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(B) Given the velocity $v = k \sqrt{x}$.Squaring both sides,we get $v^2 = k^2 x$. Comparing this with the kinematic equation $v^2 = u^2 + 2ax$,where i

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$\frac{m k^4 t^2}{8}$

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(B) Given the velocity $v = k \sqrt{x}$.Squaring both sides,we get $v^2 = k^2 x$. Comparing this with the kinematic equation $v^2 = u^2 + 2ax$,where initial velocity $u = 0$,we get $2a = k^2$,which implies acceleration $a = \frac{k^2}{2}$.Since the acceleration is constant,the displacement $x$ at time $t$ is given by $x = \frac{1}{2} a t^2 = \frac{1}{2} \left( \frac{k^2}{2} \right) t^2 = \frac{k^2 t^2}{4}$.The force acting on the particle is $F = ma = m \left( \frac{k^2}{2} \right) = \frac{m k^2}{2}$.The work done $W$ is given by $W = F \cdot x = \left( \frac{m k^2}{2} \right) \left( \frac{k^2 t^2}{4} \right) = \frac{m k^4 t^2}{8}$.

$A$ particle of mass $m$ starts moving from the origin along the $x$-axis and its velocity varies with position $x$ as $v = k \sqrt{x}$. The work done by the force acting on it during the first $t$ seconds is ...........

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