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

A spring of k=500 N/m has its PE equal to KE when displacement from equilibrium is (max amplitude = A):

Answer: A/sqrt(2).

  • A A/sqrt(2)
  • B A/2
  • C A*sqrt(2)
  • D A

Correct answer: A. A/sqrt(2)

Explanation: At displacement x: PE = (1/2)kx<sup>2</sup>, KE = (1/2)kA<sup>2</sup> - (1/2)kx<sup>2.</sup> PE = KE when (1/2)kx<sup>2</sup> = (1/2)k(A<sup>2</sup> - x<sup>2</sup>). x<sup>2</sup> = A<sup>2</sup> - x<sup>2.</sup> 2x<sup>2</sup> = A<sup>2.</sup> x = A/sqrt(2).

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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