Modelling assumptions when friction is involved

[etotheipi]

Whenever friction exists within a mechanics problem, there must be some dissipation of mechanical energy into thermal energy. However, I'm not sure how we determine which bodies can or cannot possesses thermal energy.

Suppose we consider the case of a block sliding down a rough wedge, which is fixed to the Earth. Treating the whole Earth-wedge-block system gives something along the lines of $U_{1} = U_{2} + T_{block} + E_{th}$.

There are essentially two options, either the wedge gains all of this thermal energy, or it is split between the wedge and the block. If the block is specified to be rigid, the former applies and we might then say that $E_{th} = f_{k}d$ if $f_{k}$ is the frictional force applied by the block on the wedge. If not, I'm not entirely sure how it's possible to solve for the final velocity of the block.

My guess is that it is assumed in any mechanics problem that the quote-and-quote "main" object in the question is rigid (in this example, the block), and all of this thermal energy is dissipated in other components. I wondered whether anyone could think of any exceptions? On perhaps another level, does it matter/make any difference to the answer where this thermal energy is dissipated? I'm inclined to say yes, simply because I don't believe there is any way of computing $E_{th}$ unless it ends up in one body only.

Please do let me know if this is confused. Thanks!

 

[PeroK]

On perhaps another level, does it matter/make any difference to the answer where this thermal energy is dissipated?

As far as the purely kinetic model is concerned it makes no difference how or where the energy is lost. It makes no difference whether it is thermal energy inside the bodies, thermal energy of the surrounding air or sound waves in the air - or even radiated away as EM radiation.

These dissipative processes themselves are outside the scope of the model.

As you improve your model, the form of the equations, e.g. involving coefficient of drag and dependence on the square of the velocity, may reflect a deeper knowledge of how the dissipative processes operate. But, the details of the processes themselves remain irrelevant to the kinematic equations.

 

[etotheipi]

As far as the purely kinetic model is concerned it makes no difference how or where the energy is lost. It makes no difference whether it is thermal energy inside the bodies, thermal energy of the surrounding air or sound waves in the air - or even radiated away as EM radiation.

In the instance where only the block gains thermal energy, $E_{th} = f_{k}d$. Since the mode of dissipation is then unimportant, is that to say that the total $\Delta E_{th}$ will always be $f_{k}d$ (even if, for instance, there are a pair frictional forces contributing to it)?

If we wanted to solve the system then the two routes would either be work-energy on the block $W_{grav} + W_{fr} = \Delta T$, but this would only apply if the block were rigid.

Or if the block is no longer constrained to be rigid, we could try a more general energy conservation for the entire system, in which case we need to come up with a number for $\Delta E_{th}$ somehow. It's not immediately obvious how to do this, thought it seems sensible to also call it $W_{fr}$ so that the block ends up with the same final translational speed.

 

[jbriggs444]

Or if the block is no longer constrained to be rigid, we could try a more general energy conservation for the entire system, in which case we need to come up with a number for $\Delta E_{th}$ somehow. It's not immediately obvious how to do this, thought it seems sensible to also call it $W_{fr}$ so that the block ends up with the same final translational speed.

If what you want to know is "how far does the block slide before coming to rest", then, as @PeroK pointed out, the resulting thermal energy is irrelevant.

If what you want to know is "how hot does the block get" then you are out of luck. Conservation laws will not get you there.

 

[etotheipi]

If what you want to know is "how far does the block slide before coming to rest", then, as @PeroK pointed out, the resulting thermal energy is irrelevant.

I'm only really concerned about the kinetic energy of the block at the bottom of the ramp, and where everything fits into a conservation of energy expression.

The block is acted upon by weight and friction, the wedge by friction (and weight, but this is irrelevant). In the case of the block, the frictional force is responsible for reducing its kinetic energy and perhaps increasing its thermal energy. For the wedge, which I will assume is fixed, friction only increases its thermal energy.

In this scenario, work-energy doesn't really help us so our only option is conservation of energy, $\Delta U + \Delta T + \Delta E_{th} = 0$. Then we've got to figure out what $\Delta E_{th}$ is.

