An object of mass 2.00 kg starts at rest from the top of a rough inclined plane of height 20.0 m as shown in Fig. 8-9. If the work done by the force of friction is -150 J, what is the speed of the object as it reaches the bottom of the inclined plane? Use g = 10.0 m/s².

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Quizplus · Accepted Answer

The potential energy at the top is mgh = 2 * 10 * 20 = 400 J. The work done by friction is -150 J, so the kinetic energy at the bottom is 400 - 150 = 250 J. Using KE = 0.5 * m * v^2, we solve for v: 250 = 0.5 * 2 * v^2, giving v = sqrt(250) = 15.8 m/s.

An Object of Mass 2