Answer :
Final answer:
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.
