I have seen the total energy of a system, $E$, given in two forms: $$E = K + U$$ where $K$ is the kinetic energy and $U$ is the potential energy, as well as $$E = K + U + I$$ where $I$ is the internal energy.
Is it correct to always view the internal energy as part of the potential energy? Or is this simply a matter of modeling?
Is there any way to distinguish when a non-kinetic energy is internal or potential?
Is it correct to always view the internal energy as part of the potential energy?
Yes, but only the potential energy associated with internal molecule forces, as opposed to the potential energy of the system as a whole with respect to an external frame of reference (which is the mechanical potential energy of the system, $U$ in your equation). The other part of the internal energy is the kinetic energy of the molecules of the system.
Is there any way to distinguish when a non-kinetic energy is internal or potential?
Yes. Non kinetic energy is potential energy. The question is whether the potential energy is external to the system (part of $U$ in your equation) or internal to the system (part of $I$ in your equatio).
For example, consider a ball and the Earth (the system). The ball moving through air. The gravitational potential energy associated with the ball is respect to a specific frame of reference on earth external to the system (typically the ground is chosen). On the other hand there is molecular potential energy associated with intermolecular forces of the ball which is internal to the system.
Hope this helps.
This is a serious edit to rectify my mistake of not explaining things correctly or more precisely
I must rewrite this answer again as I misdirected the OP to understand only one side of The story, Thanks to @BobD, his answer is to the point and precise, I will try again to be as precise technically as possible for me
I must be clear as already mentioned in BobD's answer that any non-kinetic energy can be generally called Potential energy except in some cases, see https://physics.stackexchange.com/a/245189/283030
For classical mechanics, Now, When you say internal energy, you really should ask internal to what?
Let us consider An example, You are observing a ball translationally falling(Not rotating) from a height on earth
When You only observe The ball, Internal energy of ball
$I_{ball}=P_{\mathbb{microscopic\ interaction\ between\ atoms \ of \ ball}}+K_{\mathbb{microscopic \ w.r.t \ Centre \ of \ mass \ of \ ball}}\tag 1$
When You only observe The Earth, Internal energy of earth $I_{earth}=P_{\mathbb{microscopic\ interaction\ between\ atoms \ of \ earth}}+K_{\mathbb{microscopic \ of \ earth \ w.r.t \ Centre \ of \ mass \ of \ earth}}\tag 2$
where $P$ and $K$ are potential and kinetic energies at microscopic levels,The microscopic interactions includes all other interactions like electrostatic and gravitational etc.
When you observe complete "ball-earth" system from outside, The internal energy
$$I= P_{\mathbb{microscopic\ interaction\ between\ atoms \ of \ ball}}+P_{\mathbb{microscopic\ interaction\ between\ atoms \ of \ earth}}+P_{\mathbb{microscopic\ interaction\ between\ atoms \ of \ ball \ and \ earth}}+K_{\mathbb{microscopic \ of \ earth \ w.r.t \ Centre \ of \ mass \ of \ earth}}+K_{\mathbb{microscopic \ of \ ball \ w.r.t \ Centre \ of \ mass \ of \ ball}} \tag 3$$
Now keeping this in mind
You can now write $$E=K_{\ Centre \ of \ mass \ of \ ball }+K_{\ Centre \ of \ mass \ of \ Earth }+I$$
which can be re-written using (3) as
$$E=K_{\ Centre \ of \ mass \ of \ ball }+K_{\ Centre \ of \ mass \ of \ Earth }+K_{\mathbb{microscopic \ of \ earth \ w.r.t \ Centre \ of \ mass \ of \ earth}}+K_{\mathbb{microscopic \ of \ ball \ w.r.t \ Centre \ of \ mass \ of \ ball}}+P_{\mathbb{microscopic\ interaction\ between\ atoms \ of \ ball}}+P_{\mathbb{microscopic\ interaction\ between\ atoms \ of \ earth}}+P_{\mathbb{microscopic\ interaction\ between\ atoms \ of \ ball \ and \ earth}}$$
This can further be reduced to
$$E=K_{\ Centre \ of \ mass \ of \ ball }+K_{\ Centre \ of \ mass \ of \ Earth }+I_{ball}+I_{earth}+P_{\mathbb{microscopic\ interaction\ between\ atoms \ of \ ball \ and \ earth}}$$
Now Let $P_{\mathbb{microscopic\ interaction\ between\ atoms \ of \ ball \ and \ earth}}=P_{EB}$ includes gravitational and non gravitational interactions$
