4
$\begingroup$

Suppose a box (which I assume to be a rigid body) with an initial velocity that starts to slide on a level surface with friction. Imagine this experiment is done in vacuum, so there is no air drag or dissipation of energy as sound waves. What I've learned is that as the box slides on the surface, the work done by friction decreases the kinetic energy of the box, and part of this energy is converted into the thermal energy of the box itself (it heats up) and part of it exits the box as other forms of energy (the floor heats up).

My textbook suggests the following formula for conservation of energy in a system: $$W_{external}=\Delta E_{mechanical} + \Delta E_{thermal} + \Delta E_{internal}$$ Where $\Delta E_{internal}$ represents the change in forms of internal energy other than thermal energy (that is accounted for separately). I deliberately took the box as the system and didn't include the floor/ground, etc. There is no potential energy, and we neglect other forms of internal energy than thermal energy. So for the "box" system, the equation above can be rewritten as: $$W_{external}=\Delta K + \Delta E_{thermal,box} ~~~~~~~~~~(1)$$

Whereas the work-energy theorem states that: $$W_{net}=\Delta K ~~~~~~~~~~(2)$$

The box is our system, and the only force that is exerted on the box and does work, is friction. Thus $W_{net}=W_{friction}=W_{external}$ (is this correct?). But if we apply this latter result on equations (1) and (2), it gives that $\Delta E_{thermal,box}=0$, meaning that the box has not heated up, which is of course, incorrect.

The only thing I could think of is that the $\Delta K$ in the work-energy theorem also includes the change in thermal energy, as itself is the kinetic energy of the molecules and atoms vibrating (or not?), whereas the $\Delta K$ in equation (1) only accounts for the change in kinetic energy of the center of mass of the box, and the change in thermal energy of the box is written as a separate term, $\Delta E_{thermal,box}$.

This is very confusing. Is there any assumption behind these formulas that I overlooked? What am I missing?

$\endgroup$
4
  • $\begingroup$ The Work Energy Theorem is pretty unrealistic in that it ignores all forms of internal energy. If you already understand where energy can go (thermal, other internal forms), why not just forget it? $\endgroup$
    – Rod Bhar
    Feb 13, 2023 at 16:05
  • $\begingroup$ @RodBhar There is no problem in ignoring it for real-world cases. What I'm interested in is that why the equations don't work out in this ideal situation (or I couldn't interpret them correctly). Also, most of the problems in textbooks, exams, etc. actually assume the object is rigid, but I still need to be able to apply the relevant equations to solve the problem. $\endgroup$
    – M. Bagheri
    Feb 13, 2023 at 16:14
  • $\begingroup$ Well, you state that the only force doing work on the box is friction. Shouldn't you include gravity? $\endgroup$
    – Rod Bhar
    Feb 13, 2023 at 16:20
  • $\begingroup$ @RodBhar Since I assumed that the floor has a level surface, the gravitational force is perpendicular to the path the box follows, and thus does no work on the box. $\endgroup$
    – M. Bagheri
    Feb 13, 2023 at 16:26

6 Answers 6

5
$\begingroup$

In physics mechanics texts a typical assumption is that we are dealing with rigid bodies which have constant internal energy (no "heating" effects), and work is defined as the change in the kinetic energy (work by a conservative force is equivalently the negative of the change in potential energy).

In thermodynamics, the definition of work is expanded to be "energy that crosses a boundary between a system and its surroundings, without mass transfer, due to an intensive property difference other than temperature". The first law of thermodynamics allows work (and heat) to change internal energy.

The mechanics definition of work is always applicable to the center of mass (CM). Some call this work "pseudo work" to distinguish it from the broader thermodynamics definition of work.

See other questions on this exchange regarding work

$\endgroup$
4
$\begingroup$

The work energy theorem states that the net work done on an object equals its change in kinetic energy. What actually happens to the kinetic energy is not covered by the theorem. Only that kinetic energy is either given to the object by net positive work, or removed from the object by net negative work (negative kinetic friction work in your example).

Hope this helps

$\endgroup$
3
$\begingroup$

The work-energy theorem does account for thermal energy, but you have to be careful.

If you think of the box as a collection of atoms that exert forces on each other, you can see that atoms from the floor bump atoms on the bottom of the box and make them all vibrate faster. The loss in kinetic energy of the box has gone into increased kinetic energy of these atoms. These bump other nearby atoms, and the kinetic energy spreads all around the box and floor.

