Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
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.
PhysicsWork Power EnergyJEE MainJEE Main 2023 (01 Feb Shift 1)
Solution:
1846 Upvotes Verified Answer
The correct answer is: 40

Work done by a constant force is given by W=ForceF·Displacementr, where, r=r2-r1

Given here, r1=5i^-2j^+k^ and r2=2i^+3j^-4k^

So, W=F·r2-r1

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

W=40 J

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.