6.1
WORK, ENERGY, AND POWER
E
XERCISES
Section 6.1 Work
12.
If the force is constant and parallel to the displacement,
W
F
r
F
r
=
=
=
=
⋅ ∆
∆
(
)(
)
.
75
12
900
N
m
J
13.
I
NTERPRET
The relevant physical quantity here is the work done by a person sliding a box.
D
EVELOP
If you push parallel to a level floor, the applied force equals the frictional force (since the acceleration
is zero),
F
f
a
=
k
.
Since the normal force equals the weight of the box:
n
mg
=
,
the applied force is
F
n
mg
a
=
=
µ
µ
k
k
Equation 6.1,
W
F
x
=
∆
,
can then be used to find the work done.
E
VALUATE
The applied force is constant and parallel to the displacement. So using Equation 6.1, we get
W
F
r
f
r
n r
mg r
a
a
=
=
=
=
=
∆
∆
∆
∆
k
k
k
kg
µ
µ
( .
)(
)( .
0 21 50
9 8 m s
m
J
/
)( .
)
2
4 8
494
=
A
SSESS
The units are correct,
1
1
1
2
J
N m
kg m s
m.
=
⋅
=
⋅
⋅
/
If the floor were frictionless
(
),
µ
k
=
0
then the work
done would be
W
a
=
0,
as expected.
14.
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
mgj
y j
xi
mg y
⋅ ∆
∆
∆
∆
=
⋅
+
=
=
(
ˆ
) (
ˆ
ˆ
)
(
)( .
kg
650
9 8
23
147
2
m s
m
kJ
/
)(
)
.
=
15.
I
NTERPRET
This problem is about the work done by gravity on the water falling from a certain height.
D
EVELOP
Since the density of water is 1000 kg/m
3
, the mass of a cubic meter of water is 1000 kg, and the force
of gravity at the Earth’s surface on a cubic meter of water is
F
mg
g
=
=
=
(
)( .
)
1000
9 8
9800
kg
m/s
N
2
vertically downward. We can then use Equation 6.1,
W
F
x
=
∆
,
to find the work done.
E
VALUATE
The displacement of the water is parallel to this, so the work done by gravity on the water is
W
F
y
g
g
=
=
=
×
∆
(
)(
)
.
9800
980
9 6
10
6
N
m
J
A
SSESS
The units are correct,
1
1
J
N m.
=
⋅
The greater the distance the water falls, the larger the amount of work
done by gravity.
16.
The average force exerted by the meteorite parallel to its penetration into the ground is
F
W
x
av
MJ/
=
=
/
∆
140
0 75
187
.
,
m
MN
=
or about 21,000 tons.
17.
I
NTERPRET
This problem is about the work done by the elevator cable on the elevator as it accelerates upward.
D
EVELOP
To give the elevator a constant upward acceleration
a
y
,
the tension in the cable must satisfy
T
mg
ma
T
m g
a
y
y
−
=
→
=
+
(
)
We can then use Equation 6.1,
W
F
x
=
∆
,
to find the work done.
E
VALUATE
Acting over a parallel displacement
∆
y
h
=
upward, the amount of work done by the cable on the
elevator is
W
T
y
m g
a
y
m g
g h
mgh
T
y
=
=
+
=
+
=
∆
∆
(
)
(
.
)
.
0 1
1 1
6
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6.2
Chapter 6
A
SSESS
The units are correct,
1
1
J
N m.
=
⋅
The greater the upward acceleration
a
y
,
the larger the amount of work
must be done by the cable. Of course, if the elevator undergoes free fall with
a
g
y
= −
,
then the tension in the cable
is zero, and no work is done on the elevator.
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 Fall '10
 Pheong
 Force, Work, wA

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