2
$\begingroup$

During my attempt of fully grasping the work energy theorem, I came across this written in my textbook:

The frictional force, which we represented as a constant force is in reality quite complicated, involving the making and breaking of many microscopic welds, which deform the surfaces and result in changes in internal energy of the surfaces (which may in part be revealed as an increase in the temperature of the surfaces). Because of the difficulty of accounting for these other forms of energy, and because the objects do not behave as particles, it is generally not correct to apply the particle form of the work–energy theorem to objects subject to frictional forces.

one oddity that comes to mind when finding the work done by kinetic friction is that the point of application is continuously moving along the surface, and thus the force of kinetic friction is exerted by a different asperity/ adhesive bond each time. So displacement of the point of application for the force of each asperity or adhesive bond and thus the work done should, by definition, be 0. This, however, is not possible as kinetic energy is clearly converted to heat

What exactly is wrong with applying the work energy theorem for rigid bodies in case of friction?

Edit

Suppose there is a pointed vertical metal rod, that is fixed to the ceiling and just presses into the rubber surface of a conveyer belt placed on the ground. When the conveyer belt is turned on, kinetic friction scts between the pointed end and the rubber surface. If constant force isnt provided, the belt will clearly stop due to the friction ( ignoring any internal friction of the machinery) so friction is clearly doing Negative work on it. The problem arises When calculating the work, as the point of contact hasnt moved in space at all, from the earth frame of reference. Whats going on here?

$\endgroup$
1
$\begingroup$

Nothing is wrong with using the work energy theorem for rigid bodies in the case of friction, but as always you need to be careful. The work energy theorem is a bit tricky.

First, you need to distinguish between “net work” and the thermodynamic work.

Thermodynamic work is a transfer of energy by any means other than heat. This is the work that you are interested in when you are looking at the conservation of energy and seeing where energy flows from and to. The thermodynamic work done on an object in Newtonian mechanics is given by $\vec F \cdot \vec d$ where $\vec F$ is the force on the object and $\vec d$ is the displacement of the object’s material at the point of application of the force. The sum of the thermodynamic work for each force acting on an object is the total thermodynamic work.

The “net work” is defined very similarly as $\vec F_{net} \cdot \vec d_{CoM}$ where $\vec F_{net}$ is the net force acting on the object and $\vec d_{CoM}$ is the displacement of the center of mass of the object. The “net work” is only useful for tracking changes in kinetic energy (KE), but provides no information about where that energy came from nor about the total thermodynamic work done on the object.

With these definitions in mind, the work energy theorem states that the change in the KE is equal to the net work: $\Delta KE = \vec F_{net}\cdot \vec d_{CoM}$. This expression holds in general, including in cases of friction. So the work energy theorem is valid, even with friction. However, it does not tell you anything about the flow of energy between objects.

Now, the thermodynamic work is more interesting. Suppose that we have a stationary table and we are sliding a rough block across the table. The block begins with some initial velocity in the positive direction and comes to a stop after a displacement $\vec d$ due to a frictional force $-\vec F$ where the - indicates that it is pointing in the negative direction. Now, the thermodynamic work done on the block is $-\vec F \cdot \vec d= -Fd$. By Newton’s 3rd law the force on the table is $\vec F$ and so the thermodynamic work on the table is $\vec F \cdot \vec 0=0$.

This is interesting. Mechanical work $Fd$ leaves the block but no mechanical work enters the table. The difference is mechanical energy that is lost and is converted into thermal energy at the interface. From that point it is a question of thermodynamics about where the thermal energy spreads.

The contact weld idea is a good “microscopic” explanation, and if you knew the location and displacement of each contact weld then you could replicate that information. However, in practice you never have that level of information, so it is better to simply use the macroscopic information you do have to analyze the energy flow.

In short, “net work” can be used with friction, but it doesn’t tell you as much as the thermodynamic work does. In either case you need to be careful to use the correct forces and displacements for the quantity you want to calculate.

Now, your quote mentions the “particle form of the work energy theorem”, implying that they may have a different expression for non-particle objects. I don’t know what that may be. The work energy theorem I refer to is $\Delta KE=\vec F_{net} \cdot \vec d_{CoM}$. Your textbook may have multiple variants.

