# Homework Help: Friction and internal energy

1. Jan 8, 2012

### chiddler

1. The problem statement, all variables and given/known data
A block slides to a stop along a table top. Which is true?
A. The work done by friction is equal to the force of friction times the distance that the block slid.
B. The work done on the block is equal to the force of friction times the distance that the block slid.
C. The work done on the table is equal to the force of friction times the distance that the block slid.
D. The work done on the block is equal to the sum of its changes in kinetic and internal energies.

Explanation (that I don't understand well): When considering the energy transfer due to friction, we must take into account internal energy change. The friction force times distance is equal to change in mechanical energy: fd = ΔKE + ΔPE, but this does not take into account the internal energy change of the block. Work by friction does not equal fd.

2. Relevant equations
Fdcos(theta) = Work = mg(uk)d

3. The attempt at a solution
My initial reaction was that this question is playing with positive and negative signs, so I was looking for the one with the proper sign. Work done by friction on the block is negative, and on the table is positive.

Been trying to reason the answer for half an hour. Stuck and frustrated.

Thanks.

2. Jan 8, 2012

### Simon Bridge

Conservation of energy - work = change in energy ... therefore, the work done on a body is equal to the change in it's energy. The work energy can be divided into kinetic and potential ... since the table is flat, there is no change in gravitational potential energy, which leaves "internal energy" ... which just means "we dunno": heat, other vibrations, whatever. Calculating it would require a very detailed model of the structure of the block and table... not supplied.

Some of the work done went into the table, noise, heat that sort of thing, some into the air and so on ... but none of that work was done on the block.

3. Jan 8, 2012

### chiddler

Thanks so much for your help.

Ok, but why are the other answers incorrect?

Work by friction is equal to m*g*u*d, yes? So why is A, B, and C incorrect?

4. Jan 8, 2012

### Simon Bridge

Can you find a reference frame in which friction and displacement are in the same direction? Basically, work is being done by everything in the list.

You'll want to be careful with C because you may want to argue that the table heats up.
It does this because it is working against the friction with the block ... otherwise it would slide.

5. Jan 9, 2012

### chiddler

OH!

So they are all incorrect because they fail to account for the internal energy change. Work by friction is both m*g*u*d AND the heat and sound, etcetera.

The table also works against friction which is why it also heats up. If it works with friction, then it would slide as you said. Like sliding a book forward by putting your hand on it!

But I don't understand the relevance of what you wrote:

" Basically, work is being done by everything in the list."

Do you mean work is being done BY everything as opposed to ON everything? If so, why is that important?

6. Jan 9, 2012

### ehild

When you speak about work, it has to be associated to a force and to a body. "Work done by force F on the block".

Work is associated with force and displacement. There is no work without displacement. The table does not move. Does friction do any work on it?

If the force of friction is F on the block and the direction of the frictional force is opposite to the displacement d of the block, the work of friction on the block is negative Wf= Fd=-μmgd.

It is not state if there are other forces doing work on the block. If there are other forces, the KE will change by the amount of the total work done on it.

Friction is a nonconservative force. If it acts on a body, the mechanical energy of the body is not conserved. To preserve the law of Conservation of Energy we add the term internal energy U to the mechanical energy. ΔKE+ΔPE+ΔU=0. But the energy of the whole system is conserved here: that of the block and the table and the surrounding air, as Simon Bridge pointed out. The lost mechanical energy of the block will be distributed as internal energy (heat) to the whole Universe.

The Work-energy Theorem is valid both for conservative and for non-conservative forces. Assuming that the only non-conservative force that acts on the block is friction, then ΔKE= Wc+Wf. The work done by a conservative force diminishes the potential energy ΔPE=-Wc. So you can rewrite the Work-energy Theorem as ΔKE+ΔPE=Wf. The work of friction is equal to the lost mechanical energy, which is not equal to the raise of internal energy of the block alone.

So which statement is right? A? "The work done by friction is equal to the force of friction times the distance that the block slid" True or false?

ehild

Last edited: Jan 9, 2012
7. Jan 9, 2012

### Simon Bridge

Well that too ... I was going to say it's important because B C and D all talk about the work "on" something. But that all works too :)

8. Jan 9, 2012

### chiddler

Yes but a different body, the block, is performing this displacement. So the body only needs to have displacement if it does work on something. If work is done on the body, then it doesn't need to have displacement.

Can it also diminish kinetic energy?

9. Jan 9, 2012

### ehild

??????? Check your Physics lecture notes.... If it says what you have written, I quit, as I learnt a totally different Physics.

Can it also diminish kinetic energy?[/QUOTE]

Negative work diminishes the kinetic energy. When the potential energy increases, the kinetic energy decreases. Think of a stone thrown up. gravity does negative work on it, he potential energy increases, the kinetic energy decreases.

ehild

10. Jan 9, 2012

### chiddler

D:!! Oh man...

Maybe I misinterpreted my previous question:

It is a "which is true" question. And this answer is NOT true: "No work is done on an object if that object remains stationary."

I thought that work can be done on an object if it is stationary just like I wrote with the friction.

Can you please tell me if I misinterpreted this? Why is what I wrote incorrect?

Very frustrated now because i've spent the past 5 hours wrestling with energy....

