500
4. Mosca vol I -Fisica-Tipler-5ta-Edicion-Vol-1.pdf
4. Mosca vol I -Fisica-Tipler-5ta-Edicion-Vol-1.pdf
4. Mosca vol I -Fisica-Tipler-5ta-Edicion-Vol-1.pdf

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4. Mosca vol I -Fisica-Tipler-5ta-Edicion-Vol-1.pdf

  • 615. ANSWERS Chapter 1 1. (c) 3. (c) 5. (n) 7. (e) 9. (n) True (b) False (c) True 11. 1.19 X 1057 13. (n) 10-8 m (b) 20 atoms 15. (a) 3 X 1010 diapers (b) 1.5 x 107 m3 (c) 0.6 mi2 17. $177 M 19. (n) 0.000040 W (b) 0.000000004 s (c) 3,000,000 W (d) 25,000 m 21. (n) C1 is in m; C 2 is in 111/S (b) CI is in m/s2 (c) C1 is in m/s2 (d) CI is in m; C 2 is ill S-I (e) C1 is in m/s; C 2 is in S-l 23. (a) 4 x 107 m (b) 6.37 X 106111 (c) 2.48 X 104 mi; 3.96 X 103 mi 25. 210 em 27. 1.28 km 29. (n) 36.0 km/h·s (b) 10.0 m/s2 (e) 88.0 ft/s (d) 26.8 m/s 31. 4050 m2 33. (n) m/s2 (b) s (c) m 35. T-I 37. IIIv2/r 41. M/L3 43. (n) 30,000 45. 47. 49. 51. 53. 55. 57. (b) 0.0062 (c) 0.000004 (d) 217,000 (a) 1.44 X 105 (b) 255 X 10-8 (c) 8.27 X 103 (d) 6.27 X 102 4 X 106 (a) 1.69 X 103 (b) 4.8 (c) 5.6 (d) 10 31.7 Y 2.0 x 1023 (n) 1.41 X 1017 kg/m3 (b) 216 m (a) 4.85 X 10-6 parsec (b) 3.08 X 106 m (c) 9.47 x 1015 m (d) 6.33 X 104 AU (e) 3.25 c·y 59. The claim is conservative as the actual weight of water used is closer to 55,000 tons. 61. (n) 11 = 3/2; C = 17.0 y/(Gm)3/2 (b) 0.510 Gm 63. 1.16 X 1019 lb Chapter 2 -------------------- 1. 0 3. It is safer to land against the wind. 5. (a) Negative (b) During the last five steps, gradually slow the speed of walking, until the wall is reached. (c) 1,-----------, O�--�-�--� t (5) A-l
  • 616. A-2 Answers 7. (a) True (b) True in one dimension 9. False 11. (a) 13. (b) 15. Yes. In any round-trip, A to B, and back to A, the average veloc­ ity is zero. 17. No. If the velocity is constant, a graph of position as a function of time is linear with a constant slope equal to the velocity. 19. (b) 21. (a) False (b) True 23. Vtop of fHght = 0; atop of fHght = -g 25. (b) 27. (e) 29. (e) 31. (c) 33. (a) 35. (d) 37. (d) 39. Velocity: (a) negative at 10 and 11; (b) positive at 13, 14, 16, and 17; (e) zero at 12 and 15 Acceleration: (a) negative at 14; (b) positive at 12 and 16; (e) zero at 1o, I" 13, 15, and 17 41. (a) Graphs (a), (f), and (i) (b) Graphs (e) and (d) (e) Graphs (a), (d), (e), ( f), (11), and (i) (d) Graphs (b), (e), and (g) (e) Graphs (a) and (i) are mutually consistent. Graphs (d) and (h) are mutually consistent. Graphs (f) and (i) are also mutually consistent. 43. (a) 54.2 m/s; (b) -123g 45. 47. 49. 51. 53. 55. 4.02 m/s2 14.2 ms (a) 0.278 km/min (b) -0.0833 km/min (e) 0 (d) 0.128 km/min (a) 2.25 h (b) 4.99 h (e) 880 km/h (d) 611 km/h (n) 4.33 y (b) 4.33 x 106y; No 35.8 m 57. (a) 0 (b) 0.333 m/s (e) -2.00 m/s (d) 1.00 m/s 59. 122 km/h; 1.04v,v 61. (a) 400 ,-�������--�� 350 300 250 :§: 200 (b) 15 s (e) 300 m (d) 100 m 63. 6 h 65. -2.00 m/s2 67. (a) 2 m " 150 100 50 O ¥-��L+�--������ o 2 4 6 8 10 12 14 16 18 20 I (s) (b) �x = (21 - 5)M + (Mj2, where x is in meters if I is in seconds. (e) v = 21 - 5, where v is in meters per second if I is in seconds. 69. (a) a,v,All = 3.33 m/s2; a,v,BC = 0; a,v,CE = -7.50 m/s2 (b) 75.0 m (c) 90 80 70 60 :§: 50 " 40 30 20 10 0 0 1 2 3 4 5 6 7 8 9 I (s) A B C 0 10 E (d) At point 0, I = 8 s, the graph crosses the time axis and so v = 0, 71. (a) 80.0 m/s (b) 400 11l (e) 40,0 m/s 73. 15.6 Ill/S2 75. (a) 4,68 s (b) 20,4 11l (e) 0.991 s and 3.09 s 77. (a) (b) 7.27 m (c) 1,73 s S >< (d) 11.9 m/s 8 7 6 5 4 3 2 0.2 0.4 0.6 0.8 1.0 1,2 1.4 1.6 1.8 t (s)
  • 617. 79. 43.6 m 81. 68.0 m/s 83. (n) 666 m (b) 13.6 m/s 85. (n) 10.4 5 (b) 27.4 5 (C) 12.8 5 87. (n) 19.0 km (b) 2 min 18 5 (c) 610 m/s 89. 40.0 cm/s; -6.88 cm/s2 91. (n) 4.76 m/s or 10.7 mi/h (b) 0.595 93. 10.9 m 95. 27.6 m 97. 4.59 km 99. 2.40 m; 1.40 5 103. (n) -25.7 m/s2 (b) 2.33 5 105. (n) 1.03 coy/y2 = 1.03 ely (b) = 2 d 107. 4.80 m/5 60 50 40 5 30 '"' 20 10 2 4 6 8 10 1 (5) 109. ! " 111. (n) 34.7 5 (b) 1.20 km (c) 1400 1200 1000 S 800 '"' 600 400 200 0 0 5 10 15 20 1 (5) 113. (n) Ll = � L (b) I = � If;n 12 14 16 25 30 35 Chapter 2 115. (n) A = 90 m 35 30 25 (j) 20 " r 15 S ;::> 10 5 0 0 0.5 1 1.5 2 2.5 1 (5) (b) x(t) = (3 m/s2)t2 + (3 111/5)1; nx = 90.0 m 117. x(t) = (� 111/S3)13 - (5 m/s)1 119. (n) 0.250 m/s per box 3 A-3 (b) v(l 5) = 0.930 m/s; v(2 5) = 3.20 m/s; v(3 5) = 6.20 m/s 121. (c) x(3 s) = 7.00 m 123. (n) v(l) = (0.1 m/s3)12 (b) 2.23 m/s 125. 12.8 m/s2; 30.5% 7 6 5 4 3 2 o o V / v vs. I / / V / 2 3 1 (5) 4 127. (n) a = wvmaxcos(wl); Because a varies sinusoidally with time, it is not constant. VOla>: (b) X = Xo + - ' [1 - cos(wI)] w
  • 618. A-4 129. (a) (b) Answers (b = 1 s) Because the numerical value of b, expressed in 51 units, is one, the numerical values of a, v, and x are the same at each instant in time. Chapter 3 1 . The magnitude o f the displacement o f a particle is less than or equal to the distance it travels along its path. 3. The displacement for any trip around the track is ZERO. Thus we see that no matter how fast the race car travels, the average velocity is always ZERO at the end of each complete circuit. 5. No. The magnitude of a component of a vector must be less than or equal to the magnitude of the vector. If the angle () shown in the figure is equal to 0° or multiples of 90°, then the magnitude of the vector and its component are equal. 7. No 9. (e) 11. (c) 13. (a) The velocity vector, as a consequence of always being in the direction of motion, is tangent to the path. (b) Ij v v x 15. (a) A car moving along a straight road while braking 17. (b) A car moving along a straight road while speeding up (c) A particle moving around a circular track at constant speed (a) (b) (c) t - UV? V2 1 -", A� = ,,- v, V1 t a = /:'vl/1t � -v21t�= v2- v1 V1 1 V2 t a = Mil/1t � Vl - - - ""V2 . V2 t;.V = V2- Vl , � -V1 t a = /:,vl/1t 19. 21. True 23. (d) 25. (a) False (b) True 27. -VA VB -VA VB 29. (a) Direction of velocity Path vector AB north BC northeast CD east DE southeast EF south (b) Direction of acceleration Path vector AB north BC southeast CD 0 DE southwest EF north (c) The magnitudes are approximately equal. 31. The droplet leaving the bottle has the same horizontal velocity as the ship. During the time the droplet is in the air, it is also moving horizontally with the same velocity as the rest of the ship. Because of this, it falls into the vessel, whjch has the same horizontal velocity. Because you have the same horizontal velocity as the ship, you see the same thing as if the ship were standing still. 33. True 35. 37. 39. The principal reason is aerodynamic drag; when moving through a fluid such as the atmosphere, the ball's acceleration will depend strongly on its velocity. 14.8 m/s R = 22.2 m; Cl' = 22.5° N
  • 619. Chapter 3 A-S 41. (11) Y 5l. D = (3 m){ + (3 m)] + (3 m)k; 0 = 5.20 m 53. v,v = (14.1 km/h)£ + (-4.1 km/h)] � 55. (b) A 57. (a) v". = (33.3 m/s)i + (26.7 m/s)] A + B (b) a,v = (-3.00 m/s2)£ + (-1.77 m/s2)J 59. v = (30 m/s)£ + [40 m/s - (10 m/s2)11J; a = (-10 m/s2)] x 61. (a) vov = (20 m/s)(- [ + J) (b) y (b) a", = (-2 m/s2)i (c) M = (600 m)(-£ + J) 63. (a) 13.1° west of north (b) 300 km/h A - B 65. 8.47°; 2.57 h 67. You should fly your plane acrOS5 the wind. x 69. (a) rAil (6 5) = (120 m)i + (4 m)J (b) VAB (6 5) = (-20 m/5)£ - (12 m/5)J (c) y (c) aAB = (-2 m/s2)J 71. 1.52 X 10-6 m/52; 1.55 X 1O-7g 73. 3.44 X 1O-3g; 6.07 X 1O-4g 75. 33.4 min-1 77. 11 = (va sin Ba)2 x 2g (d) Y 79. 33.8 m/s 81. 20.3 m/s; 36.2° x 83. 69.3° 85. (a) 18.0 111/5 B - A (b) 14.0° 87. (11) 8.14 m/s (b) 23.2 m/5 89. -63.4° 91. 209 m (e) y 93. (a) 0.452 5 (b) 22.6 m x 95. (a) 485 km 2B (b) 1.70 km/5 101. L = 2v� tan B g cos B 103. 10.8 m/s; v = (6.50 m/s)l + (-21.6 m/5)J 105. 40.5 m/5; 0.994 5 43. (b) 107. 7.41 111/S; 0.756 5; 15.9 m/5; 17.5 m/5; 25.0° 45. 109. 0.785 m A () Ax Ay lll. 4.91 m/s2; 8.50 m/52 (a) 10 m 30° 8.66 m 5 m 113. (a) y (m) (b) 5 m 45° 3.54 m 3.54 m 25 (c) 7 km 60° 3.50 km 6.06 km (d) 5 km 90° 0 5 km 20 (e) 15 km/s 150° - 13.0 km/5 7.50 km/s 15 (j) 10 m/5 240° -5.00 m/5 -8.66 m/5 (g) 8 m/s2 270° 0 -8.00 m/52 10 5 47. (a) 5.83; 31.0° (b) 122; -35.0° 0 5 10 15 (c) 5.39; B = 42.1°; <p = 236° x (m) 49. (a) v = (5 m/s)i + (8.66 m/5)J (b) v = (5 m/5)[ + (10 m/5)J; 11.2 m/s (b) A = (-3.54 m)[ + (-3.54 m)J (c) r = (14 m)l - (6 m)J
