Is Mechanical Energy Conservation Free of Ambiguity?

[kuruman]

In 1 you describe the system as including part of the table. So that would mean that friction is an internal force. Was that intentional?

My comments above about where I thought it went wrong were all focused on the block as the system, so that the friction would be an external force.

Good catch. Statement 1 is unintentional, the diagram is correct. If I meant statement 1 then friction would not have been included in the FBD. In part II when I address the problem using the first law, I use the block and part of the surface for the system and I even have a separate figure. I must have been thinking of that when I wrote (or rewrote) statement 1 when I edited the document for the ##n##th time, ##0 \lt n \lt 244.## I think I'll take a break until tomorrow. Thanks again.

 

[Mister T]

I was referring to this specific case of the block being pulled at constant speed on the rough floor which I erroneously used as to illustrate that kinetic friction does not do work.

The issue here is what one means by the word "work". There is nothing wrong with creating a work-energy relation (##\Delta K=W##) and saying that work done by the friction force is included as (at least) part of ##W##, It is a perfectly valid dynamical relation, but it will not serve as the work ##W## that appears in the 1st Law of Thermodynamics. How could it? Conservation of Energy is one of the three great conservation laws of classical physics. The work-energy relation, in the way that I've referred to it above. is derived from Newton's Laws, but the Conservation of Energy also includes an additional consideration of the mechanical equivalent of heat, something that is not part of Newton's Laws of motion. You can derive Conservation of Momentum from Newton's Laws, but you need the additional consideration of the mechanical equivalent of heat to construct the law of Conservation of Energy.

On the other hand, you can restrict the validity of the work-energy relation to particle-like objects, that is, objects with no internal structure that are hence not capable of possessing internal energy, and come up with the same work-energy relation, but now you no longer are using the same definition of the word "work_. You do, however, have a way of defining ##W## that the same as the ##W## that appears in the 1st Law of Thermodynamics. Using this definition is not correct to say that work is done by a dissipative friction force, you instead can do things like set the product of the friction force and the displacement, and equate it to the loss of mechanical energy.

Since the early 1990's you see the two different ways of defining ##W## in different calculus-based college-level introductory physics textbooks.

 

[kuruman]

I will be updating the insight soon and I thank you once more for your help to improve it.

I intend to cut the entire section "Work done against friction" and paste what is shown below. I would appreciate any comments you anyone may have before I do so. Thank you all for your inputs.

I. Work done against friction
Consider the problem of a billiard ball of radius ##R## spinning in air with initial angular speed ##\omega_0##. The ball falls straight down and lands on a a pool table. At first the ball slips and rolls but eventually it starts rolling without slipping. Find the linear speed of the ball when it starts rolling without slipping.

We ask the question what the work energy theorem has to say in this case. The change in kinetic energy on the left hand side is $$\Delta K=\frac{1}{2}I_{cm}(\omega_f^2-\omega_0^2)+\frac{1}{2}mV_{cm}^2$$where ##I_{cm}## is the moment of inertia about the center of mass and ##V_{cm}=\omega_f R##. What about ##W_{net}## on the right hand side of the W-E theorem? Friction is the only horizontal force acting on the ball while the vertical forces cancel. Therefore friction is the net force and is responsible for the gain in kinetic energy of the CM. It seems logical to conclude that ##W_{net}=\vec f_k \cdot \Delta \vec s_{cm}## because what else is there to say?. This is exactly the kind of "logic" that Arons warns against. The ##\vec f_k \cdot \Delta \vec s_{cm}## term is responsible for increasing the kinetic energy of the center of mass. However, within the confines of the W-E theorem, there is no term to account for the decrease of the spin kinetic energy of the ball. What else there is to say is that Bridgman's sentries report that additional mechanical work ##\Delta W=\text{-}f_k R\Delta \theta## is converted by friction into heat on the periphery of the system boundary. Since ##\vec f_k## in this case is in the same direction as the velocity of the CM, one ought to write the right hand side as shown below, $$\Delta K=f_k \Delta s_{cm}-f_k R\Delta \theta.$$Note that when the ball rolls with slipping, (##\Delta s_{cm}<R\Delta \theta~##) the kinetic energy decreases; when it rolls without slipping, (##\Delta s_{cm}=R\Delta \theta~##), the kinetic energy does not change.