11. Jan 9, 2012

### ehild

There are other kinds of work besides mechanical, and they involve other kind of displacement: elastic, electric, magnetic, chemical...
I do not know what was taught to you. It is possible that your teacher said to you that work can be done on an object if it stays in rest. And work is force time the distance along the force acts. How is work defined in your book or lecture notes?

Note, force does not exist by itself. It must act on something. If the force acts along a distance, that something on what it acts has to move, by the same distance. This is not so clear when the force acts on a deformable object. When a hammer hits a nail the nail penetrates into the wood, and meanwhile it exerts force on the wood at its tip. Although the whole block of wood does not look moving, a tiny piece of it moves at the tip of the nail. There is work done on the wood by the nail, and it causes deformation, and the energy of deformation converts to heat... We calculate the work done on the wood as the product of the displacement of the nail and the force exerted by the nail, but it is assumed that there is always a tiny piece of wood moved by the nail.

Or you have a spring and forces acting on it at both ends, compressing the spring.
If the forces are of equal magnitude and opposite direction, the CM of the spring stays stationary, but the forces move the ends and exert work on the spring. That work increases the elastic energy of the spring.

Force of friction is some kind of force of interaction between the block and table. There is a backward force due to friction on the moving block and a forward force of the same magnitude on the table. The forces are equal, but the work of both forces do not need to be equal. If the legs of the table could slide along the floor, the table would move because of the interaction with the moving block. There would be work done by friction on the table in that case.

If the table stays in rest, that means the resultant force on it is zero. There are two horizontal forces acting on it, the friction from the block and friction between the ground and legs. These two forces cancel. No force, no total work done on the table. If you say that the force of friction from the block does Fd work on the table, the frictional force at the legs must do the same work with opposite sign. But the force does not act along a distance at the legs. Nothing moves there. Therefore the total work can be zero only when there is no work done by friction on the table.

ehild

12. Jan 9, 2012

### Simon Bridge

Presumably you can do work on something by heating it up?

I notice that the first 4 options are implicitly about mechanical work when they talk about force times distance... the last one allows for different kinds of work - which can blur the distinction a bit. Physics questions need to be interpreted lest they become too long and pedantic.

The question does not explicitly require that all the work is mechanical.
One could argue that the friction between the block and table heats both up ... that the bigger the friction and the longer the run the more they heat up - therefore, the thermal work from the friction is proportional to the force times the distance?

This argument does not aide understanding the question though. Part of the metadata is that only one of the options is true ... a different interpretation is needed.

13. Jan 9, 2012

### chiddler

That's right! I didn't think of the table doing work back. So the net force becomes 0. Please correct me if i'm wrong: work can be done on an object that is stationary, but the net work on the object must be 0. Yes?

I'm studying for my MCAT so i'm not really in a class right now. I'm relearning all the physics I took last year by myself from a book. The book is terse because it is supposed to be review and not learning-for-the-first-time.

I'm glad you wrote this because now i'm trying to distinguish the definition of mechanical energy. My understanding: work is transfer of energy. When it it's friction, it's thermal. As ehild wrote, the work is non conservative so mechanical energy is lost. According to wiki, mechanical work is just Fd. What does this mean for friction? Friction IS mechanical work but it's just not conserved?

So more reason that A,B, and C are wrong is that those answers disregard the internal energy that is lost and therefore they are, as you wrote, strictly mechanical work. If this is true, then my previous classification of friction is incorrect. Then friction is not mechanical work? Or it is but it is separate...uh...i'm thinking of reasons for and against this argument now.

Could you please help me understand these? Can you give me a better definition of what mechanical work is? Wiki is not providing a very good description besides the mathematical definition.

14. Jan 10, 2012

### Simon Bridge

There is a net loss of (useable) energy in the system.
That does not mean that some things cannot gain energy.
In this system, the kinetic energy of the block gets converted to other forms.

Energy is always conserved.
Saying friction is not conservative just means that you cannot regain the energy lost by going the other way. If the force were conservative, eg gravity, then the work done moving depends only on the endpoints of the path.

W=Fd is not the only way of expressing work, just like U=mgh is not the only way to express potential energy. note: Fd has the same units as energy.

15. Jan 10, 2012

### ehild

Physics defines quantities like velocity, acceleration, force, energy, work...... trough mathematical formulas. I suggest to start studying Physics from a good book. Haliday Resnik Walker "Fundamentals of Physics" for example. And/or browse Hyperphysics http://hyperphysics.phy-astr.gsu.edu/hbase/wcon.html about work and energy.
We confuse you.

ehild

Last edited: Jan 10, 2012
16. Jan 10, 2012

### chiddler

I really appreciate your advice as well as your help. Perhaps my questions are becoming too lengthy and vague. I'm going to continue studying and attempt to ask more specific questions later on. I don't think that i'm weak enough in physics to warrant buying a book. This is just one chapter i'm having difficulty with. Plus, books are expensive!

Thank you very much for your time.

Last edited: Jan 10, 2012
17. Jan 10, 2012

### chiddler

I understand all that you wrote. That said, I understand the question and why the other answers are wrong. I had some difficulty understanding some of the responses I received so I think my questions became more and more vague. I'll continue studying and hopefully have more specific questions.

Thank you very much for your time you helped very much!