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A particle of mass ‘m’ moves on a straight line with its velocity varying with the distance travelled according to the equation $v = a \sqrt{x}$ , where ‘a’ is a constant. Find the total work done by all the forces during a displacement from $x = 0 to x = d$

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\frac{{ma}^{2} d}{2}

\frac{{mad}^{2}}{2}

\frac{mad}{2}

\frac{{mad}^{2}}{4}

answer is D.

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Detailed Solution

We know that

$a = \frac{dv}{ds} . \frac{ds}{dt}$

$a = \frac{dv}{ds} . v$

$ads = vdv$

$a \int_{0}^{d} d s = \int_{0}^{d} v d v$

$ad = \frac{v^{2}}{2}$

$ad = \frac{{(a \sqrt{x})}^{2}}{2} \Rightarrow d = \frac{a x}{2}$

$∴ F = ma S$

$= ma (\frac{ad}{2}) \Rightarrow \frac{{ma}^{2} d}{2}$

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