ch07-p033 - W s = – W a = –0.36 J. (d) With K f = K...

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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. 7-28 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. 7-28 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. 7-27 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.

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