B The work done by two objects on each other

[jbriggs444]

One would be a positive quantity and the other negative.

Not necessarily. @PeroK had pointed this out in post #2 in the thread.

It holds true for two objects (or interacting surfaces) with zero relative motion in the direction of the force between them. It does not hold in general.

For many situations, there is no relative motion in the direction of the interaction force. This would include ropes on pulleys, levers on loads and hands on boxes.

For other situations, there is relative motion. This would include kinetic friction, gravity and the electrostatic force.

For some situations there is ambiguity. It can depend on how closely we look at the interaction. There is relative motion between car and road. But not between tire treads and road. There is relative motion between the bulk of the combustion gasses in the cylinder and the piston. There is zero relative motion between the gas right at the piston surface and the directly adjacent piston surface.

 

[sophiecentaur]

Just because we survived some bad teachers, doesn't mean we have to propagate bad teaching.

Do you have actual evidence that removing the frequent use of by and on has worse results? I haven't suggested anything revolutionary; all I am saying is that you can avoid the confusion by changing the emphasis and the way through mechanics. You are ignoring the problems that arise with the standard approach because you are bright enough to cope with all that cognitive dissonance.

 

[sophiecentaur]

which sort of work

That's a new concept for me. Isn't work always a dot product of Force and Displacement?

In your post, you have quoted a very simple scenario of car wheels on roads but it's far too simplified to add the sort of confusion I'm talking about. You have to consider the 'sharing' of the energy delivered to the wheels into useful kinetic energy and losses in the tyres and the tarmac; slip too. The rim of the wheel definitely does work because the car engine is the only source of energy (a level road, here) and any student can understand that. After the losses, there will be a force on the road which will have a reaction which accelerates the car. You could also say that the axle 'does work' by accelerating the car forward. Your ('my moderately bright ') student would not find that confusing.

But questions are always asked like "how does friction make the car go forward when friction slows things down?". You must have been asked to deal with that one. In that chain of interactions between components there are wheel - tyre - road surface / the Earth and the definition of Work could be applied in there. Easy for you because you have confidence that your system works reliably but if 'the student' has to decide who does what to whom, that's a needless barrier with a perceived risk. You can just use the forces and avoid another arbitrary choice of signs for intermediate values of Work. I can see only problems and not any significant advantages.

 

[jbriggs444]

That's a new concept for me. Isn't work always a dot product of Force and Displacement?

Which displacement? Perhaps you should read what I have written in this thread.

And no. Generalized work includes other forms of energy transfer.

In your post, you have quoted a very simple scenario of car wheels on roads but it's far too simplified to add the sort of confusion I'm talking about.

Yet you are unwilling to address it at the requisite level of detail. You have to show an ability to do the work. A level of understanding sufficient to proceed with an intelligent conversation.

You have to consider the 'sharing' of the energy delivered to the wheels into useful kinetic energy and losses in the tyres and the tarmac; slip too.

We could consider those. We need not. If dissipative losses are negligible then we can neglect them.

Dissipative losses in kinetic friction are a key reason that the distinction between the work done by road on treads and treads on road is more than just a sign change.

The rim of the wheel definitely does work because the car engine is the only source of energy (a level road, here) and any student can understand that.

Yet you fail to understand that. Because it is false. The rim of the wheel does zero mechanical work on the road.

Time for you to work the exercise.


Let me work through the situation for the wheel using the notion of mechanical work. Recall that mechanical work is defined as the dot product of the force applied on an object and the motion of the object at the point of application of the force.

We begin by identifying the relevant interfaces across which mechanical work may be done.

1. Static friction between tread and road.
2. Linear force between hub and wheel bearings.
3. Torsion between drive axle and wheel.

1...

The static friction between tread and road is (per the givens of the exercise) 1000 N. This is a forward force on the tire. The tread at the contact patch has zero velocity. So zero total displacement over the duration of the scenario. The work done by road on wheel is zero.

The road is also motionless. The work done by wheel on road is zero.

2...

For convenience, we assume that air resistance on the tires is negligible. The 1000 N of retarding air resistance is entirely on the body of the car. Accordingly, we can conclude that there is a forward force of 1000 N (total across the four wheels) from hub on wheel bearings. The car is moving forward at 10 m/s over a duration of 1 s for a total displacement of 10 meters forward. So we have $F \cdot d = +1000 \cdot +10 = +10000 \text{ J}$ of work done by wheel on car.

