Question 101 of 210

A block of mass $m$ is kept on a platform which starts from rest with a constant acceleration $g/2$ upwards, as shown in the figure. Work done by normal reaction on block in time $t$ is

A$-\dfrac{m{g}^{2}{t}^{2}}{8}$B$\dfrac{m{g}^{2}{t}^{2}}{8}$C$0$D$\dfrac{3m{g}^{2}{t}^{2}}{8}$

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Answer: (D) $\dfrac{3m{g}^{2}{t}^{2}}{8}$

$N-mg=\dfrac{mg}{2}$
$N=\dfrac{3mg}{2}$
$s=ut+\dfrac{1}{2}a{t}^{2}$
$=0+\dfrac{1}{2}\dfrac{g}{2}{t}^{2}$
$=\dfrac{g}{4}$${t}^{2}$
${W}_{N}=\dfrac{3mg}{2}\times \dfrac{g}{4}{t}^{2}$
$=\dfrac{3m{g}^{2}{t}^{2}}{8}=\dfrac{3m{g}^{2}{t}^{2}}{8}$