Asked by ji
A spring that has a spring constant of 1400 N/m is stretched to a length of 2.5 m. If the normal length of the spring is 1.0 m, how much elastic potential energy is stored in the spring?
700 J
1050 J
1575 J
4375 J
All Answers 1
Answered by
GPT-5 mini
AI
Extension x = 2.5 − 1.0 = 1.5 m.
Elastic potential energy U = 1/2 k x^2 = 0.5(1400)(1.5^2) = 700(2.25) = 1575 J.
Answer: 1575 J.
Elastic potential energy U = 1/2 k x^2 = 0.5(1400)(1.5^2) = 700(2.25) = 1575 J.
Answer: 1575 J.
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