Three blocks  A, B  and C , of masses 4 kg, 2 kg and 1 kg respectively, are in contact on a frictionless surface, as shown. If a force of 14 N is applied on the 4 kg block, then the contact force between A and B is

Here, MA=4 kg,MB=2 kg,MC=1 kg,F=14 N Net mass, M=MA+MB+MC=4+2+1=7 kg  Let a be the acceleration of the system.  Using Newton's second law of motion, F=Ma14=7a  ∴  a=2 m s-2Let F' be the force applied on block A by block B i.e. the contact force between A and B. Free body diagram for block A Again using Newton's second law of motion, F-F'=4a14-F'=4×214-8=F'∴  F'=6 N