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ExamsJEE AdvancedPhysics

A wire is hung vertically and held fixed at the top. A 200 N weight attached to its lower end stretches the wire by 1 mm. What is the total elastic potential energy stored in the wire?

  1. 0.2 J
  2. 10 J
  3. 20 J
  4. 0.1 J

Correct answer: 0.1 J

Solution

Elastic PE = (1/2)*F*deltaₗ = (1/2)*200*0.001 = 0.1 J. The factor of 1/2 arises because the force increases linearly from 0 to F as the wire stretches.

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