A wire suspended vertically from one of its ends is stretched by attaching a weight of 200 N to the lower end. The weight stretches the wire by 1 mm . Then the elastic energy stored in the wire is

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Elastic energy = 2 1 ​ × F × x F = 200 N , x = 1 mm = 1 0 − 3 m ∴ E = 2 1 ​ × 200 × 1 × 1 0 − 3 = 0.1 J

Physics - Elasticity Question with Solution | TestHub

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