# 59x 10 11 m the force is fx 148 125xlo 8 n 232 on the

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Unformatted text preview: the same result, AKBc+ AUBC = WI,BC, = vc = V2(9.8 m/s2)(38 ni)(l-O.ll cot20) + (20.1 m/s)2.) = = Problem 51. A basketball dropped from a height of 2.40 m rebounds to Ii maximum height of 1.55 m. What fraction of the ball's initial energy is lost to nonconservative forces? Take the zero of potential energy at the floor. Problem 49. The potential energy of a spring is given by U ax2 - bx + c; where a = 5.20 N 1m, b== 3.12 N, and c 0.468 J, and where x is the overall length of = = CHAPTER 8 135 ,principle in the form of Equation 8-5 is ball is point A) a.nd a.tthe maximum of its rebound one has Wnc = t:J.UAB = mg(YB - YA). This energy since YB < YA' The initial energy was /;relative to zero of potential energy on the :, ~ 0, so the fraction lost is mg IYB - YAI+ \~:YB/YA = 1- 1.55/2.40 = 35.4%. (The fi'and "lost" replace the minus sign in the gei i.e., a <135% loss" equals a change ,"+ !::l.U. Since v = 0 where the 8.0-hour period. What fraction of the initial potential energy is lost to nonconservative forces (i.e., doe...
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## This note was uploaded on 07/27/2009 for the course PHYSICS 101 taught by Professor Wormer during the Spring '08 term at NYU Poly.

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