Spring's Potential Energy expression

AI Thread Summary
The discussion focuses on deriving the potential energy expression for a spring with a non-linear force defined as Fs = Ass + Bs, where s is the compression distance. The student initially attempted to relate the force to potential energy using a modified spring constant but was guided to use the integral of force to find the work done, which equals the change in potential energy. The correct final expression for potential energy was identified as (1/3 * As^3) + (1/3 * Bs^2). Additionally, there was a brief discussion about the appropriate categorization of the homework question, with clarification that it fits better in an introductory physics context. Understanding the relationship between work and potential energy through integration was emphasized as a key concept.
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Homework Statement


In an experiment, a student wishes to use a spring to accelerate a cart along a horizontal level track. The spring is attached to the left end of the track and produces a non-linear force of magnitude Fs = Ass + Bs, where s is the distance the spring is compressed in meters. A measuring tape, marked in centimeters, is attached to the side of the track.

Derive an expression for the potential energy ∪ as a function of the compression s. Express your answer in terms of A, B, s, and fundamental constants as appropriate.

Homework Equations


Us = 1/2 * k * (Δx)s (changed x to s in this problem)
Fs = -k * Δx (changed x to s in this problem)
ΔE = ∫F * dr

The Attempt at a Solution


So I know that we need to find an expression for potential energy and since they said to included: A, B, s. I wanted a way to include Fs into potential energy. So I set the formula for Fs equal to their version of Fs (so, (-k* Δx) = ( Ass + Bs)) and got k = (Ass + Bs) / Δs. And using this k, I plugged it into Us and got ((Ass + Bs) * Δs) / 2.

However, my teacher showed us that the way you were supposed to do it was using the formula for ΔE using the integral. But why would the ΔE be the same as ∪s? The final answer was (1/3 * As3 ) + (1/3 * Bs2).

Thanks so much for the help! (Additional question about forum, would this count as a intro physics homework or advanced? I posted in advanced but looked at the type of questions and realized that I should've probably posted here. But I want to make sure I place it in the right spot in the future.
Second Note: Saw that the other one was moved, so this is right spot. Sorry for that, promise that it won't happen again.)
 
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So you have a non linear force from the spring. Integrating it with respect to r will get you the work it does which in this case will also be equal to the change in potential energy. Think about conservation of energy. You start with potential, subtract the work done by the spring, and end up with less potential. Thus potential equals the work it can do. ##U_i-W=U_f## goes to ##U_f-U_i=-W##
 
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Ohh, wow this make a lot of sense now. Thank You @TJGilb !
 

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