JEE Main 2023PhysicsWork Power EnergyMediumNumerical

JEE Main 2023 — Work Power Energy Question with Solution

JEE Main 2023 (24 Jan Shift 2)

Question

A body of mass 1 kg begins to move under the action of a time dependent force F→=ti^+3t2j^ N, where i^ and j^ are the unit vectors along x and y axis. The power developed by above force, at the time t=2 s, will be _______W. 

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Show full solutionCorrect answer: 100
Correct answer
100

Step-by-step explanation

Given, F→=ti^+3t2j^ N and m=1 kg

Now, we can write from Newton's second law, F→=ma→=mdv→dt

Then, dv→=ti^+3t2j^dt

Integrating the above, 

∫0vdv→=∫0tti^+3t2j^dt⇒v→=t22i^+t3j^

Now, power developed by the force is P=F→·v→=ti^+3t2j^·t22i^+t3j^

=t32+3t5

At t=2 s, power is P=232+325=4+3×32=100 W

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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.