Edit (responding to question edit - please do not edit questions in ways that invalidate existing answers):

Notice that as I said above $d$ is "the displacement of the object’s material at the point of application of the force" (emphasis added). Although the point of contact is not moving, that is irrelevant, the material of the belt is moving at the point of contact so $d$ is non-zero for the belt and negative thermodynamic work is done on the belt. This results in the observed loss of mechanical energy of the belt and its conversion to heat at the contact point.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ isnt net work done just the signed sum of works by all the forces? Atleast thats how several of my textbooks refer to it $\endgroup$ – OVERWOOTCH Jul 7 at 9:01
  • $\begingroup$ @OVERWOOTCH no, the “net work” is often not equal to the sum of the thermodynamic work of all forces acting on an object. That is why I dislike the term “net work” and always put it in quotes. In this case they happen to be equal, but when you have a non-rigid object whose internal energy changes then they are typically not equal. For example, in your previous question about an ice skater pushing off the wall. The total thermodynamic work was 0 but the “net work” was not. Your ice skater example is a clear counter-example to any textbook which makes the claim that they are equal $\endgroup$ – Dale Jul 7 at 11:05
  • $\begingroup$ I would be careful with calling $\vec d$ the displacement. If you don't want to get into the integral definition by assuming a constant force parallel to the direction of motion at all points, then it would be best to call $\vec d$ the total distance traveled, since $\int\text d\vec x=\vec d$ is the total distance traveled, not the displacement. $\endgroup$ – BioPhysicist Jul 7 at 11:30
  • 1
    $\begingroup$ Also, I've never heard of net work being defined in terms of the center of mass, although it does make sense to me. Do you have a source for this? $\endgroup$ – BioPhysicist Jul 7 at 11:31
  • $\begingroup$ What about the static friction from the road which is accelerating a car? How does that fit into this analysis? $\endgroup$ – R.W. Bird Jul 7 at 14:03
0
$\begingroup$

This in addition to Dale's answer.

In newtonian mechanics for forces and work energy every object is assumed to be point object(except in rotational dynamics) unless dimensions is given. This point object is center of mass.

I think it was wrong of you to assume that friction is kinetic, it can be static too if the tip is moved along with conveyor belt

Let's say the belt is moving with a constant velocity of 5m/s. Then obviously net work done by by all forces is equal to 0 as there is no change in kinetic energy of the conveyor belt and as well as of the vertical rod.

But initially obviously friction will do work as the belt started from rest. However work done friction on rod is still 0 as no horizontal displacement of center of mass.

Now when work is done by conservative forces we only care about the displacement but when work is done by non conservative force we take the distance travelled by the center of mass.

Adding to RW Bird comment- That is the case of rotational dynamics thus we tend to take complete motion of object.

In that case Work done by static friction is 0 as there is no relative motion between point of contact and the road.It's because in pure rolling work done by friction on center of mass is −FRdθ as friction opposes the translation motion , but for the point of contact would also travel distance Rdθ because it's pure rolling, but now friction produces rotatory motion FRdθ , so it's work done on the whole wheel is 0

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ But the static force moves along with the car. $\endgroup$ – R.W. Bird Jul 7 at 15:37
  • $\begingroup$ @R.W.Bird The center of mass does move along static friction not the point of contact that is why we don't treat object as point object in rotational motion $\endgroup$ – Dirichlet Jul 7 at 15:41
  • $\begingroup$ as mentioned above, the rod is fixed, so the friction must be kinetic. Moreover, if the work done by friction is 0, what force is sucking out the energy supplied by the motor? Surely, the belt will come to a stop and lose all its k.e as sooon as the motor force vanishes $\endgroup$ – OVERWOOTCH Jul 7 at 16:32
  • $\begingroup$ @OVERWOOTCH I never said the work done by friction is 0. I said net work done by all forces is 0 as kinetic energy of belt is constant, And only the the top most point of the rod is fixed , is the rod allowed to rotate ? $\endgroup$ – Dirichlet Jul 7 at 16:35
  • $\begingroup$ its a pointed rod, that is completely fixed to the ceiling with no degree of freedom. Is the work done by friction non zero then? but the displacement of the point of application is 0, is it not? Im sorry if I appear to constantly change stances; I have no idea which conclusion is more reasonable or correct. $\endgroup$ – OVERWOOTCH Jul 7 at 16:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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