I'm slightly at a loss on how to calculate that number, since there are two frictional forces contributing to it. My "instinct" is that it will still just be $f_{k}d$, but I can't really rationalise this. If the block were rigid and all the thermal energy went to the wedge, then evidently the work done on the wedge would equal $\Delta E_{th} = f_k d$. But now we've got two objects to worry about!

This is only a problem when the frictional forces are internal to the chosen system.

 

[Dale]

I'm only really concerned about the kinetic energy of the block at the bottom of the ramp, and where everything fits into a conservation of energy expression.

Then which object heats up is irrelevant. You can correctly calculate the KE of the block without concern about where the thermal energy goes.

 

[etotheipi]

Then which object heats up is irrelevant. You can correctly calculate the KE of the block without concern about where the thermal energy goes.

But if I take the whole configuration to be a system, then what do I take to be $\Delta E_{th}$? In the case that both are capable of possessing thermal energy, I could take the block and wedge individually in turn as systems

Block: $mgh - f_{k}d = \frac{1}{2}mv^{2} + \Delta E_{th, block}$.

Wedge: $f_{k}d = \Delta E_{th, wedge}$

Adding these just causes the $f_k d$ terms to cancel,

$mgh = \frac{1}{2}mv^{2} + \Delta E_{th, tot}$

 

[PeroK]

But if I take the whole configuration to be a system, then what do I take to be $\Delta E_{th}$? In the case that both are capable of possessing thermal energy, I could take the block and wedge individually in turn as systems

Block: $mgh - f_{k}d = \frac{1}{2}mv^{2} + \Delta E_{th, block}$.

Wedge: $f_{k}d = \Delta E_{th, wedge}$

Adding these just causes the $f_k d$ terms to cancel,

$mgh = \frac{1}{2}mv^{2} + \Delta E_{tot}$

I propose a different model, where energy is dissipated as EM radiation, rather than internal heat. Let's say that for every $d$ of relative motion between the surfaces, a total energy of $E$ is radiated as EM radiation. The ultimate source of this energy is the KE of the block (or at least the reduction in relative motion of the surfaces).

This results in an effective force on the block of $f = E/d$. By experiment we find that, in fact, $E/d = \mu N$, where $\mu$ is some constant determined by the surfaces and $N$ is the normal force between the surfaces.

Then you don't have to worry about thermal energy any more. Instead, if you want to check conservation of energy, you need to measure the energy of all the EM radiation. But that doesn't matter, because whether you measure it or not the energy is lost to the block. It doesn't really matter to the model whether that energy is still in our universe or not. Even if it is irretrievably gone, the simple kinetic model still works.

 

[hutchphd]

The friction energy goes to another dimension.
It is outside what you have chosen to call your system. Of course it is physically "inside" the materials in question, but it is outside the configuration space you have chosen to describe the salient features of your system

 

[etotheipi]

This results in an effective force on the block of $f = E/d$. By experiment we find that, in fact, $E/d = \mu N$, where $\mu$ is some constant determined by the surfaces and $N$ is the normal force between the surfaces.

Ah okay, so first things first I should in theory replace $E_{th}$ with a more general term encapsulating any other medium that energy could be dissipated into.

But even if I limit that term to just thermal energy for simplicity, what I believe you are saying is that $\Delta E_{th}$ would equal the product of the frictional force and the distance moved.

That is indeed consistent with what you'd get if we take the model where only the wedge can gain thermal energy. It's a shame it's empirical but at least it makes sense.

 

[A.T.]

That is indeed consistent with what you'd get if we take the model where only the wedge can gain thermal energy.

Why only the wedge? The model says nothing about where the dissipated energy goes to.

 

[etotheipi]

Why only the wedge? The model says nothing about where the dissipated energy goes to.

My impression was that if we let the block be rigid but allow only the wedge to possesses thermal energy, then we end up with the following system

Block: $mgh - f_{k}d = \frac{1}{2}mv^{2}$.

Wedge: $f_{k}d = \Delta E_{th, wedge}$

And then if we take the system as both the block and the wedge, $\Delta E_{th, tot} = \Delta E_{th, wedge} = f_k d$. So this setup let's us calculate the thermal energy explicitly by considering work done on the wedge, since that's the only term on the RHS.

