Student Edition Solutions
Forces and the
Laws of Motion
Forces and the Laws of Motion, Practice B
Givens
1. F = 70.0 N
q = +30.0°
I
Solutions
Fx = F(cos q) = (70.0 N)(cos 30.0°) = 60.6 N
Fy = F(sin q) = (70.0 N)(sin 30.0°) = 35.0 N
2. Fy = −2.25 N
Fx = 1.05 N
2
2
Fnet = F
+
F
N
)2 + (−
2.25 N
)2
x y = (1.05
Fnet = 1.10
N2
+ 5.0
6 N2 = 6.16
N2 = 2.48 N
F
1.05 N
q = tan−1 x = tan−1 −2.25 N
Fy
q = 25.0° counterclockwise from straight down
3. Fwind = 452 N north
Fwater = 325 N west
Fx = Fwater = −325 N
Fy = Fwind = 452 N
2
2
Fnet = F
+
F
)2
+(45
)2
32
5N
2N
x y = (−
Fnet = 1.
05
N2+
05
N2 = 3.
05
N2
06
×1
2.0
4×1
10
×1
Fnet = 557 N
F
452 N
q = tan−1 y = tan−1  = −54.3°
Fx
−325 N
Copyright © by Holt, Rinehart and Winston. All rights reserved.
q = 54.3° north of west, or 35.7° west of north
Forces and the Laws of Motion, Section 2 Review
3. Fy = 130.0 N
Fx = 4500.0 N
2
2
+
F
)2
+(13
)2
Fnet = F
50
0.
0N
0.
0N
x y = (4
Fnet = 2.
07
N2+
04
N2 = 2.
07
N2 = 4502 N
02
5×1
1.6
90
×1
02
7×1
Fy
130.0 N
q = tan−1  = tan−1  = 1.655° forward of the side
Fx
4500.0 N
Forces and the Laws of Motion, Practice C
1. Fnet = 7.0 N forward
m = 3.2 kg
2. Fnet = 390 N north
m = 270 kg
F
7.0 N
 =  = 2.2 m/s2 forward
a = net
m 3.2 kg
F
390 N
 =  = 1.4 m/s2 north
a = net
m 270 kg
Section One—Student Edition Solutions
I Ch. 4–1
Forces and the Laws of Motion, Practice C
Givens
Solutions
3. Fnet = 6.75 × 103 N east
3
m = 1.50 × 10 kg
6.75 × 103 N east
F
 = 
a = net
= 4.50 m/s2 east
1.50 × 103 kg
m
I
4. Fnet = 13.5 N to the right
a = 6.5 m/s2 to the right
5. m = 2.0 kg
F et Fnet 13.5 N
m = n
=  = 2 = 2.1 kg
a
a
6.5 m/s
1
∆x = 2a(∆t)2
2∆x (2)(0.85 m)
a = 2 = 
= 6.8 m/s2
∆t
(0.50 s)2
Fnet = ma = (2.0 kg)(6.8 m/s2) = 14 N
∆x = 85 cm
∆t = 0.50 s
Forces and the Laws of Motion, Section 3 Review
a. Fnet = ma = (6.0 kg)(2.0 m/s2) = 12 N
1. m = 6.0 kg
a = 2.0 m/s2
12 N
F
 =  = 3.0 m/s2
b. a = net
4.0 kg
m
m = 4.0 kg
4. Fy = 390 N, north
Fx = 180 N, east
m = 270 kg
2 + F 2 = (180 N)2 + (390 N)2
Fnet = F
x y
Fnet = 3.
04
N2+1.5
05
N2 = 1.
05
N2 = 420 N
2×1
×1
8×1
Fy
390 N
q = tan−1  = tan−1 
Fx
180 N
420 N
F
 =  = 1.6 m/s2
a = net
270 kg
m
Forces and the Laws of Motion, Practice D
1. Fk = 53 N
F
Fk
53 N
mk = k = 
= 
= 0.23
Fn mg (24 kg)(9.81 m/s2)
m = 24 kg
g = 9.81 m/s2
2. m = 25 kg
Fs, max = 165 N
Fk = 127 N
2
g = 9.81 m/s
I Ch. 4–2
Fs,max Fs,max
165 N
 =  = a. ms = 
= 0.67
Fn
mg
(25 kg)(9.81 m/s2)
F
Fk
127 N
b. mk = k = 
= = 0.52
Fn mg
(25 kg)(9.81 m/s2)
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
q = 65° north of east
Givens
Solutions
3. m = 145 kg
a. Fs,max = msFn = msmg = (0.61)(145 kg)(9.81 m/s2) = 8.7 × 102 N
ms = 0.61
Fk = mkFn = mkmg = (0.47)(145 kg)(9.81 m/s2) = 6.7 × 102 N
mk = 0.47
g = 9.81 m/s2
m = 15 kg
I
2
2
b. Fs,max = msFn = msmg = (0.74)(15 kg)(9.81 m/s ) = 1.1 × 10 N
ms = 0.74
Fk = mkFn = mkmg = (0.57)(15 kg)(9.81 m/s2) = 84 N
mk = 0.57
g = 9.81 m/s2
m = 250 kg
c. Fs,max = msFn = msmg = (0.4)(250 kg)(9.81 m/s2) = 1 × 103 N
ms = 0.4
Fk = mkFn = mkmg = (0.2)(250 kg)(9.81 m/s2) = 5 × 102 N
mk = 0.2
g = 9.81 m/s2
m = 0.55 kg
d. Fs,max = msFn = msmg = (0.9)(0.55 kg)(9.81 m/s2) = 5 N
ms = 0.9
Fk = mkFn = mkmg = (0.4)(0.55 kg)(9.81 m/s2) = 2 N
mk = 0.4
g = 9.81 m/s2
Forces and the Laws of Motion, Practice E
1. Fapplied = 185 N at 25.0°
above the horizontal
m = 35.0 kg
Fapplied, y = Fapplied (sin q)
Fy, net = ΣFy = Fn + Fapplied, y − Fg = 0
mk = 0.27
g = 9.81 m/s2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Fapplied, x = Fapplied (cos q)
Fn = Fg − Fapplied, y = mg − Fapplied (sin q)
Fn = (35.0 kg)(9.81 m/s2) − (185 N)(sin 25.0°) = 343 N − 78.2 N = 265 N
Fk = mkFn = (0.27)(265 N) = 72 N
Fx, net = ΣFx = Fapplied, x − Fk = Fapplied(cos q) − Fk
Fx, net = (185 N)(cos 25.0°) − 72 N = 168 N − 72 N = 96 N
F
96 N
 =  = 2.7 m/s2
ax = x,net
m
35.0 kg
a = ax = 2.7 m/s2 in the positive x direction
