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So I was recently reading this answer on Stack Exchange, where someone enquired Why is the restoring force directly proportional to extension?

Now, the potential energy graph (in the first answer) rises on both sides of equilibrium position. Is this because on one side there is work done against the repulsive forces and on the other there is work done against the attractive forces? However, if there is work done against repulsive forces, leading to increase in potential energy there would also be work done in the direction of attractive forces, leading to a decrease in potential energy. Are these forces equal and opposite? If so, why is there any change in potential energy at all? If not, how can this graph be used generally?

sammy gerbil
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lee_dong-eun
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1 Answers1

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The graphs in the question you link to deal with the net interaction between the particles. Therefore, there really is only one force $F(r)$ to consider. If you displace the system from equilibrium, you only need to consider net attraction or only consider net repulsion. You don't need to consider both at the same time.

In general, when thinking about work done by a force you are doing an integral over some path $\int \mathbf F\cdot\text d\mathbf r$. What you are proposing seems to be considering the force at points not on the path of interest for the integral. For example, it would be like asking if we should consider some function $f(x)$ on the interval $[-5,0]$ for the integral $\int_0^5f(x)\,\text dx$

BioPhysicist
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