If we give both of them thermal energies, then the total amount will be split between them. This was the part I was unsure about. Going with the idea of friction as a measure of energy dissipated per unit distance, it appears that this total thermal energy will be the same in all scenarios (e.g. 100%/0%, 50%/50% etc.)

It appears to be the case that $\Delta E_{th, etc.}$ dissipated due to the action of a pair of frictional forces will be $f_{k}d$.

 

[A.T.]

It appears to be the case that $\Delta E_{th, etc.}$ dissipated due to the action of a pair of frictional forces will be $f_{k}d$.

You can also sum the work done the two frictional forces (with the proper signs) to get the mechanical energy dissipated or generated at the interface.

 

[etotheipi]

You can also sum the work done the two frictional forces (with the proper signs) to get the mechanical energy dissipated or generated at the interface.

Though wouldn't that approach just give $f_{k}d - f_{k}d = 0$?

 

[kuruman]

I propose a different model, where energy is dissipated as EM radiation, rather than internal heat. Let's say that for every dd of relative motion between the surfaces, a total energy of $E$ is radiated as EM radiation. The ultimate source of this energy is the KE of the block (or at least the reduction in relative motion of the surfaces).

Yes, a total energy of $E$ is radiated away as EM radiation but that happens eventually. I believe it is more appropriate to consider the details of power dissipation $\dfrac{dK}{dt}$.

 

[PeroK]

My impression was that if we let the block be rigid but allow only the wedge to possesses thermal energy, then we end up with the following system

Block: $mgh - f_{k}d = \frac{1}{2}mv^{2}$.

Wedge: $f_{k}d = \Delta E_{th, wedge}$

And then if we take the system as both the block and the wedge, $\Delta E_{th, tot} = \Delta E_{th, wedge} = f_k d$. So this setup let's us calculate the thermal energy explicitly by considering work done on the wedge, since that's the only term on the RHS.

If we give both of them thermal energies, then the total amount will be split between them. This was the part I was unsure about. Going with the idea of friction as a measure of energy dissipated per unit distance, it appears that this total thermal energy will be the same in all scenarios (e.g. 100%/0%, 50%/50% etc.)

It appears to be the case that $\Delta E_{th, etc.}$ dissipated due to the action of a pair of frictional forces will be $f_{k}d$.

A useful exercise would be to consider a block thrown onto the surface of a trolley that is free to roll without loss of energy.

Then compare this with a block thrown onto a table that is fixed to the Earth, effectively.

The other consideration you are missing is momentum. We do not have conservation of momentum in the case of friction on a fixed surface. Even if you deal with thermal energy to your satisfaction, you have conservation of momentum to worry about.

 

[etotheipi]

A useful exercise would be to consider a block thrown onto the surface of a trolley that is free to roll without loss of energy.

Then compare this with a block thrown onto a table that is fixed to the Earth, effectively.

I suppose with rolling there is no motion of the contact point so the work done by friction is zero, and the trolley would just carry on forever (assuming no rolling resistance).

Whilst the block thrown onto the table would eventually come to rest wrt the surface of the Earth. If we make the approximation that the moment of inertia of the Earth is so large that we can assume the Earth is fixed, we might then just say the kinetic energy of the block is dissipated via some frictional mechanism until it comes to rest.

The other consideration you are missing is momentum. We do not have conservation of momentum in the case of friction on a fixed surface. Even if you deal with thermal energy to your satisfaction, you have conservation of momentum to worry about.

If the system is the block then momentum isn't conserved since the external force is non-zero. In an inertial frame in space, momentum is definitely conserved. In the lab frame fixed wrt the table, momentum definitely isn't conserved, as you say, since the surface of the Earth is always at rest relative to us (this frame is non-inertial).

 

[PeroK]

I suppose with rolling there is no motion of the contact point so the work done by friction is zero,

That's not what would happen.

 

[PeroK]

Ps You can set up a problem for yourself: trollley of mass $M$, block of mass $m$, thrown onto trolley with initial velocity $v$ and coefficient of friction $\mu$.

What is the final velocity of the truck?

Where have you seen this equation before? So, what type of event has taken place?

 

[etotheipi]

That's not what would happen.

