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------------------------------------------------ A spring has a force constant of 20000 N/m. How far must it be stretched for its potential energy to be 10 J?

Answer in units of meters.

The spring with a force constant of 20000 N/m must be stretched approximately 0.03162 meters to have a potential energy of 10 J using the formula U = ½kx².

Explanation:

To determine how far a spring with a force constant of 20000 N/m must be stretched for its potential energy to be 10 J, we use the formula for the elastic potential energy of a spring: U = ½kx², where U is the potential energy, k is the spring constant, and x is the displacement from the equilibrium position.

By rearranging the formula to solve for x, we get x = √(2U/k). Plugging in the values, we have

x = √(2 * 10 J / 20000 N/m),

which simplifies to x = √(0.001 m²).

Finally, the square root of 0.001 m² is approximately 0.03162 m.

Therefore, the spring must be stretched approximately 0.03162 meters to have a potential energy of 10 J.