A spring has a force constant of 50000 N/m. How far must it be stretched for its poten-
tial energy to be 22 J? Answer in units of m
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To determine the displacement of the spring, we can use the formula for potential energy of a spring:
PE = (0.5) * k * x^2
Where:
PE is the potential energy of the spring (22 J in this case)
k is the force constant of the spring (50000 N/m)
x is the displacement of the spring from its equilibrium position (what we want to find)
Rearranging the formula, we get:
x^2 = (2 * PE) / k
Now we can substitute the given values and solve for x:
x^2 = (2 * 22 J) / (50000 N/m)
x^2 = 0.00088 m^2
To find x, we take the square root of both sides:
x = √(0.00088 m^2)
x ≈ 0.03 m
Therefore, the spring must be stretched approximately 0.03 meters for its potential energy to be 22 J.