ch08-p135 - 135. Let the amount of stretch of the spring be...

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135. Let the amount of stretch of the spring be x . For the object to be in equilibrium kx mg x mg k −= ¡ = 0. Thus the gain in elastic potential energy for the spring is Δ Uk xk mg k mg k e == F H G I K J = 1 2 1 22 2 2 while the loss in the gravitational potential energy of the system is −== F H G I K J = Δ Um g x m g mg k k g which we see (by comparing with the previous expression) is equal to 2 Δ U e . The reason why
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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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