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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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    I'm confused. Potential energy is related to work done by the conservative force associated with that potential energy. What is the point in talking about the work done by other forces "against" this force? – BioPhysicist Jan 25 '20 at 05:35
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    why is there any change in potential energy at all? Relative to what? Can you explain what scenario you are actually considering here? – BioPhysicist Jan 25 '20 at 05:35
  • Both attractive and repulsive forces are conservative. When we're stretching a spring, we're doing work against a conservative force right? One that arises from these forces? But displaying a molecule away from another does work against the attractive force which gets stored up as potential energy while it does work in the direction of the repulsive forces resulting in a loss of potential energy. Opposite when the molecules are brought close together – lee_dong-eun Jan 25 '20 at 06:54
  • Have you read the question in the link and the related discussion? That would make it clearer – lee_dong-eun Jan 25 '20 at 08:36
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    I have read the question and link. So you are confused about two conservative forces acting on the molecules at once, an attractive one and a repulsive one? – BioPhysicist Jan 25 '20 at 13:08
  • yes, i am concerned that by displacing a molecule from equilibrium, we do work against one force and work in the direction of the other simultaneously, so potential energy of the molecule should be zero as both are conservative forces. Obviously, I am wrong, but I cannot understand how and why. – lee_dong-eun Jan 25 '20 at 13:26
  • One thing that might be worth noting is that helical coil springs are really just a long bar that twists in opposite directions when it is extended or compressed. https://physics.stackexchange.com/a/355784/127931 – JMac Jan 25 '20 at 20:27
  • Have you thought about what happens to the gravitational and elastic potential energies of a spring-mass system hanging vertically in a planetary gravitational field? What is each and the total when the system hangs in equilibrium? What happens to each and the total when the system oscillates? Write the equations and plot some graphs. – Bill N Jan 25 '20 at 20:42

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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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  • Thank you, this was exactly what was missing. The point about not considering both attractive and repulsive forces at the same time. – lee_dong-eun Jan 26 '20 at 14:11