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Chapter 7 Problem Solutions
Problem
3. A crane lifts a 650kg beam vertically upward 23 m, then swings it eastward 18 m. How much work does the crane do?
Neglect friction, and assume the beam moves with constant speed.
Solution
Lifting the beam at constant speed, the crane exerts a constant force vertically upward and equal in magnitude to the weight
of the beam. During the horizontal swing, the force is the same, but is perpendicular to the displacement. The work done is
F
! "
r
=
(
mg
ˆ
j
)
!
(
"
y
ˆ
j
+
"
x
î
)
=
mg
"
y
=
(
650 kg
)(
9
.
8 m
/
s
2
)
(
23 m
)
=
147 kJ
.
Problem
4. You lift a 45kg barbell from the ground to a height of 2.5 m. (a) How much work do you do on the barbell? (b) You
hold the barbell aloft for 2.0 min. How much work do you do on the barbell during this time? (c) You lower the barbell
to the ground. Now how much work do you do on it?
Solution
(a) If we assume the barbell is lifted at a constant velocity by a vertical applied force equal to the weight and positive upward
parallel to the displacement, then
W
app
=
mg
!
y
=
(
45 kg
)(
9
.
/
s
2
2
.
5 m
)
=
1
.
10 kJ
.
(b) Just holding the weight
stationary, you must still exert an applied force of
mg
to balance gravity, but the displacement through which the force acts is
zero. Hence the work done on the barbell is also zero. (Actually, individual muscle fibers are continually contracting even
though the overall muscles are stationary, so internal work is being done in your muscles and you feel tired just holding a
weight.) (c) When the weight is lowered at constant velocity, the upward applied force,
mg
, is opposite to the displacement
downward,
!
y
=
"
2.5 m;
hence the work done on the barbell is negative,
W
=
!
1
.
10 kJ
.
Problem
6. A meteorite plunges to Earth, embedding itself 75 cm in the ground. If it does 140 MJ of work in the process, what
average force does the meteorite exert on the ground?
Solution
The average force exerted by the meteorite parallel to its penetration into the ground is
F
av
=
W
=
!
x
=
140 MJ
=
0
.
75 m
=
187 MN,
or about 21,000 tons.
Problem
10. An elevator of mass
m
rises a distance
h
up a vertical shaft with upward acceleration equal to onetenth
g
. How much
work does the elevator cable do on the elevator?
Solution
To give the elevator a constant upward acceleration
a
y
=
0.1 g,
the tension in the cable must satisfy
T
!
mg
=
ma
y
,
or
T
=
m
(
g
+
a
y
)
=
1
.
1
mg
.
Acting over a parallel displacement
!
y
=
h
upward, the tension does work
W
T
=
T
!
y
=
1
.
1
mgh
on the elevator.
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View Full DocumentProblem
14. Given the following vectors:
A
has magnitude 10 and points 30º above the
x
axis
B
has magnitude 4.0 and points 10º to the left of the
y
axis
C
=
5
.
6
î
!
3
.
1
ˆ
j
D
=
1
.
9
î
+
7
.
2
ˆ
j
,
compute the scalar products (a)
A
!
B
; (b)
C
!
D
; (c)
B
!
C
.
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 Winter '07
 Hicks
 Physics, Friction, Work

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