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Unformatted text preview: where the potential energy is equal to the mechanical energy. For 1 < x < 5, the potential energy is expressed as U(x) = 125(x 1) + 500 Equating this with the mechanical energy, we get 360 = 125(x 1) + 500 x = 2.12 m b) Suppose, instead, it is headed in the positive direction of the x axis when it is released at x = 6.0 m with 360 J. If the ball can reach x = 14 m, what is its speed there, and if it cannot, what is its turning point? For 6 < x < 14, the potential energy is always smaller than 360 J. Thus the ball reaches x = 14 m. At x = 14 m, the kinetic energy of the ball is K(x = 14 m) = E U(x = 14 m) = 60 J Kinetic energy of a rolling object consists of translational kinetic energy and rotational kinetic energy. The speed of the ball is K = (1/2)mv 2 + (1/2)I 2 = (1/2) mv 2 + (1/2)(mr 2 )(v/r) 2 = (7/10)mv 2 v = (10K/7m) = 6.55 m/s...
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This note was uploaded on 01/25/2011 for the course PHY 2048 taught by Professor Field during the Summer '08 term at University of Florida.
 Summer '08
 Field
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