Kkdkddfdk=−+−1220bg.In this first approach, we could work through the dKdd=0 condition (or with the special capabilities of a graphing calculator) to obtain the answer Kkkdfkmax1202. In the second (and perhaps easier) approach, we note that Kis maximum where vis maximum — which is where a=¡0equilibrium of forces. Thus, the second approach simply solves for the equilibrium position Ffkxkspring=¡=80. Thus, with k= 4000 N/m we obtain x= 0.02 m. But x= d0– dso this corresponds to d= 0.08 m. Then the methods of part (a) lead to the answer Kmax= 12.8 J ≈13 J. 107. (a) The effect of a (sliding) friction is described in terms of energy dissipated as
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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.