Question

62. Two blocks of masses 2 kg and 1 kg are in contact with each other on a frictionless table. When a horizontal force of 3.0 N is applied to the block of mass 1 kg the value of the force of contact between the two blocks is - (1) 2 N (2) 3 N (3) 5 N (4) 1 N​

Answer 1

Answer:

Considering both blocks as a system,

Total mass of the system M = 3 kg

So, net external force for the system = 3 N (as the table is frictionless)

F = Ma

=> 3 = 3a

=> a = 1 ms

−2

Now considering the block of 1 kg :

a = 1 ms

−2

(acceleration of the 1 kg block is same as the acceleration of the system)

so, F = ma

=> F = 1 kg x 1 ms

−2

= 1 N

Explanation:

answer is 1N