Question

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A spring is stretched by 60cm which transfers 18 J of energy to its elastic potential energy store. What is the spring constant?

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Answer 1

Potential energy of spring is $U = \frac{k x^2}{2}$, where $k$ is spring constant, and $x$ is distance stretched.

From the last formula, $k = \frac{2 U}{x^2} = \frac{2 \cdot 18 J}{(0.6m)^2} = 100 \frac{N}{m}$.