  • 620. A-6 115. 117. 119. Answers 31.3°; S.06 m Fourth step x I g (a) Vmin = cos 0 j 2(x tan 0 - /1) (b) vmin > 26.0 m/s = 5S.0 mi/h (e) hmax < x tan 0 Chapter 4 1. Ifan object with no net force acting on it is at rest oris moving with a constant speed in a straight line (i.e., with constant ve­ locity) relative to the reference frame, then the reference frame is an inertial reference frame. 3. No. If the net force acting on an object is zero, its acceleration is zero. The only conclusion one can draw is that the net force act­ ing on the object is zero. 5. No. Correctly predicting the direction of the subsequent motion requires knowledge of the initial velocity as well as the acceleration. 7. The mass of an object is an intrinsic property of the object whereas the weight of an object depends directly on the local gravitational field. Therefore, the mass of the object would not change and Wgrav = I11g]oca]· Note that if the gravitational field is zero then the gravitational force is also zero. 9. Your apparent weight would be greater than your true weight when observed from a reference frame that is accelerating up­ ward. That is, when the surface on which you are standing has an acceleration a such that ay is positive. 11. (a) Fn21 = 1111g (b) F,112 = 1111g (e) FnT2 = (1111 + 1112) g (d) Fn2T = (1111 + 1112) g 13. (b) 15. (e) 17. (a) y' 1 � T1 w 1;' 19. (a) True (b) False (e) False (d) False 21. (d) (b) Tf F T' 1 23. The velocity of the elevator has no effect on the person's apparent weight. 25. (a) 7S2 N; 62.6 N (b) Because there is no acceleration, the forces are the same going up and going down the incline. 27. (a) 6.00 m/S2 (b) 1 /3 (e) 2.25 m/S2 29. 31. 33. -3.75 kN (a) 4.24 m/S2 @ 45.0° from each force (b) S.40 m/s2 @ 14.6° from iFo 12.0 kg 35. (a) 4.00 m/S2 (b) 2.40 m/S2 37. (a) a = (1.50 m/s2)i + (-3.50 m/s2)j (b) v = (4.50 m/s)i + (-10.5 m/s)j (e) r = (6.75 m)i + (-15.S m)j 39. (a) 530 N (b) 119 1b 41. (a) 60.0 N (b) 57.7 N 43. 45. 47. 49. 51. 53. T2 > T1 (a) 36.9° (b) 4.0S N (e) 3.43 N; 2.40 N; 3.43 N (a) a = (0.500 m/ s2)i + (2.60 m/ S2)j (b) 1 = (-5.00 N)i + (-26.0 N)j (a) (b) (a) (b) (e) (a) (b) T = _ , _ v _ . 0 = 90°· T -7 T as 0 -7 0° 2 sin Of I ITIi1X 19.6 N 11.S kN 9.S1 kN 7.S1 kN 3.S2 kN 4.30 kN 55. 56.0 N 57. (d) 59. (a) 50S N; 50S N (b) 111g; 0 61. 552 N 63. (a) 65. (e) 67. (a) 19.6 N (b) 19.6 N (e) 25.6 N (d) 14.6 N 69. (a) 1.31 m/S2 71. 75. 77. 79. S1. S3. (b) 16.7 N; 21.3 N (a) (b) (a) (a) (b) (a) (b) (a) (b) (a) (b) a = _ _ F _ _ . F Fm1 1111 + 1112 ' 2,1 111] + 111 2 0.400 m/s2; O.SOO N g(m 2 - 1111 sin 0) gI111111 2 (1 + sin 0) a = ; T = - -'- -=--- - - - - 1111 + 111 2 1111 + 111 2 2.45 m/S2; 36.S N 1.37 m/s2; 61.4 N 1.19; The answer is the ratio of two quantities with the same units and so has no units. 39S N 36S N 5.00 cm aSkg = 4.91 m/s2; a20kg = 2.45 m/s2; T = 24.5 N
  • 621. 85. 87. 91. 93. 95. 97. 1.36 kg or 1.06 kg 4ml1Ti2g F = 2T = --- Ii'll + 111 2 (n) -100 m/52 (b) 6.13 cm (c) 35.0 ms 305 N; 1.55 kN (n) (b) (c) (d) (a) (b) (c) (d) 1.50 m/s 1.50 m 0.500 m/s 12.0 N F a = --- 1111 + 1'/'12 Fm2 F = --- net IH 1 + m 2 Frll1 T = --- m'1 + 1112 , � �F Yes . . . correct answers appear above. 99. (a) 55.0 g (b) 2.45 m/s2; 2.03 N 101. (a) !(F 2 + 2F1) (b) 3To 4C Chapter 5 1. The force of friction between the object and the floor of the truck must be the force that causes the object to accelerate. 3. (d) 5. (b) 7. As thespring is extended, the force exerted by the spring on the block increases. Once that force is greater than the maximum value of the force ofstatic friction on theblock, the block will be­ gin to move. However, as it accelerates, it will shorten the length of the spring, decreasing the force that the spring exerts on the block. As this happens, the force of kinetic friction can then slow the block to a stop, which starts the cycle over again. One inter­ esting application of this to the real world is the bowing of a vio­ lin string: the string under tension acts like the spring, while the bow acts as the block, so as the bow is dragged across the string, the string periodically sticks and frees itself from the bow. 9. (e) 11. Block 1 will hit the pulley before block 2 hits the wall. 13. (d) 15. (d) 17. For a rock, which has a relatively small surface area compared to its mass, the terminal speed will be relatively high; for a light­ weight, spread-out object like a feather, the opposite is true. Another issue is that the higher the terminal velocity is, the longer it takes for a falling object to reach terminal velocity: from this, the feather will reach its terminal velocity quickly, 19. 21. 23. 25. 27. 29. 31. 33. 35. 37. 39. 41. 43. 45. 47. Chapter 5 A-7 and fall at an almost constant speed very soon after being dropped; a rock, if not dropped from a great height, will have almost the same acceleration as if it were in free-faU for the dura­ tion of its fall, and thus be continually speeding up as it falls. (a) M/T; kg/s (b) M/L; kg/m (c) ML/T2 (d) 56.9 m/s (e) 86.9 m/s (b) (a) 15.0 N (b) 12.0 N 500 N (a) -5.89 m/s2 (b) 76.4 m (a) 49.1 N (b) 123 N 4.57° (a) 0.667 (b) 2.16 m/s2; 1.36 5 (a) 2.36 m/s2; 37.2 N (n) 0.599 (b) 9.25 m (c) 4.73 m/s (a) 2.75 m/s2 (b) 10.1 s (a) 0.965 m/s2directed up the incline (b) 0.184 N (a) 25.0° (b) 0.118 N (a) (b) The static-frictional force opposes the motion of the object, and the maximum value of the static-frictional force is pro­ portional to the normal force FN• The normal force is equal to the weight minus the vertical camponent Fv of the force F. Keeping the magnitude F constant while increasing 0 from 0 results ill a decrease in F" and thus a corresponding decrease in the maximum static-frictional force Im.x' The object will begin to move if the horizontal component FI-I of the force F exceeds Imax' An increase ill 0 results in a decrease in FI-I' As 0 increases from 0, the decrease in FN is larger than the decrease in FH, so the object is more and more likely to slip. However, as 0 approaches 90°, FH approaches 0, and no movement will be initiated. If F is large enough and if 0 increases from 0, then at some value of 0, the block will start to move. 240 .----,----,----,-----------,-----,-, 235 . .. .........�...............:. . "......+.. ....... .+ ......... . : ........ .� .. . +....... ......:.......... c · · · ,................:.. I i 230 ' L � 225 � :: "' "...... .. --!- . ... .... .... .f......... .....:............J',f! 210+. . ...... + ... .' ...,.. ..........,..... .. ..j / / ..........:.......... ..... f"--�! 205 +- -- -- � -- -+ -- -- � -- -- � -- � � -- � o 10 20 30 40 50 60 o (degrees)
  • 622. A-8 49. 51. 53. 55. 57. 59. 61. 63. 65. 67. Answers (b) 1400 1 200 '1000 � 800 ..... 600 400 200 a 10 (a) 0.238 (b) 1.40 In/s2 (n) 17.7 N (b) 1.47 In/s2; 5.88 N (e) 1.96 In/s2; 7.87 m/s2 (n) 0.163 In/s2 (b) 0.0381 In (e) -0.254 In/S2 -8.41 glapp/plipp2; 0.191 (n) -1.57 N; 83.8 N (b) 6.49 N; 37.5 N (n) -2.60 In/s2; 19.2 m (b) -2.11 m/s2; 23.7 m (n) 0.297 (b) 2.82 In/s (e) (n) 1.41 Ill/S (b) 8.50 N 69. (n) 8.33 In/s2; upward (b) 542 N; upward (e) 1.18 kN; upward 15 20 25 30 35 40 45 o (degrees) 71. T2 = [m2(LJ + L2)le;tTJ = [m2(LJ + L2) + 1I"1JLJ1(2 ;)2 73. 53.3°; 410 N 75. (n) 0.395 N (b) 0.644 77. 3.44 x 1O-3g; 6.07 x 1O-4g 79. 81. 83. 85. (n) V6( 1 )2 nc=� l +(IL;Vo} (b) a, = -ILknc (e) n = nch + IL� 12.8 Ill/S (n) 7.25 m/s (b) 0.536 87. 21.7° 89. (n) 7.832 kN (b) -766 kN 91. vm;n = 20.1 kln/h; v Olax = 56.1 kln/h 93. 2.79 x 10-4 kg/In 95. 88.2 kln/h 50 97. 99. 101. 103. 105. 107. 109. 111. 113. 115. 117. 3.31 s; 100 y(3.5 s) = 60.4 Ill; Ymax = 60.6 m @ t = 3.3 s; tlUghl = 7 s; The ball spends a little longer coming down than it does going up. 0.511 (n) 0.289 (b) 600 N 1.49 kN n= g(sin 01 - tan 00 cos 01) (n) 49.4 m/s2 (b) 4.49 s (n) 193 N (b) 51.8 N (e) The sled does not move. (d) ILk is undetermined. (e) 536 N 0.433 23.6 rev/min (n) Toward the earth's axis. (b) A stone dropped from a hand at a location on the earth. The effective weight of the stone is equal to 11last, surl' where aSI, surl is the acceleration of the falling stone (neglecting air resistance) relative to the local surface of the earth. The gravitational force on the stone is equal to rnast, ;ner' where aSI, ;ner is the acceleration of the stone rela­ tive to an inertial reference frame. These accelerations are related by as!, surf + asurf, iner = as!, iner' where asurf, iner is the acceleration of the local surface of the earth relative to the inertial frame (the acceleration of the surface due to the ro­ tation of the earth). Multiplying through this equation by J1l. and rearranging gives 1nasI, surf = 111asl. iner - tnasurf, iller' which relates the apparent weight to the acceleration due to gravity and the acceleration due to the earth's rotation. A vector addition diagram can be used to show that the magnitude of mast, surl is slightly less than that of maSI, ;ner' (e) 983 cm/s2 Chapter 6 1. (n) False (b) True (e) True 3. 5. 7. 9. 11. 13. 15. False No. TI::. e work done on any o!?Ject by any force F is defined as dW == F ' df. The direction of Fnel is toward the center of the cir­ cle in which the object is traveling, and df is taJ� &ent to the cir­ cle. No work is done by the net force because Fnel and df are perpendicular, so the dot product is zero. Because W IX x2, doubling the distance the spring is stretched will require four times as much work. (d) (n) False (b) False (e) True (n) False (b) False (n) 0.245 In (b) 120 J
