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A block of mass 2 Kg rests on another block of mass 3 Kg and the second block is on a smooth table. Coefficient of friction between the two block is 0.2. Then the largest force that can be applied on the lower block so that the system moves without sliding off of the upper block is

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14.7 N

9.8 N

4.9 N

19.6 N

answer is B.

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Detailed Solution

Maximum acceleration that can be given to upper block which prevents it from sliding off is $μ g$ (since $ma = μ mg$ )
So, the entire system should move with same acceleration $μ g$ .

Then maximum force applied $= (3 + 2) μ g = 5 \times 0.2 \times 9.8 = 9.8 N$

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