Hmm. For the two to coalesce the initial collision would have to be inelastic, so right off the bat some energy has to be dissipated somewhere. I'm guessing thermal in the trolley/block.

Then we're left with an object rolling at some speed relative to the surface of the Earth. The rolling wheel encounters no friction on a flat surface. Air resistance against the trolley, maybe.

 

[etotheipi]

Ps You can set up a problem for yourself: trollley of mass $M$, block of mass $m$, thrown onto trolley with initial velocity $v$ and coefficient of friction $\mu$.

What is the final velocity of the truck?

Where have you seen this equation before? So, what type of event has taken place?

Right I won't be a minute

 

[PeroK]

Hmm. For the two to coalesce the initial collision would have to be inelastic, so right off the bat some energy has to be dissipated somewhere. I'm guessing thermal in the trolley/block.

Then we're left with an object rolling at some speed relative to the surface of the Earth. The rolling wheel encounters no friction on a flat surface. Air resistance against the trolley, maybe.

This is the problem. We're not worried about air resistance or rolling resistance (or the mechanism of thermal energy dissipation). You are not seeing the wood for the trees.

Part of what you have said gives you a shortcut to the answer.

 

[etotheipi]

This is the problem. We're not worried about air resistance or rolling resistance (or the mechanism of thermal energy dissipation). You are not seeing the wood for the trees.

Part of what you have said gives you a shortcut to the answer.

I lied, I might be a little more than a minute...

Initially went to momentum conservation, then realized it probably wasn't what you're looking for. Either way, this gives $v = \frac{m}{M+m}u$.

I'm trying to think about how the coefficient fits into it. For the little mass

$-\mu N = m\frac{dv}{dt}$

whilst the relative velocity of the little mass wrt the trolley is $v - v_{t}$ where $v_{t}$ is also a function of time, such that

$\mu N = M\frac{dv_t}{dt}$

Then you've got something like

$\int_{v}^{v_{f}} m dv = \int_{0}^{v_{f}} -M dv_t \implies mv_{f} -mv = -Mv_{f} \implies v_{f} = \frac{m+M}{m}v$

Or

$v_{f} = \frac{m}{M+m}v$

 

[PeroK]

I lied, I might be a little more than a minute...

Initially went to momentum conservation, then realized it probably wasn't what you're looking for. Either way, this gives $v = \frac{m}{M+m}u$.

I'm trying to think about how the coefficient fits into it. For the little mass

$-\mu N = m\frac{dv}{dt}$

whilst the relative velocity of the little mass wrt the trolley is $v - v_{t}$ where $v_{t}$ is also a function of time, such that

$\mu N = M\frac{dv_t}{dt}$

Then you've got something like

$\int_{v}^{v_{f}} m dv = \int_{0}^{v_{f}} -M dv_t \implies mv_{f} -mv = -Mv_{f} \implies v_{f} = \frac{m+M}{m}v$

I must have made a mistake somewhere...

Yes, obviously that term should be inverted. The point is that we have a totally inelastic collision. It doesn't matter if the collision is by friction, hitting part of the trolley or a combination of both. The final velocity is determined by conservation of momentum.

 

[etotheipi]

Cool, thanks for your help.

At the end of the day lots of this energy stuff can be figured out with common sense, but I'm trying to be intentionally pedantic just to try and formalise a lot of the stuff I've covered over the past few years.

I'm finding energy especially is one of those topics that's a constant cycle of learning something and then realising that interpretation is limited, then needing to sort of "re-learn" a new framework, and then that same thing happening again. I guess that applies to a lot of Physics though. "If it ain't broke then don't fix it" works for a little while and then I run into problems and need to spend a lot of time just relearning the topic from the ground up in a different way.

 

[PeroK]

Cool, thanks for your help.

At the end of the day lots of this energy stuff can be figured out with common sense, but I'm trying to be intentionally pedantic just to try and formalise a lot of the stuff I've covered over the past few years.

I'm finding energy especially is one of those topics that's a constant cycle of learning something and then realising that interpretation is limited, then needing to sort of "re-learn" a new framework, and then that same thing happening again. I guess that applies to a lot of Physics though. "If it ain't broke then don't fix it" works for a little while and then I run into problems and need to spend a lot of time just relearning the topic from the ground up in a different way.

