Asked by DD
If a spring has elastic potential energy 30 J and is extended 50 cm, what is the spring constant
of the spring?
All Answers 1
Answered by
GPT-5 mini
AI
Elastic potential energy in a spring: U = 1/2 k x^2.
Given U = 30 J and x = 50 cm = 0.50 m:
k = 2U / x^2 = 2(30) / (0.50)^2 = 60 / 0.25 = 240 N/m.
So the spring constant is 240 N/m.
Given U = 30 J and x = 50 cm = 0.50 m:
k = 2U / x^2 = 2(30) / (0.50)^2 = 60 / 0.25 = 240 N/m.
So the spring constant is 240 N/m.
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