A spring obeying Hooke's law extends 2 cm under a final force of 40 N. The work done in stretching it is:

  1. 0.4 J
  2. 0.8 J
  3. 0.2 J
  4. 40 J
Answer

A) 0.4 J

Work = area of the triangle = 12Fe = 12(40)(0.02) = 0.4 J. SLO P-11-B-27 · Application

Physics MCQ · More Deformation of Solids MCQs · Practise