Answer to Question #1505 in Mechanics | Relativity for jon allen
A force of 340 N is applied horizontally to a wooden crate in order to displace it 39.1 m across a level floor at a constant velocity. As a result of this work the crate’s internal energy is increased by an amount equal to 15 percent of the crate’s initial internal energy. Calculate the initial internal energy of the crate. Answer in units of J.
Denote the initial internal energy as U. The work done by a force is A = FS = 340[N]39.1[m] = 13294J and equal to 15 percents of the internal energy: A = 0.15U. Thus U = A/0.15 = 13294/0.15 = 88626.7 J.
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