Q 11-05-151JEE MainJEE Main 2021 (25 Jul, Shift 2)Easy
A force of $F = (5y + 20)\hat{j}$ N acts on a particle. The work done by this force when the particle is moved from $y = 0$ m to $y = 10$ m is ______ J.
Numerical value type. Enter your answer.
Answer: 450
$$W = \int_0^{10}(5y + 20)\,dy = \left[\frac{5y^2}{2} + 20y\right]_0^{10} = 250 + 200 = 450\ \text{J}$$
Solution by Sreeraj P, M.Sc Physics