 

[anorlunda]

If you're going to use the ball example, then as a reader I would also want to hear about the case where the ball is initially spinning on an orthogonal axis to your example (like a spinning top on the floor). After being dropped on the table, the ball slows, then stops without moving the center of mass at all. All the kinetic energy is converted to heat at the contact patch.

The general case has some rotation in both axes.

Also in your example, you should explicitly say that the ball has slipping friction but no rolling friction. Real balls have both. When they roll without slipping they eventually stop.

 

[kuruman]

If you're going to use the ball example, then as a reader I would also want to hear about the case where the ball is initially spinning on an orthogonal axis to your example (like a spinning top on the floor). After being dropped on the table, the ball slows, then stops without moving the center of mass at all. All the kinetic energy is converted to heat at the contact patch.

All that is true, but the ball example is just a "for instance" to illustrate the shortcomings of the work-energy theorem. The case of the ball spinning around a vertical axis is the rotational analog of a block sliding across the floor and does not illustrate the shortcomings of the W-E theorem.

Also in your example, you should explicitly say that the ball has slipping friction but no rolling friction. Real balls have both. When they roll without slipping they eventually stop.

This point is well taken. I will make it clear that rolling friction is not included. Thank you for your suggestions.

 

[swampwiz]

At the very core of physics there are the fundamental forces, and kinetic & potential energy computations are exact, at least considering the quantum nature of stuff. But keeping track of every particle is of course ridiculously complex, so we seek to abstract the problem to greatly simply.

One such abstraction is the idea of mechanical energy, which deals with abstracted mechanics, with mechanical energy being macroscopic kinetic energy and abstracted potential energy such as gravitation, elastic deformation, and macroscopic electromagnetic energy. If we can model a system so that frictional loss is negligible, then we can use the conservation of mechanical energy.

Another abstraction is the idea of thermodynamic energy, with the idea that a compressible working material (typically a fluid) can expand at a certain pressure, thus yielding thermodynamic work, along with the idea of heat flow, and the notion of thermodynamic internal energy which is a measure of how much energy is stored microscopically in such a way that it cannot be realized as mechanical (or any other type of) energy, and such that it conforms to the model of entropy, which of course is, along with temperature, the all-important property in thermodynamics. Friction is defined by a loss of mechanical energy, and thermodynamics takes this into account as that frictional energy is treated as heat transferring into the system.

 

[anorlunda]

One such abstraction is the idea of mechanical energy,

That's a very bad way to say it. Total energy is conserved, not mechanical energy.

Mechanical energy (meaning kinetic energy plus potential energy) can be converted to/from thermal energy, electrical energy, chemical energy, nuclear energy, radiation energy and so on. In general, the individual kinds of energy are not conserved at all, only the total energy is conserved.

 

[swampwiz]

That's a very bad way to say it. Total energy is conserved, not mechanical energy.

Mechanical energy (meaning kinetic energy plus potential energy) can be converted to/from thermal energy, electrical energy, chemical energy, nuclear energy, radiation energy and so on. In general, the individual kinds of energy are not conserved at all, only the total energy is conserved.

Uh, that's why I had termed it an abstraction. The electrical energy that is being stored within a solid when it is stressed can be abstracted into the mechanical type of energy of elasticity. Or the electrical energy that is being generated by a generator can be thought of as a mechanical energy sink. It's applicability all depends on the abstraction,

 

[cianfa72]

That gives the contribution of the force of kinetic friction to the kinetic energy corresponding to the bulk motion of the center of mass of the object, yes. (i.e. to "net work" or "center of mass work").

Just to clarify myself (avoiding misunderstanding about namings): with respect to a 'compound system' (e.g. consisting of multiple particles or just multiple rigid bodies) is the 'net work' the same as 'center of mass work ? Said in other words: are they exactly the same just with two different names ?

