(b) The force acting on the particle is related to the potential energy by the negative of the slope:xUFxΔ=−ΔFrom the figure we have35 J 15 J10 N2 m 4 mxF−=+−.(c) Since the magnitude0xF>, the force points in the +xdirection.(d) At x= 7.0m, the potential energy is U3= 45 J which exceeds the initial total energy E1.Thus, the particle can never reach there. At the turning point, the kinetic energy is zero. Betweenx= 5 and 6 m, the potential energy is given by ( ) 15 30(5), 56.Uxxx−≤≤Thus, the turning point is found by solving 37 15 30(5)x−, which yields x= 5.7 m. (e) At x =5.0 m, the force acting on the particle is (45 15) J30 N(6 5) mxUFxΔ−The magnitude is |
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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.