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A box of mass 50kg at rest is pulled up on an inclined plane 12m long and 2m high by a constant force of 100N applied parallel to the inclined plane. When it reaches the top of the inclined plane if its velocity is 2ms^–1, the work done against friction in Joules is (g = 10ms^–2)

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answer is B.

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Detailed Solution

W_F = Work done by the applied force = 10 x 12 =1200J
ΔV = Change in P.E = mgh = 50 x 10 x 2 = 1000J
ΔK = Change in K.E. = 1/2 x 50 x 2² = 100J
Now, W_F = ΔK + ΔV + lost energy
$\large \therefore$ Lost energy = (1200-100-1000)J = 100J.

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