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# This is a question from All India Aakash Test series.Please solve the above question.I posted this question 2 times but not get any reply.Please sir,answer this question(see attachment)

Eshan
3 years ago
Dear student,

Maximum applicable frictional force under 4kg block is$4(0.3)(10)=12N$

When 16 N force is applied, the frictional force reaches its limit. Then the tension in the string starts to increase to oppose the change in state of rest. Hence tension=16N-12N=4N

Hence the frictional force acting on 2kg block to prevent slipping= 4N