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Kk
d
k
d
d
f
d
k
=−
+
−
1
2
2
0
bg
.
In this first approach, we could work through the
dK
dd
=
0 condition (or with the special
capabilities of a graphing calculator) to obtain the answer
K
k
kd
f
k
max
1
2
0
2
. In the
second (and perhaps easier) approach, we note that
K
is maximum where
v
is
maximum — which is where
a
=
¡
0
equilibrium of forces. Thus, the second approach
simply solves for the equilibrium position
Ff
k
x
k
spring
=
¡
=
80.
Thus, with
k
= 4000 N/m we obtain
x
= 0.02 m. But
x
=
d
0
–
d
so this corresponds to
d
=
0.08 m. Then the methods of part (a) lead to the answer
K
max
= 12.8 J
≈
13 J.
107. (a) The effect of a (sliding) friction is described in terms of energy dissipated as
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Energy, Friction

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