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Heat and the work done by friction

February 12, 2012

It’s been a while since I’ve written about a purely physics related topic, so I thought I would try to write up an interesting discussion I’ve been having with some s about colleagues about the work done by friction, heat, and how best to teach these concepts to introductory level students (9th graders for me).

It all stared when one of my colleagues invited us to give feedback on a test he’d written for his 9th grade honors physics class. At the top of the test, he’d written a bunch of formulas for energy, including the following:

\begin{array}{rcl}  KE&=&\frac{1}{2}mv^2\\  GPE &=&mgh\\  Heat&=&F_f\Delta x\\  \end{array}

Energy is one of those topics can easily succumb to the formula zoo, and students can often focus too much on memorizing formulas and not enough on thinking about conservation of energy. I think to combat this, my colleague presents fairly complicated, multi-step energy problems that force students to think about many different types of energy, and energy losses, and in order to help them not be totally at sea, provides them with a number equations to use. Giving out equations is a practice I understand pedagogically, but don’t do myself, mainly because I try to work to minimize the number of equations we use, and worry that starting the test off with a list of equations would leave them thinking that physics is about the equations. But maybe the equation sheet helps defuse this by worry by giving them the equations, implicitly telling them they will need more than equations to do well on the test. I should think about this more.

The thing that really got me thinking was the last equation Heat=F_F\Delta x, which sent my physics antennae off. I knew there were lots of arguments why this isn’t fully right from a physics point of view, but I couldn’t remember them at the moment, and more importantly, I’m unsure how many of these arguments are really appropriate for a first year physics course.

So here’s the argument, reconstructed from some conversations with Mark Hammond, re-reading Matter and Interactions, and Bruce Sherwood’s papers, Work and Heat in the Presence of Sliding Friction, and Pseudowork and Real Work. If you’ve read these papers, or are familiar with this argument, you can probably skip this part, since it consists mostly of me trying to repeat Sherwood’s arguments for my own understanding.

Explicitly calculating the work done by the frictional force

Let’s start with a block, pushed across a table at constant speed, by a constant force F. If the speed of the block is constant, the block must be experiencing a frictional force opposite its motion, and this force must be equal in size to F in order for the net force to be zero.

Let’s look at this from an energy perspective, assuming the block is pushed a distance \Delta x. We’ll start by considering a system consisting only of the block. The work done by the pushing force is F\Delta x. But since the kinetic energy of the system is unchanged, some other force must have done an equal amount of negative work on the block. And since the frictional force acting on the block is equal to -F, it makes sense to think the frictional force force of the table did -F\Delta x of work. But there’s only one problem with this—if you feel the bottom of the block, it will be warmer; something must have increased the thermal energy of the block. So the table must have removed less energy than what you put into the block by pushing it, and W_{table} < F_f\Delta x.

We can view this in an LOL energy chart.

Energy LOL chart for a block pushed along a table at constant speed.

Revised energy LOL diagram to account for the thermal energy increase of the block. In this case, the work done by the frictional force can no longer be equal to the work done by the hand.

But how could it be that the work done by the frictional force is less F_f\Delta x? We know the size of the frictional force must be equal to F, so the only explanation is that \Delta x is somehow smaller.

To understand this, Shrewood points us to the difference between the Work-Kinetic Energy theorem and the First Law of Thermodynamics. The Work-Kinetic Energy thoerem, usually derived in introductory courses by combining Newton’s Second Law, a=\frac{F_{net,ext}}{m}, and the constant acceleration equation, v^2=v_i^2+2a\Delta x, tells us that:

F_{net,ext}\Delta x=\Delta KE

For our object moving at a constant velocity (F_{net}=0) this equation tells us that the kinetic energy of the system won’t change. Note that this result is the same whether we choose a system consisting only of the block where the applied force would be equal to the frictional force), or of the block and the table, where the applied force would be equal and opposite to the external force of the floor on the table. In both cases, the net force on the system is zero.

The other law we should consider is the First Law of Thermodynamics, which tells us that the change in energy of the system is equal to the sum of energy transfers (work, heat, etc), across the boundary of the system. In both of our systems, we will assume that the only energy transfers are caused by work. In this case, the First Law of Thermodynamics can be written

W_{net}=\Delta E_{thermal}

For the case of the system with the table and the block (we will assume the contact table is stationary, and therefore the frictional force of the floor does no work), only the external force pushing the block is doing work. In this case, we can write:

\begin{array}{rcl}  W_{net}&=&\Delta E_{thermal}\\  F\Delta x &=& \Delta E_{thermal\; of \;block\; and\; table}\\  \end{array}

But if our system is reduced to be simply the block, the frictional force of the table acting on the block is now doing negative work, -F*\Delta x_{eff}. \Delta x_{eff} is a distance smaller than the total displacement of the center of mass \Delta x, since we are allowing for the possibility for the the points of contact between the block and the table to not move as far as the entire block.