 

[cianfa72]

Just to clarify to myself (avoiding misunderstanding about namings): with respect to a 'compound system' (e.g. consisting of multiple particles or just multiple rigid bodies) is the 'net work' the same as 'center of mass work ? Said in other words: are they exactly the same just with two different names ?

Any thought ?

 

[jbriggs444]

Just to clarify myself (avoiding misunderstanding about namings): with respect to a 'compound system' (e.g. consisting of multiple particles or just multiple rigid bodies) is the 'net work' the same as 'center of mass work ? Said in other words: are they exactly the same just with two different names ?

As I would use the terms, "net work" is computed as the sum of all external forces on a body multiplied by the motion of the center of mass and "center of mass work" is computed as a particular external force on a body multiplied by the motion of the center of mass. Multiplication in the sense of the vector dot product, of course.

Google can dig up all manner of sloppy definitions.

 

[Dale]

is the 'net work' the same as 'center of mass work ?

From what I have seen they are the same: both referring to the work energy theorem. They are not generally equal to the total thermodynamic work, which is the relevant one for the conservation of energy

Note that the way I use the term differs from the way that @jbriggs444 uses it. That means that different authors will use the term differently. So you need to check and see how a given author is using it, and not assume consistency

 

[cianfa72]

As I would use the terms, "net work" is computed as the sum of all external forces on a body multiplied by the motion of the center of mass and "center of mass work" is computed as a particular external force on a body multiplied by the motion of the center of mass.

Thus the difference between 'net work' and 'total thermodynamic work' -- taking into account that external forces alone are involved in both definitions-- is just a matter of points of application of external forces to be accounted for work calculation, I believe

 

[jbriggs444]

Thus the difference between 'net work' and 'total thermodynamic work' -- taking into account that external forces alone are involved in both definitions-- is just a matter of points of application of external forces to be accounted for work calculation, I believe

Yes, it is a matter of exactly what displacement you look at. The displacement of the center of mass or the displacement of the material at the point of application.

I suspect that @Dale might consider an electrical current flow as a contribution to total thermodynamic work without going to the bother of identifying a force and a point of application.

 

[Dale]

@kuruman hey, I just noticed the revisions. Good work! I like the “sentries” idea. I am reading the follow-ups now, but I think this revised version is solid

 

[atyy]

The main points of this insight are not correct. In the Atwood machine, the net force is conservative because the work done is path independent. In the gymnast example, the work energy theorem does not fail.

Jumping can be treated with the work energy theorem as follows:
https://openoregon.pressbooks.pub/bodyphysics/chapter/comparing-work-energy-and-energy-conservation/

 

[etotheipi]

In the gymnast example, the work energy theorem does not fail.

I think that is not quite correct. Really the work energy theorem only applies to rigid bodies (and this is also assuming the strong form of Newton III holds!). Since the theorem is always applicable to particles, we can find the total work by all forces on all particles in a general extended body via. a summation,$$W = \sum_i \int \vec{F}_i \cdot d\vec{x}_i = \sum_i \int (\vec{F}_i^{\text{ext}} + \vec{F}_i^{\text{int}}) \cdot d\vec{x}_i = \sum_i (W_i^{\text{int}} + W_i^{\text{ext}})$$here ##\vec{F}_i = \vec{F}_i^{\text{ext}} + \vec{F}_i^{\text{int}}## is a sum of both the internal forces and external forces on that particle. If we consider a rigid body, then we have the constraint that ##|\vec{x}_i - \vec{x}_j| = \text{constant}##, which means that$$|\vec{x}_i - \vec{x}_j|^2 = (\vec{x}_i - \vec{x}_j) \cdot (\vec{x}_i - \vec{x}_j) = \text{constant}$$ $$2(\vec{x}_i - \vec{x}_j) \cdot (\vec{v_i} - \vec{v}_j) = 0$$Also notice that, for any pair of particles, the total power of internal forces on each of them due to their mutual interaction (##\vec{F}_{ij} = -\vec{F}_{ji}##) is$$P = \vec{F}_{ij} \cdot \vec{v}_i + \vec{F}_{ji} \cdot \vec{v}_j = \vec{F}_{ij} \cdot (\vec{v}_i - \vec{v}_j)$$If the strong form of Newton III holds, then it is guaranteed that ##\vec{F}_{ij} = k(\vec{x}_i - \vec{x}_j)## where ##k## is some arbitrary scaling constant, in which case due to the aforementioned orthogonality, the power of these internal forces is zero. Hence the total work on the rigid body due to internal forces is zero, and the total work on the rigid body is solely due to external forces.

