CBSE Class 9 Answered

Let, ‘m’ be the mass of the man, ‘g’ be the acceleration due to gravity, ‘a’ is the acceleration of the lift upwards.

Forces on the floor by the man are:

Gravitational force downward = mg

The reaction on the floor due to the upward acceleration of the lift = ma

So, the net force on the floor by the man = mg + ma = m(g + a)

This force m(g + a) is the apparent weight of the man, which is more than the actual weight.

If the lift is moving downward with acceleration ‘a’, then the net force on the floor by the man = mg - ma = m(g - a). Here in this case the apparent weight of the man is less than his actual weight.

If the lift is moving up or down with constant velocity then the apparent weight and the actual weights are equal.