Is Mechanical Energy Conservation Free of Ambiguity?

[kuruman]

Introduction
“Close to any question that is in the textbook, there is another question that has never been answered that is interesting.”
[Stephen Wolfram, remarks to The University of Vermont physics students, September 30, 2005]
Mechanical energy conservation is the assertion that the sum of kinetic and potential energies of a system (the mechanical energy) does not change as a mass moves from point A to point B. Conventionally we may write
$$K_A+U_A=K_B+U_B~~~~~(\rm{I.1a})$$which can be rewritten in a form that does not require specifying the “zero of energy”, $$\Delta K+\Delta U=0~~~~~(\rm{I.1b})$$Equation (I1.b) has found extensive use as a problem-solving technique in cases where the SUVAT equations do not apply, e.g. a roller coaster, in which one needs to relate position and speed at point A with position and speed at point B.  A familiar textbook  derivation...

Continue reading...

 

[Dale]

This is good, it is a tough topic and lots of errors abound. 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 $\vec F_i$ acting on a system, each force displacing the point of contact by $\Delta \vec r_i$ then $W_{net} = \vec F_{net} \cdot \Delta x_{cm} \ne \Sigma_{i=1}^{N} \vec F_i \cdot \Delta \vec r_i$ 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 $\Sigma_{i=1}^{N} \vec F_i \cdot \Delta \vec r_i$ 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.

 

[kuruman]

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 ΣNi=1→Fi⋅Δ→ri\Sigma_{i=1}^{N} \vec F_i \cdot \Delta \vec r_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.

Precisely. Your compressed spring is an example of "action on a deformable body". As far as the work done on the spring is concerned, the work-energy theorem fails. As I argue in part II of this series, the first law provides clarity and the answer. Here is a spoiler preview of the argument one might use.

The system is the spring. The work $W$ that crosses the system boundary is related to the system's change in internal energy by $\Delta E_{int}=W$. If the spring is compressed by $x$, the only change in internal energy is elastic energy. Then $$\Delta E_{int}=\Delta U_{elastic}=\frac{1}{2}kx^2-0=W$$which gives the work done by the environment on the system without integration.

 

[Dale]

So in your example of the friction force, I would say that the problem isn’t the friction force but rather steps 2 and 3 where you use the same variable for the “net work” and the thermodynamic work.

 

[kuruman]

I do not distinguish "thermodynamic work" from "work", they are the same. In the example of the friction force, my initial assumption is that there are two agents in the environment that do (thermodynamic) work, the pulling force and friction. Isn't it true that, if two agents do work on a system, the "$W$" in $\Delta E_{int}=W$ is the sum of the two works, i.e. the net work? Did I miss your point?

 

[Dale]

I do not distinguish "thermodynamic work" from "work", they are the same.

Then statement 3 is false. That is where the problem comes in. Not with the friction force.

Isn't it true that, if two agents do work on a system, the "W" in ΔEint=W is the sum of the two works, i.e. the net work?

The sum of the two works is the (thermodynamic) work. It is not necessarily the “net work” (from the work energy theorem), which is precisely why I don’t like the concept of “net work”.

Did I miss your point?

I don’t know. My point is that your statement 2 and your statement 3 are incompatible. The W in your statement 2 is the thermodynamic work and the W in your statement 3 is the work-energy theorem “net work”. They are not the same thing in general.

 

[kuruman]

The sum of the two works is the (thermodynamic) work. It is not necessarily the “net work” (from the work energy theorem), which is precisely why I don’t like the concept of “net work”.

That clarifies your remarks to me. I think we assign different meanings to "net work". In statement 3, I used the word "net" as in "sum" not as in $W_{net}=\int \vec F_{net} \cdot d\vec s$. It did not occur to me that "net" could be considered a reserved word tied to the work-energy theorem. Perhaps statement 3 should have read,
"3. The sum of external works $W$ done on the system is zero because its kinetic energy is constant."
However that looked a bit awkward to me and that's why I used the particular language that I did not realizing that I sacrificed clarity. Perhaps Greg will let me edit this, but I will wait to see if more problems crop up before I ask him.

The confusion introduced by the $W_{net}$ of the work-energy theorem is precisely the reason I reject the W-E theorem in part III of the series. I think we are in full agreement here.