2. q1 = 12.0°
Fg,y = mg(cos q1) = (35.0 kg)(9.81 m/s2)(cos 12.0°) = 336 N
q2 = 25.0°
Fg,x = mg(sin q1) = (35.0 kg)(9.81 m/s2)(sin 12.0°) = 71.4 N
Fapplied = 185 N
Fapplied,x = Fapplied (cos q2) = (185 N)(cos 25.0°) = 168 N
m = 35.0 kg
Fapplied,y = Fapplied (sin q2) = (185 N)(sin 25.0°) = 78.2 N
mk = 0.27
Fy,net = ΣFy = Fn + Fapplied,y − Fg,y = 0
2
g = 9.81 m/s
Fn = Fg,y − Fapplied,y = 336 N − 78.2 N = 258 N
Fk = mkFn = (0.27)(258 N) = 7.0 × 101 N
Fx,net = ΣFx = Fapplied,x − Fk − Fg,x = max
Fapplied,x − Fk − Fg,x
ax = 
m
Section One—Student Edition Solutions
I Ch. 4–3
Givens
Solutions
168 N − 7.0 × 101 N − 71.4 N
27 N
ax =  =  = 0.77 m/s2
35.0 kg
35.0 kg
a = ax = 0.77 m/s2 up the ramp
I
3. m = 75.0 kg
a. Fx,net = max = Fg,x − Fk
q = 25.0 °
Fg,x = mg(sin q)
2
ax = 3.60 m/s
2
g = 9.81 m/s
Fk = Fg,x − max = mg(sin q) − max
Fk = (75.0 kg)(9.81 m/s2)(sin 25.0°) − (75.0 kg)(3.60 m/s2)
Fk = 311 N − 2.70 × 102 N = 41 N
Fn = Fg,y = mg(cos q) = (75.0 kg)(9.81 m/s2)(cos 25.0°) = 667 N
F
41 N
mk = k =  = 0.061
Fn 667 N
m = 175 kg
mk = 0.061
b. Fx,net = Fg,x − Fk = mg sin q − mkFn
Fn = Fg,y = mg cosq
mg sin q − mkmg cosq
F
 =  = g sin q − mkg cosq
ax = x,net
m
m
ax = g(sin q − mk cos q) = 9.81 m/s2 [sin 25.0° − (0.061)(cos 25.0°)]
ax = 9.81 m/s2 (0.423 − 0.055) = (9.81 m/s2)(0.368) = 3.61 m/s2
a = ax = 3.61 m/s2 down the ramp
4. Fg = 325 N
Fx,net = Fapplied,x − Fk = 0
Fapplied = 425 N
Fk = Fapplied,x = Fapplied (cos q) = (425 N)[cos(−35.2°)] = 347 N
q = −35.2°
Fy,net = Fn + Fapplied,y − Fg = 0
Fn = Fg − Fapplied,y = Fg − Fapplied (sin q)
F
347 N
mk = k = 
= 0.609
Fn 5.70 × 102 N
Forces and the Laws of Motion, Section 4 Review
2. m = 2.26 kg
g = 9.81 m/s2
1
1
a. Fg = 6 mg = 6 (2.26 kg)(9.81 m/s2) = 3.70 N
b. Fg = (2.64)mg = (2.64)(2.26 kg)(9.81 m/s2) = 58.5 N
3. m = 2.0 kg
q = 60.0°
g = 9.81 m/s2
a. Fx,net = F(cos q) − mg(sin q) = 0
mg (sin q ) (2.0 kg)(9.81 m/s2)(sin 60.0°)
F =  =  = 34 N
cos q
cos 60.0°
b. Fy,net = Fn − F(sin q) − mg(cos q) = 0
Fn = F(sin q ) + mg(cos q) = (34 N)(sin 60.0°) + (2.0 kg)(9.81 m/s2)(cos 60.0°)
Fn = 29 N + 9.8 N = 39 N
I Ch. 4–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Fn = 325 N − (425 N)[sin (−35.2°)] = 325 N + 245 N = 5.70 × 102 N
Givens
4. m = 55 kg
Fs, max = 198 N
Fk = 175 N
g = 9.81 m/s2
Solutions
Fs,max Fs,max
198 N
 =  = 
ms = 
= 0.37
Fn
mg
(55 kg)(9.81 m/s2)
F
Fk
175 N
mk = k = 
=  = 0.32
Fn mg (55 kg)(9.81 m/s2)
I
Forces and the Laws of Motion, Chapter Review
10. Fx,1 = 950 N
Fx,2 = −1520 N
Fy,1 = 5120 N
Fy,2 = −4050 N
Fx,net = Fx,1 + Fx,2 = 950 N + (−1520 N) = −570 N
Fy,net = Fy,1 + Fy,2 = 5120 N + (−4050 N) = 1070 N
Fnet = (Fx,net)2
+ (Fy ,net
)2 = (−570
N)2 + (
1070 N
)2
Fnet = 3.
05
N2+
06
N2 = 1.
06
N2 = 1.21 × 103 N
2×1
1.1
4×1
46
×1
F net
1070 N
q = tan−1 y,
= tan−1  = −62°
Fx,net
−570 N
q = 62° above the 1520 N force
11. F1 + F2 = 334 N
−F1 + F2 = −106 N
b.
F1 + F2 =
334 N
+(−F1 + F2) = (−106 N)

2F2 = 228 N
F2 = 114 N
F1 + 114 N = 334 N
F1 = 220 N
F1 = 220 N right for the first situation and left for the second
Copyright © by Holt, Rinehart and Winston. All rights reserved.
F2 = 114 N right for both situations
12. F = 5 N
Fx = F(cos q) = (5 N)(cos 37°) = 4 N
q = 37°
Fy = F(sin q) = (5 N)(sin 37°) = 3 N
20. m = 24.3 kg
Fnet = 85.5 N
21. m = 25 kg
a = 2.2 m/s2
F
85.5 N
 =  = 3.52 m/s2
a = net
m
24.3 kg
Fnet = ma = (25 kg)(2.2 m/s2) = 55 N
Fnet = 55 N to the right
Section One—Student Edition Solutions
I Ch. 4–5
I
Givens
Solutions
22. F1 = 380 N
a. F1,x = F1(sin q1) = (380 N)(sin 30.0°) = 190 N
q1 = 30.0°
F1,y = F1(cos q1) = (380 N)(cos 30.0°) = 330 N
F2 = 450 N
F2,x = F2(sin q2) = (450 N)[sin (−10.0°)] = −78 N
q2 = −10.0°
F2,y = F2(cos q2) = (450 N)[cos (−10.0°)] = 440 N
Fy,net = F1,y + F2,y = 330 N + 440 N = 770 N
Fx,net = F1,x + F2,x = 190 N − 78 N = 110 N
2
Fnet = (F
)2 = (1
N
)2
+(77
)2
x,
(F
y,n
et
10
0N
ne
t)+
Fnet = 1.
N2+
N2 = 6.
