A large slab of mass 5 kg lies on a smooth horizontal surface, with a block of mass 4 kg lying on the top of it. The coefficient of friction between the block and the slab is 0.25. If the block is pulled horizontally by a force of F = 6 N, then the work done by the force of friction on the slab, between the instants t = 2s to t = 3s is (g = 10 ms-2)

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2.4 J

5.55 J

4.44 J

10 J

answer is B.

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Detailed Solution

Maximum frictional force between the slab and the blockfmax = μN = μmg ; m = mass of upper block Evidently, F < fmax so, the two bodies will move together as a single unit. If ‘a’ be their combined acceleration, thena=Fm+MTherefore, frictional force acting can be obtained as f = Ma ,Using s=12at2 , find S2 and S3. W = (S3 - S2)f

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