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_ $\mathrm{N}$.
$\left[\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right]$
Solution:
When lift is at rest
$\mathrm{N}=\mathrm{mg}$
$\Rightarrow 60 \times 10=600 \mathrm{~N}$
When lift moves with downward acceleration. In frame of lift pseudo force will be in upward direction.
$N^{\prime}=M(g-a)$
$\Rightarrow 60(10-1.8)$
$N^{\prime} \Rightarrow 492 \mathrm{~N}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.