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

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$\frac{3 g}{6}$

$\frac{3 g}{7}$

$\frac{3 g}{4}$

$\frac{3 g}{5}$

answer is A.

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Detailed Solution

$N = \frac{m g}{4}$ is given

$m g - \frac{m g}{4} = m a a = \frac{3 g}{4}$

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