ch08-p118

# ch08-p118 - 11.0 m. This point may be found by...

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The total energy, therefore, is great enough to reach the point x = 0 where U = 11.0 J, with a little “left over” (11.9 J – 11.0 J = 0.9025 J). This is the kinetic energy at x = 0, which means the speed there is v = 2(0.9025 J)/(2 kg) = 0.950 m/s. It has now come to a stop, therefore, so it has not encountered a turning point. (b) The total energy (11.9 J) is equal to the potential energy (in the scenario where it is initially moving rightward) at x = 10.9756
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Unformatted text preview: 11.0 m. This point may be found by interpolation or simply by using the work-kinetic-energy theorem: K f = K i + W = 0 11.9025 + (4) d = 0 d = 2.9756 2.98 (which when added to x = 8.00 [the point where F 3 begins to act] gives the correct result). This provides a turning point for the particles motion. 118. (a) At x = 5.00 m the potential energy is zero, and the kinetic energy is K = 1 2 mv 2 = 1 2 (2.00 kg)(3.45 m/s) 2 = 11.9 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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