The potential energy function (in J ) of a particle in a region of space is given as U= (2 x 2 +3 y 3 +2 z ) . 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 :

Physics · Work, Power & Energy

JEE Main 2024 — 29 January, Shift 1 — Question 35

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

Option A:
2
Option B:
6
Option C:
4
Correct
Option D:
8

Answer: C

Step-by-step solution

Given $U=2 x^{2}+3 y^{3}+2 z$

$F_{x}=-\frac{\partial U}{\partial x}=-4 x$

At $x=1$ magnitude of $F_{x}$ is $4 N$

Answer key and solution verified before publishing.

Practise Work, Power & Energy

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2024
Paper: 29 January, Shift 1
Subject: Physics
Chapter: Work, Power & Energy
Topic: Conservative Forces and Potential Energy