 

[anorlunda]

Perhaps Greg will let me edit this

Authors can always edit their own Insights posts. You don't need help from Greg.

 

[kuruman]

Authors can always edit their own Insights posts. You don't need help from Greg.

Done. Thank you for pointing that out. I thought that because the "Edit" button was gone upon submission, I could no longer edit. I didn't realize I could always "Update".

 

[Dale]

The confusion introduced by the $W_{net}$ of the work-energy theorem is precisely the reason I reject the W-E theorem in part III of the series. I think we are in full agreement here.

Yes, hearty and complete agreement. It trips up many people and is very difficult to get a derivation which acknowledges the limitations/assumptions in the derivation.

Done. Thank you for pointing that out. I thought that because the "Edit" button was gone upon submission, I could no longer edit. I didn't realize I could always "Update".

So now your statement 3 reads "The sum of thermodynamic work W done on the system is zero because its kinetic energy is constant" which is clear but false. The relationship between the KE and work is given by the work energy theorem and it is the work energy theorem "net work" which is equal to the change in KE, not the sum of the thermodynamic work.

Again, the easy example is two equal forces compressing a spring. There the change in KE is zero as is the work energy theorem "net work" but the sum of thermodynamic work is non-zero. So it is a clear counterexample to statement 3: the KE being constant does not imply that the sum of thermodynamic work is zero.

This is why the work energy theorem is so pernicious. The problem isn't friction, it is the work energy theorem which leads you to make an incorrect statement like 3. From that wrong statement you get the obvious contradiction that you raise, but you misattribute the problem to friction when the problem is really the work energy theorem, or rather the understandable mistake that the work energy theorem's "net work" is the same thing as the sum of thermodynamic work.

 

[kuruman]

So now your statement 3 reads "The sum of thermodynamic work W done on the system is zero because its kinetic energy is constant" which is clear but false. The relationship between the KE and work is given by the work energy theorem and it is the work energy theorem "net work" which is equal to the change in KE, not the sum of the thermodynamic work.

Step 2 says
2. The first law of thermodynamics says $\Delta E_{int}=Q+W$.
This establishes the context to be the first law, not the W-E theorem in what follows. The revised step 3 says
3. The sum of thermodynamic work $W$ done on the system is zero because its kinetic energy is constant.

Perhaps there ought to be step 2.5 and a yet another modification to step 3 that read
2.5 The change in internal energy due to external mechanical work can only be mechanical, $\Delta E_{int}=\Delta K+\Delta U$. Both $\Delta K$ and $\Delta U$ are zero in this case which means $\Delta E_{int}=0$.
3. According to the first law (step 2), the total thermodynamic work $W$ done on the system is zero.

The problem isn't friction, it is the work energy theorem which leads you to make an incorrect statement like 3.

But the work-energy theorem in the absence of friction (and dissipative forces in general) is fine and is a statement of mechanical energy conservation. It's the inclusion of $\vec f_k\cdot d\vec s$ as part of $dW_{net}$ within the context of the work-energy theorem that creates the problem.

BTW @Dale, I appreciate your critical remarks that identify points of needed clarification.

 

[Dale]

2.5 The change in internal energy due to external mechanical work can only be mechanical

I think that may be generally problematic. A generator is an example of a device that converts external mechanical work into electrical work. I suppose that you can probably find examples of other devices that can convert external mechanical work to many forms of energy other than mechanical energy. In fact, in this case the mechanical work winds up changing the internal energy by increasing the thermal energy.

But the work-energy theorem in the absence of friction (and dissipative forces in general) is fine

Friction does not cause a problem for the work-energy theorem. According to the work-energy theorem, $\Delta K = \int \vec F_{net} \cdot d\vec s_{cm}$ where the quantity on the right is called the "net work". For your block example $\vec F_{net}= \vec P + \vec f=0$ and $\Delta K =0$ so the work energy theorem is satisfied even when we include the friction force.

Where you get problems is when you try to relate the "net work" to the total thermodynamic work. That problem happens regardless of whether the forces are dissipative or not. My spring example is one where the total thermodynamic work and the "net work" differ, but there is no dissipation involved. This is my issue with the work energy theorem: the "net work" is not the same as the total thermodynamic work, as most people would expect.