  • 623. 17. "" 1% 19. 20.8 kJ 21. (n) 147 J (b) 266 J 23. 10.6 kJ 25. (n) 6.00 J (b) 12.0 ] (e) 3.46 m/s 27. W = -�kx12 - laxf 29. (a) m(y) = 40 kg - (1 kg/m)y (b) 5.89 kJ 31 . (a) 4.17 N 35. 37. 39. (b) T, Fg, and Fn; Because all of these forces act perpendicu­ larly to the direction of motion of the object, none of them do any work. 1800 (a) -24 (b) -10 (e) 0 (n) 1.00 J (b) 0.213 N 43. No. Let A = i, B = 3i + 4J and C = 3i - 4] and form A . B and A · C. 45. (b) The results of (a) and (b) tell us that a is perpendicular to v and parallel (or antiparallel) to r. 47. (a) 98.1 W (b) 392 J 49. (a) v = (� m/s2)1 (b) P = 3.13I W/s (e) 9.38 W 51. 445 W 53. v = Vvl + 2gH 55. 4.71 kJ 57. (a) 392 J 59. 61. 63. (b) 2.45 m; 4.91 m/s (e) (d) (n) (b) (n) (b) (e) (a) (b) (e) (d) 24.1 J; 368 J 392 J; 19.8 m/s 0.100 m 0.141 m U(O) = (m 2 C 2 - 1111C1)g sin 0 U is a minimum at 0 = - 7T/2 and a maximum at 0 = 7T/2 U = 0 independently of 0 ( Fx = - 2 X F" is positive for x '* 0 and therefore F is directed away from the origin. U(x) decreases with increasing x. F" is negative for x '* 0 and therefore F is directed toward the origin. U(x) increases with increasing x. a 65. U(x) = - + Uo x 67. (a) F" = 4x(x + 2)(x - 2) (b) -2 m, 0, 2 m (e) Unstable equilibrium at x = -2 m; stable equilibrium at x = 0; unstable equilibrium at x = 2 111 Chapter 6 69. (a) 0 and 2 m; neutral equilibrium for x > 3 m (b) 4.a .------,_--,----,'"'"'---, 3.5 3.0 2.5 ....-., 2.0 :::; 1.5 1.0 0.5 0.0 +----".;L--i--;-.---I -1 a 1 2 3 x (m) A-9 (e) Stable equilibrium at x = 0; unstable equilibrium at x = 2 m (d) 2.00 m/s 71. (a) U(y) = -/JIgy - 2Mg(L - V y2 + d2) ! 1112 (b) y = d f 4M2 _ 1112 (e) Stable equilibrium 73. (a) 706 MJ (b) 11.8 MW 75. 0.500 m 77. (a) 34.4 N (b) T _ 4 nun / 3 f0. _ - 1.6 ..:.-- - y 79. 81. 1.68 N (e) 3.38 mJ ,?-- - (a) F(x) = rn(2x (b) W = 1 111(2X12 In the following, if I is in seconds and 111 is in kilograms, then v is in m/s, a is in m/s2, P is in W, and W is in J. (n) v = (612 - 81); a = (121 - 8) (b) P = 8rlll(9t2 - 181 + 8) (e) W = 21111� (311 - 4)2 83. 5.74 km 85. (a) x (m) -4 -3 -2 -1 a 2 3 4 (b) W (J) 6 4 2 0.5 0 0.5 1.5 2.5 3 � ::J a -1 -2 -3 -4 -5 -3 -2 -1 0 1 2 3 4 x (m)
  • 624. A- l 0 Answers 87. (b) W = (107T m)Fo if the rotation is clockwise; -(107T m)Fo if the rotation is counterclockwise. Because W '" 0 for a com­ plete circuit, F is not conservative. 89. (a) "6-1 2" Potential 0.120 ,..--,------,-----,---..,----,----.,---,------,----, (b) (c) 0.100 0.080 0.060 0.040 0.020 0.000 1-=1;k;�:±=±�rJ[1=J -0.020 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 r (nm) The minimum value is about -0.0107 eV, occurring at a separation of approximately 0.380 nm. Because the func­ tion is concave upward at this separation, this separation is one of stable equilibrium, although very shallow. -6.69 X 10-12 N; 7.49 X 10-11 N Ch a pter 7 1. (a) 3. (a) False (b) False 5. As she starts pedaling, chemical energy inside her body is con­ verted into kinetic energy as the bikepicks up speed.As she rides it up the hill, chemical energy is converted into gravitational poten­ tial energy. While freewheeling down the hill, potential energy is converted to kinetic energy, and while braking to a stop, kinetic energy is converted into thermal energy (a more random form of kineticenergy) by thefrictionalforcesactingon the bike. 7. (d) 9. No. From the work-kinetic energy theorem, no total work is being done on the rock, as its kinetic energy is constant. How­ ever, the rod must exert a tangential force on the rock to keep the speed constant. The effect of this force is to cancel the com­ ponent of the force of gravity that is tangential to the trajectory of the rock. 11. 33.6 s 13. 3.04 X 1019 J/y; = 6% 15. 1.10 x 106 Lis 17. (c) 19. 3.89 m 21. 5.05 m 23. 25.6° 25. [111g(Sin 8 + f.L cos 8)]2 U = 5 27. 29. 31. 33. 6mg (c) 6mg (a) 31.0 m (b) -31.7 J (c) 33.7 mls 2k 35. 37. 39. 41. 43. 45. 47. (a) 151 m (b) 45.3 mls (a) � mgL (b) 6mg (a) 20.2° (b) 6.39 mls V = L)2t(1 - COS 8) + �(V¥ - 3 COS 8 - 1r (a) 94.2 kJ (b) The energy required to do this work comes from chemical energy stored in the body. (c) 471 kJ (a) 104 J (b) 70.2 J (c) 33.8 J (d) 2.91 m/s (a) 7.67 m/s (b) 58.9 J (c) 0.333 49. (a) (b) (c) Wf = (13.7 N)y Emeeh = -(13.7 N)y 1.98 m/s 51. 0.875 m; 2.49 mls 53. (a) 9.00 x 1013 J (b) $2.5 X 106 (c) 28,400 y 55. 1.88 X 10-28 kg 57. 3.56 X 1014 reactions 59. 0.782 MeV 61. (a) 3.16 kg (b) 8.04 X 109 kg 63. (b) 65. 57.6 MJ 67. (a) 0.208 (b) 3.45 MJ 69. (a) From the FBD we can see that the forces acting on the box are the normal force exerted by the inclined plane, kinetic friction force, and the gravitational force (the weight of the box) exerted by the earth. (b) 0.451 m (c) l.33 J (d) 2.52 mls 71. 11.3 kW; -6.77 kW 73. (a) 1.60 kJ (b) 619 J (c) 23.4 m/s 75. (a) 147 J (b) The energy is transferred to the girder from its surround­ ings, which are warmer than the girder. As the tempera­ ture of the girder rises, the atoms in the girder vibrate with a greater average kinetic energy, leading to a larger average separation, which causes the girder's expansion.
  • 625. 77. 79. 81. 83. 87. 89. 91. 93. (n) 0.8 0.6 0.4 ::::0 0.2 ::::; 0 -0.2 -0.4 0 Yeq (b) F = -ky + Ing (c) 21ng Yrnnx = T (d) mg Yeq = T (e) 11l2g2 Wf = -- 2k (n) 17.3 m (b) 4.91 kN (c) 4.91 m/s2 (d) 13.4 kN, upward (e) 5.46 kN; 63.9° (j) 1.44 kN (a) 491 N; 981 N (b) 9.82 kW; 29.4 kW (c) 8.85° (d) 6.36 kmlL (n) 17.4 MJ (b) 1 .39 x 1010 J (c) 9.73 x 109 J (d) 1.59 MW (n) 2111gY v = M + In (b) v = 0 = 2VHL(1 - cos 0) (n) (b) (c) (n) (b) 11l2g2 K",ax = II/gh + U mg x = - + IllilX k Ing x = T + 246 245 244 g 243 ::::; 242 241 240 239 238 5.39 kJ 0 11l2g2 2111gh -- + -­ F k Potential Energy 50 100 s (m) 150 200 Chapter 8 A- l l Chapter 8 1. A doughnut. 3. (b) 5. No. Consider a I-kg block with a speed of 1 mls and a 2-kg block with a speed of 0.707 m/s. The blocks have equal kinetic energies but momenta of magnitude 1 kg·m/s and 1.414 kg-m/s, respectively. 7. Precoil = Prine = - Pbullet or Prine + Pbullet = 0 9. Conservation of momentum requires only that the net external force acting on the system be zero. It does not require the pres­ ence of a medium such as air. 11. Think of someone pushing a box across a floor. Her push on the box is equal but opposite to the push of the box on her, but the action and reaction forces act on dif ferent objects. You can only add forces when they act on the same object. 13. The problem is that the comic situations violate the conserva­ tion of momentum! To move forward requires pushing some­ thing backward, which Superman doesn't appear to be doing when flying around. In a similar manner, if Superman picks up a train and throws it at Lex Luthor, he (Superman) ought to be tossed backward at a pretty high speed to satisfy the conserva­ tion of momentum. 15. The friction of the tire against the road causes the car to slow down. This is rather subtle, as the tire is in contact with the ground without slipping at all times, so as you push on the brakes harder, the force of static friction of the road against the tires must increase. Also, of course, the brakes heat up, and not the tires. 17. Assume that the ball travels at 80 milh = 35 m/s. The ball stops in a distance of about 1 cm, so the distance traveled is about 2 cm at an average speed of about 18 m/s. The collision 0.02 m time is - - 1 - = 1 ms. 18 m s 19. (n) False (b) True (c) True 21. (a) The loss of kinetic energy is the same in both cases. (b) The percentage loss is greatest for the case in which the two objects have oppositely directed velocities of magni­ tude �v. 23. (b) is correct because all of l's kinetic energy is transferred to 2 when /1'12 = 1111, 25. The water is changing direction when it rounds the corner in the nozzle. Therefore, the nozzle must exert a force on the stream of water to change its direction, and, from Newton's 3rd law, the water exerts an equal but opposite force on the nozzle. 27. No. FeXl,net = dpIdt defines the relationship between the net force acting on a system and the rate at which its momentum changes. The net external force acting on the pendulum bob is the sum of the force of gravity and the tension in the string and these forces do not add to zero. 29. Think of the stream of air molecules hitting the sail. Imagine that they bounce off the sail elastically-their net change in mo­ mentulll is then roughly twice the change in momentum that they experienced going through the fan. Another way of look­ ing at it: initially, the air is at rest, but after passing through the fan and bouncing off the sail, it is moving backward; therefore, the boat must exert a net force on the air pushiJlg it backward, and there must be a force on the boat pushing it forward.
