# The work of friction forces - and - the total work of systems

1. Oct 9, 2005

### Bonulo

I've been given two conceptual problems that I can't quite crack, at least, I'm not sure that I'm cracking them:

Problem 1
When a box is moving across a rough surface the friction force does negative work. Can a friction force ever do positive work? If yes, give an example, if no, explain why.
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I've aserted, in my so-far solution, that friction force can never do positive work, based on rudimentary readings on the perpetuum mobile (the eternity machine). The main reason for it's unattainability is the existence of friction, thus I don't think friction can do positive work. Since it would result in an increase in kinetic energy, without having anything to transform it from. I am, however, not sure that the arguments hold. So I'm now chasing either the right arguments or the one example that will prove friction force can do positive work.

If a box is on a conveyor belt, and the conveyor belt is turned on, and thus accelerates, the box stays for a while (because of static friction), but is accelerating along with the belt. At some point the box starts sliding across the belt, and there is kinetic friction. But - since the absence of kinetic friction would cause the box to stay in it's initial position (with respect to the inertial coordinate system - the Earth), isn't the existence of kinetic friction then directed like the movement itself, thus doing positive work on the box?

Problem 2
A rope is used to pull a box, which therefore accelerates. According to Newtons 3rd Law the box pulls the rope with a force as big but with opposite direction. Is the total work done, as mentioned in the work-energy theorem, then 0? If it is, how can you change the kinetic energy of a body?
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The work-energy theorem says that W.tot=K2-K1=deltaK. But this applies to the work of one body on another body, not to the total work of a system. If I pull the box, I do the work W.tot, which it counters with -W.tot (the same size force, with the opposite direction). The total value of work will then, always, be 0, since W.tot+(-W.tot) = 0. But isn't that addition quite irrelevant? The kinetic energy of both bodies has been altered, which can easily by seen by using the work-energy theorem.

Is this some kind of trick question?

Any help would be muchly appreciated. The assignment's due tomorrow.

Last edited: Oct 9, 2005
2. Oct 9, 2005

### Bonulo

Comment on 1

The friction force is a resistive force - it is directed as to oppose relative motion - not motion. So I don't think that the solution is in the definition itself. But does the conveyor belt theory solve the problem? Or is it invalid?

3. Oct 9, 2005

### lightgrav

Comment on 2:

If a push lasts for a while, the contact Force transfers momentum through the contact Area from one object to the other. This is guaranteed by the definition of momentum, so we know we can use momentum conservation for a system.

If you push for a ways, the Force transfers Energy from one object to the other. This is not a trick question - if it did NOT work this way, we could not use the Energy conservation approach to describe mechanical systems!

Comment on 1's box-on-conveyor-belt:

The Work done by friction on the box is positive/negative according to :
a) the box's reference frame
b) the conveyor belt's reference frame
How about the Work done by friction on the conveyor belt in
a) and b)?

4. Oct 9, 2005

### Bonulo

On 2:
I don't quite follow you - if you were aiming at a clear response to the problem itself. Is the total work W zero? I know kinetic energy is changed for each object, but what do you think the question means?

On 1:
I'm just using the Earth as a frame of reference. If I answered if friction can or can't do positive work according to a not-inertial frame of reference - then I wouldn't have answered the question, would I? Since we always use inertial frames of reference.

I made a Google search on the string "friction does positive work" and found this: http://www.artsci.gmcc.ab.ca/courses/enph131oa/sem08_w05_solutions.pdf [Broken]

A professor from MacEwan Faculty of Arts & Science claims that friction, in a situation similar to mine, does positive work. But I also found some professors who claimed otherwise. Which is true?

(a co-student, to add to the confusion, says that another co-student asked our professor, and that he said friction can't do positive work - I just don't know if I can trust this 3rd hand statement to be true)

Last edited by a moderator: May 2, 2017
5. Oct 9, 2005

### Andrew Mason

Work is a scalar quantity but it is derived from the dot product of two vectors - force and distance:

$$W = \int \vec{F} \cdot d\vec{x} = \int Fdxcos\theta$$

where $\theta$ is the angle between the force and the direction of the change in position.

Negative work results when the direction of the force is opposite to the direction of motion. What direction is friction relative to the direction of motion? That should tell you whether friction always does negative work.

AM

Last edited by a moderator: May 2, 2017
6. Oct 9, 2005

### Bonulo

No. I know all that stuff already. Friction has the opposite direction of RELATIVE motion, not motion (i.e. motion relative to the inertial frame of reference). I want the .pdf-file solution to be validated or invalidated. If it's valid, friction can do positive work, if it's not, friction can't.

