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

10 J

4.44 J

5.55 J

answer is B.

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Detailed Solution

Maximum frictional force between the slab and the block

f_max = $μ$ N = $μ$ mg ; m = mass of upper block

Evidently, F < f_max

so, the two bodies will move together as a single unit. If ‘a’ be their combined acceleration, then

$a = \frac{F}{m + M}$

Therefore, frictional force acting can be obtained as f = Ma ,Using $s = \frac{1}{2} {at}^{2}$ , find S₂ and S₃.

W = (S₃ - S₂)f

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