Block A of mass m rests on the plank B of mass 3m which is free to slide on a frictionless horizontal surface. The coefficient of friction between the block and plank is 0.2. If a horizontal force of magnitude 2 mg is applied to the plank B, the acceleration of A relative to the plank and relative to the ground respectively, are

detailed solution

Correct option is D

Block A moves due to friction. Maximum value of friction can be μmAg. Therefore, maximum acceleration of A can be μmAgmA or μg=0.2g=g5. When force 2 mg is applied on lower block, common acceleration (if both move together) will bea= Net force  Total mass =2mg4m=g2Since, a = g/2 is greater than maximum acceleration of A which can be given to it by friction. Therefore, slipping will take place.aA=0.2g=g5aB=2mg-0.2mg3m=0.6g=3g5aAB=aA-aB=g5-3g5=-2g5 (in backward direction)