JEE Main 2019PhysicsWork Power EnergyEasyMCQ

JEE Main 2019 — Work Power Energy Question with Solution

JEE Main 2019 (09 Jan Shift 2)

Question

A force acts on a 2 kg object so that its position is given as a function of time as x=3t2+5. What is the work done by this force in first 5 seconds?

Choose an option

Show full solutionCorrect option: D
Correct answer
D900 J

Step-by-step explanation

Here, the position of the object is x=3t2+5, therefore the velocity of the object will be,

v=dxdt=d3t2+5dt

v=6t+0

From the work-energy theorem, all the work done by the force acting on it will be equal to the change in its kinetic energy, i.e.,

W=KEt=5 s-KEt=0 s
W=12×2×6×52-12×2×6×02
W=900 J.

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About this question

This is a previous-year question from JEE Main 2019, 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.