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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
A
2 N
B
6 N
C
8 N
D
18 N
Detailed Solution
Acceleration of the system: $a = \frac{F_{net}}{M_{total}} = \frac{14}{4+2+1} = 2$ m/$s^2$
The contact force between A and B moves the 2 kg and 1 kg blocks together with this acceleration.
$F = (2 + 1)\times 2 = 6$ N
The contact force between A and B moves the 2 kg and 1 kg blocks together with this acceleration.
$F = (2 + 1)\times 2 = 6$ N