If x is doubled in a spring with potential energy U_s = (1/2) k x^2, how does U_s change?

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Multiple Choice

If x is doubled in a spring with potential energy U_s = (1/2) k x^2, how does U_s change?

Explanation:
For a spring, the potential energy scales with the square of the displacement from equilibrium. If you double the displacement, you plug into U_s = (1/2) k x^2: U_s' = (1/2) k (2x)^2 = (1/2) k · 4x^2 = 4 · (1/2) k x^2 = 4 U_s. So the potential energy becomes four times larger. The squared dependence means doubling x fourfolds the energy, not just doubling it or keeping it the same.

For a spring, the potential energy scales with the square of the displacement from equilibrium. If you double the displacement, you plug into U_s = (1/2) k x^2:

U_s' = (1/2) k (2x)^2 = (1/2) k · 4x^2 = 4 · (1/2) k x^2 = 4 U_s.

So the potential energy becomes four times larger. The squared dependence means doubling x fourfolds the energy, not just doubling it or keeping it the same.

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