Mass on a spring from equilibrium

Thread starter mancity
Start date Dec 22, 2023
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Equilibrium Formula Spring

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The discussion centers on the use of the formula Fs = 1/2 kx^2 in relation to a mass on a spring and its equilibrium position. The confusion arises over why x is taken as 0.1m instead of 0.05m, given that 0.05m is the displacement from equilibrium. The key point is that the formula measures energy based on the spring's extension from its unstretched position, not from equilibrium. Therefore, x should represent the total extension from the unstretched state. Understanding this distinction clarifies the application of the formula in calculating spring energy.

Dec 22, 2023

mancity

Homework Statement: An object with mass m is suspended at rest from a spring with a spring constant of 200 N/m. The length of the spring is 5.0 cm longer than its unstretched length L, as shown above. A person then exerts a force on the object and stretches the spring an additional 5.0 cm. What is the total energy stored in the spring at the new stretch length?

Relevant Equations: Fs=1/2kx^2

Can someone explain that, when using the formula (Fs=1/2 kx^2) why do we use x=0.1m instead of 0.05m? Seems like a simple concept but why isn't it 0.05m (since 0.05m from equilibrium). Thanks.

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Dec 22, 2023

nasu

Homework Helper

If you use the formula ##E=\frac{1}{2}kx^2##, the reference for energy (zero energy) is the position with the spring unstretched. So, you need the extension of the spring relative to the unstretched position.

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