  • 626. A- 1 2 Answers 31. (a) 2.33 s (b) 6.74 m/s 33. (0.233 m, 0) 35. (2.00 m, 1.40 m) 37. (1.50 m, 1.36 m) 41. zem = �R 43. vcm = (3 m/s)f - (1.5 m/s)! 45. acm = (2.4 m/s2)1 47. (a) Fn = (rl'lp + 11Ib)g (b) Fn = (l11p + 2l11b)g (c) Fn = m� 49. (a) Fn = (11'lp + I11b)g (b) F = 11'1 g + In g(l + n -p b 51. VlO = (4 m/s)f 53. v' = 2vf - v! 55. -Jii-f 57. (a) 43.5 J (b) vcm = (1.50 m/s)f (c) vrcl = (3.50 m/s)f and V2•rel = (-3.50 m/s)f (d) 36.8 J (e) Kcm = 6.75 J = K - Krel 59. (a) 10.8 N·s (b) 1.34 kN 61. 1.81 MN·s; 10.602 MN 63. 230 N 65. (a) 1 = (1.08 N·s)f (directed into wall) 67. 69. 71. 73. (b) 360 N, into wall (c) (d) (a) (b) (a) (b) (a) (b) (a) (b) (c) 0.480 N·s, away from wall 3.84 N, away from wall 20.0 m/s 20% of the initial kinetic energy is transformed into ther­ mal energy, sound, and the deformation of metal. -2.00 m/s The collision was inelastic. von = (23.1 m/s)f -254 m/s 5.00 m/s 0.250 m Vlf = 0; V2f = 7.00 m/s 75. (a) 0.200vo (b) 0.400vo 77. 450 m/s 81. h = - - v2 (111, )2 8g 1112 83. 0.0529 85. 1.50 X 106 m/s 87. (a) v, = (312 m/s)f + (66.6 m/s)! (b) 5.61 km (c) 35.8 kJ 89. 0.913 91. (a) 20% of its mechanical energy is lost. (b) 0.894 93. 95. 97. 99. 101. (a) 1.70 m/s (b) 0.833 (a) 60° (b) 2.50 m/s; 4.33 m/s (a) 1.00 m/s; 1.73 m/s (b) The collision was elastic. v, = 8.66 mls; v2 = 5.00 mls In an elastic collision . = = p, 2 [11l� + 6m,11I2 + In� ] = p'f [m� . + 6111111 . 12 +. m� ] K, Kr 2 2 ? ? 2 111,1112 + 11111112 2 rl'l,I1l2 + mimi If p{ = +p" the particles do not collide. 103. (a) vcm = 0 (b) (c) (d) (e) 105. (a) (b) (c) 107. (a) (b) (c) u3 = (-5 m/s)f; Us = (3 m/s)f u; = (5 m/s)f; u� = 0.75 m/s v; = (5 m/s)f; v� = (-3 m/s)f 60.0J; 60.0 J 360 kN 120 s 1.72 km/s ao 7 = 1 + - o g ( m 1 ( 111 )) vr = gIsp ln� - - 1 - ----.!. 11Ir 70 mo 3 en 2 + ................: . .... '- E C " O +---i---;--------;------+---I o 2 (d) 28.1 109. 0.192 m/s; 31.3 mJ; 12.0 mJ 111. 0.462 m/s 4 6 8 10 113. (a) p = -(1.10 X 105 kg·km/h)f + (1.05 x 105 kg·km/h)! (b) 43.4 km/h; 46.3° west of north 115. (a) 6.26 m/s (b) 20.0 m 117. 3.72 m 119. (a) The velocity of the basketball will be equal in magnitude but opposite in direction to the velocity of the baseball. (b) Vlf = 0 (c) v2r = 2v 121. (a) 29.6 km/s (b) 8.10; The energy comes from an immeasurably small slowing of Saturn. 123. 3.00 x 105 m/s 125. (a) 0.600 m/s2 (b) 960 N 127. No. The driver was traveling at 23.3 km/h. 129. 8.85 kg 131. �r 133. (a) 0.716'1 Eo (b) 55
  • 627. 135. (a) (b) (c) v2 Ycm = 2L 12 v2 nCm = L VI (v ) F = L gt + 1 Mg V l f = 2 V (m1 + I1lb) (/112 + I11b) 139. -0.960 m/s2 141. v = (1.70 mI/2/s) VL Chapter 9 1. Because r is greater for the point on the rim, it moves the greater distance. Both turn through the same angle. Because r is greater for the point on the rim, it has the greater speed. .Both have the same angular velocity. Both have zero tangential acceleration. Both have zero angular acceleration. Because r is greater for the point on the rim, it has the greater centripetal acceleration. 39. 41. 43. 45. 47. 49. 51. 55. 57. (b) 124 rad/S2 (c) 620 rad/s (a) g sin 8 Chapter 9 A- 1 3 (b) Because the line-of-action of the tension passes through the pendulum'S pivot point, its lever arm is zero and it causes no torque. (c) g sin8 (a) (b) (c) 2i-<kMg dT = --- r2dr f R2 Tf = �MRi-<kg 3Rw M = -- 4i-<kg 56.0 kgom2 (a) 28.0 kgom2 (b) 28.0 kg·m2 2.60 kg·m2 (b) [em = 12/11 (a2 + b2) 5.41 X 10- 47 kg·m2 1 = foMR2 (H2 R2) J = 3M - + ­ x 5 20 3. � �. (a) 84.6 mJ 5. (d) (b) 347 rev/min 7. No. A net torque is required to change the rotational state of an 63. (a) 19.6 kN object. A net torque may decrease the angular speed of an ob- ject. AU we can say for sure is that a net torque will change the angular speed of an object. 9. (b) 11. (b) 13. For a given applied force, this increases the torque about the hinges of the door, which increases the door's angular accelera­ tion, leading to the door being opened more quickly. It is clear that putting the knob far from the hinges means that the door can be opened with less effort (force). However, it also means that the hand on the knob must move through the greatest dis­ tance to open the door, so it may not be the quickest way to open the door. Also, if the knob were at the center of the door, you would have to walk aroLmd the door after opening it, assuming the door is opening toward you. 15. (b) 17. (b) 19. (a) 21. True. If the sphere is slipping, then there is kinetic friction that dissipates the mechanical energy of the sphere. 23. 10.3% 25. 27. 29. 31. 6.42 (a) (b) (c) (d) (a) (b) (a) (b) 33. (d) 15.6 rad/s 46.8 rad 7.45 rev 73.0 m/s2 40.0 rad/s 0.960 m/s2; 192 m/s2 0.589 rad/s2 4.71 [ad 35. 1.04 radls; 9.92 rev/min 37. (a) 1 .87 Nom (b) 5.89 kN-m (c) 0.267 rad/s (d) 1.57 kN 65. (a) 3.62 rad/s (b) 3.62 rad/s 67. Unless M, the mass of the ladder, is zero, v, > vf. It is better to let go and fall to the ground. 69. 3.11 m/s2; T1 = 12.5 N; T2 = 13.4 N 71. 8.23 m/s 73. (a) g a = --- 2M 1 + - 5rn (b) 211lMg T = 5111 + 2M 75. (a) 72.0 kg (b) 1.37 rad/s2; TI = 294 N; T2 = 746 N 77. (a) g sin 8 a = --- 1111 1 + - 2m2 (b) T = �1111 g sin 8 1111 1 + - 21'/"12 (c) E = 1112gh (d) Ebo!lom = 1112gh (e) v = � 1111 1 + - 2m2 (j) For 8 = 0: a = T = 0 For 8 = 90°· a = - - g - - T = " 111 a and v = . ' 1111 ' 2 l ' 1 + - 2m2 For 1'1'11 = 0: a = g sin 8, T = 0, and v = V2ih
  • 628. A- 1 4 Answers 79. 81. 83. 85. 87. 89. 9l. 93. 97. 99. 101. 103. 105. 107. 111. 113. 115. 117. 119. 121. 123. 125. 0.0864 m/s2; 3.14 m/s 0.192 m/s2; 0.962 N 1.13 kJ 45.9 m 19.5° (a) a = �g sin 0 (b) I s = �Illg sin 0 (c) Omax = tan-, (3,us) v' = �v 223 J (a) 2F a = ; counterclockwise R(M + 3111) (b) (c) (a) (b) (a) (b) (c) F a = --- C M + 3m 2F aCB = - M + 3m 0.400 rad/S2; 0.200 rad/S2 4.00 N 12 v6 2 va s, = - -- , tl = - --, and v 49 ,ukg 7 ,ukg 5/7 26.6 m; 3.88 s; 5.71 m/s 2rwo v = -- 7 (a) 360 kN (b) 120 s (c) 1.72 km/s (a) v = 1.57 va (b) 4 v a !It = -- 7 ,ukg v2 (c) �x = 0.735- ° ,ukg 1 = 2mR2 0.134 m (a) 7.36 m/s2 (b) 14.7 m/S2 (c) 2.43 m/s (a) 780 kJ (b) 90.3 N·m; 150 N (c) (a) (b) (a) (b) (c) (a) (b) 1380 rev 15.0 m 15.4 rad/s 51 S3 S5 w - f4i }3; F = �Mg ,-- ----:- - - (a) v = 2MgD sin 0 I M + - (b) f = Mg sin 0 s R 1 + ­ r r2 5 1 = "7vo 127. (a) 14.7 m/s2 (b) 66.7 cm 129. 41.7 J 131. The solid line on the graph shown below shows the position y of the bucket when it is in free fall and the dashed line shows y under the conditions modeled in this problem. 20 r---�---�--�--� 15 ]: 10 ;:,., 5 0 0 0.5 1.0 1.5 2.0 t (s) 133. (a) 25.7 N (b) 3.21 kg (c) 1.10 m/s2 Chapter 1 0 1. (a) True 3. 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. (b) True (c) True 90° (a) Doubling p doubles L. (b) Doubling r doubles L. False (e) It is easier to crawl radially outward. In fact, a radially inward force is required just to prevent you from sliding outward. The hardboiled egg is solid inside, so everything rotates with a uniform velocity. By contrast, it is difficult to get the viscous fluid inside a raw egg to start rotating; however, once it is rotat­ ing, stopping the shell will not stop the motion of the interior fluid, and, for this reason, the egg may start rotating again after momentarily stopping. (b) (b) (a) The lifting of the nose of the plane rotates the angular mo­ mentum vector upward. It veers to the right in response to the torque associated with the lifting of the nose. (b) The angular momentum vector is rotated to the right when the plane turns to the right. In turning to the right, the torque points down. The nose will move downward. (b) The center of mass of the rod-and-putty system moves in a straight line, and the system rotates about its center of mass. 4.17 rev/s (a) 2.40 X 10-8 kg·m2/s (b) 5.22 X 1052; 2.29 X 1026 (c) The quantization of angular momentum is not noticed in macroscopic physics because no experiment can differen­ tiate between e = 2 X 1026 and e = 2 X 1026 + 1.