This is the important point; for deformable bodies, we often have no way of computing the work done by internal forces. Of course, hypothetically, if we could determine this number, then the work energy theorem would work fine. For rigid bodies, this internal work is exactly zero, so we can always use the work energy theorem.

This is what I believe @kuruman was explaining with the example of the gymnast. The normal contact force due to the floor does zero work, since the floor is fixed. The gravitational force in fact does negative work when the gymnast is jumping. However his/her change of kinetic energy is positive, which means there must have been positive work done by internal forces that exceeded in magnitude that done by gravity! It is of course easiest to use energy conservation, and attribute this to a transfer of chemical energy to kinetic energy.

Jumping can be treated with the work energy theorem as follows:
https://openoregon.pressbooks.pub/bodyphysics/chapter/comparing-work-energy-and-energy-conservation/

This is not the work energy theorem, this is another construct entirely. It deals with so-called centre-of-mass work (or what @Dale called 'net work' near the start of the thread), which is not real (thermodynamic) work in the sense of energy. It is derived from Newton's second law, and computes the integral of the scalar product of the force and the displacement of the centre-of-mass. That correctly gives the change in the kinetic energy of the centre-of-mass only:$$\begin{align*}

\left( \sum_i \vec{F}_i \right) = \vec{F} = m\vec{a}_{\text{cm}} \implies \int \vec{F} \cdot d\vec{x}_{\text{cm}} &= \int_{\vec{x}_1}^{\vec{x}_2} m \frac{d\vec{v}_{\text{cm}}}{dt} \cdot d\vec{x}_{\text{cm}} \\

& = \int_{t_1}^{t_2} m \frac{d\vec{x}_{\text{cm}}}{dt} \cdot \frac{d\vec{v}_{\text{cm}}}{dt} dt \\

& = \frac{1}{2} m \int_{t_1}^{t_2} \frac{d}{dt} \left( \vec{v}_{\text{cm}} \cdot \vec{v}_{\text{cm}} \right) dt \\

& = \Delta \left( \frac{1}{2}m v_{\text{cm}}^2 \right) = \Delta T_{\text{cm}}
\end{align*}$$We can imagine a cylinder rolling down a rough inclined plane. The work done by the force of friction is zero, since the friction force has zero power (the relative velocity at the point of contact is zero). Hence, the proper work energy theorem yields$$W_{\text{grav}} = \Delta T$$Here ##\Delta T = \Delta T_{\text{cm}} + \Delta T^*## is the total change in kinetic energy, which is the sum of the kinetic energy of the centre of mass and that in the frame of the centre of mass (König theorem). We can also use the centre-of-mass work concept that you linked, which would give$$\mathcal{W}_{\text{grav}} + \mathcal{W}_{\text{friction}} = \Delta T_{\text{cm}}$$here ##\Delta T_{\text{cm}}## is the change of linear kinetic energy only, and not the total change. N.B. I use ##\mathcal{W}## to distinguish it from thermodynamic work, but AFAIK there is no standard notation.

 

[cianfa72]

I personally think that the main culprit is the concept of "Net Work" introduced in the work energy theorem. There are many instances where if you have N mechanical forces F→i acting on a system, each force displacing the point of contact by Δr→i then Wnet=F→net⋅Δxcm≠Σi=1NF→i⋅Δr→i as you would expect. The quantity on the right is what matters with respect to conservation of energy and is the thermodynamic work (for a system acted on by N mechanical forces).

A good example is a spring being compressed by two equal and opposite mechanical forces. The "Net Work" is 0, but thermodynamic work is being done on the system and Σi=1NF→i⋅Δr→i is non-zero. People tend to get in trouble when they assume that "Net Work" is the thermodynamic work or is at all relevant for the conservation of energy.

