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A spring has a spring constant of 450 N/m. How much must this spring be stretched to store 49 J of potential energy?
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4 votes
A spring has a spring constant of 450 N/m. How much must this spring be stretched to store 49 J of potential energy?

User Eumcoz
by
3.4k points

1 Answer

5 votes

Answer:

W = 1/2 K x^2

x^2 = 2 * W / K = 2 * 49 J / (N/m) = .218 / m^2

x = .467 m

User Vadym Tyemirov
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3.7k points
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