What is the potential energy of a spring that is stretched 0.15 m from equilibrium and has a spring constant of 0.55 N/m?

asked Jun 28, 2021 13.7k views

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Step-by-step explanation:

EE = ½ kx²

EE = ½ (0.55 N/m) (0.15 m)²

EE = 0.62 J

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