Going back to the beginning...I believe the work-energy theorem as introduced in this insight actually establishes the following:

##\Delta K_{cm} = W_{net} = \vec F_{net} \cdot \vec \Delta x_{cm} = \Sigma_{i=1}^{N} \vec F_i \cdot \Delta \vec r_i##

##\vec F_{net}## is just the sum of all external forces acting on the system. 'Net Work' is then defined by the dot product between net force and center of mass displacement

It basically evaluates the change of kinetic energy of the 'bulk' system alone (no internal kinetic energy as measured relative to the system center of mass rest frame) using the concept of 'Net Work'. In other words from W-E perspective the system has not an internal structure for storing energy

 

[cianfa72]

We can imagine a cylinder rolling down a rough inclined plane. The work done by the force of friction is zero, since the friction force has zero power (the relative velocity at the point of contact is zero). Hence, the proper work energy theorem yields $$W_{\text{grav}} = \Delta T$$

Basically the force of friction does zero work just because it 'moves' its point of application along the point of contact during the rolling down, right ?

Here ##\Delta T = \Delta T_{\text{cm}} + \Delta T^*## is the total change in kinetic energy, which is the sum of the kinetic energy of the centre of mass and that in the frame of the centre of mass (König theorem). We can also use the centre-of-mass work concept that you linked, which would give$$\mathcal{W}_{\text{grav}} + \mathcal{W}_{\text{friction}} = \Delta T_{\text{cm}}$$here ##\Delta T_{\text{cm}}## is the change of linear kinetic energy only, and not the total change.

Thus, what we're really doing, is compute the sum of all forces (gravity + friction) acting on the cylinder -- as if they were applied to the center of mass -- and multiply it for the center-of-mass displacement.

It is worth to highlight that the 'system' taken in account is the cylinder thus gravity and friction are actually external forces.

 

[etotheipi]

Basically the force of friction does zero work just because it 'moves' its point of application along the point of contact during the rolling down, right ?

The point of application moves as the cylinder rolls down, yes, but the important part is really that the point of the cylinder in contact with the ramp has zero velocity w.r.t. the ramp (due to the rolling condition), i.e. the power ##P = \vec{F}_{\text{f}} \cdot \vec{v}_{\text{edge}} = \vec{F}_{\text{f}} \cdot \vec{0}## of the friction force on the cylinder is zero in the rest frame of the ramp.

Thus, what we're really doing, is compute the sum of all forces (gravity + friction) acting on the cylinder -- as if they were applied to the center of mass -- and multiply it for the center-of-mass displacement.

It is worth to highlight that the 'system' taken in account is the cylinder thus gravity and friction are actually external forces.

Yes, exactly

 

[vanhees71]

I think that is not quite correct. Really the work energy theorem only applies to rigid bodies (and this is also assuming the strong form of Newton III holds!).

This in turn is also not correct. If you have an arbitrary system of point particles with arbitrary interaction and/or external forces derivable from a potential, i.e.,
$$\vec{F}_i=-\vec{\nabla}_i V(\vec{x}_1,\ldots,\vec{x}_n)$$
the total energy
$$E=\sum_{i=1}^n \frac{m_i}{2} \dot{\vec{x}}_i^2+V$$
is conserved.

Proof:
$$\dot{E}=\sum_{i=1}^{n} \dot{\vec{x}}_i \cdot (m_i \ddot{\vec{x}}+\vec{\nabla}_i V)=\sum_{i=1}^{n} \dot{\vec{x}}_i \cdot (m_i \ddot{\vec{x}}-\vec{F}_i)=0.$$

 

[etotheipi]

@vanhees71 I agree, but I don't know, how this is different to what I write. When I say that work-energy theorem only applies to rigid bodies, I mean it in the sense that for rigid bodies, the total work done is identical to the total work done by external forces (because the internal forces do zero work).

