Chapter 6:
Work and Energy
.
Selected Conceptual Questions, pp.187.
Problems pp. 188: # 11, 26, 40, 44, 45,
49, 50, 57, 69, 86
Chapter 7: Impulse and Momentum
.
Selected Conceptual Questions, pp.216.
Problems pp. 217: # 11, 18, 22, 23, 34,
42, 43, 45, 52
Chapter 6:
Work and Energy
11.
REASONING
The work done by a constant force of magnitude
F
that makes an angle
θ
with a
displacement of magnitude
s
is given by
(
29
cos
W
F
s
θ
=
(Equation 6.1). The magnitudes and
directions of the pulling forces exerted by the husband and wife differ, but the displacement of
the wagon is the same in both cases. Therefore, we can express the work
W
H
done by the
husband’s pulling force as
(
29
H
H
H
cos
W
F
s
=
. Similarly, the work done by the wife can be
written as
(
29
W
W
W
cos
W
F
s
=
. We know the directional angles of both forces, the magnitude
F
H
of the husband’s pulling force, and that both forces do the same amount of work (
W
W
=
W
H
).
SOLUTION
Setting the two expressions for work equal to one another, and solving for the
magnitude
F
W
of the wife’s pulling force, we find
(
29
W
W
cos
F
s
(
29
W
H
H
cos
W
F
s
=
1 442 4 43
(
29
(
29
H
H
H
W
W
67 N
cos58
cos
or
45 N
cos
cos38
W
F
F
=
=
=
o
o
1 442 4 43
26.
REASONING
a.
The magnitude
s
of the displacement that occurs while the snowmobile coasts to a halt is
the distance that we seek.
The work
W
done by the net external force
F
acting on the snowmobile
during the coasting phase is given by
W
= (
F
cos
θ
)
s
(Equation 6.1), and we can use this expression
to obtain
s
.
To do so, we will need to evaluate
W
.
This we will do with the aid of the workenergy
theorem as given by
2
2
1
1
2
2
f
0
W
mv
mv
=

(Equation 6.3), since we know the mass
m
and the final and
initial velocities
v
f
and
v
0
, respectively.
The net external force
F
during the coasting phase is just
1