Conservative forces vs friction

[Like Tony Stark]

Homework Statement

A body of mass $2 kg$ moves with constant speed $v=(0;3.5;0)$ at a position $r_1=(0;0;0.5)$ when it reaches a region where a force $F=(10;-6;-4)$ acts on it.
Calculate the speed of the body when it reaches $r_2=(2.5;0.5;0)$

Relevant Equations

$\Delta T + \Delta V=0$

Hello
I've written that homework statement as an example to illustrate my doubt:
How can I tell if a force is conservative or not?
I've read that, if the curl of the force is 0, it's conservative. But what about the friction force ($f=\mu N$)? Its curl is also zero, but it's not conservative.

Consider the example that I've written.
Is that force conservative? How would this situation be different if the force $F$ is replaced by friction?
How should I solve the problem? With $\Delta T + \Delta V=0$?

 

[TSny]

But what about the friction force ($f=\mu N$)? Its curl is also zero,...

The curl is an operator applied to a vector field. But, the friction force cannot be expressed as a vector field in some region of space. So, it's not possible to take the curl of the friction force.

 

[Lnewqban]

The resistive nature of friction always consumes energy and turn it into heat that is wasted.

 

[jbriggs444]

Homework Statement:: A body of mass $2 kg$ moves with constant speed $v=(0;3.5;0)$ at a position $r_1=(0;0;0.5)$ when it reaches a region where a force $F=(10;-6;-4)$ acts on it.
Calculate the speed of the body when it reaches $r_2=(2.5;0.5;0)$
Relevant Equations:: $\Delta T + \Delta V=0$

Hello
I've written that homework statement as an example to illustrate my doubt:
How can I tell if a force is conservative or not?
I've read that, if the curl of the force is 0, it's conservative. But what about the friction force ($f=\mu N$)? Its curl is also zero, but it's not conservative.

Consider the example that I've written.
Is that force conservative? How would this situation be different if the force $F$ is replaced by friction?
How should I solve the problem? With $\Delta T + \Delta V=0$?

As long as you are only concerned about work done on one pass (no back-tracking) down a path that is free of loops, any pattern of force whatsoever along that path can be completed to become a conservative force field.

If your frictional force just happens (one in a jillion chance) to match an inverse-square central force field then this means that you can happily use the formulas for potential energy in an inverse-square central force field.

In the usual case, you are not going to be handed a frictional force that just happens to match an obvious conservative field with an easy-to-find potential. [Though some one dimensional first year problems do allow the possibility...]

 

[Like Tony Stark]

The resistive nature of friction always consumes energy and turn it into heat that is wasted.

And what about tension or a force applied by person? Are those conservative forces or not? How can you know?

 

[jbriggs444]

And what about tension or a force applied by person? Are those conservative forces or not? How can you know?

For the notion of a "conservative" force to make sense, one has to be contemplating a force that is defined for every point in a region. This is formalized as the notion of a vector field.

The force applied by a person is only applied at a point or, perhaps, along a path over time. That is not the same thing. If you want to know whether the force applied by the person or by whatever is conservative, you would have to fill in force values for all the places where the person did not apply any force.

But before you go to that trouble, you have to ask: why does it matter. What difference do you think it makes whether a pattern of force is conservative or not?

 

[etotheipi]

This may not be related to your question, but you might also wonder why friction - a phenomenon that can be identified with microscopic electric interactions at the interface - is non-conservative, considering that the electric force is conservative (if $d_t \vec{B} = \vec{0}$).

What you find is that you can't reasonably account for all of the microscopic degrees of freedom and write down the equations of motion for every individual particle. What we do is replace the tangential behaviour of these interactions with a single effective (and non-conservative) force, of friction. The work done by this force accounts for the conversion of mechanical energy into microscopic forms, e.g. heat.

Because this force of friction is dependent on the direction of the velocity, it should be clear that it is not conservative. That is because a conservative force field can be written as a function of the position only, $-\nabla \phi(\vec{x})$. If a body passes through the same position with two different velocities, the friction force $\vec{f}(\vec{v})$ will not be the same!