If we can figure out the work done by the internal forces of an arbitrary system of particles, then sure the work-energy theorem works again. (E.g. we could keep track of all of the potential energy changes like you did, if these interactions are purely conservative!),

If you have e.g. non-conservative internal interactions, then I think it's not going to be as straightforward to compute total internal work. A gymnast is quite a complicated system of particles, with all sorts of dissipative internal interactions for instance, so finding internal work will be difficult

 

[vanhees71]

The point is that the work-energy theorem applies not only to rigid bodies but to arbitrary systems of point masses. It's even more general than what I wrote, because what I wrote is energy conservation for arbitrary systems of point masses with all forces being derivable from a time-independent potential. The general work-energy theorem holds for any point charges and arbitrary forces acting on them:
$$\dot{T}=\sum_{i=1}^n m_i \dot{\vec{x}}_i \cdot \ddot{\vec{x}}_i=\sum_{i=1}^{n} \dot{\vec{x}}_i \cdot \vec{F}_i.$$

 

[etotheipi]

@vanhees71 it is definitely true if you look at it with that level of granularity, since for any particle ##\dot{T}_i = \dot{\vec{x}}_i \cdot \vec{F}_i##. And clearly if you use summation, it holds for the entire body.

But to do this you would need to know what ##\vec{F}_i## is exactly, for every particle. It's possible conceptually, but if I try to apply work-energy theorem to a complicated macroscopic object like a car, or something, then it's going to be impossible to work out all of the internal forces and works done by those internal forces.

That's why, for non-rigid bodies, I don't think it's so useful a construction because it relies on you knowing all of these internal works.

But for rigid bodies, you only need to worry about the external forces (which are a lot easier to keep track of), and everything will work out okay!

 

[vanhees71]

Sure, but a rigid body is only a special case. The energy-work theorem holds much more generally. E.g. it hplds also in fluid dynamics.

 

[etotheipi]

Sure, but a rigid body is only a special case. The energy-work theorem holds much more generally. E.g. it hplds also in fluid dynamics.

That is a very good point! Though we must still be careful, because sometimes we must also take into account heat flux as well as work done by surface and body forces in order to find the time derivative of energy of a fluid element!

 

[cianfa72]

but the important part is really that the point of the cylinder in contact with the ramp has zero velocity w.r.t. the ramp (due to the rolling condition), i.e. the power ##P = \vec{F}_{\text{f}} \cdot \vec{v}_{\text{edge}} = \vec{F}_{\text{f}} \cdot \vec{0}## of the friction force on the cylinder is zero in the rest frame of the ramp.

ok, thus friction force does not actually any thermodynamic work. It basically accounts for the 'transformation' of the gravity thermodynamic work (due to the force of gravity as applied to the center-of-mass ) into the kinetic energy of the cylinder w.r.t. its center-of-mass.

 

[etotheipi]

ok, thus friction force does not any thermodynamic work. It basically accounts for the 'transformation' of gravity thermodynamic work (due to the force of gravity as applied to the center-of-mass ) into the kinetic energy of the cylinder w.r.t. its cnter-of-mass.

I'm not quite sure how to interpret that; I would just say it like - friction stops the cylinder from slipping, and gravitational work is responsible for the (rotational and linear) kinetic energy gained by the cylinder.

 

[cianfa72]

friction stops the cylinder from slipping, and gravitational work is responsible for the (rotational and linear) kinetic energy gained by the cylinder.

ok sorry for my english. Anyway the main point in the process as described here is that no heat transformation takes place, I believe.

 

[Dale]

Jumping can be treated with the work energy theorem as follows:
https://openoregon.pressbooks.pub/bodyphysics/chapter/comparing-work-energy-and-energy-conservation/

That is an odd approach. There they are considering the legs as external to the system. I agree that their approach successfully uses the work energy theorem in the context of jumping.

However, personally I would prefer approaches that can work regardless of how you choose the system boundary. It has trouble if you make the choice to consider the legs as part of the system.

Frankly, I like to keep my legs attached to my torso, even when just assigning system boundaries.

