Q 11-05-195JEE MainJEE Main 2018 (16 Apr, Shift 1)Medium
A body of mass $m$ starts moving from rest along $x$-axis so that its velocity varies as $v = a\sqrt s$ where $a$ is a 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 $t$ second after the start of the motion is
Answer: (D) $\dfrac18ma^4t^2$
$v^2 = a^2s$, so the acceleration is constant:
$$\text{acceleration} = v\frac{dv}{ds} = \frac12\frac{d(v^2)}{ds} = \frac{a^2}{2}$$
Starting from rest, $v = \dfrac{a^2t}{2}$ after time $t$. By the work-energy theorem,
$$W = \frac12mv^2 = \frac12m\cdot\frac{a^4t^2}{4} = \frac18ma^4t^2$$
Solution by Sreeraj P, M.Sc Physics