So in this case, the change First Law of Thermodynamics states:

\begin{array}{rcl}  W_{net}&=&\Delta E_{thermal}\\  F\Delta x-F\Delta x_{eff} &=& \Delta E_{thermal\; of \;block\; only}\\  \end{array}

Now consider the case where the the block and table are identical (suppose we have two identical blocks sliding past one another). In this case, there will be no heat transfer between the blocks, and the the change in thermal energy of the top block will be half the total change in thermal energy we calculated for the system of the two blocks (\Delta E_{thermal,\; block}=\frac{1}{2}\Delta E_{thermal, 2 blocks}).

But when we apply this back to our analysis of the single block system using the first law of thermodynamics, we find a surprising result:

\begin{array}{rcl}  F\Delta x-F\Delta x_{eff} &=& \Delta E_{thermal\; of \;block\; only}\\  F\Delta x-F\Delta x_{eff} &=& \frac{\Delta E_{thermal\; 2 blocks}}{2}\\  \end{array}

Knowing that \Delta E_{thermal\; 2blocks}= F\Delta x, we can write:

\begin{array}{rcl}  F\Delta x-F\Delta x_{eff} &=& \frac{F\Delta x}{2}\\  \Delta x_{eff} &=& \frac{\Delta x}{2}\\  \end{array}

Which is quite a surprising result. But this does allow us to fully account for the energy in the system. For a block sliding across an identical block, the external force does work equal to F\Delta x and half of that energy input is taken from the system as work done by the frictional force, while the other half is transformed into thermal energy of the upper block. This makes sense because that other half of energy that is taken by work would go into the lower block, and result in an equal increase in thermal energy for the lower block, which is what we should expect for identical blocks.

Sherwood takes this even further in his paper, where he considers the cases where the blocks are not identical, lubricated friction and more. But the basic takeaway that the work done by the frictional force acting on an object sliding across a surface is always less than F_f\Delta x.

How this affects an example problem

Now consider the pretty classic physics problem that asks a student to calculate the work done by friction when a block is dragged across a level flood a distance, d. We already know that since the object is moving at constant velocity, the frictional force must be equal to the applied force, F.

We have previously seen that the complete answer to this problem is -F_f,d_{eff} where d_{eff} is some unknown distance less than the total displacement of the block, that depends on the material composition of the two surfaces that are interacting.

If we apply the first law of thermodynamics to the system consisting only of the block moving at constant velocity, we must also consider that the temperature of the block’s lower surface will be higher than the cold parts of the table it will encounter as it slides, and so there will also be a net transfer of heat (Q) from the block to the table.

Fd-Fd_{eff}-|Q|=\Delta E_{thermal,\; block}

If we consider the system of the table only, we get

Fd_{eff}-|Q|=\Delta E_{thermal,\;table}

and if we combine these two systems we get:

Fd=\Delta E_{thermal,\;block+table}

And this makes sense—the work done by the external force is transformed into thermal energy which is distributed in both the block and the table.

Implications and questions for intro physics students

Obviously, it’s not wise to go into this level of analysis with a first year physics course, especially a 9th grade one. So the question is how much of this do you share with students? Is the equation Heat=F_fd a valid one for introductory students? It certainly allows them to come out with right answers on most problems we can devise at level.

My thought is that it is important that students are thinking carefully about systems when they are introduced to energy—to me, it is critically important that they understand conservation not from a “energy can’t be created or destroyed” or as mgh=\frac{1}{2}mv^2, but that the total energy of a system can’t change unless there is a transfer of energy across the system boundary. Mastering this idea can pay huge dividends in later studies of chemistry and biology, and it’s one of the reasons I love energy LOL diagrams.

The old misconception zealot in me also doesn’t like the use of the word heat in this particular context—heat really should be the flow of energy between the system and surroundings due to a change in temperature, but I wonder if this distinction is too much for introductory students, and if introducing this distinction is counter-productive at this level.

And the idea that the work done by friction is equal to F_fd is useful for a ton of problems at the introductory level. If you decide to be strict, and not call F_fd the work done by friction, what is a better term that doesn’t open up the entire can of worms above?

