JEE Main 2017PhysicsWork Power EnergyEasyMCQ

JEE Main 2017 — Work Power Energy Question with Solution

JEE Main 2017 (02 Apr)

Question

A time dependent force F=6t acts on a particle of mass 1 kg. If the particle starts from the rest, the work done by the force during the first 1 sec will be:

Choose an option

Show full solutionCorrect option: B
Correct answer
B4.5 J

Step-by-step explanation

m=1 kg
F=6t
∴  a=Fm=6t
dvdt=6t
∴ ∫0v0dv=∫016t·dt
∴  v0=6×122=3 m s-1
Δw=kf-ki=12×1×9-0=4.5 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 2017, 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.