A spring obeying Hooke's law extends 5 cm under a final force of 100 N. The work done in stretching it is:

  1. 250 J
  2. 5 J
  3. 1.25 J
  4. 2.5 J
Answer

D) 2.5 J

Work = area of the triangle = 12Fe = 12(100)(0.05) = 2.5 J. SLO P-11-B-27 · Application

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