7 Comments leave one →
  1. February 12, 2012 9:17 pm

    The distinction regarding the use of the word “heat” is not, I believe, too much for introductory students, as long as it’s presented properly. It is important to note that heat is the transfer of energy due to a difference in temperature between the system and its surroundings (not “due to a change in temperature” which may reinforce the misconception that heat and thermal energy are the same). Introductory students are right with you when you talk about a system gaining (or losing) energy because the surroundings are warmer (or cooler) than the system. This concept seems totally commonsensical to them. Heat defined this way just seems right.

    Heat defined as something that happens when work is done on a system and the surroundings (or the system, or both) change temperature gets terribly, terribly confusing.

    Potential student response: Which direction was the flow of energy? If the table got warmer, shouldn’t it warm up the box (thereby transferring energy into the system, not out of it)? Wait, is THIS why the box is warmer? Do we need an extra arrow pointing into the system in the diagram? No, wait the box’s heat increased… no, the table’s did… no, what? Why are you screaming “thermal energy, not heat” at me now?

    So “Heat = F_f \Delta{x}” I really don’t like at all, because it mixes up what heat really is.

    I do, however, tell my first year students that they can calculate the amount of energy removed from the system by friction by putting the object exerting the friction force outside the system and simply pretending that the system’s thermal energy doesn’t increase. This is clearly wrong, so why does it work? It works because our model of sliding solid-on-solid friction is a pretty rough model of what is really going on. To get the real answer, we need a more refined model of friction (insert advertisement for second year physics here).

    So we use a crummy model of friction, add some fiction, and we somehow get a good answer. Once one does this fictional frictional calculation, you can make yourself feel better by just putting both objects interacting via friction into the system, and show an increase in the thermal energy equal to the amount of friction work done in the first calculation. But note that you did not REALLY get \Delta E_{therm} from F_f \Delta{x}! While, this is proper, given my caveat, you can’t say how much of that thermal energy ended up in either object (except for the identical box-on-box situation you noted above).

  2. Meechan permalink
    October 20, 2017 4:46 pm

    Thank you for writing this, as I am currently wrestling with how much to go into it with my first-year calculus-based college physics class. Interestingly, I never learned the friction model properly in school (I have a PhD in Physics), so this topic was apparently sort of glossed over and was never adequately covered, unless I just don’t remember it.
    I never use the word “heat” in the context of friction, but identifying the system properly, discussing the point contacts between objects, deriving the expressions being careful with all the subscripts etc. takes so much of the lecture time and ends up confusing the students that it’s a challenge to give it a proper treatment.

  3. Lonegreat permalink
    May 17, 2019 2:01 pm

    Exactly… I was with this problem and thinking and thinking but u have helped me to shed some light on it ….Thank you

  4. metacarpo10 permalink
    November 17, 2020 12:54 pm

    I didn’t really understand the argument. When moving the block on the table, you are making work that is F.Dx, and friction is making -F.Dx, I would think that this work done by friction is exactly the dissipation of energy you are putting on the system in the form of heat, or even better, transforming kinectic energy into thermal energy, so the net kinectic energy does not change. If you day the work done by friction is less (in absolute terms) then the work you are doing on the book, then the book should be accelerating! What am I thinking wrong here?

    • November 17, 2020 1:56 pm

      The key point in the argument is that the point(s) of contact between the block and the table is not moving as far as the block itself, because those points of contact are somewhat flexible and they stick to the table, so they won’t move as far as the block itself moves. This is the \Delta x_{eff}, the effective distance that the points of contact move. Because of this, the work done on the block by the person will be greater than the negative work done by friction. The difference between these two quantities goes into raising the thermal energy of the block.

      • January 31, 2022 12:22 pm

        This just isn’t true. All of the work done by friction dissipates into heat in the surfaces in contact. Just let go of the block and observe the deceleration: Delta K = Ff delta x will hold.

        It’s not mathematically complicated, it just doesn’t play well by by either mechanical or thermodynamic principles, because it bridges the macro-micro gap. Ordered mechanical KE is passed to entropic particle KE through the electrical interactions of the surface atoms.

  5. January 31, 2022 12:29 pm

    You’ve gone off the deep end a bit here…
    Heat = Ff delta x is numerically correct, it’s just a bad way to notate it.
    No heat is exchanged between the surfaces, they are both heated by the friction process symmetrically (Q = Q, temp changes depend on specific heats).
    Since this is not how Q works otherwise in thermodynamics, and the work in question isn’t the p delta V type, I think it’s best to avoid mentioning either to new students and present it as follows:

    The work done by friction *dissipates* macroscopic KE into microscopic thermal KE, heating up both surfaces. Since work done by friction is inherently negative,
    delta K = Wf (< 0)
    delta Eth = – Wf or | Wf |
    The increase in thermal energy is true for both surfaces, and calling it this avoids any confusion surrounding Q or Wth.

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