N2 = 770 N
2×104
5.9
×105
0×105
F net
110 N
q = tan−1 x,
= tan−1  = 8.1° to the right of forward
Fy,net
770 N
770 N
F
 =  = 0.24 m/s2
b. a = net
3200 kg
m
m = 3200 kg
anet = 0.24 m/s2 at 8.1° to the right of forward
24. m = 0.150 kg
a. Fnet = −mg = −(0.150 kg)(9.81 m/s2) = −1.47 N
vi = 20.0 m/s
g = 9.81 m/s2
b. same as part a.
a. Fn = mg = (5.5 kg)(9.81 m/s2) = 54 N
26. m = 5.5 kg
q = 12°
b. Fn = mg(cos q) = (5.5 kg)(9.81 m/s2)(cos 12°) = 53 N
q = 25°
c. Fn = mg(cos q) = (5.5 kg)(9.81 m/s2)(cos 25°) = 49 N
q = 45°
d. Fn = mg(cos q) = (5.5 kg)(9.81 m/s2)(cos 45°) = 38 N
Fn = mg(cos q) = (5.4 kg)(9.81 m/s2)(cos 15°) = 51 N
29. m = 5.4 kg
q = 15°
g = 9.81 m/s2
Fn = Fg = mg = (95 kg)(9.81 m/s2) = 930 N
35. m = 95 kg
Fs,max = 650 N
Fk = 560 N
g = 9.81 m/s2
Fs,max 650 N
ms =  =  = 0.70
Fn
930 N
F 560 N
mk =  k =  = 0.60
F n 930 N
Fn = mg(cos q)
36. q = 30.0°
2
mg(sin q) − Fk = max , where Fk = mkFn = mkmg(cos q)
2
mg(sin q) − mkmg(cos q) = max
a = 1.20 m/s
g = 9.81 m/s
sin q
ax
ax
mg(sin q) − ma
mk = x =  − 
= tan q − 
cos q
g(cos q)
g(cos q)
mg(cos q)
1.20 m/s2
mk = (tan 30.0°) − 
= 0.577 − 0.141 = 0.436
(9.81 m/s2)(cos 30.0°)
I Ch. 4–6
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
g = 9.81 m/s2
Givens
Solutions
37. m = 4.00 kg
Fapplied,x = Fapplied(cos q) = (85.0 N)(cos 55.0°) = 48.8 N
Fapplied = 85.0 N
Fapplied,y = Fapplied(sin q) = (85.0 N)(sin 55.0°) = 69.6 N
q = 55.0°
Fn = Fapplied,y − mg = 69.6 N − (4.00 kg)(9.81 m/s2) = 69.6 N − 39.2 N = 30.4 N
ax = 6.00 m/s2
max = Fapplied,x − Fk = Fapplied,x − mkFn
48.8 N − (4.00 kg)(6.00 m/s2)
Fapplied,x − max
 = 
mk = 
30.4 N
Fn
2
g = 9.81 m/s
I
48.8 N − 24.0 N 24.8 N
mk =  =  = 0.816
30.4 N
30.4 N
38. Fapplied = 185.0 N
Fapplied,x = Fapplied (cos q) = (185.0 N)(cos 25.0°) = 168 N
q = 25.0°
Fapplied,y = Fapplied (sin q) = (185.0 N)(sin 25.0°) = 78.2 N
m = 35.0 kg
Fy,net = Fn + Fapplied,y − mg = 0
mk = 0.450
Fn = mg − Fapplied,y = (35.0 kg)(9.81 m/s2) − 78.2 N = 343 N − 78.2 N = 265 N
g = 9.81 m/s2
Fk = mkFn = (0.450)(265 N) = 119 N
Fx,net = max = Fapplied,x − Fk
49 N
F
168 N − 119 N
 =  =  = 1.4 m/s2
ax = x,net
m
35.0 kg
35.0 kg
a = ax = 1.4 m/s2 down the aisle
39. Fg = 925 N
Fapplied,x = Fapplied (cos q) = (325 N)(cos 25.0°) = 295 N
Fapplied = 325 N
Fapplied,y = Fapplied (sin q) = (325 N)(sin 25.0°) = 137 N
q = 25.0°
Fy,net = Fn + Fapplied,y − Fg = 0
mk = 0.25
Fn = Fg − Fapplied,y = 925 N − 137 N = 788 N
2
Fk = mkFn = (0.25)(788 N) = 2.0 × 102 N
g = 9.81 m/s
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Fx,net = max = Fapplied,x − Fk = 295 N − 2.0 × 102 N = 95 N
925 N
Fg
m =  = 2 = 94.3 kg
9.81 m/s
g
95 N
F
 =  = 1.0 m/s2
ax = x,net
m
94.3 kg
mg
(6.0 kg)(9.81 m/s2)
Fn =  =  = 68 N
cos q
cos 30.0°
40. m = 6.0 kg
q = 30.0°
F = Fn sin q = (68 N)(sin 30.0°) = 34 N
g = 9.81 m/s2
41. m = 2.0 kg
∆x = 8.0 × 10
Because vi = 0 m/s,
−1
m
1
∆x = 2a∆t2
∆t = 0.50 s
2∆x (2)(0.80 m)
a = 2 = 
= 6.4 m/s2
∆t
(0.50 s)2
vi = 0 m/s
Fnet = ma = (2.0 kg)(6.4 m/s2) = 13 N
Fnet = 13 N down the incline
Section One—Student Edition Solutions
I Ch. 4–7
Givens
Solutions
42. m = 2.26 kg
b. Fg = mg = (2.26 kg)(9.81 m/s2) = 22.2 N
∆y = −1.5 m
g = 9.81 m/s2
I
43. m = 5.0 kg
a = 3.0 m/s
FT − Fg = ma
2
FT = ma + Fg = ma + mg
2
g = 9.81 m/s
FT = (5.0 kg)(3.0 m/s2) + (5.0 kg)(9.81 m/s2) = 15 N + 49 N = 64 N
FT = 64 N upward
44. m = 3.46 kg
b. Fg = mg = (3.46 kg)(9.81 m/s2) = 33.9 N
g = 9.81 m/s2
45. F1 = 2.10 × 103 N
a. Fnet = F1 + F2 = 2.10 × 103 N + (−1.80 × 103 N)
F2 = −1.80 × 103 N
Fnet = 3.02 × 102 N
m = 1200 kg
F
3.02 × 102 N
 =  = 0.25 m/s2
anet = net
m
1200 kg
2
anet = 0.25 m/s forward
∆t = 12 s
vi = 0 m/s
1
1
b. ∆x = vi ∆t + 2a∆t 2= (0 m/s)(12 s) + 2(0.25 m/s2)(12 s)2
∆x = 18 m
c. vf = a∆t + vi = (0.25 m/s2)(12 s) + 0 m/s
vf = 3.0 m/s
mk = 0.050
Fg = 645 N
g = 9.81 m/s2
vf = 0 m/s
Fk = mkFn = (0.050)(645 N) = 32 N
645 N
Fg
m =  = 2 = 65.7 kg
9.81 m/s
g
−32 N
−F
a = k =  = −0.49 m/s2
65.7 kg
m
2
2
(0 m/s)2 − (7.0 m/s)2
vf − vi
∆x = 
= 
(2)(−0.49 m/s2)
2a
1
∆x = 5.0 × 10 m
I Ch. 4–8
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
46. vi = 7.0 m/s
Givens
Solutions
47. Fg = 319 N
a. Fapplied, x = Fapplied(cos q) = (485 N)[cos(−35°)] = 4.0 × 102 N
Fapplied = 485 N
Fapplied,y = Fapplied(sin q) = (485 N)[sin(−35°)] = −2.8 × 102 N
