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Question

# A block of mass m is placed on another block of mass M, which itself is lying on a horizontal surface. The coefficient of friction between the two blocks is μ1 and that between the block of mass M and the horizontal surface is μ2. What maximum horizontal force can be applied to the lower block so that the two blocks move without separation?

A
(M+m)(μ2μ1)g
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B
(Mm)(μ2μ1)g
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C
(Mm)(μ2+μ1)g
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D
(M+m)(μ2+μ1)g
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Solution

## The correct option is D (M+m)(μ2+μ1)g The FBDs of the upper block and of the system (upper+lower block) are shown above. The force applied should be such that frictional force acting on the upper block m should not be more than the limiting friction i.e f1≤μ1N=μ1mg Let the system move with a common acceleration a, then for the whole system, F−f=(M+m)a where f=μ2N′=μ2(M+m)g ⇒F−μ2(M+m)g=(M+m)a ⇒F=(M+m)(a+μ2g).....(1) For block of mass m, f1=ma≤μ1mg ⇒a≤μ1g......(2) Therefore, from (1) and (2), we get F≤(M+m)(μ1+μ2)g Option D is correct.

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