Now you think of the box as an object and reclassify the energy into $\Delta E_{mechanical} + \Delta E_{thermal} + \Delta E_{internal}$. The box atoms used to have an average sideways velocity, and individual $\Delta v$'s from the average. The total energy was

$$E = \sum_i \frac{1}{2} \space m_i \space 2|(\vec{v}_{avg}+\vec{v}_i)|^2$$

$$= \sum_i \frac{1}{2} \space m_i \space\bigl(v_{avg}^2 + 2(\vec{v}_{avg}\cdot\vec{v}_i) + v_i^2 \bigr)$$

If you average the center term over time, it sums to $0$. Likewise at any instant, the average of a large number of velocities in random directions is equally distributed in all directions. So again the center term averages to $0$.

$$E = E_{mechanical} + E_{thermal}$$

You get this before and after the box slides to a stop.

$$\Delta K = \Delta E = \Delta E_{mechanical} + \Delta E_{thermal}$$

You can further divide the thermal energy

$$\Delta E_{thermal} = \Delta E_{thermal, box} + \Delta E_{thermal, floor}$$

In our example, $\Delta E_{interal} = 0$. An example where it would be non-$0$ would be if the heat induced melting or a chemical reaction. This would change the separation of atoms and rearrange chemical bonds. You would need quantum mechanics to correctly calculate the energy in bonds, but you can see that accounting for the energy in the equations above would be straight forward.

$\endgroup$
3
$\begingroup$

The OP running into the issue that the work-energy principle in the form $$ W_{\textrm{external}} = \Delta K_{\textrm{center of mass}} $$ is only correct kinematically and not microscopically and that when dealing with dissipative forces, the best thing to do is to put both objects interacting via that force in the system. The OP has the right idea in the penultimate paragraph

The only thing I could think of is that the $\Delta K$ in the work-energy theorem also includes the change in thermal energy, as itself is the kinetic energy of the molecules and atoms vibrating (or not?), whereas the $\Delta K$ in equation (1) only accounts for the change in kinetic energy of the center of mass of the box, and the change in thermal energy of the box is written as a separate term, $\Delta E_{thermal,box}$.


Here's the analysis I would do. First, take the system = box. The initial condition is that the box is sliding with speed $v$ and therefore kinetic energy $K=\frac{1}{2}mv^2$. The net external force on the box is $F_{\textrm{kinetic friction by surface}}=f_K$, and so, the object slows to a stop, kinematically---that is, via $$ 0=v_f = v_i - 2\frac{f_K}{m}t $$ given the right amount of time ---and the change in kinetic energy of the box is $\Delta K = -\frac{1}{2}mv^2$. Note crucially that we have derived this kinematically/dynamically and not by making an energy argument.

Now, where does the energy go? For that, we take a different viewpoint and set the system = box + ground. In that case, there are no external forces acting, and so $$ 0 = \Delta E_{\textrm{system}} = \Delta K_{\textrm{box}} + \Delta K_{\textrm{surface}} + \Delta E_{\textrm{box, thermal}} + \Delta E_{\textrm{surface, thermal}} + \Delta E_{\textrm{other}}\,, $$ where $\Delta E_{\textrm{other}}=0$ includes other irrelevant energies. We know that the kinetic eneriges don't change, and so we're left with $$ 0 = \Delta E_{\textrm{box, thermal}} + \Delta E_{\textrm{surface, thermal}} = \Delta E_{\textrm{thermal}}\,. $$ The box and surface, by rubbing against each other, have generated thermal energy in the other. The thermal energy of both objects increases, and not by heating (which is energy flow across a temperature difference, but this occurs even when they have the same temperature).