 

[kuruman]

I think that may be generally problematic.

Generally yes, but statement 2.5 is not a general statement. It refers to a block being pulled on a rough surface at constant speed. The change in internal energy can only be in the kinetic form because the potential form is irrelevant. The hiding places for mechanical energy have been exhausted and the kinetic form does not change (in this case) by assumption. Of course you and I both know that there is additional internal energy change in the form of thermal energy. The purpose here is to paint the "work done by friction" into a corner and label $\vec f_k\cdot d\vec s$ the change in thermal internal energy $\Delta E_{therm}$. That is argued in part II of this series.

Where you get problems is when you try to relate the "net work" to the total thermodynamic work.

I hope the "you" above is generic and does not refer specifically to @kuruman. The stuff I present in the Introduction of part I, is a reference point of where the general consensus is now, i.e. the status quo. In parts II and III, yet to be released, I challenge the status quo and present alternatives to it. Specifically, in part III I argue in favor of discarding the W-E theorem and replacing it with equation (I.2)$$\Delta K=\int_A^B\vec F_{net}\cdot d\vec s.$$without associating the word "work" with any part of the equation explicit or implied.

My spring example is one where the total thermodynamic work and the "net work" differ, but there is no dissipation involved. This is my issue with the work energy theorem: the "net work" is not the same as the total thermodynamic work, as most people would expect.

I am with you all the way. Dissipation is not the only instance where the W-E theorem fails. Action on or by deformable bodies is another. Both are mentioned in part I of the series and revisited in part II. Your spring example belongs to the latter instance. Post #3 shows how to find the thermodynamic work done on the spring without using the W-E theorem.

In most cases people "agree to disagree". Here I think we "disagree to agree".

 

[Dale]

Post #3 shows how to find the thermodynamic work done on the spring without using the W-E theorem.

It had better since in general you cannot find thermodynamic work with using the work energy theorem.

The purpose here is to paint the "work done by friction" into a corner

Well, I guess I will have to see the future installments. I think the thing that needs to be painted into a corner is the work energy theorem.

The work done by friction is a perfectly valid concept and it is perfectly compatible with both the concept of total thermodynamic work and “net work”. So I guess I don’t get why you are corner-painting it.

 

[kuruman]

Well, I guess I will have to see the future installments.

I guess you will have to. To give you a preview, my argument in part II in terms of the first law is that $\vec f_k \cdot d\vec s$ does not belong on the right hand side of the law as work but on the left hand side as change in thermal energy (part of the overall change in internal energy), i.e. $d E_{therm}=\text{-}\vec f_k \cdot d\vec s.$ Thus, the first law (always in the case of the block being pulled at constant speed by force $P$) is $$\Delta E_{int}=W_P$$ $$\text{-}\vec f_k \cdot \Delta\vec s_{cm}=\vec P\cdot \Delta\vec s_{cm}~~~~~~(1)$$Equation (1) says that the mechanical work that crosses the system boundary is converted by kinetic friction into internal (thermal) energy and there is none left for kinetic energy change. By saying that kinetic friction does work, I guess you would write the total work equation as $$0=\vec f_k \cdot \Delta\vec s_{cm}+\vec P\cdot \Delta\vec s_{cm}~~~~~~(2)$$Equation (2) is algebraically equivalent to (1) and, as a statement of the work energy theorem, it might make sense. As a statement of the first law it is pure nonsense because it says that the change in internal energy of the block (left hand side) is zero when we know that it is not. So I see a problem with both the work energy theorem and the idea that kinetic friction does work.

 

[bagasme]

Work-energy theorem has flaws on work done against friction and deformed body, and neither the mechanical energy conservation address that 'perfectly', right?

 

[Dale]

As a statement of the first law it is pure nonsense because it says that the change in internal energy of the block (left hand side) is zero when we know that it is not.

Not really. With respect to the first law the internal energy increases via the $Q$ term. If you trace the flow of mechanical energy you see that mechanical energy goes into the block by the pulling force and out of the block by the friction force, but where does the energy go then? It does not go into the table because $\Delta s=0$ for the table. So that mechanical energy goes into the contact patch and does not come out because it is changed to thermal energy at the contact patch. Then that thermal energy flows (heat) into both the block and the table as $Q$.

