ch08-p039 - 39. From Fig. 8-50, we see that at x = 4.5 m,...

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(b) The force acting on the particle is related to the potential energy by the negative of the slope: x U F x Δ =− Δ From the figure we have 35 J 15 J 10 N 2 m 4 m x F =+ . (c) Since the magnitude 0 x F > , the force points in the + x direction. (d) At x = 7.0m, the potential energy is U 3 = 45 J which exceeds the initial total energy E 1 . Thus, the particle can never reach there. At the turning point, the kinetic energy is zero. Between x = 5 and 6 m, the potential energy is given by ( ) 15 30( 5), 5 6. Ux x x 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) J 30 N (6 5) m x U F x Δ− 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.

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