Q 11-05-076JEE MainJEE Main 2024 (29 Jan, Shift 1)Easy
The potential energy function (in J) of a particle in a region of space is given as $U = (2x^2 + 3y^3 + 2z)$. Here $x$, $y$ and $z$ are in metre. The magnitude of $x$-component of force (in N) acting on the particle at point $P(1, 2, 3)\ \text{m}$ is:
Answer: (C) 4
$$F_x = -\frac{\partial U}{\partial x} = -4x$$
At $x = 1\ \text{m}$, $F_x = -4\ \text{N}$, so its magnitude is $4\ \text{N}$.
Solution by Sreeraj P, M.Sc Physics