A Hooke's law spring is compressed 12.0 cm from equilibrium, and the potential energy stored is 36.0 J. What compression (as measured from equilibrium) would result in 100 J being stored in this case?Select one:a.The correct answer is not given.b.14.1 cmc.13.6 cmd.20.0 cm
Question
A Hooke's law spring is compressed 12.0 cm from equilibrium, and the potential energy stored is 36.0 J. What compression (as measured from equilibrium) would result in 100 J being stored in this case?Select one:a.The correct answer is not given.b.14.1 cmc.13.6 cmd.20.0 cm
Solution
To solve this problem, we can use the formula for potential energy stored in a spring, which is given by:
PE = 1/2 * k * x^2
where PE is the potential energy, k is the spring constant, and x is the displacement from equilibrium.
First, we can solve for the spring constant k using the given values:
36.0 J = 1/2 * k * (12.0 cm)^2
Rearranging for k gives:
k = 2 * 36.0 J / (12.0 cm)^2 = 50 J/cm^2
Now, we can use this value of k to find the displacement x that would result in 100 J of potential energy:
100 J = 1/2 * 50 J/cm^2 * x^2
Solving for x gives:
x = sqrt(2 * 100 J / 50 J/cm^2) = 20 cm
So, the compression that would result in 100 J being stored is 20.0 cm. Therefore, the correct answer is d. 20.0 cm.
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