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A spring with a spring constant of 200 N/m stretches by 0.03 m. What is the potential energy of the spring?

User Nikshep
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1 Answer

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Answer: 0.09 J

Explanation: K = 200 N/m , 1/2 X 200 N/m X (0.03 M)^2 = 0.09 J

User Igorgue
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