Lift Acceleration | MCQs

Explanation

Let's denote the weight of the man in the stationary lift as W and the weight in the accelerating lift as W'.

We know that W' / W = 2/3.

When the lift is accelerating downward with acceleration 'a', the apparent weight (W') is given by:

W' = W - ma

where m is the mass of the man.

Since W' / W = 2/3, we can write:

(W - ma) / W = 2/3

Substitute W = mg (where g is the acceleration due to gravity):

(mg - ma) / mg = 2/3

Simplify:

g - a = (2/3)g

a = g - (2/3)g

a = (1/3)g

a = g/3