Chapter 7 – Student Solutions Manual
3. (a) From Table 2-1, we have
v
v
. Thus,
a
2
0
2
2
=
+
Δ
x
(
)
(
)
(
)
2
2
7
15
2
0
2
2.4
10
m/s
2 3.6
10
m/s
0.035 m
2.9
10 m/s.
v
v
a x
=
+
Δ
=
×
+
×
=
×
7
(b) The initial kinetic energy is
(
)(
)
2
2
27
7
0
1
1
1.67
10
kg
2.4
10 m/s
4.8
10
J.
2
2
i
K
mv
−
−
=
=
×
×
=
×
13
The final kinetic energy is
(
)(
)
2
2
27
7
1
1
1.67
10
kg
2.9
10 m/s
6.9
10
J.
2
2
f
K
mv
−
−
=
=
×
×
=
×
13
The change in kinetic energy is
Δ
K
= 6.9
×
10
–13
J – 4.8
×
10
–13
J = 2.1
×
10
–13
J.
17. (a) We use
to denote the upward force exerted by the cable on the astronaut. The
force of the cable is upward and the force of gravity is
mg
downward. Furthermore, the
acceleration of the astronaut is
g
/10 upward. According to Newton’s second law,
F
–
mg
=
mg
/10, so
F
= 11
mg
/10. Since the force
G
F
G
F
and the displacement
G
d
are in the same
direction, the work done by
G
is
F
2
4
11
11 (72 kg)(9.8 m/s
)(15 m)
1.164
10
J
10
10
F
mgd
W
Fd
=
=
=
=
×
which (with respect to significant figures) should be quoted as 1.2
×
10
4
J.
(b) The force of gravity has magnitude
mg
and is opposite in direction to the
displacement. Thus, using Eq. 7-7, the work done by gravity is
2
4
(72 kg)(9.8 m/s
)(15 m)
1.058
10
J
g
W
mgd
= −
= −
= −
×
which should be quoted as – 1.1
×
10
4
J.
(c) The total work done is
W
. Since the
astronaut started from rest, the work-kinetic energy theorem tells us that this (which we
round to 1
) is her final kinetic energy.
=
×
−
×
=
×
1164
.
10 J
1.058
10 J
1.06
10 J
4
4
3
1
.
×
10 J
3
(d) Since
K
mv
=
1
2
2
,
her final speed is

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