So friction can be consistently treated just like any other force both for the work energy theorem and for the first law of thermo. All that needs to be done is to recognize that kinetic friction involves a loss of mechanical energy at the boundary which energy is converted to thermal energy.

I.e. that statement 2.5 you were proposing is incorrect for the block. If you get rid of that then the observed behavior follows: mechanical power does enter the contact patch and does not leave it and the same amount of thermal energy is produced. The thermal energy transfers as heat.

So I see a problem with both the work energy theorem and the idea that kinetic friction does work.

How about instances where the kinetic friction clearly does do positive work? For example, a box, initially at velocity 0, is set onto a moving conveyor belt. If the kinetic friction force between the box and the belt does not do work on the box then how does the box gain KE?

 

[andresB]

I have to re-read it with more care casue I have several doubts. But, is it the point of the insight post that (assuming no exchange of heat) the work-energy theorem as usually thaught is not correct and can't be used to find mechanical quantities?

 

[kuruman]

Work-energy theorem has flaws on work done against friction and deformed body, and neither the mechanical energy conservation address that 'perfectly', right?

That is my position.

 

[andresB]

I can't say anything about deformable bodies.

But, let's look at the following example: A point particle is throw upward along the surface of inclined plane with friction. Is it incorrect to use the work-energy theorem to find how high the particle goes until it stops?

 

[kuruman]

I can't say anything about deformable bodies.

But, let's look at the following example: A point particle is throw upward along the surface of inclined plane with friction. Is it incorrect to use the work-energy theorem to find how high the particle goes until it stops?

If you are asking whether you will get the correct answer if you use the W-E theorem in this case, the answer is yes as long as you accept that kinetic friction does mechanical work. If you are asking whether it is conceptually correct, I would say "no". @Dale may have a different answer to this.

There is a better way to think about such problems (actually two ways) that bypass the W-E theorem. These I present in parts II and III. I have not publish these yet because I want to see the reactions to part I and perhaps clarify certain points. I have arranged with @Greg Bernhardt to post this on March 9.

 

[andresB]

If you are asking whether you will get the correct answer if you use the W-E theorem in this case, the answer is yes as long as you accept that kinetic friction does mechanical work. If you are asking whether it is conceptually correct, I would say "no". @Dale may have a different answer to this.

That's very strange. Though I guess I'll have to wait until you complete your series of insight posts in order to actually give an informed opinion.

 

[Dale]

I can't say anything about deformable bodies.

But, let's look at the following example: A point particle is throw upward along the surface of inclined plane with friction. Is it incorrect to use the work-energy theorem to find how high the particle goes until it stops?

For this situation it is correct because you do not care about total energy, only kinetic energy. The work-energy theorem is fine for that. It does not matter what kind of force or forces are involved in producing the net force.

Where you get into trouble with the work energy theorem is if you are interested in anything other than KE. E.g. total thermodynamic work.

 

[jbriggs444]

If you are asking whether you will get the correct answer if you use the W-E theorem in this case, the answer is yes as long as you accept that kinetic friction does mechanical work.

Looking at kinetic friction as just another external force on a body then it is immediately obvious that kinetic friction can do mechanical work.

Looking at kinetic friction as a third law force pair that acts on each of two mating and relatively moving surfaces, it is less immediately obvious that kinetic friction absorbs mechanical work. For me, the useful take-away from this viewpoint is that the amount of work absorbed is an invariant quantity. It does not depend on one's choice of reference frame.

 

[Dale]

For me, the useful take-away from this viewpoint is that the amount of work absorbed is an invariant quantity. It does not depend on one's choice of reference frame.

Yes, that is an important and interesting point.

