What is the apparent weight of a man of mass 60 kg who is standing on a lift which is moving up with a uniform speed?

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CONCEPT:

The apparent weight of a person:

(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

(iii) 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:

Given m = 60 kg and a = 6 ms-2

As we know that weight, W = mg

Here, when the lift is going downwards with acceleration, then the apparent weight of the is,

W = m(g - a)

Where ‘m’ is the mass of the boy, ‘a’ is the acceleration and ‘g’ is the acceleration due to gravity.

Therefore,