JEE Main 2015PhysicsWork Power EnergyHardMCQ

JEE Main 2015 — Work Power Energy Question with Solution

JEE Main 2015 (10 Apr Online)

Question

A block of mass m=0.1 kg is connected to a spring of unknown spring constant k. It is compressed to a distance x from its equilibrium position and released from rest. After approaching half the distance x2 from the equilibrium position, it hits another block and comes to rest momentarily, while the other block moves with velocity  3 m s-1. The total initial energy of the spring is:

Choose an option

Show full solutionCorrect option: A
Correct answer
A0.6 J

Step-by-step explanation

By mechanical energy conservation between compression positions x and x2

12kx2=12kx22+12mv2

12kx2-12kx24=12mv2

12kx234=12mv2

v=3kx24m=3kmx2

On collision with a block at rest

∵ Velocities are exchanged ⇒ elastic collision between identical masses.

∴v=3=3kmx2

⇒6=3km  x

⇒x=6m3k

∴ The initial energy of the spring is

U=12k x2=12k×36m3k=6m

U=6×0.1=0.6 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 2015, 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.