The weight of a man in an elevator descending with an acceleration of 4.9 ms⁻² will become?

Answer: Half

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

Let:

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

$W^{'} = m (g - a)$

Substitute values:

$W^{'} = m (9.8 - 4.9) = m (4.9)$ $W^{'} = 1 \times m g$

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

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