Work done by friction of a box

[alingy1]

Using Knight's Physics : A Strategic Approach (3rd edition):

I am on the section concerning dissipative forces. A box is pulled with a rope on a surface with friction at a constant speed. The book says Wtension=Change in thermal energy.
By Newton's First Law, the book says: Change in thermal energy=TΔs=kinetic friction times Δs.
Then it says:
"You might wonder why we didn’t simply calculate the work done by friction. The rather subtle reason is that work is defined only for forces acting on a particle.
[...]
There is work being done on individual atoms at the boundary as they are pulled this way and that, but we would need a detailed knowledge of atomic-level friction
forces to calculate this work. The friction force fk is an average force on the object as a whole; it is not a force on any particular particle, so we cannot use it to calculate work. Further, increasing thermal energy is not an energy transfer from the book to the surface or from the surface to the book. Both book and surface are gaining thermal energy at the expense of the macroscopic kinetic energy."

The book seems to be contradicting itself. Everywhere on the web, I see problems where we find the work done by friction with formula W=FΔs.

Why is Wtension okay but not Wfriction? Why are there so many questions with Wfriction if it is wrong?

 

[alingy1]

The end result gives the same thing as the formula for Wfriction.

 

[CWatters]

I think this is a roundabout way for them to say that you can't calculate the friction force from first principles. It can only be done by experiment. In other words...

In the example of a box being pulled by a rope they can't calculate the work done by friction directly using ...

Work = kinetic friction * Δs = μ*m*g*Δs

..because μ is unknown and can't be calculated from first principles. The only way to calculate μ is to do an experiment which involves measuring the tension in the rope. They could then calculate μ but since the velocity is constant it's just as easy to use tension directly...

Work = tension * Δs

If you could model the surfaces at the molecular level then you could probably calculate μ from first principles without having to measure the tension.

 

[lendav_rott]

what if the object was so heavy that you couldn't pull it hard enough for it to start moving? How would you be able to calculate the effort you put into it, the object didn't move.

 

[alingy2]

CWatters: μ is usually given in problems.
lendav_rott: I do not see your point. Are you asking another question? Or are you trying to make me think? You do no work in that case. But thermal energy should increase at the surface of the object and the floor. How does that relate?

 

[CWatters]

what if the object was so heavy that you couldn't pull it hard enough for it to start moving? How would you be able to calculate the effort you put into it, the object didn't move.

If the object doesn't move then you have done no work on it... because Δs = 0.

 

[alingy2]

Sorry! I posted on my brother's account "alingy2".

 

[CWatters]

CWatters: μ is usually given in problems.

Well yes, but where do you think they get it from?

 

[lendav_rott]

Pardon me I was thinking about the whole system not just the object observed. My thoughts went something like:
If the box is too heavy and the mechanism pulling on the rope is not exerting enough force to even make the box overcome its stationary friction there is no way there would be no work done at all. I didn't pay enough attention to the original text.

 

[alingy2]

I just don't get why they say we can't use W=Ffriction*delta s. This would allow us to find change in thermal energy.

 

[alingy2]

Cwatters: experimental data of course! But how does that link to W=Ffriction* delta s not being usable? I am so lost :(

 

[CWatters]

I just don't get why they say we can't use W=Ffriction*delta s. This would allow us to find change in thermal energy.

I agree it would but to calculate Ffriction you need to know μ and you can't just look up μ for every box. You can look up typical values of μ for different materials in contact with different materials but even then you will only find typical values or perhaps a range of values for μ.

If you want to know the exact work done using that method you would need to measure μ for your particular box on your particular surface. That would require you to do the experiment and measure the tension in the rope. You could then calculate u and the work done but it's far easier to just calculate the work done from the tension directly.

 

[CWatters]

If the box is too heavy and the mechanism pulling on the rope is not exerting enough force to even make the box overcome its stationary friction there is no way there would be no work done at all.

Is there a typo in the above?

Imagine you hang the free end of the rope over the edge of the table. You could put a pulley on the edge of the table if you like. Then on the free end you add a weight that puts some tension in the rope but not enough tension to move the box. Nothing moves so no work is done by anything on anything else.

 

[lendav_rott]

Is there a typo in the above?

Imagine you hang the free end of the rope over the edge of the table. You could put a pulley on the edge of the table if you like. Then on the free end you add a weight that puts some tension in the rope but not enough tension to move the box. Nothing moves so no work is done by anything on anything else.