For the example of the block, the block is pulled with force $\vec P$ against friction force $\vec f = -\vec P$ at a constant velocity $\vec v$ (in the frame of the floor). The mechanical power going into the block by the pulling force is $p_1=\vec P \cdot \vec v$ and the mechanical power going into the block by the friction force is $p_2=\vec f \cdot \vec v = -\vec P \cdot \vec v$ and the mechanical power going into the floor by friction is $p_3=-\vec f \cdot 0=0$. The mechanical power provided externally is $p_1=\vec P \cdot \vec v$, the power gained by the block is $p_1+p_2 = 0$, the mechanical power gained by the floor is $p_3=0$, and the mechanical power converted to heat at the interface is $-p_2-p_3=-(-\vec P \cdot \vec v) - 0 = \vec P \cdot \vec v$

Now, in a frame moving at velocity $\vec u$ with respect to the floor we have the mechanical power into the block by the pulling force is $p'_1=\vec P \cdot (\vec v - \vec u)$ and the mechanical power into the block by the friction force is $p'_2=\vec f \cdot (\vec v - \vec u) = -\vec P \cdot (\vec v - \vec u) = -p'_1$ and the mechanical power into the floor by friction is $p'_3 = -\vec f \cdot (-\vec u)= -\vec P \cdot \vec u$. So now the mechanical power provided externally is $p'_1=\vec P \cdot (\vec v - \vec u)$ the mechanical power gained by the block is $p'_1+p'_2 = 0$, the mechanical power gained by the floor is $p'_3=-\vec P \cdot \vec u$, and the mechanical power converted to heat at the interface is $-p_2-p_3=\vec P \cdot (\vec v - \vec u)+\vec P \cdot \vec u=\vec P \cdot \vec v$.

So, overall energy is conserved in both frames. The input power either goes to mechanical power into the floor or heat at the interface, and the amount of mechanical power converted to heat at the interface is (Galilean) invariant.

 

[kuruman]

This is a place holder. I tried saving an edited post and the LaTeX code got all screwed up. I will repost after I get it fixed.

 

[Dale]

This is a place holder. I tried saving an edited post and the LaTeX code got all screwed up. I will repost after I get it fixed.

If you want I can revert it to the pre-screwed up version.

 

[hutchphd]

Is that like a super power?? I could occasionally make use of that in my life!

 

[kuruman]

If you want I can revert it to the pre-screwed up version.

Thank you for your kind offer, but that version is now obsolete. I have seen the error of my ways and I have become convinced that there is nothing wrong with the idea that $\vec f_k \cdot d\vec s_{cm}$ can be considered as the work done by friction. Of course this retraction necessitates finding the fault in the logic of the step by step reasoning that was supposed to show the opposite. I think the fault is in step 6, "The internal energy of the block increases because its temperature rises, $\Delta E_{int}> 0.$" The statement itself is true but irrelevant. As you have shown in #25, the power accounting is OK therefore the energy accounting must be OK; there is no energy loss or gain per unit time anywhere. The temperature of the block rises some time later as heat is transported to the block through the standard channels, no because it is not accounted for. The heat is generated at the bottom interface and where it goes thereafter is a separate issue.

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

 

[jbriggs444]

Thank you for your kind offer, but that version is now obsolete. I have seen the error of my ways and I have become convinced that there is nothing wrong with the idea that $\vec f_k \cdot d\vec s_{cm}$ can be considered as the work done by friction.

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").

But it does not correspond to the total energy contributed to the object due to the application of the force of kinetic friction, either with or without inclusion of the thermal energy dissipated due to the kinetic friction.

[I thought we'd just gone through this discussion at length in the context of a car on a road and static friction on tires]

 

[Dale]

Is that like a super power?? I could occasionally make use of that in my life!

Yes, if you need it just ask a moderator. We cannot rescue anything you didn’t submit, but if you need to go back to a previous version you submitted we can do that.

 

[kuruman]

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").
But it does not correspond to the total energy contributed to the object due to the application of the force of kinetic friction, either with or without inclusion of the thermal energy dissipated due to the kinetic friction.

[I thought we'd just gone through this discussion at length in the context of a car on a road and static friction on tires]

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.

 

[jbriggs444]

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.

For a rigid, non-rotating block you do indeed escape the complications that led that thread into its terminal impasse.

 

[kuruman]

For a rigid, non-rotating block you do indeed escape the complications that led that thread into its terminal impasse.

Impasse or recalcitrance? (This is the last time I will mention that thread.)

 

[Dale]

f course this retraction necessitates finding the fault in the logic of the step by step reasoning that was supposed to show the opposite. I think the fault is in step 6, "

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.