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 [figure]
- 8 N
- 18 N
- 2 N
- 6 N
Answer
(4) 6 N
Solution
Common acceleration: a = FmA + mB + mC = 144+2+1 = 2 m s-2
The contact force between A and B moves B and C together:
NAB = (mB + mC)a = (2+1) 2 = 6 N
Answer: (4)