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 :

6 N 8 N 18 N 2 N

6 N

Acceleration : a = \(\frac{F_{net}}{M_{total}}\) Fnet = 14 N ; Mtotal = 4 + 2 + 1 = 7 kg ⇒ a = 2 m/s2 Fcontact = (2 + 1) a = 6 N