# Consider having 2-D Cartesian coordinate xy-plane, a block connected to a spring

Braylon Lester 2022-07-19 Answered
Let's consider having 2-D Cartesian coordinate xy-plane, a block connected to a spring (placed horizontally on negative part of x-axis and having natural length L) sitting at the origin point with the spring at natural length.It is allowed to oscillate back and forth about the origin. From the equation that relates the work done by conservative force and the potential energy, $U\left({x}_{a}\right)-U\left({x}_{b}\right)=-{\int }_{{x}_{b}}^{{x}_{a}}{F}_{x}\text{d}x$. If we define the zero point at x=0, we can express its potential energy for $0, with
$U\left(x\right)-U\left(x=0\right)=-{\int }_{0}^{x}-kx\text{d}x⇒U\left(x\right)=\frac{1}{2}k{x}^{2}$
while for −L<x<0, we have (the restoring force switch sign as it now points to the positive x- direction):
$U\left(x\right)-U\left(x=0\right)=-{\int }_{0}^{x}kx\text{d}x⇒U\left(x\right)=-\frac{1}{2}k{x}^{2}$
My question: Is my idea and this approach for the energy of the system correct? Because I saw from some reference books, they simply say that it is just $\frac{1}{2}k{x}^{2}$ for both of the conditions.
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## Answers (1)

neobuzdanio
Answered 2022-07-20 Author has 13 answers
For $-L, a spring force of $+kx$ would point in the negative direction (since k>0 and x<0.) This would not be a restoring force, as it would point away from the origin.
On the other hand, a force of $F=-kx$ for all x points in the negative direction for positive x (F<0 and x>0), but points in the positive direction for negative x (F>0 and x<0). This is the correct function to model the force, and leads to a potential energy of $\frac{1}{2}k{x}^{2}$ for all x
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