A person standing on a spring balance inside a stationary lift measures

Question:

A person standing on a spring balance inside a stationary lift measures

$60 \mathrm{~kg}$. The weight of that person if the lift descends with uniform

downward acceleration of $1.8 \mathrm{~m} / \mathrm{s}^{2}$ will be

N. $\left[\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right]$

Solution:

$\mathrm{Mg}-\mathrm{N}=\mathrm{Ma}$

$\mathrm{N}=\mathrm{M}(\mathrm{g}-\mathrm{a})$

$\mathrm{N}=60(10-1.8)$

$\mathrm{N}=60 \times 8.2=492 \mathrm{~N}$

$\mathrm{~N}=492$