  • 629. 29. (a) 0.331 (b) Because experimentally C < 2/5 = 0.4, the mass density must be greater near the center of the earth. 31. 10.1 rad/s 33. T = FRk 35. (n) 24k (b) -24J (c) -51( 39. B = 4J + 3k 45. (a) 54.0 kgom2/s (b) w increases as the particle approaches the point and de­ creases as it recedes. 47. (a) 1.33 x 10-5 kgom2/s (b) 1.33 x 10-5 kgom2/s (c) 1.33 x 10-5 kgom2/s (d) 8.83 x 10-5 kgom2/s; -6.17 X 10-5 kgom2/s 49. (n) 4.00 Nom (b) (0.192 rad/s2)t 51. (a) Tnet = Rg(m 2 sin e -m,) (b) L = VR(;2 + 111, + 111 2 ) g(m 2 sin e -111,) (c) a = =--- -=- ----'- I - + 111, + 111 2 R 2 55. (a) 5.00 rev/s (b) 622 J (c) The energy comes from your internal energy. 57. 59. 61. 63. 65. 67. 69. 71. 9.67 mm/s (n) (b) (c) 54.7° (n) (b) Lo = rolrlvo Ko = � II/V2 0 v2 3 T = F = 111�; W = --mv2 e ,.o 2 0 3.46 X 10-47 kgom2 1.99 meV; 5.98 meV; 12.0 meV 82.5 m/s mv mMvd v = ---' w = em M + 111 ' tzMU(M + 11'1) + Mmd2 (0.5 M + 0.8111)(�Md2 + 0.64md2)g v = 0.32dm2 (n) K Vem = M (b) 4K M (c) 2K -- M (d) X = � e 73. 0.349 75. 12 rad/s; 10.8 J 77. (n) 18.1 Jos (b) 0.414 rad/s (c) 15.2 s (d) 0.0791 Jos 79. (n) r =-(47. 7 kgom2/s)k (b) T = (15.9 Nom)k 81. (a) 243 JoS (b) 306 J 83. (a) No, L decreases. Chapter R (b) Its kinetic energy is constant. (c) Vo (The kinetic energy remains constant.) 85. Yes. 87 v = ew , I(U -(2) . , 2L v 91. (n) 0.228 rad/s (b) 0.192 rad/s 93. 4.47 x 1022 Nom 95. 12.5 rad/s 97. (a) 26.5 rad/s (b) r = (0.303 kgom2/s)e(1.4'S-')1 Chapter R 1. The friend in the car. A- 1 5 3. Yes. If two events occur at the same time and place in one reference frame they occur at the same time and place in aU ref­ erence frames. (Any pair of events that occur at the same time and at the same place in one reference frame are called a space­ time coincidence.) 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 31. 33. (a) 1 + (8.61 x 10-11) 6.00 ns (n) 6.63 m (b) 12.6 m (a) 599 m (b) 13.4 km (a) 129 y (b) 87.6 Y (a) 0.600 m (b) 2.50 ns 0.800e (a) 4.50 x 10-10% (b) 0.142 ms 25.0 min; 25.0 min 60.0 min 0.400c; event B can precede event A provided II > 0.400e (a) 11.3 y (b) 40.0 Y (a) 1.005 (b) 1.155 (c) 1.667 (d) 7.089 (a) 0.155Eo (b) 1.29Eo (c) 6.09Eo 35. 2.97 GeV 39. (b) 0.866c (c) 0.999c
  • 630. A- 1 6 Answers 41. (a) 0.794% (b) 68.7% 43. (a) 0.943 45. (a) 617 eV (b) 79.6 eV (e) 7.96 eV 47. (a) 0.745 (b) 5.00 ft (e) No. In Keisha's rest frame, the back end of the ladder will clear the doorbefore the front end hits the wall of the shed, while in Ernie's rest frame, the front end will hit the wall of the shed while the back end has yet to reach the door. Chapter 1 1 1. (a) False (b) True 3. (d) 5. (a) 7. The gravitational field is proportional to the mass within the sphere of radius r and inversely proportional to the square of r, i.e., proportional to ,.3/r2 = 1". 9. (d) 11. 1.08 X 1011 Ms 13. (a) 2.78 h (b) 19.3 X 1042 kgom2/s; 7.85 X 1042 kgom2/s; 0.703% (e) 4.80 X 10-4 rad/s 15. 84.0 Y 17. (a) 1.59 X 1011 m (b) 2.71 X 1010 m; 2.91 X 1011 m 19. (a) 90° (b) 0.731 AU 21. (a) 1.90 X 1027 kg (b) 0.282 m/s2; 0.0356 m/s2 23. (a) 8.18 X 104 s (b) 1.22 X 109 m 25. 1.99 X 1030 kg 27. lOw, where w is your weight on earth. 29. 2.27 X 104 m/s 31. 1.43 33. (a) 7.37 m (b) 0.0319 mm 35. 0.605 37. 39. 41. (a) 2.27kg (b) It is the inertial mass of 111 2 , 109 m GMmo W = -- R 43. 6.94 km/s 45. (a) (b) --> GMmo , Foutside = -- r - Z - r GMrl"lo GMmo U(r) = ---; U(R) = --- r R GMmo (d) U(r) = U(R) = -- R - 47. 49. 51. 53. 55. 57. 59. 61. 63. 65. (e) 2.38 km/s R 19.4 km/s (a) 62.7 MJ (b) 17.4 kWoh (e) $139 (a) 7.31 h (b) 1.04 GJ (e) 8.72 X 1012 J-s 11.1 GJ g = (4 N/kg)1 (a) � Gm < Gm < g = U I + U' (b) v'2 Gm g = 2 U (a) g = (-1.67 X 10-11 N/kg)1 (b) g = (-8.34 X 1O-12 N/kg)1 (e) (a) (b) (a) (b) (e) 2.48 m M = 1 CU � 2GM [ ( Xo ) g = -- In -- - U Xo - L 0 0 3.20 X 10-9 N/kg GrnM, (b) F = 3.61a2 (e) 0 Co :L)]I 71. (a) mg F = - r g R (b) FN = C�- mUlZ}- (e) The change in mass between you and the center of the earth as you move away from the center is more impor­ tant than the rotational effect. 73. g(x) = GC7T �oR3 )[�- 8(x �1R)Z] 75. Ul = �47T ;oG 77. 0.104 mm/s 79. (a) --> GMm [ � l' F = ----;r- 1 - {d2 + �2y/2 i -, GMm , (b) F(R) = -0.821 J?2i
  • 631. 81 . 249 Y 83. (n) (b) W = IIlgR�(� _ _ _ 1 _) . RE RE + h 85. 8.96 x 107 m 87. 1.70 Mm (GM 91. v = 1 .64 y ----;- - n - GM 93. For r < RI, g(r) =0; For r > R" g(r) = - , ; - r- GM(r3 - Rt) For R, < r < R2, g(r) = r2(R� _ R�) 2GA 95. g = - }- . 0.12 -,--,--.,---,--.,----,--.,---,-----, 0.10 0.08 :s 0.06 bl) 0.04 002 o +--+-+--+-�-+-��� o 2 3 4 5 6 7 8 I I Rl R2 GMmo (Xo + L/2 ) 97. (b) U = --- In L Xo - L/2 99. 33.5 pN 101. (n) The gravitational force is greater on the lower robot, so if it were not for the cable its acceleration would be greater than that of the upper robot, and they would separate. In opposing this separation the cable is stressed. (b) 220 km Chapter 1 2 ---------------------- 1 . (n) False (b) True (c) True (d) False 3. No. The definition of the center of gravity does not require that there be any material at its location. 5. Th.is technique works because the center of mass must be di­ rectly under the balance point. Hence the intersection of the two lines must be at the center of mass. 7. (b) 9. (c) 11. The tensile strengths of stone and concrete are at least an order of magnitude lower than their compressive strengths, so you want to build compressive structures to match their properties. 13. (b) 200 N/m 15. 318 N Chapter 1 2 A- 1 7 1 7. (b) Taking long strides requires a larger coefficient of static friction because 0 is then large. (c) If ILs is small, that is, there is ice on the surface, 0 must be small to avoid slipping. 19. . _ (�n2b - 7TnR2 + 7TR3 � ) 21. (xcg' Ycg) - ab _ 7TR2 ' 2 b 23. 692 N; 900; 2.54 kN; No block is required to prevent the mast from moving. 25. 0.728 m 1 V3 27. F2 = "2 W; FI = - 2 - W 29. (n) 5.00 III (b) 4.87 m � MgV � h( - 2- R- - - J -- l) A A 31. FI = h - R i + Mgj 33. (n) F = (30.0 N)[ + (30.0 N)J (b) F = (35.0 N)l + (45.0 N)J 35. (n) N- F = Mg - F - - n h (b) FC•h = F (c) _ )2R - h Fe." - F h 37. (n) 6.87 N (b) 1.65 Nom (c) -8.26 N; 15.1 N 39. 636 N; 21.50 41. (n) 70.7 N (b) 1.77 m (c) 3.54 m (d) 497 N 43. 7 net = (69.3 N)b - (40.0 N)n 45. 0 = � ( V3b - n) 35.7 m - 30.4x 47. Y = 3.57 m - (294m-l)x 12.0 10.0 8.0 I 6.0 ;0-, 4.0 2.0 0 0.0 0.2 0.4 0.6 x (111) 0.8 1 .0 1.2
  • 632. A- 1 8 Answers 49. 51. 53. 55. 57. 61. 63. 65. 69. 71. 73. 75. 77. 79. 81. 83. 85. h = f.LsL tan e sin e 21-1 f.Ls = - L - t - an - e - s - il - l - e 59.00 (a) 41.6 N (b) 0.136% 5.010 (a) 1.82 x 106 N/m2 (b) 6.62 mJ 0.686 It will not support the elevator. FL = 117 N; FR = 333 N WI = 1.50 N; w 2 = 7.00 N; W 3 = 3.50 N 0.148 f.Ls < 0.500 f.Ls = 1 (cot e - 1) (a) 147 N (b) 3.62 m The block will tip before it slides. f.Ls < 0.500 (a) The stick remains balanced as long as the center of mass is between the two fingers. For a balanced stick the normal force exerted by tbe finger nearest the center of mass is greater than that exerted by the other finger. Consequently, a larger static-frictional force can be exerted by the finger closer to tbe center of mass, which means the slipping occurs at tbe otber finger. (b) The finger farthest from the center of mass will slide in­ ward until the normal force it exerts on the stick is suffi­ ciently large to produce a kinetic-frictional force exceed­ ing the maximum static-frictional force exerted by the other finger. At that point the finger that was not sliding begins to slide, the finger that was sliding stops sliding, and the process is reversed. When one finger is slipping the other is not. 87. (a) 23.0 m/s (b) 29.1 m/s 89. (c) Cs = 1.142 m; Clo = 1.464 m; Cwo = 2.594 m 91. 93. (d) Increasing N in the spreadsheet solution suggests that the sum of the individual offsets continues to grow as N in­ creases without bound. The series is, in fact, divergent and the stack of bricks has no maximum offset or length. 566 N mg R - r Fn = 2111g; F = -- ; Fw = mg---: = = = = cos e YR(2r - R) Chapter 1 3 1. (e) 3. (d) 5. Nothing. The fish is in neutral buoyancy, so the upward accel­ eration of the fish is balanced by the downward acceleration of the displaced water. 7. (b) 9. It blows over the ball, reducing the pressure above the ball to below atmospheric pressure. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 31. 33. 35. False The buoyant force acting on the ice cubes equals the weight of the water they displace (i.e., B = wr = PrVrg). When the ice melts, the volume of water displaced by the ice cubes will oc­ cupy the space previously occupied by the submerged part of the ice cubes. Therefore the water level remains constant. Because the pressure increases with depth, the object will be compressed and its density will increase. Thus it will sink to the bottom. The drawing shows the beaker and a strip within the water. As is readily established by a simple demonstration, the surface of the water is not level while the beaker is accelerated, showing that there is a pressure gradient. That pressure gradient results in a net force on the small element shown in the figure. From Bernoulli's principle, the opening above which the air flows faster will be at a lower pressure than the other one, which will cause a circulation of air in tbe tWlllel from opening 1 toward opening 2. It bas been shown tbat enough air will cir­ culate inside the tunnel even with the slightest breeze outside. 0.673 kg 103 kg 29.8 inHg 230 N 198 atm (a) 14.8 kN (b) 0.339 kg 0.453 m pga3 F = - 8 37. 4.36 N 39. (a) 11.1 x 103 kg/m3 (b) lead 41. 800 kg/m3; 1.11 43. 250 kg/m3 45. 3.89 kg 47. 2.46 x 107 kg 49. 491 kN 51. (a) 9.28 cm/s (b) 0.331 cm (c) 8.31 cm, in reasonable agreement witb everyday experience. 53. (a) 12.0 m/s (b) 133 kPa 55. (c) The volume flow rates are equal. (a) 4.58 L/min (b) 763 cm2 57. 144 kPa 59. (a) 21.2 kg/s (b) 636 kg·m/s (c) 899 kg'm/s; 899 N 61. (a) x = 2Yh(H - 1-1) (b) h = 1H ::': �YH2 - x2 63. (b) Ptop = Pot'" - pgd 65. 1.43 mm 67. 93.4 mi/h; Since most major league pitchers can throw a fast­ ball in the low-to-mid-90s, this drag crisis may very well play a role in the game.
