28. We take the original height of the box to be the y= 0 reference level and observe that, in general, the height of the box (when the box has moved a distance ddownhill) is sin 40yd=−°.(a) Using the conservation of energy, we have KUK Umvmgykdii+=+¡++00121222.Therefore, with d= 0.10 m, we obtain v= 0.81 m/s. (b) We look for a value of d≠0 such that K= 0. KU KUmgy kd+¡+=++000122.Thus, we obtain mgdkdsin 40
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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.