elastic potential energy

Question

Spring of k = 200 N/m compressed by 0.1 m stores energy:

RankGuru · Accepted Answer

The elastic potential energy ($U$) stored in a spring compressed or stretched by distance $x$ from its equilibrium position is:
$U = \frac{1}{2} k x^2$
Given:
- $k = 200 	ext{ N/m}$
- $x = 0.1 	ext{ m}$
$U = \frac{1}{2} (200) (0.1)^2 = 100 	imes 0.01 = 1 	ext{ Joule}$.

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