q = −35°
Fy,net = Fn + Fapplied,y − Fg = 0
mk = 0.57
Fn = Fg − Fapplied,y = 319 N − (−2.8 × 102 N) = 6.0 × 102 N
∆x = 4.00 m
Fk = mkFn = (0.57)(6.0 × 102 N) = 3.4 × 102 N
g = 9.81 m/s2
Fx,net = max = Fapplied,x − Fk = 4.0 × 102 N − 3.4 × 102 N = 6 × 101 N
vi = 0 m/s
319 N
Fg
m =  = 2 = 32.5 kg
9.81 m/s
g
F
6 × 101 N
 =  = 2 m/s2
ax = x,net
m
32.5 kg
I
1
∆x = vi ∆t + 2ax ∆t 2
Because vi = 0 m/s,
∆t =
2m
/s =
a = 
2∆x
(2)(4.00 m)
2
x
mk = 0.75
2s
b. Fk = mkFn = (0.75)(6.0 × 102 N) = 4.5 × 102 N
Fk > Fapplied,x ; the box will not move
48. m = 3.00 kg
q = 30.0°
∆x = 2.00 m
∆t = 1.50 s
g = 9.81 m/s2
vi = 0 m/s
1
a. ∆x = vi ∆t + 2a∆t 2
Because vi = 0 m/s,
2∆x (2)(2.00 m)
2
a = 2 = 
= 1.78 m/s
∆t
(1.50 s)2
b. Fg,x = mg(sin q) = (3.00 kg)(9.81 m/s2)(sin 30.0°) = 14.7 N
Fg,y = mg(cos q) = (3.00 kg)(9.81 m/s2)(cos 30.0°) = 25.5 N
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Fn = Fg,y = 25.5 N
max = Fg,x − Fk = Fg,x − mkFn
14.7 N − (3.00 kg)(1.78 m/s2)
Fg,x − ma
mk = x = 
25.5 N
Fn
14.7 N − 5.34 N
9.4 N
mk =  =  =
25.5 N
25.5 N
c. Fk = mkFn = (0.37)(25.5 N) =
d. vf2 = vi2+ 2ax∆x
0.37
9.4 N
m/s
vf = vi 2+
2ax ∆
x = (0
)2+(2)
(1
.7
8m
/s
2)(2
.0
0m
) = 2.67 m/s
49. vi = 12.0 m/s
∆t = 5.00 s
vf = 6.00 m/s
vf = vi + a∆t
vf − vi 6.00 m/s − 12.0 m/s −6.0 m/s
a =  =  = 
∆t
5.00 s
5.00 s
a = −1.2 m/s2
Fk = −ma
F
−ma −a 1.2 m/s2
mk = k =  =  = 2 = 0.12
Fn
mg
g
9.8 m/s
Section One—Student Edition Solutions
I Ch. 4–9
I
Givens
Solutions
50. Fg = 8820 N
vi = 35 m/s
Fg
8820 N
m =  = 2 = 899 kg
g 9.81 m/s
∆x = 1100 m
vf 2= vi 2 + 2a∆x = 0
g = 9.81 m/s2
−vi 2
−(35 m/s)2
a = 
=  = −0.56 m/s2
2∆x (2)(1100 m)
2
Fnet = ma = (899 kg)(−0.56 m/s2) = −5.0 × 10 N
51. mcar = 1250 kg
mtrailer = 325 kg
2
a = 2.15 m/s
a. Fnet = mcara = (1250 kg)(2.15 m/s2) = 2690 N
Fnet = 2690 N forward
b. Fnet = mtrailera = (325 kg)(2.15 m/s2) = 699 N
Fnet = 699 N forward
52. m = 3.00 kg
Fs, max = msFn
q = 35.0°
mg(sin q) = ms[F + mg(cos q)]
ms = 0.300
mg[sin q − ms(cos q)]
mg(sin q) − msmg(cos q)
F = 
= 
ms
ms
g = 9.81 m/s2
(3.00 kg)(9.81 m/s2)[sin 35.0° 0.300 (cos 35.0°)]
F = 
0.300
(3.00 kg)(9.81 m/s2)(0.574 0.246)
(3.00 kg)(9.81 m/s2)(0.328)
F =  
0.300
0.300
F=
53. m = 64.0 kg
32.2 N
At t = 0.00 s, v = 0.00 m/s. At t = 0.50 s, v = 0.100 m/s.
F = ma = (64.0 kg)(0.20 m/s2) = 13 N
At t = 0.50 s, v = 0.100 m/s. At t = 1.00 s, v = 0.200 m/s.
vf − vi 0.200 m/s − 0.100 m/s
a =  =  = 0.20 m/s2
1.00 s − 0.50 s
tf − ti
F = ma = (64.0 kg)(0.20 m/s2) = 13 N
At t = 1.00 s, v = 0.200 m/s. At t = 1.50 s, v = 0.200 m/s.
a = 0 m/s2; therefore, F = 0 N
At t = 1.50 s, v = 0.200 m/s. At t = 2.00 s, v = 0.00 m/s.
vf − vi 0.00 m/s − 0.200 m/s
a =  =  = −0.40 m/s2
2.00 s − 1.50 s
tf − ti
F = ma = (64.0 kg)(−0.40 m/s2) = −26 N
I Ch. 4–10
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
0.100 m/s − 0.00 m/s
vf − v
a = i =  = 0.20 m/s2
0.50 s − 0.00 s
tf − ti
Givens
Solutions
54. Fapplied = 3.00 × 102 N
Fnet = Fapplied − Fg sin q = 0
Fg = 1.22 × 104 N
g = 9.81 m/s2
Fapplied
sin q = 
Fg
3.00 × 102 N
Fapplied
q = sin−1  = sin−1 1.22 × 104 N
Fg
I
q = 1.41°
Forces and the Laws of Motion, Standardized Test Prep
3. Fx,1 = 82 N
Fx,net = Fx,1 + Fx,2 = 82 N − 115 N = −33 N
Fx,2 = −115 N
Fy,net = Fy,1 + Fy,2 = 565 N − 236 N = 329 N
Fy,1 = 565 N
2
)2 = (−
N
)2
+(32
)2
Fnet = (F
x,
(F
33
9N
ne
t)+
y,n
et
Fy,2 = −236 N
Fnet = 1.
N2+
N2 = 1.
N2 = 3.30 × 102 N
09
×103
1.0
8×105
09
×105
Fy,net
329 N
q = tan−1  = tan−1  = −84° above negative x-axis
Fx,net
−33 N
q = 180.0° − 84° = 96° counterclockwise from the positive x-axis
5. m = 1.5 × 107 kg
Fnet = 7.5 × 105 N
vf = 85 km/h
vi = 0 km/h
6.
7.5 × 105 N
F
 = 
= 5.0 × 10−2 m/s2
a = net
1.5 × 107 kg
m
(85 km/h − 0 km/h)(103 m/km)(1 h/3600 s)
vf − v
∆t = i = 
5.0 × 10−2 m/s2
a
∆t = 4.7 × 102 s
Apply Newton’s second law to find an expression for the acceleration of the truck.
ma = Ff = mkFn = mkmg
Copyright © by Holt, Rinehart and Winston. All rights reserved.
a = mkg
Because the acceleration of the truck does not depend on the mass of the truck, the
stopping distance will be ∆x regardless of the mass of the truck.
7.
2
vi
+2a∆
vf = x = 0
−vi 2
a = 
2∆x
The acceleration will be the same regardless of the intial velocity.
a1 = a2 = mkg (see 6.)