There is no relative motion in the forward direction. So the work done by car on wheel through this interface is the opposite: -10000 J.

3...

To simplify the analysis, I will pretend that only one wheel is driven. We could total across four driven wheels with various loads, but that would complicate matters with no benefit to understanding.

We are given that the radius of the wheel assembly is 0.5 meters. The required torque ($\tau$) to achieve 1000 N at the rim is 500 N m. In order to traverse 10 meters in one second, we are going to need 20 radians of rotation ($\Delta \theta$). The work done by torque is given by $\tau \cdot \Delta \theta = 500 \text{ Nm}\cdot 20 \text{ rad} = 10000 \text{ J}$. For drive axle on wheel the torque is in the same direction as the rotation of the wheel. So +10000 J of mechanical work is done.

[If you like, I can walk you through a derivation of $W = \tau \cdot \Delta \theta$ from first principles -- integration and the definition of mechanical work]

For wheel on drive axle, the torque is opposite and -10000 J of mechanical work is done.

In summary, the mechanical energy flows over the duration of the scenario are 10000 J from drive axle to wheel and 10000 J from wheel to car body.

 

[Herman Trivilino]

Not necessarily. @PeroK had pointed this out in post #2 in the thread.

That seems like a different point to me.

When you refer to the work done "on" something, or the work done "by" something aren't you always referring to a positive quantity?

My point is that if a positive amount of work $W$ is done on A in some interaction with B, then a negative amount of work $-W$ is done by A in the same interaction.

 

[sophiecentaur]

If dissipative losses are negligible then we can neglect them.

That's so true but answering questions about brakes, skidding and slipping and the dreaded word "friction" need answers involving more than the ideal.

Yet you fail to understand that. Because it is false. The rim of the wheel does zero mechanical work on the road.

Here's more confusion. In the frame of the car (we can choose?) the work is done on the road But you are right; the Earth doesn't move so that's the best reference frame. bI was being sloppy about the rim doing work; It's the couple (wheel rim and axel) that does work

I appreciate that the workings of the transmission system is analysed in terms of torque. The work done in the cylinder starts off as linear motion.

In summary, the mechanical energy flows over the duration of the scenario are 10000 J from drive axle to wheel and 10000 J from wheel to car body.

You put that well.

If dissipative losses are negligible then we can neglect them.

Now here's a bit of a problem. Your idealised model with no losses is fine and has much fewer apparent paradoxes but, I'm sure you will remember countless questions about friction (slipping) and internal losses in tyres and tarmac. The reason I got started on all this was because of the confused phrases about friction doing the work etc. etc.. I guess I'll just have to get over it but, when you read the next post that's confused about work being done by friction and "how is that possible?", you'll notice the cognitive dissonance and understand that there real difficulties because, basically, by the' by or on' question.

Cheers

 

[jbriggs444]

That's so true but answering questions about brakes, skidding and slipping and the dreaded word "friction" need answers involving more than the ideal.

We are still working toward a meeting of the minds. Your insistence on adding complications at every turn is not helpful in achieving that goal.

Here's more confusion. In the frame of the car (we can choose?)

There you go again. Another complication.

Yes, we can choose to work in a different reference frame. I was trying to avoid that complexity until we had some agreement on the basics.

If we shift to the rest frame of the car, the wheel assembly does zero work on the car body. But now the wheel assembly does positive work on the retreating road. The energy balance through the wheel assembly remains intact, of course. 10000 J in and 10000 J out.

We could shift to the Earth centered inertial frame and consider an eastbound car at the equator. Now we wheel is doing 470000 J on the car body and -46000 J on the Earth. The energy balance is still intact with 10000 J in and 10000 J out.

A westbound wheel would do -450000 J on the car body and +460000 on the Earth. Still for a total of 10000 J in and 10000 J out.

the work is done on the road But you are right; the Earth doesn't move so that's the best reference frame. I was being sloppy about the rim doing work; It's the couple (wheel rim and axel) that does work

First of all, it is not correct to say that the Earth does not move. Its movement is conditional on the reference frame that one chooses. A key lesson from Newtonian mechanics is that there is no such thing as absolute rest.

In addition, the wheel, rim and axle are a rigid assembly. No dissipation. No energy production. Zero total work done on the assembly [in this case anyway].

If you want to talk about a couple, you will need to be clear about which couple you are talking about.

