Internal forces converting kinetic into potential energy (vice versa)

[fog37]

Hello,
Trivial question: a system is isolated and all its internal forces are conservative. Because of Newton's 3rd law, all internal forces are pairwise and the net internal force is always zero (regardless of the forces being conservative or not) hence the system's total momentum is conserved. If all the internal forces are conservative, then also the system's mechanical energy is conserved. But it is possible for some kinetic energy to convert into potential energy and vice versa. How do the internal conservative forces manage to do work to transform kinetic energy into potential energy and vice versa? Shouldn't the net work done by the internal forces be zero?

Example: Isolated system = Earth + apple.
$F_{Ea}=$ force of Earth on apple.
$F_{aE}=$ force of apple on Earth $=-F_{Ea}$.
$F_{net}=0$.
$E_{mech_{total}}=KE_{tot}+U_{g_{tot}}=constant$
Net work $W_{net}=F_{net}\dot \Delta s=0$. The work done by $F_{Ea}$ is equal and opposite to the work done by $F_{aE}$.

Thanks!

 

[Chestermiller]

Summary:: Internal conservative forces

Hello,
Trivial question: a system is isolated and all its internal forces are conservative. Because of Newton's 3rd law, all internal forces are pairwise and the net internal force is always zero (regardless of the forces being conservative or not) hence the system's total momentum is conserved. If all the internal forces are conservative, then also the system's mechanical energy is conserved. But it is possible for some kinetic energy to convert into potential energy and vice versa. How do the internal conservative forces manage to do work to transform kinetic energy into potential energy and vice versa? Shouldn't the net work done by the internal forces be zero?

Example: Isolated system = Earth + apple.
$F_{Ea}=$ force of Earth on apple.
$F_{aE}=$ force of apple on Earth $=-F_{Ea}$.
$F_{net}=0$.
$E_{mech_{total}}=KE_{tot}+U_{g_{tot}}=constant$
Net work $W_{net}=F_{net}\dot \Delta s=0$. The work done by $F_{Ea}$ is equal and opposite to the work done by $F_{aE}$.

Thanks!

To move them apart, positive work has to be done on either (or both). The force you have to apply to either is in the same direction as its displacement.

 

[kuruman]

In the example you mentioned, if you consider the Earth + apple as your system, the mechanical energy of that system is $ME=K+U$. If the apple and the Earth move closer together $K$ increases at the expense of $U$ but the sum stays constant because there is no force external to the system that does work. The work $W$ is work done by external forces on the system. In this case if the Earth and apple are isolated, no work is done on this system. And yes, the work done by the internal forces is zero for the reasons you mentioned. In that case, the relevant equation is $$\Delta K_{\text{Earth}}+\Delta K_{\text{apple}}+\Delta U_{\text{Earth-apple}}=0.$$When you talk about the work done on the two components separately, you are changing the system to two separate systems on which external forces do work. Then you have two equations, $$\Delta K_{\text{Earth}}=W_{\text{by apple on Earth}}~~\text{and}~~\Delta K_{\text{apple}}=W_{\text{by Earth on apple}}$$By comparing with the previous equation, you see that $$\Delta U_{\text{Earth-apple}}=-(W_{\text{by apple on Earth}}+W_{\text{by Earth on apple}})$$which is the definition of change in potential energy between the two. See how it works?

The moral of the story here is that you have to be clear about what system you are talking about when you do.

 

[A.T.]

Net work $W_{net}=F_{net}\dot \Delta s=0$.

This looks like work done by external forces, not by the internal forces you are asking about.

 

[nasu]

Net work $W_{net}=F_{net}\dot \Delta s=0$. The work done by $F_{Ea}$ is equal and opposite to the work done by $F_{aE}$.

Thanks!

The forces are equal but the displacements are not, in general. For this reason the net work is NOT equal to some net force time some displacement, in general. Only for rigid systems thus may be true. So in general, the work of internal forces can change the kinetic energy. If the internal forces are conservative the change in KE is equal to the change in PE with opposite sign. Even for your example, the Earth did not move by the same distance as the apple does. So the net work of internal forces is not zero. The KE increases (net work is positive) and PE decreases. The displacement of the Earth is so small that usually is neglected.

 

[fog37]

Thank you everyone. This is exactly the discussion I was looking for. Just to make sure I am on your same page (sorry for the long post. Hopefully it is useful to others beside me), I am referring to a perfectly isolated system:

a) the net external mechanical work $W_{net\ by\ external \ forces}=0$
b) no energy of any kind can enter or leave the system through its boundary via heat, sound, radiation

As far as external forces go, it would seem that ok for some of of the external forces to be conservative and some to be nonconservative but I would argue that an external force cannot be of conservative type because that would imply a potential energy between a member of the system and an entity outside the system. Is that an incorrect?

In my scenario of a closed system, the only forces applied to the system are internal forces. Some of internal forces can be nonconservative (ex: frictional forces) and some conservative. In my case, all the internal forces are conservative. It remains that the net internal, conservative force $F_{net\ due \ to \ all \ internal \ conservative \ forces}=0$ as well as $F_{net\ due \ to \ all \ internal \ nonconservative \ forces}=0$ in virtue of Newton's 3rd law hence $F_{net \ all \ internal \ forces}=0$.

