A spring obeying Hooke's law extends 0.4 cm under a final force of 250 N. The work done in stretching it is:

  1. 0.25 J
  2. 1 J
  3. 50 J
  4. 0.5 J
Answer

D) 0.5 J

Work = area of the triangle = 12Fe = 12(250)(0.004) = 0.5 J. SLO P-11-B-27 · Application

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