Possibly one interesting detail that as come out of this is: If we let the second object move, then the frictional force accelerates it (no loss of energy directly). But, the force decelerates the incoming object (which is where the energy loss comes from).

If you imagine the friction with the wedge accelerating the wedge-Earth a tiny amount, then that interaction incurs no energy loss. That's why you don't double count the friction force and get twice the energy loss.

 

[A.T.]

Though wouldn't that approach just give $f_{k}d - f_{k}d = 0$?

No, because the displacement is different for the two contact patches, if they are in relative motion (sliding).

 

[etotheipi]

No, because the displacement is different for the two contact patches, if they are in relative motion (sliding).

I'm not sure I follow. We might consider making the block arbitrarily small, the point of application of both frictional forces (block on wedge and wedge on block) would be the same point. Then the displacements of both forces should be equal.

 

[A.T.]

I'm not sure I follow. We might consider making the block arbitrarily small, the point of application of both frictional forces (block on wedge and wedge on block) would be the same point. Then the displacements of both forces should be equal.

No, the displacements are different, because the physical points of force application (the material parts of each body were the force is applied) are moving relative to each other.

 

[etotheipi]

No, the displacements are different, because the physical points of force application (the material parts of each body were the force is applied) are moving relative to each other.

I've realized I've drawn cylinders instead of a block (please ignore this part, I didn't intend for any rolling, just sliding!). Though I can't see why the two displacements would be any different.

 

[A.T.]

Though I can't see why the two displacements would be any different.

Displacements are integrals of velocity, but you can also consider the velocities and thus the instantaneous power, instead of work. Since the velocities of the contact patches are different, the equal but opposite kinetic friction forces are not doing equal but opposite work/time (power). The discrepancy represents the dissipated energy.

 

[etotheipi]

Since the velocities of the contact patches are different

I might need to come back to this in the morning in the case that I'm missing something obvious since it seems I'm stuck in a bit of a pantomime-style "Oh yes it is!", "Oh no it isn't!" situation, but I really struggling to figure this out.

To me it seems there is only one common point of contact between our arbitrarily small block and the wedge, and the velocity of the contact point is obviously equal to the velocity of the block. At all points in the motion, the two opposite frictional forces act at the same point (the contact point). Both undergo the same displacement by a vector $\vec{d}$ as measured in the Earth frame in which the wedge is fixed.

I do apologise if I'm missing the point!

 

[jbriggs444]

I do apologise if I'm missing the point!

The work done by wedge on block is determined by the force of wedge on block multiplied by the motion of the block material at the contact point.

The work done by block on wedge is determined by the force of block on wedge multiplied by the motion of the wedge material at the contact point.

The forces are equal and opposite, of course. That's Newton's third law. But no matter what frame of reference we choose, the motion of the block material and the motion of the wedge material at the contact point will differ. Otherwise we are not dealing with kinetic friction.

[One should draw sliding objects as rectangles rather than as circles to avoid the implication that rolling and static friction are involved]

 

[etotheipi]

The forces are equal and opposite, of course. That's Newton's third law. But no matter what frame of reference we choose, the motion of the block material and the motion of the wedge material at the contact point will differ. Otherwise we are not dealing with kinetic friction.

Ah, okay. I was working under the assumption that work done by a force in a certain frame of reference was the dot product of the force and the displacement of the point of application of the force.

The two bodies are certainly in relative motion. Though the work done as calculated by the definition I gave as opposed to the $P = \vec{F} \cdot \vec{v}$ calculation are then apparently different. I don't understand how this can be the case.

 

[jbriggs444]

I was working under the assumption that work done by a force in a certain frame of reference was the dot product of the force and the displacement of the point of application of the force.

Indeed, that is a common misunderstanding. It is the displacement of the target material at the point of application that matters.

The point of application cannot be in a fixed position relative to both of two slipping surfaces.

 

[etotheipi]

Indeed, that is a common misunderstanding. It is the displacement of the target material at the point of application that matters.

The point of application cannot be in a fixed position relative to both of two slipping surfaces.

That makes life a lot more complicated!

