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We note that
x
c
is also the point where the applied and spring forces “balance.”
(e) At
x
c
we find
K = K
max
= 0.090 J.
33. (a) This is a situation where Eq. 728 applies, so we have
Fx
=
1
2
kx
2
¡
(3.0 N)
x
=
1
2
(50 N/m)
x
2
which (other than the trivial root) gives
x
=
(3.0/25) m = 0.12 m.
(b) The work done by the applied force is
W
a
=
Fx
= (3.0 N)(0.12 m) = 0.36 J.
(c) Eq. 728 immediately gives
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Unformatted text preview: W s = – W a = –0.36 J. (d) With K f = K considered variable and K i = 0, Eq. 727 gives K = Fx – 1 2 kx 2 . We take the derivative of K with respect to x and set the resulting expression equal to zero, in order to find the position x c which corresponds to a maximum value of K : x c = F k = (3.0/50) m = 0.060 m....
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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, Force, Work

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