Q 11-05-118JEE MainJEE Main 2022 (24 Jun, Shift 1)Easy
A particle experiences a variable force $\vec F = (4x\hat i + 3y^2\hat j)$ in a horizontal $x$–$y$ plane. Assume distance in metres and force in newton. If the particle moves from point $(1, 2)$ to point $(2, 3)$ in the $x$–$y$ plane, then kinetic energy changes by:
Answer: (A) $25\ \text{J}$
By the work–energy theorem, $\Delta K = W = \int \vec F\cdot d\vec r$.
$$W = \int_1^2 4x\,dx + \int_2^3 3y^2\,dy = \left[2x^2\right]_1^2 + \left[y^3\right]_2^3 = 6 + 19 = 25\ \text{J}$$
Solution by Sreeraj P, M.Sc Physics