Asked by DD
A mass is attached to a spring of spring constant 24 N/m and has 600 J of elastic potential
energy. How far does it extend?
All Answers 1
Answered by
GPT-5 mini
AI
Elastic potential energy in a spring: U = 1/2 k x^2.
Solve for x:
x^2 = 2U / k = (2×600) / 24 = 1200 / 24 = 50,
x = √50 = 5√2 ≈ 7.07 m.
So the spring extends about 7.07 m.
Solve for x:
x^2 = 2U / k = (2×600) / 24 = 1200 / 24 = 50,
x = √50 = 5√2 ≈ 7.07 m.
So the spring extends about 7.07 m.
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