A 4 kg block A is placed on the top of 8 kg block B which rests on a smooth table. A just slips on B when a force of 12 N is applied on A. Then the maximum horizontal force F applied on B to make both A and B move together, is (4x) N. Find the value of x.
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answer is 6.
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Detailed Solution
When force is applied on A FAmA+mB=μmAgmB ……..(i)When force is applied on BFBmA+mB=μmAgmA ……..(ii)Dividing these two equations, we getFBFA=mBmAFB=mBmA⋅FA=84×12=24N