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:
$\mathrm{Mg}-\mathrm{N}=\mathrm{Ma}$
$\mathrm{N}=\mathrm{M}(\mathrm{g}-\mathrm{a})$
$\mathrm{N}=60(10-1.8)$
$\mathrm{N}=60 \times 8.2=492 \mathrm{~N}$
$\mathrm{~N}=492$
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