JEE Main 2024PhysicsWork Power EnergyEasyMCQ

JEE Main 2024 — Work Power Energy Question with Solution

JEE Main 2024 (29 Jan Shift 1)

Question

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:

Choose an option

Show full solutionCorrect option: C
Correct answer
C4

Step-by-step explanation

Given U=2x2+3y3+2z

The expression for the x-component of the force can be calculated as follows: 

Fx=-∂U∂x=-4x

Hence, at the point 1, 2, 3 m, the value of the force is given by

Fx=-4×1 N=4 N

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About this question

This is a previous-year question from JEE Main 2024, covering the Work Power Energy chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.