Question

What should be the acceleration ${}^{\prime }{a}^{\prime }$ of the box shown in figure so that the block of mass $m$ exerts a force $\frac{mg}{4}$ on the floor of the box ? (where $g$ is acceleration due to gravity)

• A. $\frac{3g}{4}$
• B. $\frac{g}{4}$
• C. $\frac{2g}{4}$
• D. $\frac{4g}{4}$

Answer 1

The correct option is A $\frac{3g}{4}$

As lift is moving down
Net Normal reaction = $N=m(g-a)$
Given $N=\frac{mg}{4}$
$\frac{\text{/}mg}{4}=\text{/}m(g-a)$
$a=\frac{3g}{4}\left(\text{downwards}\right)$