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A wire 4 m long and 0.3 mm in diameter is stretched by a force of 800 gram-weight (take g = 9.8 m/s²). If the resulting extension in its length is 1.5 mm, calculate the elastic potential energy stored in the wire.

1. 5.88 * 10⁻³ J
2. 1.18 * 10⁻² J
3. 2.94 * 10⁻³ J
4. 11.76 * 10⁻³ J

Correct answer: 5.88 * 10⁻³ J

Solution

The restoring force grows linearly from 0 to F as the wire stretches, so the work stored is the area of the force-extension triangle: U = (1/2)*F*x. With F = 0.800 kg * 9.8 = 7.84 N and x = 1.5 mm = 1.5*10⁻³ m, U = 0.5 * 7.84 * 1.5*10⁻³ = 5.88*10⁻³ J. The wire's length and diameter are not needed once the force and extension are known.

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