$P_{EB}=P_{EB \ gravitational}+P_{EB \ non-gravitational}$
This $P_{EB \ gravitational}$ is called gravitational potential energy $U$ for this case
the final equation becomes
$$E=K_{\ Centre \ of \ mass \ of \ ball }+U+I_{ball}+I_{earth}+P_{EB \ non-gravitational}+K_{\ Centre \ of \ mass \ of \ Earth }$$
For change analysis, this becomes
$$\Delta E=\Delta K_{\ Centre \ of \ mass \ of \ ball }+\Delta U+\Delta I_{ball}+\Delta I_{earth}+\Delta P_{EB \ non-gravitational}+\Delta K_{\ Centre \ of \ mass \ of \ Earth }\tag 4$$
Now in most of mechanics problem
The term
$$I_{earth}+P_{EB \ non-gravitational}+K_{\ Centre \ of \ mass \ of \ Earth }$$ is assumed to be constant
hence the term $$\Delta I_{earth}+\Delta P_{EB \ non-gravitational}+\Delta K_{\ Centre \ of \ mass \ of \ Earth }=0$$
which gives out the final result
$$\Delta E=\Delta K_{\ Centre \ of \ mass \ of \ ball }+\Delta U+\Delta I_{ball}\tag 5$$
Here note that $\Delta I_{ball}$ cannot be solely included in potential energy category, Because it includes both $\Delta P_{\mathbb{microscopic\ interaction\ between\ atoms \ of \ ball}}$ and $\Delta K_{\mathbb{microscopic \ w.r.t \ Centre \ of \ mass \ of \ ball}}$
The change in internal energy of ball can be done using changing its shape, which changes the potential part of Internal energy, while it can also be changes by heating, which changes kinetic as well as potential part of Internal energy
Which answers you question..
Note:In The complete analysis of kinetic energy, I have used $K_{total}=K_{C.O.M}+K_{w.r.t \ C.O.M}$
In general, Energy is not as well-defined as it may seem. In thermodynamics, we commonly attribute a certain type of energy as the "Internal Energy" of a system, which we don't bother to categorize; it's simply the energy of the system. However, in other areas of physics, we rarely encounter this term.
For the most classical assumptions, potential energy is a conservative field $U(\textbf{r})$ associated with a force $\textbf{F}$ such that $\textbf{F}=-\nabla U$. Nothing more than that, and this is how you demonstrate that the quantity $K+U$ must remain constant during a possible motion (simply integrate Newton's second law over time). In this sense, Internal Energy would be any negligible energy in the analysis of the system; it is confined and unchanging within the system and can thus be disregarded. It's interesting to note that $K+U$ is constant, not $K+U+I$; after all, if $I$ itself is constant, this additional information is irrelevant. Even in philosophical terms, Potential Energy is that which has the potential to become motion (in a very Aristotelian view, motion is the act of potentiality as such), meaning it can transform into kinetic energy through some process (for example, a ball you drop in the air has the potential to attain high kinetic energy when nearing the ground).
Returning to Thermodynamics, you'll see internal energy playing a role when transitioning one statistical state to another, such as heating your gas sample or changing its magnetic properties. In a sense, you can think of it as the kinetic energy of electrons in the gas atoms, but this analogy doesn't work well and there's no need to think of it that way. You write $K+U+I$ when you want to clearly separate $K$ (particle motion energy), $U$ (energy from a "classical" force), and $I$ (intrinsic particle energy, determining thermodynamic state characteristics).
Internal energy is typically referred to the energy that is stored inside an object or contained. Potential energy or what I would consider kinetic energy, is what that object can produce. For example you have a motor with a capacitor that has stored internal energy. The potential energy that motor has could be based on the force or torque of the spindle on the end of the motor that would turn a gear. So therefore when you need to know what the potential energy of a motor is to turn a specific gear, you would not include the energy that the motor itself uses to turn itself that it holds internally. You would only use the formula based on the force output at the end of the spindle to turn the gears. I hope that makes sense and I didn't confuse things worse for you LOL.
