22(16.0 J)12.6 m/s.0.200 kgKvm===(c) At the turning point, the speed of the particle is zero. Let the position of the right turning point be.RxFrom the figure shown on the right, we find Rxto be 16.00 J 024.00 J 16.00 J7.67 m.7.00 m8.00 mRRRxxx−−=¡=(d) Let the position of the left turning point be.LxFrom the figure shown, we find Lxto be 16.00 J 20.00 J9.00 J 16.00 J1.73 m.1.00 m3.00 mLLLx=¡=40. In this problem, the mechanical energy (the sum of Kand U) remains constant as the particle moves. (a) Since mechanical energy is conserved,
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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.