The potential energy function (in J) of a particle in a region of space is given as U = (2x² + 3y³ + 2z). Here x, y and z are in meter. The magnitude of x -component of force (in N) acting on the particle at point P(1, 2, 3) m is: (a) 4 (b) 2 (c) 8 (d) 6

Answer: (a). Fx = -(∂ U)/(∂ x) = -4x At x = 1: Fx = -4 N, magnitude 4 N. So the answer is (a) 4.

The potential energy function (in J) of a particle in a region of… | JEE Main 2024 (29 Jan, Shift 1)

Q5JEE Main 2024 · 29 JAN · SHIFT 1Medium

Potential Energy & Equilibrium

The potential energy function (in $J$ ) of a particle in a region of space is given as $U = \left(2x^2 + 3y^3 + 2z\right)$ . Here $x, y$ and $z$ are in meter. The magnitude of $x$ -component of force (in $N$ ) acting on the particle at point $P(1, 2, 3)\text{ m}$ is:

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Answer: (a)

$F_x = -\frac{\partial U}{\partial x} = -4x$
At $x = 1$ : $F_x = -4\text{ N}$ , magnitude $4\text{ N}$ .
So the answer is (a) 4.

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