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So far what I had understood about potential energy is that it is defined for a system of particles (at least two particles) with forces acted by the particles on each other of same magnitude and opposite in direction as, $$-W_\mathrm{conservative} = ΔU.$$

For instance in case of gravitational potential energy, we only consider work done on small body as there is negligible work done on the earth. But in cases such as the following example, the concept is not completely clear. Consider a massless spring, the left side of which is attached to the wall, and there is a block attached to the right side of the spring. If we consider the block $\cup$ spring as our system, then the the net work done by internal forces would be zero. So what is increasing the potential energy of the system as we stretch the spring?

As per one of the answers to this question, it seems we need to consider the earth to be a part of the system and the spring just stores potential energy similar to the gravitational field while not really being the part of the system.

Is this reasoning correct? If not, please help me clarify the concept.

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  • $\begingroup$ >For example in case of gravitational potential energy ,we only consider work done on small body as there is negligible work done on the earth. This statement is not very clear for me $\endgroup$ – Buraian Mar 14 at 14:26
  • $\begingroup$ For example there is a small block placed on the ground and then it is raised by the height 'h', then work done by gravitational force is mgh but same gravitational force is acting on the earth but earth's mass is tremendously greater than block so there would be negligible work done by gravitational force on the earth , so mgh is the the only work done on the earth + block system ; that's what I mean ;Is it correct ? $\endgroup$ – Yash Agrawal Mar 14 at 14:36
  • $\begingroup$ I mean if you denote work as $ \vec{F} \cdot ds$ , then the earth is not really be pushed down by any distance so the work done on earth must be zero $\endgroup$ – Buraian Mar 14 at 14:37
  • $\begingroup$ that's what i am also saying $\endgroup$ – Yash Agrawal Mar 14 at 14:38
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If we consider block + spring as our system then the the net work done by internal forces would be zero , So what is increasing the potential energy of the system then ?

See this is the whole point behind the harmonic oscillator , you don't need any 'external' to drive the object to keep it's oscillating behaviour. You give the harmonic oscillator a little push and, due to the nature of the set up, the force law is such that that the body keeps moving without ever stopping (in the ideal case).

As per 'what' is increasing the potential energy, you can say that the internal forces of the system are responsible for it. The internal forces do no work, but it needn't be that they can't convert PE and KE back and forth.

I think it may help to learn the distinction between gravitational potential and gravitational potential energy see here

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  • $\begingroup$ if internal forces do no work , how potential energy of system can be increased ? "but it needn't be that they can't convert PE and KE back and forth." - this isn't very clear to me. $\endgroup$ – Yash Agrawal Mar 14 at 14:43
  • $\begingroup$ If KE of block increases then PE of spring decreases, hence you rise one then the other drops which makes the net work on system zero @YashAgrawal $\endgroup$ – Buraian Mar 14 at 14:48
  • $\begingroup$ Ok I understand your point. $\endgroup$ – Yash Agrawal Mar 14 at 18:26
  • $\begingroup$ But my real concern is that, potential energy is always defined for a pair of objects with the some mutual forces between them , So what is that mutual force pair in spring cases due to which potential energy arises in the system ? $\endgroup$ – Yash Agrawal Mar 14 at 18:30
  • $\begingroup$ You don't need a 'pair' to define potential energy, for the earth + object case, it is just a coincidence (or I don't know, maybe work of god) that it does. The potential energy is a property of the geometry of the system in the classical mech cases we deal with ( don't cite me on this) in the sense that it depends on how everything is put together geometrically $\endgroup$ – Buraian Mar 14 at 18:40
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You need to provide the potential energy by doing work on the spring-mass system to stretch or compress the spring.

If you stretch or compress the system by a distance x, the restoring force on the block by the spring will be $-k x$. Hence in order to stretch/compress the spring, the force that you apply has to be equal and opposite to this force and that will be $+k x$. If you pull the spring from the mean position = 0 to a position x, then the work done by you is $\dfrac{1}{2}$ k $x^2$. This work done gets stored as the potential energy and when you release it, the mass will start moving. At that point there is no longer any external force and the potential energy starts getting converted to kinetic energy. This conversion goes back and forth. The source of the energy however, is the external force.

