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A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Compute the(a) Work done by the applied force in 10 s(b) Work done by the friction in 10 s(c) Work done by the net force on the body in 10 s(d) Change in kinetic energy of the body in 10 s and interpret your results. |
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Answer» Here, m = 2 kg; u = 0, μ = 0.1, applied force, F = 7 N Force of friction, f = μ mg = 0.1 x 2 x 9.8 = 1.96 N Net force under which body moves, F1 = F - f = 7 - 1.96 = 5.04 N Hence, acceleration produced, a = F/m = 5.04/2 = 2.52 ms-2 Now, s = ut + 1/2 at2 = 0 x 10 + 1/2 x 2.52 x 102 = 126 m (a) Work done by applied force in 10 s = 7 x 126 = 882 J (b) Work done by friction in 10 s = 1.96 x 126 = 246.96 J (c) Work done by net force in 10 s = 5.04 x 126 = 635.04 J (d) Since the initial kinetic energy of the body (at rest) was zero and the final K.E. acquired by the body in 10 s under the net force is 635.04 J, the change in K.E. in 10 s is 635.04 J Work done by the applied force in 10 s is more than the work done by the net force. It means work has been done to overcome the friction equal to the work done by the friction the 10 seconds. |
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