You can see that that's the whole story. So where's the inconsistency in the OP? Let's start from Newton's 2nd Law for a system of particles, i.e., $$ m_i\vec{a}_i = \vec{F}_{\textrm{on }i} = \sum_{j \neq i}\vec{F}_{j{\textrm{ on }i}} + \vec{F}_{\textrm{external on }i}\,, $$ so that the net force on the entire system of particles is $$ \vec{F}_{\textrm{sys}} = \sum_i \vec{F}_{\textrm{on }i} = \sum_ i \sum_{j \neq i}\vec{F}_{j{\textrm{ on }i}} +\sum_i \vec{F}_{\textrm{external on }i} = \sum_i \vec{F}_{\textrm{external on }i}\,. $$ This sum is just the sum of external forces because the internal forces cancel out due to Newton's 3rd Law. Via Newton's 2nd Law, then, $$ \vec{F}_{\textrm{external}} = \sum_i \vec{F}_{\textrm{external on }i} = \sum_i \vec{F}_{\textrm{on }i} = \sum_i m_i\vec{a}_i = \left(\sum_j m_j\right)\sum_i \frac{m_i}{\sum_j m_j}\vec{a}_i = M\vec{a}_{\textrm{com}}\,, $$ Thus, we can compute the acceleration of the center of mass by focusing on the net external force on the system. Integrating this equation along a path taken by the center-of-mass yields the ``work-energy theorem'' in the form $$ W_{\textrm{net external along com}} = \Delta K_{\textrm{com}}\,, $$ but crucially this ignores all of the internal details of the system where the objects are exerting forces on each other, thereby changing potential energies in the context of conservative interactions or changing the thermal energy in the context of dissipative interactions like friction. This "center-of-mass work-energy theorem", involves a quantity called the pseudo work or center-of-mass work, and it is useful for computing kinematic results, but it is not a statement of conservation of energy and should never be used as such!

To get the the form of the work-energy theorem that corresponds to conservation of energy, we need to sum up the works done on the individual particles by integrating along each path taken by *each individual particle. That is, $$ W_{\textrm{on }i} = \sum_j W_{j\textrm{ on}i} + W_{\textrm{external on }i} = W_{\textrm{internal on }i} + W_{\textrm{external on }i}\,, $$ so that \begin{align} W_{\textrm{net on system}} &= \sum_i W_{\textrm{internal on }i} + \sum_iW_{\textrm{external on }i} \\&= W_{\textrm{internal}} + W_{\textrm{external}} \\& = -\Delta U +\sum_i \Delta K_i \\& = -\Delta U + \Delta K \,, \end{align} which is the statement of conservation of energy for a system of particles. (We are ignoring dissipative forces, because that requires a non-microscopic view, and here we are summing over the individual particles in the system.) Note crucially that $W_{\textrm{external}}$ is not the same as $W_{\textrm{external along com}}$! Alternatively, we could break up $\Delta K$ as $$ \Delta K_{\textrm{com}} + \Delta K_{\textrm{relative to com}}\,, $$ and attempt to understand the parts of the work that leads to change of the center of mass motion and to changes in the relative motion.


Finally, then, in equation (1) in the OP, the $W$ must correspond to $W_{\textrm{external}}$, which sums up the individual works done on all the particles. But this can't be the whole story, because the particles in the box are pushing on the particles in the surface, too, and so there is loss of energy to the surface. This could be taken up inside $W_{\textrm{external}}$, but then there is really no way to compute the left- and right-hand sides of the equation (see mmesser314's answer for how to think about this). Equation (2) in the OP could be correct if it's the center-of-mass equation, in which case we would change the name of $W$ to $W_{\textrm{external along com}}$ of maybe just $W_{\textrm{com}}$, showing explicitly that we are integrating along the path of the center of mass.


Here are some papers that really hammer this stuff home:

Pseudowork and real work, i.e., distinguishing between a kinematic relation and a conservation of energy relation: https://aapt.scitation.org/doi/10.1119/1.13173

A taxonomy of different works: https://aapt.scitation.org/doi/10.1119/1.16878

The work-energy theorem and developing a correct thermodynamic conservation of energy principle: https://aapt.scitation.org/doi/10.1119/1.19182

Examples of where you have to be careful when frictional forces are present (I really particularly like this one): https://aapt.scitation.org/doi/10.1119/1.13775

$\endgroup$
2
$\begingroup$

Does work-energy theorem account for thermal energy?

Yes, if by "account" we mean a mechanical model of many particles, which takes into account effect of all working forces on the invisible kinetic energy of microscopic motions. However, this general work-energy theorem does not know the concept of potential energy, and thus in case of interacting systems with potential energy, does not state anything on potential or internal energy of such a system.

My textbook suggests the following formula for conservation of energy in a system: $$W_{external}=\Delta E_{mechanical} + \Delta E_{thermal} + \Delta E_{internal}$$

There is the problem. This is generally incorrect and misconceived, as there is no valid concept of "thermal energy of the box" that is not at the same time a contribution to "internal energy of the box". And in addition, this wrong equation statement is missing also increase of energy through acceptance of heat.

