A wire hanging vertically is stretched by 1 mm when a 20 N weight is hung from its lower end. How much energy does the wire store?
Correct answer C. 0.01 J
Explanation
Elastic energy stored is ½ × force × extension = ½ × 20 N × 0.001 m = 0.01 J, assuming the wire obeys Hooke's law.
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