Two blocks of mass m₁ = 6 kg and m₂ = 3 kg are as shown in the diagram. Coefficient of friction between m₁ and m₂ and between m₁ and the surface is 0.5 and 0.4, respectively. The maximum horizontal force that can be applied to the mass m₁ so that they move without separation is

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81 N

61 N

41 N

101 N

answer is C.

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Detailed Solution

If the blocks move together without any relative motion between them, then their acceleration is given by

$a = \frac{F - 0 .4 (3 + 6) \times 10}{(3 + 6)} m / s^{2} = \frac{F - 36}{9} m / s^{2}$

Pseudo force on 3 kg block $= 3 a = \frac{F - 36}{3} N$

The 3 kg block will not slip on 6 kg block $i.e., If = 3 a = \frac{F - 36}{3} \leq 0 .5 \times 3 \times 10$

$\Rightarrow F_{max} = 81 N$

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