Elevator Motion | MCQs

Explanation

The weight of a man in an elevator descending with an acceleration of 4.9 m/s² will become:

Let:

• Actual weight $W=mg$

• Acceleration due to gravity g=9.8m/s^2

• Acceleration of the elevator a=4.9m/s^2

When an elevator descends with acceleration $a$, the apparent weight becomes:

W′=m(g−a)

Substitute values:

W′=m(9.8−4.9)=m(4.9)
W′=1​/2×mg

So, the apparent weight becomes half of the actual weight.

Explanation

If the lift is accelerated downward, the apparent weight experienced by an object inside the lift is m(g−a).