 

[Dale]

It basically evaluates the change of kinetic energy of the 'bulk' system alone (no internal kinetic energy as measured relative to the system center of mass rest frame) using the concept of 'Net Work'. In other words from W-E perspective the system has not an internal structure for storing energy

Yes, which is why it can differ from the total thermodynamic work.

The more I consider it, the more I like the term “center of mass work”. It is more descriptive and makes it less likely to get confused between the work energy theorem and the conservation of energy.

 

[Dale]

The general work-energy theorem holds for any point charges and arbitrary forces acting on them:

That equation is correct but note that it is not what is referred to as the work energy theorem in this article. What he is talking about is (in your notation):

$$\dot{T}=\dot{\vec{x}}_{COM} \cdot \sum_{i=1}^{n} \vec{F}_i$$

 

[atyy]

If we can figure out the work done by the internal forces of an arbitrary system of particles, then sure the work-energy theorem works again. (E.g. we could keep track of all of the potential energy changes like you did, if these interactions are purely conservative!),

The point is that the work-energy theorem applies not only to rigid bodies but to arbitrary systems of point masses. It's even more general than what I wrote, because what I wrote is energy conservation for arbitrary systems of point masses with all forces being derivable from a time-independent potential.

However, personally I would prefer approaches that can work regardless of how you choose the system boundary. It has trouble if you make the choice to consider the legs as part of the system.

Yes, I would prefer to stress the generality of the work-energy theorem in Newtonian mechanics, rather than inferring its failure from macroscopic examples. For the conversion of internal energy to mechanical energy, there are classical models. An example of a simple classical model is two hard spheres attached to each other by a compressed spring and held in check by a constraint, with the motion initiated by removal of the constraint. We should stress that Newtonian mechanics with conservative forces is in use even for regimes that "common sense" would consider microscopic, such as molecular dynamics simulations.

Of course, ultimately there is a failure of the work-energy theorem, but that is because there is the failure of the whole Newtonian concept of force (and definite position and momentum) in quantum mechanics. Here I would still stress the primacy of mechanics, and say that although force is no longer a concept, energy and momentum conservation remain valid. In fact, I would stress that at this fundamental level, all "forces" (in the generalized sense of electromagnetic force, weak force, strong force in quantum theory) are "conservative". There are no dissipative forces at the fundamental level.

Philosophically, most physicists would say that it is the conservative forces of mechanics (classical or quantum) that undergirds the energy conservation of the first law of thermodynamics. However, in thermodynamics there is the additional idea that because of our ignorance and lack of control of microscopic motion, there is a subjective division of energy into work and heat.

As for the second law of thermodynamics, I would say that philosophically, we still believe in the primacy of mechanics, and the second law of thermodynamics arises from the initial conditions of the universe.

So overall, I think the approach in this insight fails the work-energy theorem too early, and introduces thermodynamics as too fundamental. Of course, thermodynamics is fundamental in the sense that we don't expect it to fail, but it is not fundamental in the sense that it is more general than mechanics. Mechanics is fundamental, and thermodynamics emerges from mechanics together with coarse graining and the initial conditions of the universe.

 

[atyy]

However, personally I would prefer approaches that can work regardless of how you choose the system boundary. It has trouble if you make the choice to consider the legs as part of the system.

I found this article which has a simple model with many of the features in the scenario of the gymnast jumping off the floor.

Springbok: The physics of jumping
The Physics Teacher 39, 109 (2001); https://doi.org/10.1119/1.1355171
Robert J. Dufresne, William J. Gerace, and William J. Leonard
https://srri.umass.edu/sites/srri/files/dufresne-2001spj/index.pdf

The next article also looks relevant, although it's not peer reviewed.

Simple Model of a Standing Vertical Jump
Chris L. Lin
https://arxiv.org/abs/2007.08706

 

[cianfa72]

That equation is correct but note that it is not what is referred to as the work energy theorem in this article. What he is talking about is (in your notation):

$$\dot{T}=\dot{\vec{x}}_{COM} \cdot \sum_{i=1}^{n} \vec{F}_i$$

Yes, exactly. I believe the main source of confusion in this thread is due to naming.

Work-Energy theorem and center-of-mass work theorem are really two different things...