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

Given : MA=4 kg, MB=2 kg, MC=1 kg
Step 1 : Acceleration of all the blocks
From figure, we see that the blocks A, B and C will move with a common acceleration a.
By Newton's Second Law, Acceleration=Fnet External/Mtotal
⇒a=F/(MA+MB+MC)
⇒a=14N/(4kg+2kg+1kg)
⇒a=14N/7kg=2m/s^2
Step 2 : Contact force between A and B
Let FAB be the contact force between A and B and FBC be the contact force between B and C.
Considering block B and C together as a system, the net external force is FAB.
By Newton's Second Law,
FAB=(MB+MC)a
⇒FAB=(2kg+1kg)2m/s^2=6N
Considering block C alone, the net force is FBC.
By Newton's Second Law,
FBC=MCa
⇒FBC=1kg2m/s^2=2N
Therefore, the contact force between A and B is 6N.