−vi,12 −vi,22
 = 
2∆x
2∆x2
vi,22∆x
1
 where vi,2 = 2vi,1
∆x2 = 
vi,12
2
∆x
∆x2 = 
=
vi,12
1
v
2 i,1
1

4
∆x
Section One—Student Edition Solutions
I Ch. 4–11
Givens
Solutions
10. m = 3.00 kg
Because the ball is dropped, vi = 0 m/s
∆y = −176.4 m
1
∆y = 2ay ∆t 2
Fw = 12.0 N
2
I
ay = −g = −9.81 m/s
11. m = 3.00 kg
2∆y
∆t =  =
ay
(2)(−176.4 m)

−9.81
m/s =
2
6.00 s
12.0 N
F
 =  = 4.00 m/s2
ax = w
3.00 kg
m
Fw = 23.0 N
∆t = 6.00 s (see 10.)
12. ay = −g = −9.81 m/s2
1
1
∆x = 2ax ∆t2 = 2(4.00 m/s2)(6.00 s)2 = 72.0 m
vy = ay ∆t = (−9.81 m/s2)(6.00 s) = −58.9 m/s
∆t = 6.00 s (see 10.)
vx = ax ∆t = (4.00 m/s2)(6.00 s) = 24.0 m/s
ax = 4.00 m/s2 (see 11.)
2
)2
+(−
m/s
)2
v = vx2+
v
4.
0m
/s
58
.9
y = (2
v = 57
s2
+347
s2 = 40
m2/
s2 = 63.6 m/s
6m
2/
0m
2/
50
16. m = 10.0 kg
Fnet,y = Fn + F sin q − mg = 0
F = 15.0 N
Fn = mg − F sin q
q = 45.0°
Fnet,x = F cos q − Fk = F cos q − mkFn
mk = 0.040
Fnet,x = F cos q − mk(mg − F sin q)
2
g = 9.81 m/s
F cos q − mk(mg − F sin q)
F
 = 
ax = net,x
m
m
(15.0 N)(cos 45.0°) − (0.040)[(10.0 kg)(9.81 m/s2) − (15.0 N)(sin 45.0°)]
ax = 
10.0 kg
10.6 N − 3.5 N 7.1 N
ax =  =  = 0.71 m/s2
10.0 kg
10.0 kg
a = ax = 0.71 m/s2
I Ch. 4–12
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
10.6 N − (0.040)(98.1 N − 10.6 N) 10.6 N − (0.040)(87.5 N)
ax =  = 
10.0 kg
10.0 kg
Problem Workbook Solutions
Forces and the
Laws of Motion
4
Additional Practice A
Givens
Solutions
a.
1.
b.
FEarth-on diver
Fair resistance-on diver
FEarth-on diver
2.
II
Fscale-on-sack
Fchef-on-sack
FEarth-on-sack
3.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Ffloor-on-toy-vertical
Ffloor-on-toy-horizontal
Fhandlebars-on-toy
FEarth-on-toy
Additional Practice B
1. mw = 75 kg
mp = 275 kg
g = 9.81 m/s2
The normal force exerted by the platform on the weight lifter’s feet is equal to and
opposite of the combined weight of the weightlifter and the pumpkin.
Fnet = Fn − mwg − mpg = 0
Fn = (mw + mp)g = (75 kg + 275 kg) (9.81 m/s2)
Fn = (3.50 × 102 kg)(9.81 m/s2) = 3.43 × 103 N
Fn = 3.43 × 103 N upward against feet
Section Two — Problem Workbook Solutions
II Ch. 4–1
Givens
Solutions
2. mb = 253 kg
Fnet = Fn,1 + Fn,2 − mbg − mwg = 0
mw = 133 kg
2
g = 9.81 m/s
The weight of the weightlifter and barbell is distributed equally on both feet, so the
normal force on the first foot (Fn,1) equals the normal force on the second foot (Fn,2).
2Fn,1 = (mb + mw)g = 2Fn,2
m
(253 kg + 33 kg) 9.81 2
s
(mb + mb)g
 = 
Fn,1 = Fn,2 = 
2
2
(386 kg)(9.81 m/s2)
Fn,1 = Fn,2 =  = 1.89 × 103 N
2
Fn,1 = Fn,2 = 1.89 × 103 N upward on each foot
3. Fdown = 1.70 N
Fnet = 4.90 N
Fnet2 = Fforward2 + Fdown2
2
2
Fforward = F
−
Fdo
)2
−(1.
N
)2
w
.9
0N
70
net
n = (4
Fforward = 21
N2 = 4.59 N
.1
II
4. m = 3.10 × 102 kg
2
Fx,net = ΣFx = FT,1(sin q1) + FT,2(sin q2) = 0
g = 9.81 m/s
Fy,net = ΣFy = FT,1(cos q1) + FT,2(cos q2) + Fg = 0
q1 = 30.0°
FT,1(sin 30.0°) = −FT,2[sin (−30.0°)]
q2 = − 30.0°
FT,1 = FT,2
FT,1(cos q1) + FT,1(cos q2) = −Fg = mg
FT,1(cos 30.0°) + FT,1[cos (−30.0°)] = (3.10 × 102 kg)(9.81 m/s2)
(3.10 × 102 kg)(9.81 m/s2)
FT,1 = 
(2)(cos 30.0°)[cos(−30.0°)]
FT,1 = FT,2 = 1.76 × 103 N
Copyright © by Holt, Rinehart and Winston. All rights reserved.
As the angles q1 and q2 become larger, cos q1 and cos q2 become smaller. Therefore,
FT,1 and FT,2 must become larger in magnitude.
II Ch. 4–2
Holt Physics Solution Manual
Givens
Solutions
5. m = 155 kg
Fx,net = FT,1(cos q1) − FT,2(cos q2) = 0
FT,1 = 2FT,2
2
g = 9.81 m/s
q1 = 90° − q2
Fy,net = FT,1(sin q1) + FT,2(sin q2) − mg = 0
1
FT,1[(cos q1) − (cos q2)] = 0
2
2 (cos q1) = cos q2 = cos(90° − q1) = sin q1
2 = tan q1
q1 = tan−1(2) = 63°
q2 = 90° − 63° = 27°
F ,1
FT,1(sin q1) + T(sin
q2) = mg
2
mg
FT,1 = 
1
(sin θ1) +  (sin θ2)
2
(155 kg)(9.81 m/s2)
(155 kg)(9.81 m/s2) (155 kg)(9.81 m)
 = 
FT,1 = 
(sin 27°) = 
0.89 + 0.23
1.12
(sin 63°) + 
2
FT,1 = 1.36 × 1.36 × 103 N
II
FT,2 = 6.80 × 102 N
Additional Practice C
1. vi = 173 km/h
vf = 0 km/h
[(0 km/h)2 − (173 km/h)2](103 m/km)2(1 h/3600 s)2
vf 2 − vi2
a = 
= 
(2)(0.660 m)
2∆x
∆x = 0.660 m
a = −1.75 × 103 m/s2
m = 70.0 kg
F = ma = (70.0 kg)(−1.75 × 103 m/s2) = −1.22° × 105 N
g = 9.81 m/s2
Fg = mg = (70.0 kg)(9.81 m/s2) = 6.87 × 102 N
Copyright © by Holt, Rinehart and Winston. All rights reserved.
The force of deceleration is nearly 178 times as large as David Purley’s weight.