The energy of the spring mass system that is equal to $\dfrac{1}{2} k A^2$, where A is the amplitude of the oscillation is thus provided from an external source. This can be done in many different ways. However you have to provide the system with kinetic or potential energy initially so that it has some energy. This can then convert between kinetic and potential energy.

The same situation (w.r.t. work done and potential energy) also occurs when you place a body at a height in a gravitational field. That work-energy phenomenon also follows the same logic. Work is done by an external force is lifting the object to the height and this is then stored as the gravitational potential energy ($mg \cdot \text{height}$).

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For the elastic potential, the wall plays the role of a big celestial body in the case of gravitational potential.

In both cases, one of the objects are supposed to be big enough to be taken as fixed. And the other moves according to the field.

The earth alternates periods of greater distance from the sun, (higher potential) and smaller kinetic energy, and others of smaller distances and greater kinetic energy.

The spring also oscillates with a trade-off between potential and kinetic energy. The elastic field $\frac{1}{2}k\Delta x^2$ of the spring can be compared to the gravitational field $-\frac{GM}{r}$.

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  • $\begingroup$ Sir , What you mean by "elastic field" ? Isn't it just spring ? $\endgroup$ – Yash Agrawal Mar 15 at 10:18
  • $\begingroup$ It is a field in the meaning that for each $\Delta x$ there is a value of potential associated. Let's say that the field is a property of the spring. $\endgroup$ – Claudio Saspinski Mar 15 at 12:59
  • $\begingroup$ Thanks this answer confirmed my reasoning on the question. $\endgroup$ – Yash Agrawal Mar 15 at 13:05
  • $\begingroup$ Can you please comment whether potential energy is defined for a single particle or not ? $\endgroup$ – Yash Agrawal Mar 15 at 13:07
  • $\begingroup$ It is defined for a particle in a field, where the potential energy is a function of the position. $\endgroup$ – Claudio Saspinski Mar 15 at 14:25
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I was able to clear my misconception , So I am just writing so anybody else could get helped ;

So first of all , I got to know that only "internal" conservative forces changes potential energy as: $$-W_{internal} = ΔU \tag{1}$$ Note:- I am considering non-conservative internal forces to be zero.

Note when we write Work energy theorem , the $W_{net}$ includes both $W_{internal}$ and $W_{external}$. $$W_{internal} + W_{external} = ΔKE$$

By using $(1)$ $$W_{external} = ΔU + ΔKE \tag{2}$$

If $W_{external}$ is zero then we say total energy of system is conserved.

Now let's consider spring as our system in which let $W_{internal}$ mean work done by internal particles of spring on each other due to elastic forces and the block applies the force $kx$ on the spring which is $W_{external}$. As spring is massless this implies $ΔKE$ = 0 ,So we could write work energy theorem for the spring as $$W_{internal} + W_{external} = 0$$ Using (1) $$W_{external} = ΔU_{spring}$$

By calculation , $W_{external}$ turn outs to be $kx^2/2$. $$ΔU_{spring} = kx^2/2$$

Now if I were to consider Block ∪ Spring as my system then: $$W_{block on the spring} + W_{spring on the block} = 0$$

Considering no external force on the block ; $$W_{internal} = ΔKE$$ By using (1) $$ ΔKE + ΔU = 0 $$

So my statement "If we consider the block ∪ spring as our system, then the the net work done by internal forces would be zero" was wrong as there are internal forces of spring which are responsible for potential energy !!

I would also like to clear out that in a field the source remains fixed , So potential energy of the "source + particle" system is wrongly attributed as potential energy of particle as source has $ΔKE = 0$.

Reference : - HC Verma : Concept of physics - Volume 1

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