A body of mass M starts moving from rest along x-axis so that its velocity varies $v = a \sqrt{s}$ Where a is constant and s is the distance covered by the body. The total work done by all the forces acting on the body in the first second after the start of the motion is

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$\frac{1}{6} M a^{4} t$

$\frac{1}{8} M a^{4} t^{2}$

$4 M a^{4} t^{2}$

$\frac{1}{2} M a^{4} t^{2}$

answer is C.

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Detailed Solution

$v = a \sqrt{s} \Rightarrow \frac{d s}{d t} = a \sqrt{s} \Rightarrow \frac{d s}{\sqrt{s}} = a d t \Rightarrow 2 \sqrt{s} = a t \Rightarrow s = \frac{a^{2} t^{2}}{4} F = M (v \frac{d v}{d s}) = M (a \sqrt{s}) (\frac{a}{2 \sqrt{s}}) = \frac{M a^{2}}{2} ∴ W o r k d o n e = \frac{M a^{2}}{2} \times \frac{a^{2} t^{2}}{4} = \frac{1}{8} M a^{4} t^{2}$