A wooden crate of mass 14 kg is sitting on a horizontal cement floor. Then, one

Question

A wooden crate of mass 14 kg is sitting on a horizontal cement floor. Then, one end of a horizontal pulley is attached to the crate. The rope in the pulley loops around twice, so that the pulley provides a force 4 times bigger than the tension in the rope. Now the rope is given a continuous tension of 50 N, until the crate has been dragged a distance of 9.5 m across the floor. What is the kinetic energy K (in J) of the crate when it is just 3.4 m from its starting point? The kinetic coefficient of friction between the wood and the cement is 0.57.

Explanation / Answer

Given data Mass of the wooden crate is, m = 14 kg Tension in the rope is, T = 50 N Force exerted by the pulley is, F = 4 T                                                     = 4 (50 N)                                                     = 200 N Coefficient of friction between the wood and the cement is, = 0.57 Distance moved by the crate is, d = 3.4 m Solution: From the work energy theorem, Wtot = KE KE = (F *s ) - ( mg * s)         = (200 N * 3.4 m) - (0.57 * 14 kg *9.8 m/s2 * 3.4 m)         = 680 J - 265.89 J         = 414 J         = 414 J

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