Q.

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, the work done by the force of friction on the slab, between the instants t = 2 s and t = 3 s, is (g = 10 ms-2)

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a

2.4 J

b

5.55 J

c

4.44 J

d

10 J

answer is B.

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Detailed Solution

Maximum frictional force between the slab and the block fmax=μN =μmg =14× 4 × 10 = 10N. Evidently, f < fmax. So, the two bodies will move together as a single unit. If a be their combined acceleration, thena=Fm+M=64+5=23ms−1. Therefore, factional force acting can be obtained as f=Ma=23×5N=103N. Using s=12at2.s(2)=12×23(2)2=43 and s(3)=12×23(3)2=3. Therefore, work done by friction = =1033−43=509=5.55 J
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