First slide

Connected bodies

Question

Three blocks A,B and C of masses 4 kg, 2 kg and 1kg 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 :

Easy

A

6N
B

8 N
C

18 N
D

2 N

Solution

Acceleration of system = $\frac{F_{net}}{M_{total}} = \frac{14}{4 + 2 + 1} = 2 m / s^{2}$
The contact force between 4 kg and 2 kg block will move 2 kg and 1 kg block with the same acceleration so
F_contact = (2+ 1) a = 3(2) = 6 N

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