A spring of constant 400 N/m stores 2 J of elastic energy. Its extension is:

  1. 0.01 m
  2. 0.2 m
  3. 0.1 m
  4. 0.05 m
Answer

C) 0.1 m

x = 2Ek = 2(2)400 = 0.1 m. SLO P-11-B-28 · Application

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