A spring has a spring stiffness constant, k, of 440 N/m. How much must the spring be stretched to store 25 J of potential energy?

Question

Physics

Izaiah Andrade

27 September, 22:52

A spring has a spring stiffness constant, k, of 440 N/m. How much must the spring be stretched to store 25 J of potential energy?

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Answers (1)

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Gary Spencer · Answer 1

E p = 25 J, k = 440 N/m

E p = 1/2 k x²

25 J = 1/2 · 440 N/m x²

25 J = 220 N/m · x²

x² = 25 : 220 = 0.113636

x = √0.113636

x = 0.337 m

The string must be stretched 0.337 m.