Q 11-05-083JEE MainJEE Main 2024 (9 Apr, Shift 1)Easy
A particle of mass $m$ moves on a straight line with its velocity increasing with distance according to the equation $v = \alpha\sqrt x$, where $\alpha$ is a constant. The total work done by all the forces applied on the particle during its displacement from $x = 0$ to $x = d$ will be:
Answer: (D) $\dfrac{m\alpha^2d}{2}$
By the work–energy theorem, $W = \Delta K$. At $x = 0$, $v = 0$; at $x = d$, $v^2 = \alpha^2d$:
$$W = \frac12m\alpha^2d$$
Solution by Sreeraj P, M.Sc Physics