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⚛️ Physics  ·  Work, Energy and Power  ·  NEET & JEE

A spring of constant 200 N/m is compressed by 10 cm. The elastic potential energy stored is:

Answer: 1 J.

  • A 0.5 J
  • B 1 J
  • C 2 J
  • D 4 J

Correct answer: B. 1 J

Explanation: PE = (1/2)kx² = 0.5·200·(0.1)² = 1 J.

Energy Conservation: KE ⇄ PE on a SwingA: PE maxKE = 0B: KE maxPE minC: PE maxKE = 0

As a pendulum swings, total mechanical energy (KE+PE) stays constant: at the highest points A and C, all energy is potential (zero speed); at the lowest point B, all of that energy has converted into kinetic energy (maximum speed).

Concept context

Work-energy theorem, potential energy, conservation of energy, collisions.

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