I assume by motion of the 'block material' or 'wedge material' you mean at a microscopic level.

How is it then possible to calculate anything that involves kinetic friction? Since in this case $W=f_{k} d$ no longer has any meaning.

 

[jbriggs444]

I assume by motion of the 'block material' or 'wedge material' you mean at a microscopic level.

No. I mean at an ordinary macroscopic level.

If you have a tire skidding down the pavement, the material of the tire and the material of the road are moving relative to one another. We do not need to delve into the microscopic details of that motion in order to calculate the energy that is dissipated. Instead, we do as @A.T. has suggested:

1. Pick a frame of reference.
2. Using this frame of reference, multiply the motion of the road material by the force of tire on road.
3. Using this same frame of reference, multiply the motion of tire material by force of road on tire.
4. Add the results together.

That's how much mechanical energy has been dissipated. No matter what frame of reference you choose, the result will be the same. The motion of the point of application relative to the frame of reference is irrelevant.

 

[A.T.]

That makes life a lot more complicated!

You seem to be making things more complicated than necessary.

 

[etotheipi]

No. I mean at an ordinary macroscopic level.

If you have a tire skidding down the pavement, the material of the tire and the material of the road are moving relative to one another. We do not need to delve into the microscopic details of that motion in order to calculate the energy that is dissipated. Instead, we do as @A.T. has suggested:

1. Pick a frame of reference.
2. Using this frame of reference, multiply the motion of the road material by the force of tire on road.
3. Using this same frame of reference, multiply the motion of tire material by force of road on tire.
4. Add the results together.

That's how much mechanical energy has been dissipated.

Okay, I finally see what you mean. My head sort of hurts, mainly because I've been living a lie.

Essentially we're saying that zero work is done on the wedge. That's... really odd.

Then everything you say makes sense, since the change in thermal energy would naturally be the total work done by friction which would indeed, in this case, be $f_k d$.

 

[jbriggs444]

Essentially we're saying that zero work is done on the wedge. That's... really odd.

If the wedge doesn't move then no work has been done on the wedge.

If you change to a different reference frame then friction may do work on the wedge. That work can be positive or negative depending on which way the wedge is moving in that reference frame.

Neither Work nor Energy are invariant quantities.

The energy dissipated by kinetic friction is an invariant quantity, however.

 

[etotheipi]

It does throw up a few new questions, the answers to which you might have already given earlier in the thread; I'd need to have a closer look at some of the posts above, since I presume the answer lies in my strange conviction that we must include a thermal energy term.

If the system is only the wedge, the first law of thermodynamics becomes $Q + W = \Delta E = 0$; that is to say the change in thermal energy of the wedge is zero.

My instinct is that the eventual change in thermal energy of the wedge must be due to a transfer of heat from the block to the wedge during its travel, and not due to the block working on the wedge.

If the wedge doesn't move then no work has been done on the wedge.

You and @A.T. have fully convinced me, it's just a little strange since I'd never thought of it like that before. My "mantra" was force multiplied by displacement of point of application. This, in hindsight, always works when the actual material at the point of contact is moving with the force (i.e. someone pushing a rigid body around an axis), but evidently doesn't work in scenarios like these.

Displacement of the material at the contact point. Got it.

Sorry about this, I am conscious of how annoying I'm being...

 

[jbriggs444]

My "mantra" was force multiplied by displacement of point of application. This, in hindsight, always works when the actual material at the point of contact is moving with the force.

Yes indeed. One place where this really becomes important is with rotation. Wheels on roads and such.

 

[Dale]

But if I take the whole configuration to be a system, then what do I take to be ΔEthΔEth\Delta E_{th}?

For the whole system that is just $f_k d$. That is all that you need to know to calculate the KE of the block using conservation of energy.

In the case that both are capable of possessing thermal energy, I could take the block and wedge individually in turn as systems

That is a different problem. If you are ONLY interested in the KE of the block, as you said earlier, then you wouldn’t do this approach. So this is the same scenario but a different problem. Now, you are interested in the division of thermal energy between the two objects. That is no longer just a mechanics problem, but a thermodynamics problem too.