Now yes, there might be some very specific cases where the force of friction might be able to be written in this form during a certain interval, e.g. a constant friction force acting on a body that travels in a straight line in one direction (without turning around). But I don't think it is at all helpful to tie this to the formalism of conservative vector fields.

 

[hutchphd]

The work done by this force accounts for the conversion of mechanical energy into microscopic forms, e.g. heat.

I know this is standard rhetoric for "friction" , but I abhor it. Can we say instead "mechanical energy lost to degrees of freedom outside our mathematical description" or something better than this. The energy is not "converted" nor "lost"...it makes this a pedagogical minefield IMHO

 

[etotheipi]

I know this is standard rhetoric for "friction" , but I abhor it. Can we say instead "mechanical energy lost to degrees of freedom outside our mathematical description" or something better than this. The energy is not "converted" nor "lost"...it makes this a pedagogical minefield IMHO

I'm not sure I understand your objection. Why isn't the internal/thermal energy of a body within our mathematical description? When the total energy of a body can be written as the sum of a part related to the ordered, macroscopic motion and another part related to the microscopic degrees of freedom, why is it unreasonable to say that in our accounting, energy has been converted between those two contributions?

Sometimes you can ignore where the used-to-be mechanical energy goes, e.g. for the purposes of dynamics, but that doesn't help you to figure out by how much the object heats up, for instance.

 

[hutchphd]

I'm not sure I understand your objection.

I object to the definition of the force of friction as something other than a convenient calculational fiction. For the beginning student it seems "just another force" like gravity or electric

why is it unreasonable to say that in our accounting, energy has been converted between those two contributions?

So long as it is apparent to the student that simply our accounting ledger has changed, We choose to put it in another column because we can no longer keep track of it not because it has converted form somehow.

I remind you the OP wanted to define field operation on essentially the "friction field". I think that confusion is not completely unwarranted

Incidentally I thought most of your description quite good...my major problem is the usual "grumpy old guy" objection that starts out "why do we always teach it this way ...

 

[Lnewqban]

And what about tension or a force applied by person? Are those conservative forces or not? How can you know?

Friction is the manifestation of the work done by a force, it is a reaction, a result, an effect.
It is a process that collects and dissipates mechanical energy mainly in form of heat and vibration.
We discuss friction force as the reaction to the force that does the work.
It can be useful, if you are after heat generation, like when initiating a fire while camping by energic friction of pieces of dry wood.

If you want to keep the same amount of potential and kinetic energy within a closed system, conserving the total mechanical energy, you need to avoid processes that transform that energy into heat, permanent deformation of material, vibration, turbulence of fluid and any other from which you can’t recover the invested mechanical energy.

Copied from
https://courses.lumenlearning.com/physics/chapter/7-4-conservative-forces-and-potential-energy/

Conservative force: a force that does the same work for any given initial and final configuration, regardless of the path followed.
Conservation of mechanical energy: the rule that the sum of the kinetic energies and potential energies remains constant if only conservative forces act on and within a system.

 

[haruspex]

So long as it is apparent to the student that simply our accounting ledger has changed, We choose to put it in another column because we can no longer keep track of it not because it has converted form somehow.

I agree, and the boundary is fuzzy. We could start with a body of fluid all moving at the same velocity, and happily regard this as KE. Now half goes the other way... still KE? Now ten parcels, each going in a different direction...
I'm sure this can all be nicely described with enthalpy or whatever, and the 'conversion' to thermal energy becomes gradual, but there is no clear boundary.

 

[etotheipi]

Thanks for clarifying, I see where you are coming from. My initial, simplistic, view was that a rigid body has 6 macroscopic degrees of freedom, and that any further kinetic degrees of freedom could then be lumped into the 'thermal' term. For deformable bodies it is way more complicated, since you can still have macroscopic internal movements of e.g. individual water parcels, which wouldn't be classified as translational or rotational kinetic energy, but would also not be thermodynamic energy.

I'm too much of a noob in statistical mechanics to know how it's treated in the thermodynamic limit, but I agree that this is where the answer lies.