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A spring with spring constant 120 N/m is extended 35 m. What is the elastic potential energy stored?
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Elastic potential energy U = 1/2 k x^2.
U = 1/2 (120 N/m) (35 m)^2 = 60 × 1225 = 73,500 J.
U = 1/2 (120 N/m) (35 m)^2 = 60 × 1225 = 73,500 J.
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