Going with the idea of friction as a measure of energy dissipated per unit distance, it appears that this total thermal energy will be the same in all scenarios (e.g. 100%/0%, 50%/50% etc.)

That would be the thermodynamics part of the problem, but I highly doubt that it will be a simple split. I would assume that, for similar materials in both sides that the slower the block is moved the closer to 50/50 split you get, but the faster you go the larger fraction of the total energy will go into the ramp.

 

[Dale]

How is it then possible to calculate anything that involves kinetic friction? Since in this case W=fkdW=fkdW=f_{k} d no longer has any meaning.

It is actually easier to work with power: $P=f_k \cdot v$. Note that by definition the $v$ is different for the two sides during kinetic friction. This results in some amount of mechanical power that goes into the contact patch that does not go out. This is the mechanical power that is converted to thermal power.

 

[etotheipi]

That is a different problem. If you are ONLY interested in the KE of the block, as you said earlier, then you wouldn’t do this approach. So this is the same scenario but a different problem. Now, you are interested in the division of thermal energy between the two objects. That is no longer just a mechanics problem, but a thermodynamics problem too.

Right, sure. So I suppose the conclusion is that we can't find the final KE of a non-rigid block if we consider the block as a single system.

Whilst we can if we take the wedge-block system and note that the change in thermal energy is the negative of the total work done by internal kinetic frictional forces.

The algebra's a bit weird though. For the block-wedge system

$mgh = \Delta T + \Delta E_{th} = \Delta T + f_k d$

But for the block and wedge individually,

Block: $mgh - f_{k}d = \Delta T + \Delta E_{th, block}$
Wedge $0 = \Delta E_{th, wedge}$

Since the two models have to be consistent, we can only then conclude that $\Delta E_{th, block} = 0$. That can't be right, though, since we've somehow determined that the change in thermal energy of the block + the wedge is zero, whilst we know it to be $f_k d$.

 

[jbriggs444]

Right, sure. So I suppose the conclusion is that we can't find the final KE of a non-rigid block if we consider the block as a single system.

What do you mean by "KE of a non-rigid block"?
What are you using as inputs to an attempted calculation of "KE of a non-rigid block?"
What constraints are you imposing on yourself by "considering the block as a single system"?

I do not understand what you are going on about.

 

[Dale]

Right, sure. So I suppose the conclusion is that we can't find the final KE of a non-rigid block if we consider the block as a single system.

No. I am not sure where that conclusion came from.

 

[etotheipi]

What do you mean by "KE of a non-rigid block"?
What are you using as inputs to an attempted calculation of "KE of a non-rigid block?"
What constraints are you imposing on yourself by "considering the block as a single system"?

I do not understand what you are going on about.

I'm just trying to approach it via the first law of thermodynamics, such that any external work or heat will change the total energy content of a chosen system.

Granted 'non-rigid' is poor terminology, I mean to say that I am allowing the block to possesses thermal energy. As opposed to a rigid body which will only ever possesses mechanical energy.

So if I draw a system boundary around only the block, friction becomes an external force and the total energy of the block becomes its kinetic plus its thermal energy. Then, the work done by friction plus the work done by gravity should equal the total change in the block's energy.

It's fine if we take the block and the wedge as one system since then we have no external work and we can apply conservation of energy in one line.

 

[jbriggs444]

I'm just trying to approach it via the first law of thermodynamics, such that any external work or heat will change the total energy content of a chosen system.

Granted 'non-rigid' is poor terminology, I mean to say that I am allowing the block to possesses thermal energy. As opposed to a rigid body which will only ever possesses mechanical energy.

So if I draw a system boundary around only the block, friction becomes an external force and the total energy of the block becomes its kinetic plus its thermal energy. Then, the work done by friction plus the work done by gravity should equal the total change in the block's energy.

It's fine if we take the block and the wedge as one system since then we have no external work and we can apply conservation of energy in one line.

None of that defines what you mean by "kinetic energy of a non-rigid block".

 

[etotheipi]

None of that defines what you mean by "kinetic energy of a non-rigid block".

I'm referring to translational plus rotational, though in this case just translational. I'm aware the decomposition into macroscopic (mechanical) and microscopic (thermal) KE is somewhat arbitrary, though it seems useful.