6.1
WORK, ENERGY, AND POWER
EXERCISES
Section 6.1 Work
12.
If the force is constant and parallel to the displacement,
WFrF
r
==
=
=
v
v
⋅∆
∆
()
.
75
12
900
Nm
J
13.
INTERPRET
The relevant physical quantity here is the work done by a person sliding a box.
DEVELOP
If you push parallel to a level floor, the applied force equals the frictional force (since the acceleration
is zero),
Ff
a
=
k
. Since the normal force equals the weight of the box:
nm
g
=
, the applied force is
Fn
m
g
a
µµ
kk
Equation 6.1,
WF
x
=∆
, can then be used to find the work done.
EVALUATE
The applied force is constant and parallel to the displacement. So using Equation 6.1, we get
W
F
r
f
rn
rm
g
r
aa
===
=
=
∆∆
∆
∆
k
kg
( .
)(
)( .
021 50
98ms
m
J
/)
(
. )
2
4 8
494
=
ASSESS
The units are correct,
11
1
2
JN
mk
g
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
v
v
Fr m
g
j y
j x
i m
g
y
∆
∆
∆
+
(
ˆ
ˆˆ
)
(
.
kg
650
9 8
23
147
2
ms
m
kJ
(
)
.
=
15.
INTERPRET
This problem is about the work done by gravity on the water falling from a certain height.
DEVELOP
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
Fm
g
g
=
(
.
)
1000
9 8
9800
kg
m/s
N
2
vertically downward. We can then use Equation 6.1,
x
, to find the work done.
EVALUATE
The displacement of the water is parallel to this, so the work done by gravity on the water is
y
gg
=
×
∆
(
)
.
9800
980
9 6 10
6
J
ASSESS
The units are correct,
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
FW
x
av
MJ/
/
∆
140
0 75
187
.,
mM
N
=
or about 21,000 tons.
17.
INTERPRET
This problem is about the work done by the elevator cable on the elevator as it accelerates upward.
DEVELOP
To give the elevator a constant upward acceleration
a
y
, the tension in the cable must satisfy
Tm
gm
a
ga
yy
−=
→
=
+
We can then use Equation 6.1,
x
, to find the work done.
EVALUATE
Acting over a parallel displacement
∆
yh
=
upward, the amount of work done by the cable on the
elevator is
W Tym
ga ym
g
g
h
m
g
h
Ty
==+
=+
=
(
.
)
.
01
6