The work-energy theorem in mechanics states $$ \text{work of all forces, internal and external, acting on particles in the system = } $$ $$ \text{ = change of kinetic energy of the system} $$ $$ W_{total} = \Delta K_{total}. $$

We can't apply this theorem directly to the microscopic work of friction forces on the box decelerating on a table, because we can't easily express all the friction forces and total work $W_{total}$ done on all particles of the box $$ W_{total} = \sum_{a} \int_{\mathbf r_{a,i}}^{\mathbf r_{a,f}} \mathbf F_{-a} \cdot d \mathbf r_a $$ using just net force $\mathbf F$ and displacement of the block $\Delta\mathbf x$. There are too many particles, they move chaotically and they experience varying unknown forces $\mathbf F_{-a}$, so this total work is unknown exactly. We can only say it has to be negative when the box decelerates, since the table heats up and the box has to lose some energy in the process.

However, because the box is solid, we can apply the work-energy theorem to its surrogate system, a mass point of the same mass, experiencing net friction force of magnitude $\mathbf F$, while it gets displaced on the table by $\Delta \mathbf{x}$. Such a mass point is a purely mechanical system, and thus the work done by total force $\mathbf F$ on this point equals change in its kinetic energy $K_{point}$:

$$ W_{point} = \mathbf F \cdot \Delta \mathbf{x} = \Delta \bigg(\frac{1}{2}Mv^2\bigg). $$

However, notice that this work $W_{point}$ is not the same as the actual total work $W_{total}$ in the fully detailed description, and $K_{point}$ is not $K_{total}$. The work $W_{point}$ ignores the work that increases internal energy of the box, and $K_{point}$ ignores kinetic energy of particles due to chaotic microscopic motions that contributes to total kinetic energy $K_{total}$.

There is also the 1st law of thermodynamics which applies to the box as well. For a system at rest, it states:

$$ \text{work of external forces acting on the system + net heat added to that system = } $$ $$ \text{ = change of internal energy of that system}. $$

We can apply this to the box, if we get into the (non-inertial) frame of the box itself, where the box stays at rest. In this frame, the box is rubbed by the table and thus the table does work on the box. There is initially no heat exchange unless a temperature gradient develops between the box and the table (which it can). All the work done by the table and heat added by the table to the box transforms into internal energy of the box.

$\endgroup$
1
$\begingroup$

I read two enlightening papers by professor Bruce Sherwood that give many details which can answer the questions mentioned above. I'll link to those papers at the end. Here are the points to mention:

1. What is the work-energy theorem?

There are some ambiguities in defining this theorem:

  • Does it apply to compound systems?
  • What about non-rigid bodies?
  • Does it include internal works and energies? etc.

I realized that a lack of precise definition for this theorem leads to abusing it. My textbook (Halliday and Resnick) defines the "Work-Kinetic energy theorem" this way:

change in the kinetic energy of a particle = net work done on the particle

The energy of a particle is merely kinetic, so it makes sense to say that all the work done on a particle contributes to its kinetic energy. In fact, this theorem can be mathematically derived using Newton's 2nd law:

$$ W_{net} = \int_{r_1}^{r_2} F_{net}\cdot dr = \int_{r_1}^{r_2} ma\cdot dr =m\int_{v_1}^{v_2} \frac{dv}{dt} \frac{dr}{dv} \cdot dv =m\int_{v_1}^{v_2} v \cdot dv = \Delta(\frac{1}{2}mv^2) $$

But what about a system of multiple particles? It can be shown using Newton's 2nd law that for a system of particles:

$$ \sum_i \vec{F}_{external,i} = M \vec{a}_{com} $$

$M$ denotes the total mass of the system. This formula only includes external forces, because internal forces cancel each other out and don't count. Integrating this formula with respect to the path the center of mass takes gives:

$$ W_{\text{net, external, along center of mass}} = \Delta K_{\text{center of mass}} ~~~~~~~~~~(1)$$

In this formula, the work for every force is calculated by integrating it along the COM path, not the actual path that the point of application of that force takes. This means that the left-hand side of this equation does not represent the actual work done on the system, and the right-hand side does not represent the total kinetic energy of the system, so we can't really call it energy; it's just $\frac{1}{2}Mv_{\text{com}}^2$. Basically, this is not an energy formula, so the left-hand side is known as "pseudowork" or "center-of-mass work", but in any case this formula holds for all systems and bodies, be it compound, particle-like, rigid or non-rigid. For rigid bodies, however, we can say $\Delta K_{\text{center of mass}} = \Delta K_{\text{translational}}$.

However, when we consider particles, formula (1) becomes an energy formula (the left-hand side is the real work done and the right-hand side is the real change in kinetic energy), as explained above. This is because of the fact that the center of mass for a particle is equivalent to the point of application of a force on that particle.

