A man inside lift feels _______, when lift was going down with uniform acceleration. 

Concept:

Apparent weight of a person inside a lift

(i) When the lift moves upward with acceleration a. Then the net upward force on the person is-

• R - mg = ma
•  Apparent weight, R = mg + ma = m (g + a)
• So, when a lift accelerates upwards, the apparent weight of the person inside it increases.

(ii) When the lift moves downwards with acceleration a. Then the net downward force on the person is-

• Mg - R = ma
•  Apparent weight, R = mg - ma = m (g - a)
• So, when a lift accelerates downwards, the apparent weight of the person inside it decreases.

(iii) When the lift is at rest or moving with uniform velocity v downward/upward. The acceleration a = 0, then the net force on the person is-

• R - mg = m x 0 = 0
• R = mg
•  Apparent weight = Actual weight

(iv) When the lift falls freely. If the supporting cable of the lift breaks, the lift falls freely under gravity. Then a = g. The net downward force on the person is -

• R = m(g - g) = 0.
• Thus, the apparent weight of the man becomes zero. This is because both the man and the lift are moving downwards with same acceleration ‘g’ and so there are no forces of action and reaction between the person and the lift. Hence a person develops a feeling of weightlessness when he falls freely under gravity.

Explanation:

When the lift moves downward with acceleration a -

• Then the net upward force on the person is-
• mg - R = ma
•   Apparent weight, R = mg - ma = m (g - a)

This is a reason why one feels lighter in a lift when the lift just begins to go down because our body starts moving in same direction as gravitational force hence there is less opposition on our body hence net force exerted on our body is less as compared to when lift was going up. Thus, when a lift accelerates downward, the apparent weight of the person inside it decreases.