NEET Physics Question - Laws of Motion - Motion of Connected Bodies

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 Image related to

18 N

2 N

8 N

6 N

Solution:

The total mass of the system:

$M = 4 + 2 + 1 = 7 \text{ kg}$

The acceleration of the system:

$a = \frac{\text{Total Force}}{\text{Total Mass}} = \frac{14}{7} = 2 \text{ m/s}^2$

Now, considering block B and C together (mass = $2 + 1 = 3$ kg), the force required to accelerate them:

$F_{AB} = (3 \times 2) = 6 \text{ N}$

Thus, the contact force between A and B is 6 N.

Rankers Physics

Solution:

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