  • 633. 69. 0.0137; 0.0115 71. The net force is zero. Neglecting the thickness of the table, the atmospheric pressure is the same above and below the surface of the table. 73. 75. 77. 79. 81. 83. 1061 kg/m3 65.7% If you are floating, the density (or specific gravity) of the liquid in which you are floating is immaterial as you are in transla­ tional equilibrium under the influence of your weight and the buoyant force on your body. Thus the buoyant force on your body is your weight in both (a) and (b). rn V = -- 0.96pw 11.8 cm 1 m is a reasonable diameter for the pipeline. 85. hA = 12.6 m; hs = 9.78 m 87. (a) 64.6% (b) 10.7 kN (c) 17.9 m/s2 Chapter 1 3 89. 3.31 X 10-3 mmHg or 3.31 /LmHg 91. 1.37 93. (a) 70.0 m3 (b) 7.47 m/s2 95. (c) 0.126 km-1 97. (a) 33.9 kN (b) 39.8 kN; 36.1 kN 99. (c) 11 = Vii - - - Vig t ( A, Y 2A1 (d) 1 h 46 min A- 1 9
  • 634. A-20 Answers Chapter 1 4 1. 0; 47T2j2A 3. (a) False (b) True (e) True 5. (a) 7. False 9. Assume that the first cart is given an initial velocity v by the blow. After the initial blow, there are no external forces acting on the carts, so their center of mass moves at a constant velocity v12. The two carts will oscillate about their center of mass in simple harmonic motion where the amplitude of their velocity is v12. Therefore, when one cart has velocity vl2 with respect to the center of mass, the other will have velocity -v/2. So, the velocity with respect to the laboratory frame of reference will be +v and 0, respectively. Half a period later, the situation is reversed; so, one will appear to move as the other stops, and vice-versa. 11. True 13. Examples of driven oscillators include the pendulum of a clock, 31. (a) 7.85 m/s; 24.7 m/s2 (b) -6.28 m/s; -14.8 m/s2 33. (a) 0.313 Hz (b) 3.14 s (e) x = (40 cm)cos[(2 s-1)1 + 8] 35. 22.5 J 37. (a) 0.368 J (b) 3.84 cm 39. 1.38 kN/m 41. (a) 6.89 Hz (b) 0.145 s (e) 0.100 m (d) 4.33 m/s (e) 187 m/s2 (j) 36.3 ms; 0 43. (a) 682 N/m (b) 0.417 s (e) 1.51 m/s (d) 22.7 m/s2 a bowed violin string, and the membrane of any loudspeaker. 45. (a) 3.08 kN/m 15. Because f' varies inversely with the square root of 1n, taking into account the effective mass of the spring predicts that the frequency will be reduced. 17. (d) 19. (b) 21. 87T 23. (a) 3.00 Hz (b) 0.333 s (e) 7.00 cm (d) 0.0833 s; Because v < 0, the particle is moving in the nega­ tive direction at I = 0.0833 s. 25. (a) x = (25 cm)cos[(4.19 s-1)1] (b) v = -(105 cm/s)sin[(4.19 s-1)1] (e) a = -(439 cm/s2)cos[(4.19 s-1)1] 27. (a) x = (27.7 cm)cos[(4.19 s-1)1 - 0.445] (b) v = -(116 cm/s)sin[(4.19 s-1)1 - 0.445] (e) a = -(486 cm/s2)cos[(4.19 s-1)1 - 0.445] 29. (a) 10 ",",--,--�--,------,------,--,----�...,..., (b) If (s) 1 2 3 4 I; 6 2 -2 -6 -10 +-�����L-�-+-�� 0 1 2 3 4 5 6 7 8 I (s) LlX (s) (cm) 0 12.93 1 1 17.07 1 2 17.07 1 3 12.93 / 47. 49. 51. 53. 55. 57. 59. (b) 4.16 Hz (e) 0.240 5 (a) 0.438 m/s (b) 0.379 m/s; 120 m/s2 (e) 95.5 ms 0.262 s 10.1 kJ (a) 0.997 Hz (b) 0.502 s (e) 0.294 N (a) 46.66 cm (b) 0.261 s (e) 0.767 m/s (a) 0.270 J (b) -0.736 J (e) 1.01 J (d) 0.270 J (a) 1.90 cm (b) 0.0542 J (e) ::':0.224 J (d) 0.334 J 61. 12.2 5 63. 11.7 s 65. T = 27T I L jg(l - sin 0) 67. 1.10 5 69. 0.504 kg·m2 71. (b) 3.17 s 73. 21.1 cm from the center of the meter stick 77. (a) 1.63572 m (b) 14.5 mm, upward 79. 13S
  • 635. 81. 85. 87. 89 91. 93. 95. 97. 99. 3.14% (a) 0.314 (b) -3.13 X 10-2 percent (a) 1.57% (c) 0.430£0 (a) 1.01 Hz (b) 2.01 Hz (c) 0.352 Hz (a) 4.98 cm (b) 14.1 rad/s (c) 35.4 cm (d) 1.00 rad/s (a) 0 (b) 4.00 m/s (a) 14.1 cm; 0.444 s (b) 23.1 cm; 0.363 s (c) (14.1 cm)sin[(14.1 S-l)t]; (23.1 cm)sin[(17.3 S-l)t] (a) v = -(1.2 m/s)sin [(3 rad/s)t + �] (b) -0.849 m/s (c) 1.20 m/s (d) 1.31 s (a) The normal force is identical to the tension in a string of length r that keeps the particle moving in a circular path and a component of mg provides, for small displacements 00 or 52' the linear restoring force required for oscillatory motion. (b) The particles meet at the bottom. Because 51 and 52 are both much smaller than r, the particles behave like the bobs of simple pendulums of equal length and, therefore, have the same periods. 101. 1.62 s 103. 3.86 X 1O-7 N'm/rad 105. g' is closer tog than is goo. Thus the error is greater if the clock is elevated. 107. (a) Ak f-L = -,--- - - - 5 (11'11 + m)g (b) A is unchanged. £ is unchanged since £ = !kN. w is re­ duced by increasing the total mass of the system and T is increased. 109. (b) 2.04 cm/s2 113. (a) x = 0 (b) v = x � 5 . uj� (c) Xf = Xo 11'1 P m + 11'1 b P 115. (a) 10TT-��-�-�-�-- 8 :; 6 4 0.5 1.0 1.5 2.0 2.5 3.0 x/a (b) Xo = a or ao = 1 Chapter 1 5 (c) U(xo + c:) = Uo[l + f3 + (1 + f3)-1] (d) 82 U(Xo + 8) = constant + Uo? a- 119. 6.44 X 1013 rad/s 121. 7.78 .Jff 123. (a) 0.0478 (b) 0.00228 127. (a) A-2 1 0.6 ,-- -- -- -- -- -- -- -- ---,- -- -- ----,-- -- -- � -- - / ----" . . .. . . ... � �::+ ................... .-j.............................. j .. ................... +...................... ,..·......···..·.. 7 ·.... . . .. .. · . · . .. .. .. . . + / .. .. ·.. .. .. .. .. .. ··· .. -;/' ...... ::J 0.2 + ........................,..............................ii .. . ............ . . !...... � ..............:;7' C; 7 ........................ .+- ....................... . 1 0.1 .... � 0.0 0.0 0.5 1.0 1.5 (b) r = ro; k = 2f320 (c) w = 2f3IQ /-;;; Chapter 1 5 r (nm) 2.0 2.5 3.0 1. The speed of a transverse wave on a rope is given by v = �where F is the tension in the rope and f-L is its linear den­ sity. The waves on the rope move faster as they move up be­ cause the tension increases due to the weight of the rope below. 3. True 5. The speed of the wave v on the bullwhip varies with the ten­ sion F in the whip and its linear density f-L according to v = �. As the whip tapers, the wave speed in the tapered end increases due to the decrease in the mass density, so the wave travels faster. 7. No; Because the source and receiver are at rest relative to each other, there is no relative motion of the source and receiver and there will be no Doppler shift in frequency. 9. The light from the companion star will be shifted about its mean frequency periodically due to the relative approach to and recession from the earth of the companion star as it re­ volves about the black hole. 11. (a) True (b) False (c) False 13. There was only one explosion. Sound travels faster in water than air. Abel heard the sound wave in the water first, then, surfacing, heard the sound wave traveling through the air, which took longer to reach him. 15. vy 4 6 8 X (cm)
  • 636. A-22 Answers 17. Path C. Because the wave speed is highest in the water, and more of path C is underwater than A or B, the sound wave will spend the least time on path C. 19. (a) 78.5 m (b) 69.7 m (c) 70.5 m . . . about 1% larger than our result in part (b) and 11% smaller than our first approximation in (a). 21. 270 m/s; 20.6% 23. 1.32 km/s 25. 19.6 g 27. (a) 265 m/s (b) 15.0 g 29. (b) 40.0 N 33. The lightning struck 680 m from the ball park, 58.4° W (or E) of north. 39. (a) y(x,t) = A sin k(x - vt) (b) y(x,t) = A sin 27TG- ft) (c) C 1 ) y(x,t) = A sin 27T i - yt (d) 27i y(x,t) = A sinA (x - vt) (e) y(x,t) = A sin 27Tf (;- t) 41. 9.87 W 43. (a) The wave is traveling in the -x direction.; 5.00 m/s (b) 10.0 cm; 50.0 Hz; 0.0200 s (c) 0.314 m/s 45. (a) 6.82 J (b) 44.0 W 47. (a) 79.0 mW (b) Increasingfby a factor of 10 would increase Pa" by a factor of 100. Increasing A by a factor of 10 would increase Pa" by a factor of 100. Increasing F by a factor of 104 would in­ crease v by a factor of 100 and Pa" by a factor of 100. (e) Depending on the adjustability of the power source, in­ creasingfor A would be the easiest. 49. (a) 0.750 Pa (b) 4.00 m (e) 85.0 Hz (d) 340 m/s 51. (a) (b) 53. (a) (b) 55. (a) (b) (e) 57. (a) (b) 3.68 X 10-5 m 8.27 X 10-2 Pa The displacement s is zero. 3.68 fLm 138 Pa 21.7 W1m2 0.217 W 50.3 W 2.00 m (e) 4.45 x 10-3 W1m2 59. (a) 20.0 dB (b) 100 dB 61. 90.0 dB 65. (a) 100 m (b) 0.126 W 67. (a) 100 dB (b) 50.3 W (e) 2.00 m (d) 96.5 dB 69. (a) 81.1 dB (b) 80.0 dB; Eliminating the two least intense sources does not reduce the intensity level significantly. 71. 87.8 dB 73. 57.0 dB 75. (a) 260 m/s (b) 1.30 m (e) 262 Hz 77. (a) 1.70 m (b) 247 Hz 79. 153 Hz 81. 1021 Hz or a fraction increase of 2.06%; Because this fractional change in frequency is less than the 3% criterion for recognition of a change in frequency, it would be impossible to use your sense of pitch to estimate your running speed. 83. 349 mi/h 85. 7.78 kHz 87. 15.0 km west of P 89. (a) f' = (1 - u,/v)(l - usIV)-l fo 91. 1.33 m/s 93. (a) 824 Hz (b) 849 Hz 95. 184 m 97. -2.07 x 10-5 nm; 99 2.25 x 108 m/s 99. 2.25 x 1018 m/s . . . where the upper arrow means the 8 is an exponent. 101. 20.8 cm 103. 3.42 m/s 105. 529 Hz; 474 Hz 107. 7.99 m 109. (a) 55.1 NIm2 (b) 3.46 W1m2 (e) 0.109 W 111. 77.0 kN 113. 115. 117. 119. 204 m 24.0 cm (b) Vo = Jf; (e) As seen by an observer at rest, the pulse remains at the same position because its speed along the chain is the same as the speed of the chain. With respect to a fixed point on the chain, the pulse travels through 360°. (b) 2.21 s
