Q 11-05-124JEE MainJEE Main 2022 (25 Jul, Shift 1)Easy
A body of mass $0.5\ \text{kg}$ travels on a straight line path with velocity $v = (3x^2 + 4)\ \text{m s}^{-1}$. The net work done by the force during its displacement from $x = 0$ to $x = 2\ \text{m}$ is
Answer: (B) $60\ \text{J}$
$v(0) = 4\ \text{m s}^{-1}$, $v(2) = 16\ \text{m s}^{-1}$. By the work–energy theorem:
$$W = \frac12(0.5)(16^2 - 4^2) = 0.25\times240 = 60\ \text{J}$$
Solution by Sreeraj P, M.Sc Physics