Yes, you are right. I was trying to challenge that idea with an idiotic example, something like: Imagine you are trying to move a refridgerator weighing 20 times as much as you. After countless attempts in trying to move it, you can't be as ripe and raring to go as you were before. If there was no work done, then you wouldn't be exhausted either. That is what I meant when I said that I was accidentally thinking about the whole system not the object observed.

 

[PhanthomJay]

Cwatters: experimental data of course! But how does that link to W=Ffriction* delta s not being usable? I am so lost :(

It is very usable, that sentence from Knight is an unorthodox way of looking at work. I wonder what Knight would say about the work done on a box sliding on a horizontal surface with friction and coming to rest in a certain distance. No other forces acting in the horizontal direction. Perhaps they'd call it frictional pseudo work? I don't know.

 

[Doc Al]

Perhaps they'd call it frictional pseudo work?

Exactly! Or sometimes it's called center of mass work, but it's not "work" in the same sense as used in the first law of thermodynamics. It's more a consequence of Newton's laws, not energy.

I think Knight is making a subtle and important point, but it might be too subtle!

 

[PhanthomJay]

Exactly! Or sometimes it's called center of mass work, but it's not "work" in the same sense as used in the first law of thermodynamics. It's more a consequence of Newton's laws, not energy.

I think Knight is making a subtle and important point, but it might be too subtle!

Great point, thanks.

 

[alingy1]

Doc Al: Could you give me a precise definition of pseudo work? I'm still having trouble differentiating why it is not a consequence of energy, but stuff is making more sense to me slowly!

 

[Doc Al]

Doc Al: Could you give me a precise definition of pseudo work? I'm still having trouble differentiating why it is not a consequence of energy, but stuff is making more sense to me slowly!

The term 'pseudowork' has never really caught on, though I have seen it in a few books and papers. A more common term for the same thing is 'center of mass work'.

If you take a net force acting on an object (like friction) and multiply it by the displacement of the object's center of mass, you get a quantity that looks like a work term but is better called pseudowork (or "center of mass" work)--what it determines is not the real work done on the object, but the change in the KE of the center of mass of the object. This is usually called the "Work-Energy" theorem:
$$F_{net}\Delta x_{cm}=\Delta (\frac{1}{2}m v_{cm}^2)$$
Despite the name, this is really a consequence of Newton's 2nd law, not a statement of energy conservation.

"Real" work is the work that appears in the first law of thermodynamics (conservation of energy) and depends on the details of how the force is applied and the movement of the point of contact.

In Knight's example of the box being dragged, the tension acting on the box does real work, since it is a force acting through a displacement of the point of contact. Since the KE of the box isn't changing (constant speed) that work must be going somewhere: The internal energy increase due to the friction. But calculating the real work done by friction is too complicated, since it is not a single force acting at a single point. But you can surely calculate the 'center of mass' work due to the net friction force, and that's what most books have you do anyway.

Here are a couple of examples which you may find interesting where Fnet*Δx certainly exists, but the real work is zero.

One: An accelerating car. Friction of the road on the tires certainly accelerates the car, but it does no work on the car (assume no slipping). It's not the road that provides energy for the car, it's the gasoline and the engine! Of course, you can use pseudowork (if you know the friction force and air resistance) to calculate the change in KE using the work-energy theorem described above.

Here's another example: You jump in the air. Your KE obviously increased, but did the ground do work on you? No! Your point of contact with the ground did not move, so the ground did no work on you. So where did the energy come from? From the chemical energy in your muscles, of course, not from the ground. Of course, you can consider the average force that the ground exerts on you and use it to calculate your KE--but that's an application of pseudowork again, not real work.

Let me know if this helps at all.

 

[alingy1]

Awesome! Then, the concept of pseudowork allows to evaluate cases where the particle model does not work (a particle cannot exert force on itself).

Then, for Knight's case, if real work and center of mass work are different, does that make the formula deltaEthermal=friction force times displacement wrong? This formula basically calculates center of mass work.

 

[PhanthomJay]

Total energy of a system is always conserved. In equation form,
$\Delta K + \Delta U +\Delta E_o = 0$ where the first two terms represent the change in kinetic and potential energies and the last term is the change in all other forms of energy like heat, chemical, electrical, and everything else. In your particular example in the original post, there is no change in K and U, therefore all the change must be in E_o, where heat added into the system from friction is balanced by the energy lost from the tension force which might be a form of say chemical energy if some person is doing the pulling. It is often convenient to represent energy changes by Work, whether that work is real or pseudo. The work associated with an energy change is the negative of that energy change, F.d.