  • 637. Chapter 1 6 1. t = 1 �I--------T--,..-----'If-------r-----!RI-------+------""'----' 1 = 2 1 � 1 = 3 r l -- ,- -. -- � R -- � -- r- -+ -- .- -. -- � -. 3. (c) 5. (b) 7. (a) 9. since v C/. T, increasing the temperature increases resonant frequencies. 11. No; the wavelength of a wave is related to its frequency and speed of propagation (, = vlj). The frequency of the plucked string will be the same as the wave it produces in air, but the speeds of the waves depend on the media in which they are propagating. Since the velocities of propagation differ, the wavelengths will not be the same. 13. When the edges ofthe glass vibrate, sound waves areproduced in the air in the glass. The resonance frequency of the air columns depends on the length of the air column, which de­ pends on how much water is in the glass. 15. (b) 17. The pitch is determined mostly by the resonant cavity of the mouth, and the frequency of sounds he makes is directly pro­ pOI·tional to their speed. Since vHe > va;r (see Equation 15-5), the resonance frequency is higher if helium is the gas in the cavity. 19. Pianos are tuned by ringing the tuning fork and the piano note simultaneously and tuning the piano string until the beats are far apart (i.e., the time between beats is very long). If we assume that 2 s is the maximum detectable period for the beats, then one should be able to tune the piano string to at least 0.5 Hz. 21. 34.0 Hz; Because v C/. T, the frequency will be somewhat higher in the summer. 23. 7.07 em 25. (a) 90.0° (b) V 2A 27. (a) 0 (b) 210 (c) 410 29. (a) � , (b) � , 31. (a) 60.0 cm (b) 2'71" - 5 (c) 24.0 mls 33. 4726 Hz; 9452 Hz 35. (b) Y (c) 0.500 mls 37. 1.81; 51.5" 39. (a) 0.279 m (b) 1.22 kHz (c) 111 3 8m (rad) 0.432 4 0.592 5 0.772 6 0.992 7 1.354 8 undefined (d) 0.0698 rad 41. 1.98 rad or 113° 43. (a) 70.5 Hz Chapter 1 6 A-23 x (s) -1 = 0.0 -I = 0.53 1 = 1.05 (b) The person on the street hears no beat frequency as the sirens of both ambulances are Doppler shifted up by the same amount (approximately 35 Hz). 45. (a) 2.00 m; 25.0 Hz (b) Y3(x,l) = (4 mm)sin kx coswl, where k = '71"m-1 and w = 50'71"S-1 47. (a) 521 mls (b) 2.80 m; 186 Hz (c) 372 Hz; 558 Hz 49. 141 Hz 51. (a) 31.4 cm; 47.7 Hz (b) 15.0 mls (c) 62.8 cm
  • 638. A-24 Answers 53. (a) 4 .... 4 2 2 E 0 E 0 "" "" "., "., -2 -2 -4 0 0.5 1.0 1.5 2.0 2.5 2.0 2.5 x lm) 4 2 E 0 "" '" -2 -4 . 0 0.5 1.0 1.5 2.0 2.5 x (m) (b) 12.6 ms (c) Since the string is moving either upward or downward when y(x) = 0 for all x, the energy of the wave is entirely kinetic energy. 55. (a) 70.8 Hz (b) 4.89 Hz (c) 35 57. 452 Hz; It would be better to have the pipe expand so that vlL, where L is the length of the pipe, is independent of temperature. 59. (a) 80 cm (b) 480 N (c) You should place your finger 9.23 cm from the scroll bridge. 61. (a) 75.0 Hz (b) The harmonics are the 5th and 6th. (c) 2.00 m 63. (a) 0.574 glm (b) 1.29 g/m; 2.91 g/m; 6.55 glm 65. (a) The two sounds produce a beat because the th.irdharmonic of the A string equals the second harmonjc of the E string, and the original frequency of the E string is sligh.tly greater than 660 Hz. If fE = (660 + Llf)Hz, a beat of 2Llf will be heard. (b) 661.5 Hz (c) 79.6 N 69. 76.8 N; 19.2 N; 8.53 N 71. (a) Nlfa (b) Llx/N (c) 27TNILlX (d) N is uncertain because the waveform dies out gradually rather than stopping abruptly at some time; hence, where the pulse starts and stops is not well defined. 73. (a) 3.40 kHz; 10.2 kHz; 17.0 kHz (b) Frequencies near 3400 Hz will be most readily perceived. 75. ! A 77. 6.62 m 79. (a) 1.90 cm; 3.59 mls (b) 0; 0 (c) 1.18 cm; 2.22 m/s (d) 0; 0 81. (a) At resonance, standing waves are set up in the tube. At a displacement antinode, the powder is moved about; at a node the powder is stationary, and so it collects at the nodes. (b) 2fD (c) If we let the length L of the tube be 1.2 m and assume that Va;, = 344 mls (the speed of sound in air at 20°C), then the 10th harmonic corresponds to 0 = 25.3 cm and a driving frequency of 680 Hz. (d) Iff = 2 kHz and vHe = 1008 mls (the speed of sound in helium at 20°), then 0 for the 10th harmonic in helium would be 25.3 em, and 0 for the 10th harmonic in air would be 8.60 cm. Hence, neglecting end effects at the driven end, a tube whose length is the least common multiple of 8.60 cm and 25.3 cm (218 cm) would work well for the measurement of the speed of sound in either air or helium. 83. (a) The pipe is closed at one end. (b) 262 Hz (c) 32.4 cm 85. (a) Y'(X,t) = (0.01 m)Sin[(�m-')x - (407TS-')t ]; ) /2 (x,t) = (0.01 m)Sin[(�m-I)x + (407T S-I)t]; (b) 2.00 m (c) vy{l m,t) = -(2.51 m/s)sin(407T s-')t (d) ay(l m,t) = -(316 m/s2)cos(407T s-')t 87. Y,es(x,t) = 0.1 sin(kx - wt) 89. (b) 203 Hz 91. (a) What you hear is the fundamental mode of the tube and its overtones. A more physical explanation is that the echo of the finger snap moves back and forth along the tube with a characteristic time of 2L/c, leading to a series of clicks from each echo. Since the clicks happen with a fre­ quency of e/2L, the ear interprets this as a musical note of that frequency. (b) 93. (a) (b) 95. (a) (b) (c) 97. (a) 1 .5 38.6 cm Since no conditions were placed on its derivation, this ex­ pression is valid for all harmonics. 1.54% vy(x,t) = -w,AI sin wIt sin k,x - w2A2 sin w 2 t sin k2x dK = �fL[w� Ai sin2 WIt sin2 k,x + 2w,w2A,A2 sin wIt sin k,x sin w2t sin k 2x + wiAi sin2 W 2 t sin2 k2xJdx K = ! l11wiAi sin2 WIt + ! l11w�A� sin2 wi -1.5 L-----�----�----�----�----�----�----� (b) f(27T) = 1 which is equivalent to the Liebnitz formula.
  • 639. 99. (b) 0.014 .,------r--------,,------,---------, 0.012 0.010 t ························ ······,········r .-- c............. , L-- ...............................,....·················.···· ···· 1 / 0 008 · · · · ·· · · ·· 1 <i 0.006 / ........; . .......................... ,................................;. ....................... . 1 0.00. / /- ··· . · . · . · . · ..............,...............................�................................;...........················· · ··1 0.002 17 0.000 +------i----�c----+-----I o 50 100 1/ 150 200 (c) The frequency heard at any time is l/Llt", so because Llt" increases over time, the frequency of the culvert whistler decreases.; 7.65 kHz Chapter 1 7 1. (n) False (b) False (c) True (d) False 3. Mert's room was colder. 5. From the ideal-gas law we have P = nRTN. In the process de­ picted, both the temperature and the volume increase but the temperature increases faster than does the volume. Hence the pressure increases. 7. True 9. Kav increases by a factor of 2; Kav is reduced by a factor of � . 11. False 13. Since 107 » 273, it does not matter. 15. (b) 17. (d) 19. The ratio of the rms speeds is inversely proportional to the square root of the ratio of the molecular masses. The kinetic energies of the molecules are the same. 21. Because the temperature remains constant, the average speed of the molecules remains constant. When the volume decreases, the molecules travel less distance between collisions, so the pressure increases because the frequency of collisions increases. 23. The average molecular speed of He gas at 300 K is about 1.4 km/s, so a significant fraction of He molecules have speeds in excess of earth's escape velocity (11.2 km/s), and thus "leak" away into space. Over time, the He content of the atmosphere decreases to almost nothing. 25. (n) 3.61 X 103 K (b) 225 K (c) If Ven" > !ve or T 2:: 25T"m' H2 molecules escape. Therefore, the more energetiC Hz molecules escape from the upper at­ mosphere. (d) 164 K; 10.3 K; If we assume that the temperature on the moon with an atmosphere would have been approximately 1000 K, then all O2 and H2 would have escaped during the time since the formation of the moon to the present. 27. (a) 1.24 km/s (b) 310 m/s 29. 31. 33. 35. 37. 39. 41. Chapter 1 8 (c) 264 m/s (d) O2, CO2, and H2 should be found on Jupiter. 1063°C (n) 8.40 cm (b) 107°C -319°F (n) 54.9 torr (b) 3704 K -40°C = -40°F -183°C; -297°F (n) B = 3.94 X 103 K; Ro = 3.97 X 10-3 fl (b) 1.31 kO (c) -389 fl/K; -433 fl/K A-2S (d) The thermistor is more sensitive (i.e., has greater sensitiv­ ity, at lower temperatures). 43. 1.79 mol; 1.08 X 1024 molecules 45. -83.2glips 47. (n) 3.66 X 103 mol (b) 60.0 mol 49. 10.0 atm 51. 1.19 kg/m3 53. 2.56 N 55. (n) 276 m/s (b) 872 m/s 57. 499 km/s; 2.07 X 10-16 J 61. K/LlU = 7.95 X 104 65. (n) 0.142 s (b) 0.143 s 67. (n) 122 K (b) 244 K (c) 1.43 atm 69. 111 mol; 55.5 mol 71. 711'1H 73. 400.49 K 75. (n) 4.10 X 10-26 m (b) 4.28 nm; The mean free path is larger by approximately a factor of 1000. 77. (a) 48.9% (b) 70.6% Chapter 1 8 1. LlTB = 4LlT" 3. (c) 5. Yes, if the heat adsorbed by the system is equal to the work done by the system. 7. Won + Qm = LlE;nt; For an ideal gas, LlE;nt is a hmction of T only. Since W = 0 and Q = 0 in a free expansion, LlE;nt = 0 and T is constant. For a real gas, LlE;nt depends on the density of the gas because the molecules exert weak attractive forces on each other. In a free expansion, these forces reduce the average kinetic energy of the molecules and, consequently, the temperature. 9. The temperature of the gas increases. The average kinetic en­ ergy increases with increasing volume due to the repulsive in­ teraction between the ions.
