A spring of spring constant 100 N/m is stretched by 0.1 m (within the limit of proportionality). The elastic potential energy stored is:

  1. 1 J
  2. 5 J
  3. 0.25 J
  4. 0.5 J
Answer

D) 0.5 J

Ep = 12kx^2 = 12(100)(0.1)^2 = 0.5 J. SLO P-11-B-28 · Application

Physics MCQ · More Deformation of Solids MCQs · Practise