Question

What should be the acceleration a of the box shown in Fig. 5.54 so that the block of mass m exerts a force $\frac{mg}{4}$ on the floor of the box?

• A. $\frac{3g}{4}$
• B. $\frac{5g}{4}$
• C. $\frac{3g}{2}$
• D. $\frac{g}{2}$

Answer 1

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

Let box is accelerated downward with acceleration '$a$' so from the FBD of the block,

$\therefore$ $mg-N=ma$

$N=\frac{mg}{4}$ is given

$\Rightarrow mg-\frac{mg}{4}=ma$

$\Rightarrow \frac{3mg}{4}=ma$

$\Rightarrow a=\frac{3g}{4}$