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tiL12214_appd_633-642.indd Page 635 09/10/10 8:35 AM user-f463
tiL12214_appd_633-642.indd Page 635 09/10/10 8:35 AM user-f463 Volume/207/MHDQ243/tiL12214_disk1of1/0073512214/tiL12214_pagefile

1 _
Example 3.4, p. 66 KE = mv 2

wh _
w = 150 lb P = 2

t
m _
1 _
h = 15 ft = (1.0 kg)(4.4 )
2

__
t = 10.0 s = (150 lb)(15 ft) 2 s

2)
(
2
P = ? 10.0 s = (1.0 kg) 19.4
1 _
m _

_ _ 2
(150)(15) ft . lb

= s

10.0 s
kg ⋅ m
1 _ _

_ = (1.0)(19.4) × m

ft . lb

= 225 2 s 2
s
_ = 9.7 N ⋅ m
ft . lb
225

_
s
= = 9.7 J

_
ft . lb

_
s
550
_
ft . lb

_ CHAPTER 4
s

hp
s
_ _ Example 4.2, p. 91
ft . lb
9 _

= 0.41 × × hp T C = 20° T F = T C + 32°

s ft . lb 5
T F = ?
= 0.41 hp 9 _

= 20° + 32°
5
Example 3.6, p. 68 _
180 º

m = 5.00 kg W = Fd = 5 + 32°
g = 9.8 m/s 2 W = mgh = 36° + 32°
h = 5.00 m m _ = 68°
2)
(
W = ? = (5.00 kg) 9.8 (5.00 m)
s
kg . m
_ Example 4.6, p. 95

= (5.00)(9.8)(5.00) × m m = 2 kg
s 2
Q = 1.2 kcal
= 245 N . m
ΔT = 20°C
= 250 J C = ?
Q
_
Q = mc ΔT ∴ c =
Example 3.8, p. 69 m ΔT

m = 100.0 kg W = KE __
1.2 kcal

=
v = 6.0 m/s 1 _ 2 (2 kg)(20.0C°)

KE = mv

2
_
1.2

W = ? = _ kcal

1 _ m _ 2 (2)(20.0) kgC°

= (100.0 kg)(6.0 )

2 s _
kcal

= 0.03
(
2)
1 _ m _ 2 kgC°

= (100.0 kg) 36

2 s
kg ⋅ m
1 _ _

× m
= (100.0)(36)

2 s 2 CHAPTER 5
= 1,800 N ⋅ m Example 5.3, p. 121
v _

= 1,800 J f = 2,500 Hz v = λf ∴ λ =
f
v = 330 m/s
m _
Example 3.10, p. 73 λ = ? 330
s

= _
1 _

2
m = 1.0 kg V f = √ 2,500
gh
g = 9.8 m/s 2 s
√ (
2 )
m _

= 2 9.8 (1.0 m)
330 m _ _
h = 1.0 m s = _ s
×

KE = ? m _ 2,500 s 1

√ s 2 = 0.13 m or 13 cm

= 2(9.8)(1.0) × m

√ 2
m _
= 19.6

2
s
m _
= 4.4

s
D-3 APPENDIX D Solutions for Follow-Up Example Exercises 635