If we were to consider the real work done on a system (i.e. the real input/output of energy to a system by means other than heat which is due the a difference in temperature), we'd have to integrate each force along the path that the point of application of that force takes, but we can't set this equal to $\Delta K_{system}$, because some of this input of energy might be stored as potential, thermal or internal (other than thermal) energies.

So finally, calling the famous formula $W = \Delta K$ the "work-energy theorem" is somehow confusing in a sense that it feels like it's working with real energy and work, which is only the case for particles. For compound systems, as professor Sherwood suggests, it might be better to call it the "CM equation", because it can only be applied to center-of-mass quantities.

2. First law of thermodynamics

The version of "work-energy theorem" that can work with real work and change in energy, is included in the first law of thermodynamics (here we consider positive $W$ as adding energy to the system): $$ W + Q = \Delta E $$ This time $W$ is the real work. When there is no difference in temperature, as in the sliding box problem, energy transfer to and from the system does not involve heat, so we can omit the $Q$ term (actually the heated box can transfer heat to the floor, but we just neglect it). We end up with the same equation my textbook suggested ($\Delta E_{internal}$ denotes the changes in other forms of internal energy than heat): $$W_{external} = \Delta E_{mechanical} + \Delta E_{thermal} + \Delta E_{internal}$$ It is important to notice that we accounted for thermal energy in the term $\Delta E_{thermal}$, so $\Delta K$ in $\Delta E_{mechanical}$ represents the change in purely mechanical kinetic energy and we don't need to care if "thermal energy" is modeled as "kinetic energy of atoms and molecules". Now for the original sliding box problem, taking the box as our system, we can write: $$ W_{external} = \Delta K + \Delta E_{thermal,box} $$ $$ \Longrightarrow \Delta E_{thermal,box} = W_{external} - \Delta K $$ Where $ W_{external} = W_{friction} $, the real work done by friction on the box. For simplicity, I take the frictional force $f$ to be constant. Now, if the point of application of this frictional force moves a distance $d_{friction}$, then the real work friction does is $W_{friction} = -fd_{friction}$, where the negative sign is because of fact that the direction of $f$ is the opposite of the direction of the movement of the point of application. So we can write:

$$ \Delta E_{thermal,box} = -fd_{friction} - \Delta K ~~~~~~~~~~(2) $$

Taking the box as a rigid body, we can write the center-of-mass equation: $$ W_{\text{net, external, along COM}} = W_{\text{friction, along COM}} = \Delta K_{com} $$ $$ \Longrightarrow -fd_{com} = \Delta K_{com} ~~~~~~~~~~(3)$$

Where $d_{com}$ is the distance traveled by the center of mass of the box. Since the box is considered to be rigid, we can say that $\Delta K_{com} = \Delta K$, where $\Delta K$ is the total change in kinetic energy of the system. Again, here $\Delta K$ denotes changes in purely mechanical kinetic energies, and we don't include thermal energy in it. Finally we combine equations (2) and (3):

$$ \Delta E_{thermal,box} = f\cdot(d_{com} - d_{friction}) $$

So, if the effective distance the point(s) of application of the frictional force moves equals the distance the center of mass travels, the box won't heat up. Imagine, instead of friction stopping the box, someone tries to stop the box traveling in the deep space by pushing against the box with his hand (without the hand slipping on the surface of the box). The point of application of this opposing force (where the hand touches the box) moves the same distance as the COM, until the box stops, without any change in its thermal energy.

If we take friction to be the force required to break the cold-weldings that form between two surfaces, then $d_{friction}$ would be the total distance these welded spots move before they break, because they are the points of application of the frictional force. So, the reason the box heats up is that $d_{frictional} < d_{com}$. The second paper linked below has some nice explanations for this. Finally, if we had $d_{frictional} = d_{com}$, then the box wouldn't heat up, but the thermal energy of the floor would increase by an amount of $fd_{com}$ where $d_{com}$ is the distance the COM of the box moves.

Some models for friction suggest that the heating up effect is because of the vibrations of the cold-welded spots after tearing, which means this can't happen in rigid bodies. I assumed that the box is rigid so we can take $\Delta K_{com} = \Delta K_{total}$, So, these results are essentially sort of an approximation. What's important though, is that the confusion around the "work-energy theorem" and its application for friction is cleared.


Papers by professor Bruce Sherwood:

And his blog post on "pseudowork".

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.