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2 _
m _
1 _
1
2

Scientific notation is used here to simplify a large number and (a) KE = mv = (1,000.0 kg) ( 25.0 )
to easily show three signifi cant fi gures. The answer could 2 2 s

2
(
2)
1 _
m _

likewise be expressed as 113 kJ. = (1,000.0 kg) 625
2 s
3.9. (a) W = Fd kg·m 2
1 _

= (1,000.0)(625) _

= (10 lb)(5 ft) 2 s 2
= (10)(5) ft × lb = 312.5 kJ = 313 kJ
= 50 ft·lb (b) W = Fd = KE = 313 kJ

(b) The distance of the bookcase from some horizontal (c) KE = W = Fd = 313 kJ
reference level did not change, so the gravitational
1 _

potential energy does not change. 3.13. KE = mv 2
2

3.10. The force (F) needed to lift the book is equal to the weight (w) of 1 _ m _ 2

= (60.0 kg)( 2.0 )

the book, or F = w. Since w = mg, then F = mg. Work is defi ned 2 s
(
as the product of a force moved through a distance, or W = Fd. 1 _ m _ 2
2)

= (60.0 kg) 4.0

The work done in lifting the book is therefore W = mgd, and 2 s

( )
(a) W = mgd m _ 2

= 30.0 × 4.0 kg ×
2
= (2.0 kg)(9.8 m/s )(2.00 m) s 2
= 120 J
kg·m
_

= (2.0)(9.8)(2.00) × m 1 _

s 2 KE = mv 2

kg·m 2 2
_

= 39.2 2

1 _
m _
s 2 = (60.0 kg)( 4.0 )

2 s
= 39.2 J = 39 J
2)
1 _ ( m _ 2

(b) PE = mgh = 39 J = (60.0 kg) 16
2
s
(c) PE lost = KE gained = mgh = 39 J m _ 2
= 30.0 × 16 kg ×
(or) s 2

v = √ 2gh = √ (2)(9.8 m/s 2 )(2.00 m) = 480 J

2
2
= √ 39.2 m /s Thus, doubling the speed results in a fourfold increase in

= 6.26 m/s kinetic energy.
1 _ 2 1 _ 2 1 _ 2
( )

KE = mv = (2.0 kg)(6.26 m/s) 3.14. KE = mv
2 2 2
1 _
1 _
( )
2
2

= (2.0 kg)(39 m /s ) = (70.0 kg)(6.00 m/s) 2
2
2
2
2
kg·m
_ 2 = (35.0 kg)(36.0 m /s )

= (1.0)(39)
s 2 m _ 2
= 35.0 × 36.0 kg ×

= 39 J s 2
= 1,260 J
3.11. Note that the gram unit must be converted to kg to be consistent
1 _

with the definition of a newton-meter, or joule unit of energy: KE = mv 2

1 _ 2 1 _ 2 2
( )

KE = mv = (0.15 kg)(30.0 m/s)
1 _
2 2 = (140.0 kg)(6.00 m/s) 2

1 _
( ) 2 2 2
= (0.15 kg)(900 m /s ) = (70.0 kg)(36.0 m /s )
2
2
2
2
m _
kg·m
1 _
( ) _ 2 = 70.0 × 36.0 kg ×

= (0.15)(900) s 2 s 2
2
= 67.5 J = 68 J = 2,520 J

3.12. The km/h unit must first be converted to m/s before fi nding the Thus, doubling the mass results in a doubling of the kinetic

kinetic energy. Note also that the work done to put an object in energy.
motion is equal to the energy of motion, or the kinetic energy that

3.15. (a) The force needed is equal to the weight of the student. Th e
it has as a result of the work. The work needed to bring the object

English unit of a pound is a force unit, so
to a stop is also equal to the kinetic energy of the moving object.
W = Fd
Unit conversion: = (170.0 lb)(25.0 ft)
m _ m _
(

)
_ _ = 4,250 ft·lb
s _
s _
(
km
km

= 25.0 m/s
1 = 0.2778 ∴ 90.0 0.2778 )

h _ h _
km
km

h h
652 APPENDIX E Solutions for Group A Parallel Exercises E-10