The obvious place to look is the expanding gasses in the combustion chamber. Add up the work done on the piston and the work done on the cylinder and one can see that positive total work is being done by those gasses. The rest of the drive train has [ideally] zero losses. Mechanical work flows down the line to the drive axle and the wheel assembly.

The interface between car body and atmosphere involves dissipative friction. That interface is a mechanical energy sink. Work goes in and mostly thermal energy comes out.

In the exercise I posed, there was a headwind. In addition to the mechanical energy dissipated into thermal energy, there is a loss in total atmospheric kinetic energy. The atmosphere slows down as the car plows through it.

But since you not yet worked through the exercise, you may not have discovered that fine point in the energy balance.

I appreciate that the workings of the transmission system is analyzed in terms of torque. The work done in the cylinder starts off as linear motion.

Good. So we do not have to worry about a disagreement about the work associated with torque and rotation.

Now here's a bit of a problem. Your idealised model with no losses is fine and has much fewer apparent paradoxes but, I'm sure you will remember countless questions about friction (slipping) and internal losses in tyres and tarmac. The reason I got started on all this was because of the confused phrases about friction doing the work etc. etc.. I guess I'll just have to get over it but, when you read the next post that's confused about work being done by friction and "how is that possible?", you'll notice the cognitive dissonance and understand that there real difficulties

Real difficulties. Yes. I recognize that. Perhaps you recall the discussion here around the Brennan torpedo and where work is being done in that system. That was a lively discussion.

you'll notice the cognitive dissonance and understand that there real difficulties because, basically, by the' by or on' question.

Here we disagree. I believe that problem with student's understandings of work have little to do with the words "on" and "by". If anything, I see most difficulties associated with a failure to properly identify the relevant displacement.

One can trace confusion over kinetic friction and work to that difficulty. The displacements of the two interacting objects are different.

One can trace confusion about work done on cars to that difficulty. The displacements of the treads and of the car are different.

 

[jbriggs444]

That seems like a different point to me.

When you refer to the work done "on" something, or the work done "by" something aren't you always referring to a positive quantity?

As I use the terminology, the result can be negative.

Perhaps I should explain the way I use the terms. I use both terms in a single phrase to characterize one side of a particular interaction.

If I exert a retarding force of 1000 N on an object displacing forward through 10 m then I have done -10000 J of work on the object. This is the work done "by" my hand "on" the object.

If this 1000 N is instead the result of kinetic friction from a long piece of sandpaper then the mechanical work done "by" the sandpaper "on" the object is still -10000 J.

However, if I speak about the work done "by" the object "on" the sandpaper then the result is 0 J since the sandpaper is not moving. In this case, the total of the two works is -10000 J. This discrepancy will manifest as thermal energy.

My point is that if a positive amount of work $W$ is done on A in some interaction with B, then a negative amount of work $-W$ is done by A in the same interaction.

No, that is not correct.

 

[A.T.]

all I am saying is that you can avoid the confusion

Avoiding confusion by being less explicit about what you mean? Is this a joke?

 

[Herman Trivilino]

If I exert a retarding force of 1000 N on an object displacing forward through 10 m then I have done -10000 J of work on the object. This is the work done "by" my hand "on" the object.

And the work done by the object on your hand would be +10 000 J.

If this 1000 N is instead the result of kinetic friction from a long piece of sandpaper then the mechanical work done "by" the sandpaper "on" the object is still -10000 J.

However, if I speak about the work done "by" the object "on" the sandpaper then the result is 0 J since the sandpaper is not moving. In this case, the total of the two works is -10000 J. This discrepancy will manifest as thermal energy.

But note that that thermal energy is shared by both the object and the sandpaper, the total being 10 000 J, but with no way of knowing how much of that thermal energy is in the sandpaper and how much is in the object.

So if we look at the work done on the object you claim it's -10 000 J, yet the object doesn't lose a total of 10 000 J of energy. It looses 10 000 J of mechanical energy, gains an unknown amount of thermal energy, and therefore looses an unknown amount of energy.

If this is a result of the way you are defining work, there is nothing wrong with that as a perfectly valid dynamical relation. But it is not the work as we define it in the 1st Law of Thermodynamics.

 

[jbriggs444]

And the work done by the object on your hand would be +10 000 J.

Yes. In this case the force pair is not dissipative. The two works are equal and opposite.

But note that that thermal energy is shared by both the object and the sandpaper, the total being 10 000 J, but with no way of knowing how much of that thermal energy is in the sandpaper and how much is in the object.

