JEE Main 2023PhysicsWork Power EnergyEasyNumerical

JEE Main 2023 — Work Power Energy Question with Solution

JEE Main 2023 (01 Feb Shift 1)

Question

A small particle moves to position 5i^-2j^+k^ from its initial position 2i^+3j^-4k^ under the action of force 5i^+2j^+7k^ N. The value of work done will be ______ J.

Enter your answer

Show full solutionCorrect answer: 40
Correct answer
40

Step-by-step explanation

Work done by a constant force is given by W=ForceF→·Displacementr→, where, r→=r→2-r→1

Given here, r1→=5i^-2j^+k^ and r→2=2i^+3j^-4k^

So, W=F→·r→2-r→1

=(5i^+2j^+7k^)·(5i^-2j^+k^)-(2i^+3j^-4k^)

⇒W=40 J

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Work Power Energy chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

About this question

This is a previous-year question from JEE Main 2023, 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.