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A cube of soft rubber with side 5 cm has one face fixed; the opposite face is sheared parallel to the fixed face through an angle of 0.01 rad by a tangential force of 40 N. Calculate the work done by the external agent and the elastic energy stored.

1. 1 * 10⁻² J
2. 2 * 10⁻² J
3. 1 * 10⁻³ J
4. 5 * 10⁻³ J

Correct answer: 1 * 10⁻² J

Solution

The top face shifts by x = (0.01 rad)*(0.05 m) = 5e-4 m; the elastic work (and stored energy) is (1/2)*F*x = (1/2)*40*5e-4 = 1e-2 J.

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