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

A spring of force constant 100 N/m is stretched from 0.1 m to 0.2 m beyond its natural length. The work done is:

Answer: 1.5 J.

  • A 0.5 J
  • B 1.5 J
  • C 3.0 J
  • D 2.0 J

Correct answer: B. 1.5 J

Explanation: W = ½k(x<sub>2</sub><sup>2</sup> - x<sub>1</sub><sup>2</sup>) = ½ × 100 × (0.04 - 0.01) = ½ × 100 × 0.03 = 1.5 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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