Yes. Agreed.

So if we look at the work done on the object you claim it's -10 000 J, yet the object doesn't lose a total of 10 000 J of energy. It looses 10 000 J of mechanical energy, gains an unknown amount of thermal energy, and therefore looses an unknown amount of energy.

Yes. Agreed.

If this is a result of the way you are defining work, there is nothing wrong with that as a perfectly valid dynamical relation.

Good. Then we have no fundamental disagreement. We are just nattering about terminology.

But it is not the work as we define it in the 1st Law of Thermodynamics.

You are apparently talking about thermodynamic work while I am talking about mechanical work.

Can you define "thermodynamic work" precisely and explain how it relates to kinetic friction?

If it were me, I would invent an abstract system to embody the interface across which the kinetic friction acts. This abstract system has a net surplus of "thermodynamic work" supplied from its two sides. It supplies an equivalent amount of heat which egresses (in an unknown ratio, as you point out) out the two sides.

 

[Herman Trivilino]

You are apparently talking about thermodynamic work while I am talking about mechanical work.

I didn't know that you were making a distinction.

 

[sophiecentaur]

Avoiding confusion by being less explicit about what you mean? Is this a joke?

I think you have been too interested in getting across the theory (and no-one is disputing that it's right) and have not grasped that using Work as a way into understanding makes the whole thing harder. What do you mean by "explicit"? These sort of basics are very often best taught in steps and avoiding the difficulty which Work can initially introduce would be an advantage. I have never suggested that Work shouldn't;t be taught.

If you only see intellectual problem from your personal viewpoint then you will find it hard to understand how many other people see things. From a perfectly logical point of view it can be said that people who want to learn should just accept being taught the way you were taught and that should be enough.
I am always coming across the stumbling block that arises when Work is introduced too early in learning some topics. If you just tell me that they should get over it and approach it 'your way' from the start, then it's basically 'end of conversation'.

Yes. In this case the force pair is not dissipative. The two works are equal and opposite.

the force pair is not dissipative if the displacements are the same amplitude. (That, at least is true for mechanical interactions.

 

[sophiecentaur]

You are apparently talking about thermodynamic work while I am talking about mechanical work.

Can you define "thermodynamic work" precisely and explain how it relates to kinetic friction?

I didn't know that you were making a distinction.

This sub-conversation rather makes my point for me. How would someone who is confused when trying to appreciatewhat goes on in a driven car and what happens to the Energy, be in a position to appreciate the meaning of those two contributions? It only clouds the basic issue of the thread which it the order in which all these concepts should best be explained.

There's a difference between vectors, which 'cancel' and scalars which merely have signs but don't actually cancel in the same way.

 

[jbriggs444]

This sub-conversation rather makes my point for me. How would someone who is confused when trying to appreciate what goes on in a driven car and what happens to the Energy, be in a position to appreciate the meaning of those two contributions? It only clouds the basic issue of the thread which it the order in which all these concepts should best be explained.

As @A.T. suggests, the way to combat confusion is with clarity. Not vague handwaving.

There's a difference between vectors, which 'cancel' and scalars which merely have signs but don't actually cancel in the same way.

More handwaving. Show me a scalar and I'll show you a vector.

 

[jbriggs444]

the force pair is not dissipative if the displacements are the same amplitude. (That, at least is true for mechanical interactions.

What does "amplitude" mean in this context?

 

[jbriggs444]

I didn't know that you were making a distinction.

I'd already distinguished between "center of mass" work and mechanical work. I'd alluded to something else called generalized work. I think that generalized work (a transfer of mechanical energy without a transfer of mass) is the same thing as thermodynamic work, but am still awaiting your definition.

 

[sophiecentaur]

What does "amplitude" mean in this context?

I was trying for a way of avoiding the word 'equal' because then you'd say they're vectors. Isn't it true that non-dissipative implies no relative motion?

Edit: but this is all well above the intended level of the thread and is another reason to try to avoid Work, early on. This is a very common approach to initial learning / teaching many topics in Physics, actually.

 

[Herman Trivilino]

This sub-conversation rather makes my point for me. How would someone who is confused when trying to appreciatewhat goes on in a driven car and what happens to the Energy, be in a position to appreciate the meaning of those two contributions?