2. m = 2.232 × 106 kg
2
a. Fnet = manet = Fup − mg
g = 9.81 m/s
Fup = manet + mg = m(anet + g) = (2.232 × 106 kg)(0 m/s2 + 9.81 m/s2)
anet = 0 m/s2
Fup = 2.19 × 107 N = mg
b. Fdown = mg(sin q)
F
Fup − Fdown mg − mg(sin q)
 = 
 = 
anet = net
m
m
m
9.81 m/s2
anet = g(1 − sin q) = (9.81 m/s2)[1.00 − (sin 30.0°)] =  = 4.90 m/s2
2
anet = 4.90 m/s2 up the incline
3. m = 40.00 mg
= 4.00 × 10−5 kg
g = 9.807 m/s2
anet = (400.0)g
Fnet = Fbeetle − Fg = manet = m(400.0) g
Fbeetle = Fnet + Fg = m(400.0 + 1)g = m(401)g
Fbeetle = (4.000 × 10−5 kg)(9.807 m/s2)(401) = 1.573 × 10−1 N
Fnet = Fbeetle − Fg = m(400.0) g = (4.000 × 10−5 kg)(9.807 m/s2)(400.0)
Fnet = 1.569 × 10−1 N
The effect of gravity is negligible.
Section Two — Problem Workbook Solutions
II Ch. 4–3
Givens
Solutions
4. ma = 54.0 kg
The net forces on the lifted weight is
Fw,net = mwanet = F ′ − mwg
mw = 157.5 kg
2
anet = 1.00 m/s
2
g = 9.81 m/s
where F′ is the force exerted by the athlete on the weight.
The net force on the athlete is
Fa,net = Fn,1 + Fn,2 − F ′ − mag = 0
where Fn,1 and Fn,2 are the normal forces exerted by the ground on each of the athlete’s feet, and −F′ is the force exerted by the lifted weight on the athlete.
The normal force on each foot is the same, so
Fn,1 = Fn,2 = Fn
and
F ′ = 2Fn − mag
Using the expression for F ′ in the equation for Fw,net yields the following:
mwanet = (2Fn − mag) − ma g
2Fn = mw(anet + g) + mag
mw(anet + g) + ma g (157.5 kg)(1.00 m/s2 + 9.81 m/s2) + (54.0 kg)
 = 
Fn = 
2
2
II
(157.5 kg)(10.81 m/s2) + (54.0 kg)(9.81 m/s2)
Fn = 
2
1702 N + 5.30 × 102 N 2232 N
Fn =  =  = 1116 N
2
2
Fn,1 − Fn,2 = Fn = 1116 N upward
5. m = 2.20 × 102 kg
Fnet = manet = Favg − mg
anet = 75.0 m/s
Favg = m(anet + g) = (2.20 × 102 kg)(75.0 m/s2 + 9.81 m/s2)
g = 9.81 m/s2
Favg = (2.20 × 102 kg)(84.8 m/s2) = 1.87 × 104 N
2
∆t = 2.5
vf − v
(1.0 m/s − 0.0 m/s)
anet = i =  = 0.40 m/s2
2.5 s
∆t
vi = 0 m/s
Fnet = manet = FT − mg
6. m = 2.00 × 104 kg
FT = manet + mg = m(anet + g)
vf = 1.0 m/s
FT = (2.00 × 104 kg)(0.40 m/s2 + 9.81 m/s2)
g = 9.81 m/s2
FT = (2.00 × 104 kg)(10.21 m/s2) = 2.04 × 105 N
FT = 2.04 × 105 N
7. m = 2.65 kg
Fx,net = FT,1(cos q1) − FT,2(cos q2) = 0
q1 = q2 = 45.0°
2
anet = 2.55 m/s
2
g = 9.81 m/s
FT,1(cos 45.0°) = FT,2(cos 45.0°)
FT,1 = FT,2
Fy,net = manet = FT,1(sin q1) + FT,2(sin q2) − mg
FT = FT,1 = FT,2
q = q1 = q2
FT(sin q) + FT(sin q) = m(anet + g)
2FT(sin q) = m(anet + g)
II Ch. 4–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Favg = 1.87 × 104 N upward
Givens
Solutions
m(anet + g) (2.65 kg)(2.55 m/s2 = 9.81 m/s2)
 = 
FT = 
(2)(sin 45.0°)
2 (sin q)
(2.65 kg)(12.36 m/s2)
FT =  = 23.2 N
(2)(sin 45.0°)
FT,1 = 23.2 N
FT,2 = 23.2 N
vf 2 − vi2
(0.550 m/s)2 − (0.00 m/s)2
anet = 
=  = 9.76 × 10−2 m/s2
2∆x
(2)(1.55 m)
8. m = 20.0 kg
∆x = 1.55 m
Fnet = manet = (20.0 kg)(9.76 × 10−2 m/s2) = 1.95 N
vi = 0 m/s
vf = 0.550 m/s
Fmax = mmaxg = FT
9. mmax = 70.0 kg
m = 45.0 kg
Fmax = (70.0 kg)(9.81 m/s2) = 687 N
g = 9.81 m/s2
Fnet = manet = FT − mg = Fmax − mg
687 N
F ax
anet = m
− g =  − 9.81 m/s2 = 15.3 m/s2 − 9.81 m/s2 = 5.5 m/s2
m
45.0 kg
II
anet = 5.5 m/s2 upward
10. m = 3.18 × 105 kg
Fnet = Fapplied − Ffriction = (81.0 × 103 − 62.0 × 103 N)
Fapplied = 81.0 × 103 N
Fnet = 19.0 × 103 N
Ffriction = 62.0 × 103 N
19.0 × 103 N
F
 = 
anet = net
= 5.97 × 10−2 m/s2
3.18 × 105 kg
m
11. m = 3.00 × 103 kg
Fnet = manet = Fapplied(cos q) − Fopposing
3
Fapplied = 4.00 × 10 N
Copyright © by Holt, Rinehart and Winston. All rights reserved.
q = 20.0°
Fopposing = (0.120) mg
g = 9.81 m/s2
Fapplied(cos q) − (0.120) mg
anet = 
m
(4.00 × 103 N)(cos 20.0°) − (0.120)(3.00 × 103 kg)(9.81 m/s2)
anet = 
3.00 × 103 kg
3.76 × 103 N − 3.53 × 103 N
2.3 × 102 N
= 
anet = 
3
3.00 × 10 kg
3.00 × 103 kg
anet = 7.7 × 10−2 m/s2
12. mc = 1.600 × 103 kg
3
For the counterweight: The tension in the cable is FT.
mw = 1.200 × 10 kg
Fnet = FT − mwg = mwanet
vi = 0 m/s
For the car:
2
g = 9.81 m/s
Fnet = mcg − FT = mcanet
∆y = 25.0 m
Adding the two equations yields the following:
mcg − mwg = (mw + mc)anet
(mc − mw)g (1.600 × 103 kg − 1.200 × 103 kg)(9.81 m/s2)
 = 
anet = 
1.600 × 103 kg + 1.200 × 103 kg
mc + mw
(4.00 × 102 kg)(9.81 m/s2)
anet = 
= 1.40 m/s2
2.800 × 103 kg
Section Two — Problem Workbook Solutions
II Ch. 4–5
Givens
Solutions
vf = 2a
m/s
)2
ne
+vi2 = (2
)(
1.
40
2)(2
5.
