Problem 4D

Copyright © by Holt, Rinehart and Winston. All rights reserved.
Givens Solutions
6. Fapplied = 2.50 × 10 2 N
m = 65.0 kg
q = 18.0°
anet = 0.44 m/s 2
7. m = 65.0 kg
g = 9.81 m/s 2
Fk = 24 N
q = 18.0°
8. F applied = 3.00 × 10 2 N
q = 20.0°
mk = 0.250
9. Fapplied = 590 N
Fdownhill = 950 N
ms = 0.095
q = 14.0°
10. a net = 1.20 m/s 2
Fapplied = 1.50 × 10 3 N
q =−10.0°
ms = 0.650
g = 9.81 m/s 2
F net = m a net = F applied − mg(sin q) − F k
F k = F applied − mg(sin q) − ma net
Fk = 2.50 × 10 2 N − (65.0 kg)(9.81 m/s 2 )(sin 18.0°) − (65.0 kg)(0.44 m/s 2 )
Fk = 2.50 × 10 2 N − 197 N − 29 N = 24 N = 24 N downhill
Fnet = m anet = mg(sin q) − Fk anet = g(sin q) − ⎯ Fk
⎯ = (9.81 m/s
m
2 24
N
)(sin 18.0°) − ⎯⎯ = 3.03 m/s
65.0
kg
2 − 0.37 m/s 2 = 2.66 m/s 2
a net = 2.66 m/s 2 downhill
Fx,net = Fapplied(cos q) − Fk = 0
Fy,net = Fn − mg + Fapplied(sin q) = 0
F k = m kF n
Fn = ⎯ Fapplied(
cos q) (3.00 × 10
⎯ = = 1130 N
mk
2 N)[cos(−20.0°)]
⎯⎯⎯
0.25°
m = ⎯ Fn + Fapplied(sin q)
⎯ =
g
m = ⎯ 1130
N − 103N
1030N
⎯ 2 = ⎯
9.81
m/s
9. 81
m/s
2
1130 N + (3.00 × 10
⎯ = 105 kg
2 N)[sin(−20.0°)]
⎯⎯⎯⎯
9.81 m/s 2
Fnet = Fapplied + Fs,max − Fdownhill = 0
Fs,max = ms Fn = ms mg(cos q)
ms Fn = Fdownhill − Fapplied Fn = ⎯ Fdownhill − Fapplied 950 N − 590
N
⎯ = ⎯⎯ = ⎯
ms
0.095
360
N
⎯ = 3800 N
0.095
Fn =
Fn
m = ⎯⎯ = = 4.0 × 10
g(cos
q)
2 3800 N
⎯⎯⎯
kg
(9.81 m/s 2 3800 N perpendicular to and up from the ground
)(cos 14.0°)
Fx,net = Fapplied(cos q) − Fs,max = 0
Fs,max = msFn Fn = ⎯ Fapplied(
cos q)
⎯ = = 2.27 × 10
ms
3 N
Fn = 2.27 × 10 3 (1.50 × 10
N, upward
3 N)[cos(−10.0°)]
⎯⎯⎯
0.650
F y,net = m a net = F n − mg + F applied(sin q)
m(anet + g) = Fn + Fapplied(sin q)
m = ⎯ Fn + Fapplied(
sin q) 2.27 × 10
⎯ =
anet
+ g
3 N + (1.50 × 10 3 N)[sin(−10.0°)]
⎯⎯⎯⎯⎯
1.20 m/s 2 + 9.81 m/s 2
m = = ⎯ 2
2.27 × 10 3
.01
× 10
N
⎯ 2 = 183 kg
11.01
m/
s
3 N − 2.60 × 10 2 N
⎯⎯⎯
11.01 m/s 2
Section Five—Solution Manual V Ch. 4–7
V