There have been efforts in the last 15-20 years in many of the mainstream college-level introductory physics textbooks to make the definition of work the same in both dynamics and thermodynamics. It's an uphill battle, mainly because many of us who learned different formalisms have difficulty re-learning the newer formalism. Some people can be very stubborn and close-minded when it comes to re-learning something that took them years or even decades to master.

 

[jbriggs444]

I was trying for a way of avoiding the word 'equal' because then you'd say they're vectors. Isn't it true that non-dissipative implies no relative motion?

No relative motion in the direction of the force. You want the dot product of the relative motion and the force to be zero.

Edit: but this is all well above the intended level of the thread and is another reason to try to avoid Work, early on.

The original intent of this thread is to discuss the work done by two objects on each other. A detailed examination of the situation is directly on point.

A digression into the utility of the words "on" and "by" in physics pedagogy is arguably beyond the scope of this thread.

This is a very common approach to initial learning / teaching many topics in Physics, actually.

Certainly, we often paint a simple picture to begin with and fill in complexities and details later. There is nothing wrong with that. But if a student comes to us and asks why the work done by the road on the car can be different from the work done by the car on the road [as in the original post in this thread], how are we to answer if we are forbidden from discussing the matter in detail?

For me, I want to be able to reconcile a formalism with the physical reality. To do that, I need a formalism rather than a handwave. I'd be open to a formalism that focuses on the interaction rather than on the participants in the interaction. But then one loses the ability to talk about the work done by fictitious forces.

 

[Herman Trivilino]

I think that generalized work (a transfer of mechanical energy without a transfer of mass) is the same thing as thermodynamic work, but am still awaiting your definition.

I think that such a definition is beyond what I can put in a forum post. It wouldn't do justice to the question asked, and would just result in an extended back and forth discussion. It is instead something to be found in a textbook or a published article. You can find good definitions in many of the college-level introductory physics textbooks written in the last 15-20 years.

I refer you to, and strongly recommend, the paper by Arnold B. Arons that starts on page 1063 of the American Journal of Physics [Am. J. Phys. 67 (12), December 1999]. If you can't get access to the article let me know and I can send you the PDF via private message.

Here is an excerpt from that paper that may answer your question:

How must work be calculated? Experience shows
that the useful operational definition of work must be developed
along the lines discussed by P. W. Bridgman in his
careful, insightful, and thorough analysis in The Nature of
Thermodynamics:

Turn now to an examination of the W of the First Law.
This W means the total mechanical work received by the
region inside the boundary from the region outside [or
delivered to the region outside from the region inside]. As
in the case of [heat transfer] Q, this work is done across
the boundary, and the evaluation of W demands the posting
of sentries at all points of the boundary, and the summing
of their contributions. In the simple cases usually
considered in elementary discussions, the work received
by the inside from the outside is of the simple sort typified
by the motion of stretched cords or of simple linear piston
rods. Our sentry can adequately report this sort of thing in
terms of finite forces acting at points and finite displacements.
In general, however, there will be contact of the
material outside over finite regions of the boundary, and
we become involved in the stresses and strains of elasticity
theory.

[Bridgman goes on to mention the ‘‘infelicities" that result
when we apply the notion of work to the sliding of two
bodies on each other with friction.]

To emerge as a conserved quantity, work must be calculated
by summing the product of forces and their corresponding
displacements over the periphery or boundary of the system
which has been defined. For example, for a compressed
spring, we must calculate the integral over the displacement
of the end of the spring and not over the displacement of its
center of mass.

 

[sophiecentaur]

But if a student comes to us and asks why the work done by the road on the car

Yes; a problem. However, that sort of question will mostly, only be asked when previous teaching has made the student thing that it's essential to consider the work situation. We all know that this Work thing is harder than one would think so why use it when not absolutely necessary, or at least point out that there are many situations in which it really doesn't help. It's having a tool that introduces more confusion than doing a job another way.

I guess there are parallels with Photons and Electrons being used far too early by students who somehow feel those two concepts are necessary to explain basic macroscopic phenomena. Nonsense like "Why did they get the sign of the current wrong?" is an understandable quandary and pointless hurdle for many students to deal with.

The problem with trying to resolve frequently held misconceptions with students can result in their leaving with the wrong lesson being learned. (They woke up up half way through the lesson)

 

[jbriggs444]

Yes; a problem. However, that sort of question will mostly, only be asked when previous teaching has made the student thing that it's essential to consider the work situation. We all know that this Work thing is harder than one would think so why use it when not absolutely necessary, or at least point out that there are many situations in which it really doesn't help. It's having a tool that introduces more confusion than doing a job another way.