0m
)+(0m
/s
t∆y
vf = 8.37 m/s
a. Fnet = Fapplied − mg(sin q) = 2080 N − (409 kg)(9.81 m/s2)(sin 30.0°)
13. m = 409 kg
Fnet = 2080 N − 2010 N = 70 N
d = 6.00 m
Fnet = 70 N at 30.0° above the horizontal
q = 30.0°
2
g = 9.81 m/s
Fapplied = 2080 N
vi = 0 m/s
70 N
F
 =  = 0.2 m/s2
b. anet = net
409 kg
m
anet = 0.2 m/s2 at 30.0° above the horizontal
1
1
c. d = vi∆t + anet ∆t 2 = (0 m/s)∆t + (0.2 m/s2)∆t2
2
2
∆t =
II
14. amax = 0.25 m/s2
Fmax = 57 N
0 m)
(
(20).(26.m0
/s) = 8 s
2
57 N
F ax
= 2 = 2.3 × 102 kg
a. m = m
amax 0.25 m/s
b. Fnet = Fmax − Fapp = 57 N − 24 N = 33 N
Fapp = 24 N
33 N
F
 = 
anet = net
= 0.14 m/s2
2.3 × 102 kg
m
15. m = 2.55 × 103 kg
3
a. Fx,net = ΣFx = max,net = FT(cos qT) + Fwind
FT = 7.56 × 10 N
Fx,net = (7.56 × 103 N)[cos(−72.3°)] − 920 N = 2.30 × 103 N − 920 N = 1.38 × 103 N
qT = −72.3°
Fy,net = ΣFy = may,net = FT(sin qT) + Fbuoyant + Fg = FT(sin qT) + Fbuoyant − mg
Fbuoyant = 3.10 × 104 N
Fy,net = (7.56 × 103 N)[sin(−72.3°)] = 3.10 × 104 N − (2.55 × 103 kg)(9.81 m/s2)
Fwind = −920 N
Fy,net = −7.20 × 103 N + 3.10 × 104 N − 2.50 × 104 = −1.2 × 103 N
g = 9.81 m/s2
2
Fnet = (F
)2 = (1
03
N
)2
+(−
03
N
)2
x,
(Fy,n
et
.3
8×1
1.
2×1
ne
t)+
Fnet = 1.
06
N2+
06
N2
90
×1
1.4
×1
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Fnet = 3.
06
N2 = 1.8 × 103 N
3×1
Fy, net
−1.2 × 103 N
 = tan−1 
q = tan−1 
Fx, net
1.38 × 103 N
q = −41°
Fnet = 1.8 × 103 N at 41° below the horizontal
1.8 × 103 N
F
 = 
b. anet = net
2.55 × 103 kg
m
∆y = −45.0 m
vi = 0 m/s
anet = 0.71 m/s2
c. Because vi = 0
1
∆y =  ay,net ∆t2
2
1
∆x =  ax,net ∆t2
2
ax,net

∆x = 
ay,net
−45.0 m
∆x =  = 52 m
tan(−41°)
II Ch. 4–6
Holt Physics Solution Manual
anet(cos q)
∆y = 
anet(sin q)
∆y
∆y = 
tan q
Additional Practice D
Givens
Solutions
1. m = 11.0 kg
Fk = mkFn = mkmg
mk = 0.39
Fk = (0.39) (11.0 kg)(9.81 m/s2) = 42.1 N
g = 9.81 m/s2
2. m = 2.20 × 105 kg
ms = 0.220
g = 9.81 m/s2
3. m = 25.0 kg
Fs,max = msFn = msmg
Fs,max = (0.220)(2.20 × 105 kg)(9.91 m/s2) = 4.75 × 105 N
Fs,max = msFm
Fapplied = 59.0 N
Fn = mg(cos q) + Fapplied
q = 38.0°
Fs,max = ms[mg(cos q) = Fapplied] = (0.599)[(25.0 kg)(9.81 m/s2)(cos 38.0° + 59.0 N]
ms = 0.599
g = 9.81 m/s2
Fs,max = (0.599)(193 N + 59 N) = (0.599)(252 N) = 151 N
Alternatively,
Fnet = mg(sin q) − Fs,max = 0
II
Fs,max = mg(sin q) = (25.0 kg)(9.81 m/s2)(sin 38.0°) = 151 N
4. q = 38.0°
Fnet = mg(sin q) − Fk = 0
g = 9.81 m/s2
Fk = mkFn = mkmg(cos q)
mkmg(cos q) = mg(sin q)
sin q
mk =  = tan q = tan 38.0°
cos q
Copyright © by Holt, Rinehart and Winston. All rights reserved.
mk = 0.781
5. q = 5.2°
Fnet = mg(sin q) − Fk = 0
2
g = 9.81 m/s
Fk = mkFn = mkmg(cos q)
mkmg(cos q) = mg(sin q)
sin q
mk =  = tan q = tan 5.2°
cos q
mk = 0.091
Section Two — Problem Workbook Solutions
II Ch. 4–7
Givens
Solutions
6. m = 281.5 kg
q = 30.0°
F net = 3mg(sin q) − ms(3mg)(cos q) − Fapplied = 0
Fapplied = mg
(3)(sin 30.0°) − 1.00
3(sin q) − 1.00
3mg(sin q ) − mg
ms =  =  = 
(3)(cos 30.0°)
3mg(cos q)
3(cos q )
1.50 − 1.00
0.50
ms =  = 
(3)(cos 30.0°) (3)(cos 30.0°)
ms = 0.19
7. m = 1.90 × 105 kg
ms = 0.460
Fnet = Fapplied − Fk = 0
Fk = mkFn = mkmg
2
g = 9.81 m/s
Fapplied = mkmg = (0.460)(1.90 × 105 kg)(9.81 m/s2)
Fapplied = 8.57 × 105 N
II
8. Fapplied = 6.0 × 103 N
mk = 0.77
Fnet = Fapplied − Fk = 0
Fk = mkFn
2
g = 9.81 m/s
Fapplied 6.0 × 103 N
 =  = 7.8 × 103 N
Fn = 
mk
0.77
Fn = mg
F
7.8 × 103 N
m = n = 
= 8.0 × 102 kg
g
9.81 m/s2
9. Fapplied = 1.13 × 108 N
ms = 0.741
Fnet = Fapplied − Fs,max = 0
Fapplied
1.13 × 108 N
 = 2 = 1.55 × 102 kg
m= 
msg
(0.741)(9.81 m/s
10. m = 3.00 × 103 kg
q = 31.0°
Fnet = mg(sin q) − Fk = 0
Fk = mkFn = mkmg(cos q)
2
g = 9.81 m/s
mkmg(cos q) = mg(sin q)
sin q
mk =  = tan q = tan 31.0°
cos q
mk = 0.601
Fk = mkmg(cos q ) = (0.601)(3.00 × 103 kg)(9.81 m/s2)(cos 31.0°)
Fk = 1.52 × 104 N
Alternatively,
Fk = mg(sin q) = (3.00 × 103 kg)(9.81 m/s2)(sin 31.0°) = 1.52 × 104 N
II Ch. 4–8
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Fs,max = msFn = msmg
Additional Practice E
Givens
Solutions
1. Fapplied = 130 N
2
Fnet = manet = Fapplied − Fk
anet = 1.00 m/s
Fk = mkFn = mkmg
mk = 0.158
manet + mkmg = Fapplied
2
g = 9.81 m/s
m(anet + mkg) = Fapplied
130 N
Fapplied
 = 
m= 
2
1.00 m/s + (0.158)(9.81 m/s2)
anet + mkg
130 N
130 N
m = 
= 51 kg
2
2 = 
1.00 m/s + 1.55 m/s
2.55 m/s2
2. Fnet = −2.00 × 104 N
Fnet = manet = mg(sin q) − Fk
Fk = mkFn = mkmg(cos q)
q = 10.0°
m[g(sin q) − mkg(cos q)] = Fnet
mk = 0.797
−2.00 × 104 N
F et
m = n
= 
2
(9.81 m/s )[(sin 10.0°) − (0.797)(cos 10.0°)]
g[sin q − mk(cos q)]
2
g = 9.81 m/s
II
−2.00 × 104 N
−2.00 × 104 N
m = 
= 2
2
(9.81 m/s )(0.174 − 0.785) (9.81 m/s )(−0.611)
m = 3.34 × 103 kg
Fn = mg(cos q) = (3.34 × 103 kg)(9.81 m/s2)(cos 10.0°) = 3.23 × 104 N
Copyright © by Holt, Rinehart and Winston. All rights reserved.