  • 640. A-26 11. 13. 15. (a) (a) (b) (c) (d) (e) (f) (g) (d) Answers False False True True True True True 17. If V decreases, the temperature decreases. 19. Theheatcapacity of a substance is proportional to thenumber of degrees of freedom per molecule associated with the molecule. Since there are 6 degrees of freedom per molecule in a solid, and only 3 per molecule (translational) for a monatomic liquid, you would expect the solid to have the higher heat capacity. 21. 1.63 min, an elapsed time that seems to be consistent with experience. 23. ep = (1.01%)ew.,., 25. (a) 10.5 MJ (b) 121 W 27. 7.48 kcal 29. 48.8 mg 31. 365°C 33. 20.8°C 35. 453 kg 37. (a) O°C (b) 125 g 39. (a) 4.94°C (b) No ice is left. 41. (a) 2.99°C (b) 199.8 g (c) The answer would be the same. 43. 618°C 45. 2.21 kJ 47. 176°C 49. 53.7 J 51. 53. 55. 57. 59. 61. 63. (a) 6.13 W (b) 19.0 min (a) 405 J (b) 861 J (a) 507 J (b) 963 J �PoVo (a) 555 J (b) 555 J (a) 55.7 g/mol (b) Fe (a) 0; 6.24 kJ; 6.24 kJ (b) 8.73 kJ; 6.24 kJ; 2.49 kJ (c) 2.49 kJ 65. 59.6 L 67. �Cp = -!iNk 69. Cv,w.'er = 5Nk 71. (a) 465 K (b) 387 K 73. (a) 300 K; 7.80 L; 1.14 kJ; 1.14 kJ (b) 208 K; 5.41 L; 574 J; 0 75. (a) 263 K (b) 10.8 L (c) -1.48 kJ (d) 1.48 kJ 79. -142 J 81. QO-;A = 8.98 kJ; QA-;B = 13.2 kJ; QB-;C = -8.98 kJ; Qc-;o = -6.56 kJ; Wcycle = 6.62 kJ 83. (a) p , , P2 - -, " P4 " - - -' 4' P3 - - -,- ,- - - , , , Vj V4 V2 85. 180 kJ 87. (a) 65.2 K; 81.2 K (b) 1.62 kJ (c) 2.22 kJ 89. (a) 65.2 K; 81.2 K (b) 2.65 kJ (c) 3.25 kJ 91. (a) 9.20 X 10-2 J/kg'K (b) 0.0584 J/kg 93. 47.6 kPa; 51.5 K; 71.2 K; 148 kPa 95. (a) 2.49 kJ (b) 3.20 kJ 97. 171 K 99. (a) W = 0; Q = 3.74 kJ T,; T' c , 3 V3 V (b) �U = 3.74 kJ; Q = 6.24 kJ; W = 2.50 kJ 101. 110 - - - - - -10 " � __ __ __ __ __ __ __ __ __ __ __ L-- 103. 4RT 105. 396 K 107. (a) 1Po 20353.5 771.5 (b) diatomic 3028.5 3048.5 t (s) (c) In the isothermal process, T is constant, and the transla­ tional kinetic energy is unchanged. In the adiabatic process, T3 = 1.32To, and the translational kinetic energy increases by a factor of 1.32.
  • 641. 109. (a) 93.5 kPa (b) 6266 K; 1.30 MPa (c) 56.7 kPa 111. (b) flU = 4621 J, a result in good agreement with the result of Problem 106. Chapter 1 9 1. Friction reduces the efficiency of the engine. 3. Increasing the temperature of the steam increases the Carnot effi­ ciency, and generally increases the efficiency of any heat engine. 5. (c) 7. (d) 9. Note that A--7B is an adiabatic expansion. B--7C is a constant volume process in which the entropy decreases; therefore heat is released. C--7D is an adiabatic compression. D--7A is a con­ stant volume process that returns the gas to its original state. The cycle is that of the Otto engine (see Figure 19-3). 11. 5 � _ _ ---, 3 1 V 13. p C "-----�A �---------V 15. 56.5% 17. (a) 1.66 x 1017 W (b) 5.66 x 1014 JIK·s (c) 3.09 x 1013 J/K·s 19. 29.8 kJ/K 21. (a) 500 J (b) 400 J 23. (a) 40.0% (b) 80.0 W 25. (a) 2 tJ 1 4 10 20 30 40 50 V (L) 27. Chapter 20 W1....2 = 0; Ql....2 = 3.74 kJ W2....3 = 4.99 kJ; Q2....2 = 12.5 kJ W3....4 = 0; Q3....4 = -7.48 kJ W4....1 = 2.49 kJ; Q4....1 = -6.24 kJ (b) 15.4% 2 1.5 ]' - - �- � � , 1 , P.., 4, 2 0.5 - - - - , - - - .... ...._400 K A-27 , 3.... - - - , - -300 K , 0 0 10 20 30 40 50 60 V (L) 13.1% 29. (a) 600 K; 1800 K; 600 K (b) 15.4% 31. (a) 5.16%; The fact that this efficiency is considerably less than the actual efficiency of a human body does not con­ tradict the Second Law of Thermodynamics. The applica­ tion of the second law to chemical reactions such as the ones that supply the body with energy have not been discussed in the text. (b) 35. (a) (b) (c) (d) Most warm-blooded animals survive under roughly the same conditions as humans. To make a heat engine work with appreciable efficiency, internal body temperatures would have to be maintained at an unreasonably high level. 33.3% 33.3 J 66.7 J 2.00 37. Let the first engine be run as a refrigerator. Then it will remove 140 J from the cold reservoir, deliver 200 J to the hot reservoir, and require 60 J of energy to operate. Now take the second engine and run it between the same reservoirs, and let it eject 140 J into the cold reservoir, thus replacing the heat removed by the refrigerator. If 82, the efficiency of this engine, is greater than 30%, then Qh2' the heat removed from the hot reservoir by this engine, is 140 J1(1 - 82) > 200 J, and the work done by this engine is W = 82Qh2 > 200 J. The end result of all this is that the second engine can run the refrigerator, replacing the heat taken from the cold reservoir, and do additional mechanical work. The two systems working together then convert heat into me­ chanical energy without rejecting any heat to a cold reservoir, in violation of the second law. 39. (a) 33.3% (b) If COP > 2, then 50 J of work will remove more than 100 J of heat from the cold reservoir and put more than 150 J of heat into the hot reservoir. So running engine (a) to oper­ ate the refrigerator with a COP > 2 will result in the trans­ fer of heat from the cold to the hot reservoir without doing any net mechanical work in violation of the second law. 41. (a) 100°C (b) Ql....2 = 3.12 kJ; Q2....3 = 0; Q3....1 = -2.91 kJ (c) 6.73% (d) 35.5% 43. (a) 5.26 (b) 3.l9 kW (c) 4.81 kW
  • 642. A-28 Answers 45. (a) 173 kJ (b) 121 kJ 47. t.Su = 2.40 JIK 49. (a) 11.5 JIK (b) Since the process is not quasi-static, it isnonreversible and the entropy of the universe must increase. 51. 1.22 kJIK 53. 55. 57. 59. 61. 63. 65. 67. (a) 0 (b) 267 K (a) 244 kJ/K (b) -244 kJ/K (c) t.Su > 0 (a) - 117 J/K (b) 137 J/K (c) 20.3 J/K 1.97 kJ/K (a) 0.417 J/K (b) 125 J (a) 20.0 J (b) 66.7 J; 46.7 J (a) 51.0% (b) 102 kJ (c) 98.0 kJ 113 W/K 69. (a) Process (1) is more wasteful of mechanical energy. Process (2) is more wasteful of total energy. (b) 1.67 JIK; 0.833 JIK 71. 313 K 73. 10.0 W 75. (a) 253 kPa (b) 462 K (c) 6.96 kJ; 25.9% 77. (a) 253 kPa (b) 416 K (c) 6.58 kJ; 34.8% 79. 180 J 83. =10478 - Chapter 2 0 1. The glass bulb warms and expands first, before the mercury warms and expands. 3. (c) 5. (a) With increasing altitude P decreases; from curve OF, T of the liquid-gas interface diminishes, so the boiling temper­ ature decreases. Likewise, from curve OH, the melting temperature increases with increasing altitude. (b) Boiling at a lower temperature means that the cooking time will have to be increased. 7. The thermal conductivity of metal and marble is much greater than that of wood; consequently, heat transfer from the hand is more rapid. 9. (c) 11. In the absence of matter to support conduction and convection, radiation is the only mechanism. 13. (a) 15. The temperature of an object is inversely proportional to the maximum wavelength at which the object radiates (Wein's dis­ placement law). Since blue light has a shorterwavelength than red light, an object for which the wavelength of the peak of thermal emission is blue is hotter than one which is red. 17. 18.1 mWl(m·K) 19. 2.90 nm 21. (a) t.AIA y = � (b) y = 2at.T 23. 217°C 25. 15.4 x 10-6 K-l 27. 5.24 m 29. 0.255 mm 31. (a) The clock runs slow. (b) 8.21 s 33. 3.68 x 10-12 N1m2 35. (a) 90°C (b) 82°C (c) 170 kPa 37. (b) (pr+ �2}3Vr - 1) = 8Tr 39. 2.07 kBtu/h 41. (a) leu = 962 W; fAl = 569 W (b) 1.53 kW (c) 0.0523 K/W 43. (a) Conservation of energy requires that the thermal current through each shell be the same. 21TkL f = I ( I ) (T2 - TI) n 1'1 1'2 45. 47. 9.35 X 10-3 m2 49. 1598°C 51. 2.10 km 53. 5767 K 55. 57. 59. 1.18 cm f3exp - 13th � (b) < � f3t11 1.26 X 1010 kW; <0.002% 61. 132 W ignoring the cylindrical insulation; 142 W taking the in­ sulation into account. 63. L2 = LI; W2 = (1 - 2at.T)wI; E2 = EI(l - 2at.T) 65. (a) 0.698 cm/h (b) 11.9 d 67. (b) 40.5 min 600 ,-------------------------. �o -- - - -- - - - - - - - ---- - - - - - - � 400 � h 300 200 100 O +----,-----,----,----.--� o 500 1000 1500 t (h) 2000 2500