Question

Arrangement of two block system is as shown. The
net force acting on 1 kg and 2 kg blocks are (assuming
the surfaces to be frictionless) respectively
​

Answer 1

Thus the net fore on 1 Kg and 2 Kg is 2 N and 4 N respectively.

Explanation:

Both block move together.

a = F net / Total mass = 6 / 1 + 2

a = 6 / 3 = 2 m/s^2

For 2 Kg block

N = 2 x a = 2 x 2 = 4 N ( net force on 2 Kg block)

For 1 Kg block

6 N -----> | 1 Kg | <------ 4 N

Net force on 1 Kg = 6 - 4 = 2 N ( Right side)

Net force on 2 Kg = 2 N and for 2 Kg = 4 N

Thus the net fore on 1 Kg and 2 Kg is 2 N and 4 N respectively.

Answer 2

The net fore on 1 Kg and 2 Kg is 2 N and 4 N respectively.

Explanation:

Both block move together.

a = F net / Total mass = 6 / 1 + 2

a = 6 / 3 = 2 m/s^2

We know that

For 2 Kg block

F = ma

6 - f = ma

f = 6 - 1 x 2 = 6 - 2 = 4 N ( net force on 2 Kg block)

For 1 Kg block

Net force on 1 Kg = 6 - 4 = 2 N

Net force on 2 Kg = 2 N x 2 Kg = 4 N

Thus the net fore on 1 Kg and 2 Kg is 2 N and 4 N respectively.