3. Fnet = 6.99 × 103 N
Fnet = manet = mg(sin q) − Fk
q = 45.0°
Fk = mkFn = mkmg(cos q)
mk = 0.597
m[g(sin q) − mkg(cos q)] = Fnet
6.99 × 103 N
F et
m = n
= 
(9.81 m/s2)[(sin 45.0°) − (0.597)(cos 45.0°)]
g[sin q − mk(cos q)]
6.99 × 103 N
6.99 × 103 N


m = 
=
(9.81 m/s2)(0.707 − 0.422) (9.81 m/s2)(0.285)
m = 2.50 × 103 kg
Fn = mg(cos q) = (2.50 × 103 kg)(9.81 m/s2)(cos 45.0°) = 1.73 × 104 N
4. m = 9.50 kg
Fnet = manet = Fapplied − Fk − mg(sin q)
q = 30.0 °
Fk = mkFn = mkmg(cos q)
Fapplied = 80.0 N
2
anet = 1.64 m/s
2
g = 9.81 m/s
mkmg(cos q) = Fapplied − manet − mg(sin q)
Fapplied − m[anet + g (sin q)]
mk = 
mg(cos q)
80.0 N − (9.50 kg)[1.64 m/s2 + (9.81 m/s2)(sin 30.0°)]
mk = 
(9.50 kg)(9.81 m/s2)(cos 30.0°)
Section Two — Problem Workbook Solutions
II Ch. 4–9
Givens
Solutions
80.0 N − (9.50 kg)[1.64 m/s2 + 4.90 m/s2)
80.0 N − (9.50 kg)(6.54 m/s2)
mk = 
= 
2
(9.50 kg)(9.81 m/s )(cos 30.0°)
(9.50 kg)(9.81 m/s2)(cos 30.0°)
17.9 N
80.0 N − 62.1 N
= 
mk = 
(9.50 kg)(9.81 m/s2)(cos 30.0°)
(9.50 kg)(9.81 m/s2)(cos 30.0°)
mk = 0.222
5. m = 1.89 × 105 kg
5
Fnet = manet = Fapplied − Fk
Fapplied = 7.6 × 10 N
Fk = Fapplied − manet = 7.6 × 105 N − (1.89 × 105)(0.11 m/s2) = 7.6 × 105 N − 2.1 × 104 N
anet = 0.11 m/s2
Fk = 7.4 × 105 N
6. q = 38.0°
Fnet = manet = mg(sin q) − Fk
mk = 0.100
Fk = mkFn = mkmg(cos q)
2
g = 9.81 m/s
manet = mg[sin q − mk(cos q)]
anet = g[sin q − mk(cos q)] = (9.81 m/s2)[(sin 38.0°) − (0.100)(cos 38.0°)]
anet = (9.81 m/s2)(0.616 − 7.88 × 10−2) = (9.81 m/s2)(0.537)
II
anet = 5.27 m/s2
Acceleration is independent of the rider’s and sled’s masses. (Masses cancel.)
7. ∆t = 6.60 s
Fnet = manet = mg(sin q) − Fk
q = 34.0°
Fk = mkFn = mkmg(cos q)
mk = 0.198
manet = mg[sin q − mk(cos q)]
2
g = 9.81 m/s
anet = g[sin q − mk(cos q)] = (9.81 m/s2)[(sin 34.0°) − (0.198)(cos 34.0°)]
vi = 0 m/s
anet = (9.81 m/s2)(0.559 − 0.164) = (9.81 m/s2)(0.395)
anet = 3.87 m/s2
vf = 25.5 m/s2 = 92.0 km/h
II Ch. 4–10
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
vf = vi + anet∆t = 0 m/s + (3.87 m/s2)(6.60 s)
Study Guide Answers
Forces and the
Laws of Motion
Changes in Motion, p. 19
1. The diagram should show two forces: 1) Fg (or mg)
pointing down; 2) an equal and opposite force of the
floor on the box pointing up.
2. The diagram should show four forces: 1) Fg (or mg)
pointing down; 2) an equal and opposite force of the
floor on the box pointing up; 3) F pointing to the right,
parallel to the ground; 4) Fresistance pointing to the left,
parallel to the ground.
3. The diagram should show four forces: 1) Fg (or mg)
pointing down; 2) F pointing to the right at a 50° angle
to the horizontal; 3) a force equal to Fg minus the vertical
component of the force F being applied at a 50° angle;
and 4) Fresistance to the left, parallel to the ground.
Newton’s First Law, p. 20
1. Fnet = F1 + F2 + F3 = 0
2. String 1: 0, −mg
4. F1 = 20.6 N
String 3: F3 cos q2 , F3 sin q2
3. Fx net = −F2 cos q1 + F3 cos q2 = 0
F2 = 10.3 N
Fy net =
−F2 sin q1 + F3 sin q2 + F1 = 0
F3 = 17.8 N
String 2: −F2 cos q1, F2 sin q1
Newton’s Second and Third Laws, p. 21
1. Fs on b and Fb on s ; Fg on s and Fs on g; Ffr,1 and −Ffr,1;
Ffr,2 and −Ffr,2.
2. Fs on b, Fb on s , −Ffr,1
5. Fy,box = Fs on b − mg = 0
6. Fx,sled = Ma = F cos q − Ffr,1 − Ffr,2
3. Fg on s , Fs on g ; Fb on s , Ffr,1, F, Ffr,2
Copyright © by Holt, Rinehart and Winston. All rights reserved.
4. Fx,box = ma = −Ffr,1
7. Fy,sled = Fg on s + F sin q − Fb on s − Mg = 0
III
Everyday Forces, p. 22
1. 44 N
3. a. 21 N, up the ramp
2. 31 N
4. a. 18 N, down the ramp
b. yes
b. yes
Mixed Review, pp. 23–24
1. a. at rest, moves to the left, hits back wall
b. m2 a
b. moves to the right (with velocity v), at rest, neither
c. F − m2 a = m1a
c. moves to the right, moves to the right, hits front wall
m1
d. 
F
m1 + m2
2. a. mg, down
b. mg, up
c. no
d. yes
F
3. a. a = 
m1 + m2
F − Fk
4. a. a = 
m1 + m2
b. m2 a − Fk
c. F − m2 a − Fk = m1a − Fk
m1
d. 
(F − Fk)
m1 + m2
Section Three—Study Guide Answers
III–5