ch08-p073 - launched from the ground at t = with speed 9.35...

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73. (a) By mechanical energy conversation, the kinetic energy as it reaches the floor (which we choose to be the U = 0 level) is the sum of the initial kinetic and potential energies: K = K i + U i = 1 2 (2.50 kg)(3.00 m/s) 2 + (2.50 kg)(9.80 m/s 2 )(4.00 m) = 109 J. For later use, we note that the speed with which it reaches the ground is v = 2 K/m = 9.35 m/s. (b) When the drop in height is 2.00 m instead of 4.00 m, the kinetic energy is K = 1 2 (2.50 kg)(3.00 m/s) 2 + (2.50 kg)(9.80 m/s 2 )(2.00 m) = 60.3 J. (c) A simple way to approach this is to imagine the can is
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Unformatted text preview: launched from the ground at t = with speed 9.35 m/s (see above) and ask of its height and speed at t = 0.200 s, using Eq. 2-15 and Eq. 2-11: y = (9.35 m/s)(0.200 s) – 1 2 (9.80 m/s 2 )(0.200 s) 2 = 1.67 m, v = 9.35 m/s – (9.80 m/s 2 )(0.200 s) = 7.39 m/s. The kinetic energy is K = 1 2 (2.50 kg) (7.39 m/s) 2 = 68.2 J. (d) The gravitational potential energy U = mgy = (2.5 kg)(9.8 m/s 2 )(1.67 m) = 41.0 J...
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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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