236_Physics ProblemsTechnical Physics

236_Physics ProblemsTechnical Physics - 238 P8.54 Potential...

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238 Potential Energy P8.54 The gain in internal energy due to friction represents a loss in mechanical energy that must be equal to the change in the kinetic energy plus the change in the potential energy. Therefore, −= + µθ θ k mgx K kx mgx cos sin 1 2 2 and since vv if == 0, K = 0. Thus, −° = ° µ k 2 00 9 80 37 0 0 200 100 0 200 2 2 00 9 80 37 0 0 200 2 .. c o s . s i n af a f a f a f a f a f and we find k = 0115 . . Note that in the above we had a gain in elastic potential energy for the spring and a loss in gravitational potential energy. P8.55 (a) Since no nonconservative work is done, E = 0 Also K = 0 therefore, UU = where Um g x i = sin bg and Uk x f = 1 2 2 2.00 kg k = 100 N/m FIG. P8.55 Substituting values yields
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