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Q.

A block B of mass 5 kg is placed on a slab A of mass 20 kg which lies on a frictionless surface as shown in the figure. The coefficient of static friction between the block and the slab is 0.4 and that of kinetic friction is 0.2. If a force F = 25 N acts on B, the acceleration of the slab will be (g = 10 m/s²)

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a

0.4 ms^–2

b

0.5 ms^–2

c

1 ms^–2

d

zero

answer is B.

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Detailed Solution

$Maximum frictional force = 0.4 \times 5 \times g = 20 N Appliled force = 25 N So there is a relative motion . Frictional force then kinetic friction f = 0.2 \times 5 \times g = 10 N Force on the slab = 10 N acceleration ot the slab = 10 = 20 a a = 0.5 m / s^{2}$

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