A wire is hung vertically from one of its ends. A load of 200 N is attached to the free (lower) end, which stretches the wire by 1 mm. What is the elastic potential energy stored in the stretched wire?

0.2 J
10 J
20 J
0.1 J

Correct answer: 0.1 J

Solution

The elastic potential energy stored equals (1/2)*F*x = (1/2)*200*0.001 = 0.1 J, because the wire is stretched quasi-statically and the average force during stretching is F/2.

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