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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 two blocks is μ1 and that between the block of mass M and 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
Here, the force applied should be such that force acting on the upper block of m should not be more than the force of friction (=μ1mg) acting on it. Let the system moves with acceleration a. Then
Fμ2(M+m)g=(M+m)a .. (1)
For block of mass m:
μ1mg=ma or a=μ1g ... (2)
From Eqs. (1) and (2), we get;
F=μ2(M+m)g+(M+m)μ1g
=(M+m)g(μ1+μ2)
Hence, the correct answer is option (d).

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