The potential energy in a spring is given by the formula:
\[ P = \frac{1}{2} k x^2 \]
To solve for \( k \), we can rearrange the equation:
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Start with the original equation: \[ P = \frac{1}{2} k x^2 \]
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Multiply both sides by 2 to eliminate the fraction: \[ 2P = k x^2 \]
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To isolate \( k \), divide both sides by \( x^2 \): \[ k = \frac{2P}{x^2} \]
So, the correct equation for \( k \) that Lupe should use is:
\[ k = \frac{2P}{x^2} \]
This corresponds to the option:
k = StartFraction 2 P Over x squared EndFraction.