Q 11-05-177JEE MainJEE Main 2019 (9 Jan, Shift 2)Easy
A force acts on a $2\ \text{kg}$ object so that its position is given as a function of time as $x = 3t^2 + 5$. What is the work done by this force in first 5 seconds?
Answer: (D) $900\ \text{J}$
$v = \dfrac{dx}{dt} = 6t$, so $v = 0$ at $t = 0$ and $v = 30\ \text{m/s}$ at $t = 5\ \text{s}$.
By the work-energy theorem:
$$W = \tfrac12 m v^2 - 0 = \tfrac12 (2)(30)^2 = 900\ \text{J}$$
Solution by Sreeraj P, M.Sc Physics