However, as nasu points out, the work done by each individual conservative force, even if the forces are equal in magnitude and opposite in direction, can be different because the displacements can be different depending on the mass of the system's member.

In the case of the system "Apple-Earth", the gravitational interaction implies the existence of the two forces $F_{Apple\ due \ to\ Earth}$ and $F_{Earth\ due \ to \ Apple}$. When the apple at rest is drop from a certain height, both forces do positive work $W_{F_{Apple \ due \ to \ Earth}}= \Delta KE_{Apple} >0$ and $W_{F_{Earth \ due \ to \ Apple}}= \Delta KE_{Earth }>0$ but
$$ W_{F_{Apple \ due \to \ Earth}} > W_{F_{Earth \ due \ to \ Apple}}$$ because $M_{Earth}>>0$ with $\Delta KE_{Earth}$ is very small. Overall, $$\Delta KE_{total}=\Delta KE_{Apple}+\Delta KE_{Earth} >0$$

The total positive change $\Delta KE_{total}= - \Delta U_{Earth \ -Apple} >0$ leading to $$\Delta E_{mech}=\Delta KE_{total} + \Delta U_{Earth \ -Apple} = 0$$
Great!

SCENARIO 2: Now a single nonzero external force $F_1$ acts on the system (specifically on the Apple) producing nonzero work $W_{net \ due \ to \ F_1}= \Delta KE_{total} \neq 0$. The system's mechanical energy is now changed:

$$\Delta E_{mech}=\Delta KE_{total} + \Delta U_{Earth \ Apple} \neq 0$$
What happens to the potential energy $U_{Earth \ -Apple}$ while or after $F_1$ is acting? I believe $$\Delta U_{Earth \ -Apple} \neq 0$$
because the external force $F_1$ changes the configuration "Earth-Apple" increasing or decreasing the mutual distance between Earth and the Apple.
So there is a $\Delta KE_{total}$# due to the external force and an implicit $\Delta U_{Earth \ -Apple}$ that can be either positive or negative. If $\Delta U_{Earth \ -Apple}>0$, it means that the internal conservative forces will do negative work on the system to reduce the system's kinetic energy... Eventually $$\Delta E_{mech}=\Delta KE_{total} - \Delta KE_{due \ to \ conservative \ forces }\neq 0$$
which means that the overall change in the system's mechanical energy has two terms. The first term (associated to the external force) increases it and the 2nd term (associated to the internal conservative forces) decreases it.

 

[kuruman]

If you have a two-component system, each component has its own kinetic energy but it takes two to have a meaningful potential energy. The issues you raise are addressed in the following insight
https://www.physicsforums.com/insights/is-mechanical-energy-conservation-free-of-ambiguity/

 

[fog37]

If you have a two-component system, each component has its own kinetic energy but it takes two to have a meaningful potential energy. The issues you raise are addressed in the following insight
https://www.physicsforums.com/insights/is-mechanical-energy-conservation-free-of-ambiguity/

Yes! Thank you for the link. I will read it carefully.

Indeed, the term $U_{Earth \ -Apple}$ is the potential energy possessed by the two-component system. If the system included only a single object, potential energy should not be mentioned.

In general, any type of potential energy involves two or more components that together have that potential energy in virtue of their reciprocal positions.

When an ideal (massless) compressed spring+ block form the system, there is not gravitational potential energy in the system since the spring is massless. However, there is elastic potential energy $U_{spring \ -block}$ stored in the "spring-block" system.

If the system was just the spring, the spring should not have elastic potential energy, I guess. But a compressed spring is in a different configuration than when it is uncompressed and when the spring is compressed it stores "internal" potential energy between its own parts.

kuruman, would you consider the system formed by just an ideal compressed spring as having elastic potential energy?

 

[jbriggs444]

kuruman, would you consider the system formed by just an ideal compressed spring as having elastic potential energy?

An ideal spring cannot exist in a compressed state in the absence of external forces. It has only one valid state: relaxed
Scratch that. Take a straight rod made of spring steel. Bend it into a hoop and bind the ends together. Idealize by making the steel massless. Ask whether the hoop has elastic potential energy.

 

[nasu]

SCENARIO 2: Now a single nonzero external force $F_1$ acts on the system (specifically on the Apple) producing nonzero work $W_{net \ due \ to \ F_1}= \Delta KE_{total} \neq 0$. The system's mechanical energy is now changed:

The change in total kinetic energy is equal to the total work done, by both internal and external forces. You are making it more complicated than it is. If you are confused about potential energy just use the work-energy theorem. You don't have to worry about conservative-nonconservative or internal-external forces.

 

[kuruman]

kuruman, would you consider the system formed by just an ideal compressed spring as having elastic potential energy?

Assuming that the spring is held compressed by external forces (see post #9 by @jbriggs444) , there will be a change in its internal energy relative to the uncompressed stable state. If you insist on considering a compressed spring as a single-component system, you have to understand that the potential energy $\frac{1}{2}kx^2$ stored in the spring is really due to the elastic deformation of the spring as the relative distances of all the possible pairs of particles that make up the spring change.