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Re: work done by friction



Leigh,
I have not analyzed this in detail, but let me open a possible solution
to any apparent energetic paradox:

Perhaps this is another example of an interaction in which the system as
defined must lose energy to the environment.

Also, I see nothing in your calculation that the requires that the force
of interaction be friction - it could be any mechanism. Perhaps you are
showing that that mechanism (even if friction) must involve a
dissipation of energy to the surroundings. (recall the "capacitor
conundrum").

Bob

Bob Sciamanda
Physics, Edinboro Univ of PA (em)
trebor@velocity.net
http://www.velocity.net/~trebor
----- Original Message -----
From: Leigh Palmer <palmer@SFU.CA>
To: <PHYS-L@lists.nau.edu>
Sent: Wednesday, October 27, 1999 8:32 PM
Subject: Re: work done by friction


Having read only a couple of the items in this thread, and approaching
the problem from the point of view of a compulsive Blefuscan, let me
offer my analysis of the two-brick problem. Brick A has mass MA,
initial
temperature Ti, and heat capacity CA. It moves initially at velocity
VAi
in the x-direction. Brick B has mass MB, initial temperature Ti, and
heat
capacity CB. Block B moves initially with velocity VBi in the
x-direction. I will specify that MA < MB, and I will choose a frame of
reference such that

MA * VAi + MB * VBi = 0

The two bricks now interact only through a frictional force of
magnitude
f (which we shall take to be constant), and after the interaction they
separate with velocities VAf and VBf, respectively*. After a suitable
equilibration time the temperatures of the bricks are, respectively,
TA and TB. By conservation of momentum

MA * VAf + MB * VBf = 0

By conservation of energy

1 2 1 2 1 2 1
2
- MA*VAf + - MB*VBf + CA*(TA - Ti) + CB*(TB - Ti) = - MA*VAi + -
MB*VBi
2 2 2 2

I now have three equations in four unknowns, the final velocities and
temperatures of the bricks.

I could get one more independent equation from, e.g.,
impulse-momentum:

| MA * (VAf - VAi) | = f * deltat,

but in writing it I introduce another unknown, so the problem is still
indeterminate. Nevertheless, I can still calculate the work done on
either block by the friction force. The answer will be indeterminate,
however! Moreover, the meaning of "work" will be found to be somewhat
ambiguous, while the meaning of "energy" is well defined.

First, we consider the mechanical work WA which must have been done on
brick A to change its kinetic energy. By the work-energy theorem:

1 2 2
WA = - MA * ( VAf - VAi )
2

What external force acted on brick A? Hypothetically only the
frictional
force acted, so it must have done the mechanical work.

The reaction to the frictional force on brick A is the equal and
opposite frictional force which acts on brick B. That force does an
amount of work WA on brick B which must be given by

1 2 2
WB = - MB * ( VBf - VBi )
2

I note now that |WA| > |WB| (both WA and WB are negative, and I leave
the proof to the reader). It is difficult to imagine how brick A could
have scraped a longer path over Brick B than vice-versa in any really
meaningful way. Thus the product of f times that scrape distance can't
be the work in question here, since that product is of the same
magnitude and sign on both blocks!

So much for *mechanical* work. If sliding friction can do mechanical
work it must be satisfied with doing an indeterminate amount of it.
Scarcely a useful concept in my view.

But there's still thermodynamic work to consider. We see that this
quantity depends on vagaries of the structure of the bricks, that the
internal energy changes are not determined by the mechanical
specifications alone, and that still more information must be had to
predict the changes in internal energy. The internal energy changes
themselves are perfectly well defined once one final temperature is
known, but friction does an indeterminate amount of themodynamic work.
I
note in passing that, unlike the mechanical work done by friction,
this
work is always positive.

We have a word, "work", which refers to a quantity we can't define
uniquely, and for which definitions giving conflicting results are
known, and we ask the question "Can friction do work?" Ridiculous!
This
question is in the category I've abused before; once one gives an
answer
it becomes a false canon, a candidate for use on a multiple choice
exam.

I propose that whenever friction is actively involved in a problem (as
opposed to the static friction of, say, a tire on the road) one should
approach the problem dynamically rather than energetically. The latter
approach will almost always present difficulty when the two
interacting
surfaces are both in motion relative to the frame in which the
observation is being made. Of course one can probably cobble a
quantitative solution energetically for each specific case, but does
it
make the physics clearer? Probably it does not.

Leigh

*If the bricks do not separate then VAf = VBf = 0.