Potential Energy Homework: Solving F(r) = -k(r-a)

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SUMMARY

The discussion focuses on deriving the potential energy function U(r) from the central force equation F(r) = -k(r-a), where 'a' represents the natural length of the spring and 'k' is the spring constant. The relationship -dU/dr = F(r) leads to the integration resulting in U(r) = (kr²/2) - kar. The user expresses confusion regarding the parabolic shape of the graph, expecting an asymptotic function instead, but clarifies that both states (r > a and r < a) yield parabolic potential energy with the vertex at r = a.

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oldspice1212
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Homework Statement


The net force on the mass is the central force F(r) = -k(r-a). Find the potential energy U(r).
a is the spring of natural length and k is spring constant k.

Homework Equations


-dU/dx = F(x)

The Attempt at a Solution


F(r) = - \frac{ dU }{ dr } implies -k(r-a) = - \frac{ dU }{ dr }
So I did this and then integrating both sides I got
\frac{ kr^2 }{ 2} - kar = U(r) but this doesn't make much sense when I graph it out, as it gives me a parabola, and I would be expecting a asymptotic type function?
 
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Well think about what happens when r > a and r < a. Both states have potential energy and would therefore be parabolic with the vertex at r = a

P. S. I can't see your pics because I'm on my phone
 
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I didn't upload any pics, was just making sure, and yes that makes sense, thank you :)
 
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oldspice1212 said:
I didn't upload any pics, was just making sure, and yes that makes sense, thank you :)
 
Last edited:
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Yes, that's must easier to see! Thank you so much!
 

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