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6.64:
a) From Eq. (6.7),
∫
∫
-
=
-
-
=
-
=
=
2
1
2
1
2
1
.
1
1
1
1
2
2
x
x
x
x
x
x
x
x
x
k
x
k
x
dx
k
dx
F
W
The force is given to be attractive, so
0
<
x
F
, and
k
must be positive. If
1
2
1
1
1
2
,
x
x
x
x
<
,
and
0
<
W
.
b) Taking “slowly” to be constant speed, the net force on the object is zero,
so the force applied by the hand is opposite
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Unformatted text preview: x F , and the work done is negative of that found in part (a), or ( 29 2 1 1 1 x x k-, which is positive if 1 2 x x . c) The answers have the same magnitude but opposite signs; this is to be expected, in that the net work done is zero....
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