That sounds plausible in the abstract. Can you make it concrete?

For instance, we have a car on a road. The car starts with some velocity and is subject to a constant force. We are asked for the final velocity. We could apply the concept of work, multiply force by distance to get the change in the car's kinetic energy. Add the change to the car's original kinetic energy, get the final kinetic energy and extract the final velocity.

Your alternative, I suppose, would be to find the right SUVAT equation: $v^2 = u^2 + 2as$. That is the one SUVAT equation that I never bothered to memorize because I prefer the notion of "work".

So far, you seem to have a viable proposal. But what if have a large reservoir of compressed gas that is used to slowly push a piston across a known displacement in a small cylinder. The piston is further connected to a crankshaft, the details of which are irrelevant. We are asked for the energy supplied to the piston by the gas.

Now the SUVAT equations are no longer applicable. The most convenient way to proceed is through the notion of mechanical (or thermodynamic) work.

 

[sophiecentaur]

For instance, we have a car on a road. The car starts with some velocity and is subject to a constant force. We are asked for the final velocity. We could apply the concept of work, multiply force by distance to get the change in the car's kinetic energy. Add the change to the car's original kinetic energy, get the final kinetic energy and extract the final velocity.

I challenge you to find many A level students who would avoid Suvat and use work for that.Any problem involving vehicles is full of unstated extra effects. I remember that nagging feeling about 'constant accleration' in any car question and that was even before I had learned to drive.

That is the one SUVAT equation that I never bothered to memorize because I prefer the notion of "work".

Really? SUVAT is just one of those things that are taught very early on. Did you already have an understanding of mechanical work? I can understand you having a preference and I prefer it too. But that's another issue.

But what if have a large reservoir of compressed gas that is used to slowly push a piston across a known displacement in a small cylinder.

What indeed? Where would that sort of problem sit on the timeline of learning mechanics, though?

That sounds plausible in the abstract.

Yes; it is abstract because it's about how easily (or not) people can learn this stuff. If you acknowledge that you are aware of (other) people's problems with all the questions I have quoted above then you would need to address them and not give the 'right answers'. Those right answers assume familiarity with a whole lot of other stuff which (asleep in class etc.) they don't keep in their heads.

Logically, the only way to respond to the sort of naive question that's often posted on PF would be to start with a health warning: "Unless you actually understand a whole lot of mechanics already, then don't attempt to understand the following:".
PS "and don't you come back no more no more"

 

[jbriggs444]

Any problem involving vehicles is full of unstated extra effects. I remember that nagging feeling about 'constant accleration' in any car question and that was even before I had learned to drive.

Of course you are right. First year physics is full of idealizations and simplifications. We usually ignore air resistance. We assume frictionless surfaces, rails and axles. We are dealing with students who have not been exposed to calculus, so we avoid non-constant forces.

Really? SUVAT is just one of those things that are taught very early on.

Almost all of the SUVAT equations are intuitively obvious. Nothing to really memorize. You just write them down when needed. For me, the only one that does not immediately flow from the pen is $v^2 = u^2 + 2as$. But that equation obviously the same thing as the work energy theorem: $E_f = E_i + F \cdot s$ with a factor of $\frac{2}{m}$ thrown in. One less thing to memorize.

For me, understanding is fun while memorization is hard. I have a decent memory, but an ample supply of laziness.

Did you already have an understanding of mechanical work?

It is hard now for me to recall what my understanding was like back then. I certainly understood the "center of mass" work that was being taught. That was enough to get 100% on all of the tests.

But the understanding felt flawed. What we were taught did not include the possibility of external forces exerted on moving parts within non-rigid bodies. It was obviously not accounting for the complete energy flow. I had the private idea of mechanical work, but no formal coursework to provide confirmation or a name to go with the notion.

That was frustrating and uncomfortable for me. It made me want more detail from my physics instructors. Not less.

 

[sophiecentaur]

It made me want more detail from my physics instructors.

I was always too lazy to perform well in tests but my retention has always been a lot better than people who always got better exam grades; many of them went 'high flying ' and promptly forgot all that interesting g Physics.
But my comments in this thread were not aimed at 100%'ers. I suspect we are a bit in quadrature about this issue.

For me, understanding is fun but memorization is hard.

Yeah well . . . your memorisation was clearly subconscious (you lucky man).