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

A 2 kg body is dropped from a height of 5 m (g = 10 m/s²). Its kinetic energy just before striking the ground is:

Answer: 100 J.

  • A 50 J
  • B 100 J
  • C 20 J
  • D 200 J

Correct answer: B. 100 J

Explanation: KE = mgh = 2·10·5 = 100 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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