University Physics with Modern Physics with Mastering Physics (11th Edition)

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7.85: a) For the given proposed potential F kx x U dx dU + - = - ), ( , so this is a possible potential function. For this potential, k F U 2 ) 0 ( 2 - = , not zero. Setting the zero of potential is equivalent to adding a constant to the potential; any additive constant will not change the derivative, and will correspond to the same force. b) At equilibrium, the force is zero; solving 0 = + - F kx for x gives k F x = 0 . k F x U 2 0 ) ( - = , and this is a minimum of U , and hence a stable point. c) d) No; 0 tot = F at only one point, and this is a stable point. e) The extreme values of
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Unformatted text preview: E x U = ± ) ( , where ± x are the extreme points of the motion. Rather than solve a quadratic, note that k F k F x k 2 2 2 1 ) (--, so E x U = ± ) ( becomes , 2 2 1 2 2 k F k F x k F k F k F x k ± =-=- -± ± . 3 k F x k F x-= =-+ f) The maximum kinetic energy occurs when ) ( x U is a minimum, the point k F x = found in part (b). At this point k F k F k F U E K 2 2 2 2 ) ( ) ( =--=-= , so mk F v 2 = ....
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