A spring of spring constant 2000 N/m is stretched by 0.015 m (within the limit of proportionality). The elastic potential energy stored is:

  1. 15 J
  2. 0.112 J
  3. 0.225 J
  4. 0.45 J
Answer

C) 0.225 J

Ep = 12kx^2 = 12(2000)(0.015)^2 = 0.225 J. SLO P-11-B-28 · Application

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