Aw. I've asked everywhere now, and get the same, rudimentary responses or hints all the time. This is a basic university physics course, is it supposed to be this hard? :/

Last edited: Oct 9, 2005
7. Oct 9, 2005

### Andrew Mason

I would say that the paper you have linked to is wrong or, at best, is unclear in identifying the frame of reference in which force and displacement is to be measured.

Static friction force, like any force, comes in pairs: an applied force and a static friction reaction force. They are always equal and opposite. The static friction force is a reaction to the applied force. If you remove the applied force, the static friction force goes to zero.

The force that the box exerts on my hand when I push on the box (with no relative motion between my hand and the box) is a similar reaction force. It is the force of my hand that does positive work on the box as measured in the initial rest frame of the box. The reaction force of the box on my hand does negative work on my body.

It is incorrect to say that the reaction force of the box on my hand does positive work on the box. But that is what this paper appears to be saying.

AM

Last edited by a moderator: May 2, 2017
8. Oct 9, 2005

### Bonulo

Is it? The paper says that a block that's dropped unto a conveyor belt will be acted upon by a kinetic friction force - from the surface of the belt - which seeks to keep the block on the belt, thus increasing its speed (and kinetic energy) relative to any point on the planet Earth. Is that wrong?

9. Oct 9, 2005

### Erik_at_DTU

hmm..

I really think the answer is NO (friction can't do positive work)

10. Oct 9, 2005

### Bonulo

Huh? All of a sudden? But why? You're challenged to shoot down the .pdf-solution. If you can't, then your answer's at best imperfect with respect to argument-sufficiency.

11. Oct 9, 2005

### Erik_at_DTU

the pdf-solution

relative to the belt, the block does not increase its kinetic energy.

12. Oct 9, 2005

### Erik_at_DTU

msn?

do you have msn bonulo?

Mine is erikandr(at)gmail.com

13. Oct 9, 2005

### Bonulo

No, but why is that relation even relevant?

14. Oct 9, 2005

### Andrew Mason

In the case of the box landing on the moving conveyor (or a moving box landing on a stopped conveyor), the force of friction causes the box and belt to eventually move together - an inelastic collision - so energy is lost to the system. The energy that is lost is heat, due to friction. So friction does negative work in the sense that it causes a loss of mechanical energy.

If the conveyor belt was unpowered the box would slow the conveyor. Although it would speed up the box, the loss of energy of the conveyor would be greater than the gain in energy of the box - due to friction loss.

AM

15. Oct 9, 2005

### Cyrus

I dont see whats wrong with the paper talking about the box on the conveyor belt. One should be careful in stating that the static friction force TRANSMITS positive work to the box. The friction is not the the source of the energy, but it transmits the energy that the motor sends to the conveyor belt to account for the increase in mass moving at a constant velocity. Would you agree with that AM?

16. Oct 9, 2005

### Andrew Mason

I don't think it is quite correct to describe the force as 'transmitting' energy. Friction provides a mechanical linkage between two objects. That linkage allows two objects to apply equal and opposite forces to each other. This allows mechanical energy to pass from one object to the other. But it is a linkage that does not conserve mechanical energy. Kinetic friction always reduces the mechanical energy of a system.

AM

17. Oct 9, 2005

### Cyrus

Yes, but im talking static friction, which allows the block to accelerate up to speed v_conveyor belt when it falls onto it.

If the block simply falls straight down onto the belt, it has zero velocity in the x direction. So when it falls onto the belt, it will pick up speed until the speeds match. So there has to be a net force acting on the block to cause it to accelerate. Thats what the static friction does. And that force acts for some amount of time, and over some distance. So it supplies some amount of work to the box.

Last edited: Oct 9, 2005
18. Oct 9, 2005

### Dr.Brain

The work done is simply given by $\int F.ds$ . Work is done by the foce and we consider the displacement as the distance moved by the point of application of force , limiting c0ondition being that the body is rigid and each point move the same distance. Because friction is peculiarly known to oppose motion, in general motions , this force does negative work . Consider your feet when walking , you push the ground backwards the friction force directs itself forward and as a result you move forward. What do you think in this case? Does the friction do negative work? ... Consider a case of rolling without slipping body on a rough surface , so when it is 'accelerating- rotationally' . the friction directs itself forward to give an opposing torque , but the wheel keeps moving forward till it comes to rest , so for that time does friction does +ve work? ...think.

BJ

19. Oct 10, 2005

### Cyrus

Im sorry Dr. Brain, I dont understand what your saying. Are you talking about a wheel thats rolling to a rest? That is kinetic friction slowing it down and it does negative work, but thats not what im talking about. Im dealing with static friction.

Thats not true in general.

Last edited: Oct 10, 2005
20. Oct 10, 2005

### Andrew Mason

There has to be some motion of the box relative to the belt. That means the friction is kinetic. I don't see how static friction could supply the accelerating force.

AM