Q 11-05-047JEE MainEasy
A $4$ kg body moves along a straight line with velocity $v = 3t$ m/s, where $t$ is in seconds. Find the work done by the net force on it during the first $2$ s, in joules.
Numerical value type. Enter your answer.
Answer: 72
$v(0) = 0$ and $v(2) = 6$ m/s. By the work-energy theorem:
$$W = \frac{1}{2}(4)(6^2) - 0 = 72\ \text{J}$$
Solution by Sreeraj P, M.Sc Physics