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A force of 520 n keeps a certain spring stretched 0.20 m. (a) what is the potential energy of the spring when it is stretched 0.20 m? (b) what is its potential energy when it is compressed 5.0 cm? (young, 7.15)

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Final answer:

The question involves calculating elastic potential energy of a spring when stretched and compressed, using the formula U = 1/2 kx², where U is the potential energy, k is the spring constant, and x is the displacement. Without the spring constant, a hypothetical example was given assuming k as 520 N/m, resulting in 10.4 J when stretched by 0.20 m and 0.65 J when compressed by 0.05 m.

Step-by-step explanation:

The question concerns elastic potential energy of a spring in two different scenarios: stretched and compressed states. To find the potential energy stored in a spring, we can use the formula for elastic potential energy, which is U = ½ kx², where U is the elastic potential energy, k is the spring constant, and x is the displacement of the spring from its equilibrium position. Unfortunately, the spring constant is not provided in the student's question, but for a hypothetical spring with a constant of 520 N/m (assuming the given force applied corresponds to the spring constant), the potential energy when it is stretched by 0.20 m would be U = ½ × 520 N/m × (0.20 m)² = 10.4 J. A similar calculation can be made for when the spring is compressed by 5.0 cm (0.05 m), resulting in U = ½ × 520 N/m × (0.05 m)² = 0.65 J.

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