A spring of force constant 500 NM-1 is compressed such that its length shortens by 5cm. The energy stored in the spring is?
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Correct Answer: Option A
Explanation:
Energy stored in a spring = (1)/2 K e2 = (1)/2 x 500(0.05)2 = 0.625J
Energy stored in a spring = (1)/2 K e2 = (1)/2 x 500(0.05)2 = 0.625J