Re: work done by friction

[Jim Green <JMGreen@SISNA.COM>]

> So no work is done but the first law of thermodynamics still holds.
> If we take the system made up by A and B and write the first law as W
> + Q  = delta U, then the left hand side of the equation has two zero
> terms in it and on the right hand side there is a book-keeping
> transaction from external kinetic energy (macroscopic) to thermal
> energy (microscopic) (not heat).

I am noting this thread very late I take it, but I did see the word "heat"
-- and used correctly so I feel compelled to comment:

1) In the posts that I _have_ read, I don't see a clear statement of work
done on _what_.  If one says that "no work is done" one must specify which
system is doing anything on any other system.  To say that Block A and
Block B do no work makes me ask on _what_???  Is it being asserted that
Block A does no work on Block B?  This is not correct!  If Block B gets
warmer then work is being done on it!  To say that the system of Block A
_and_ Block B does not work, one must note what is in "contact" with that
system.  If nothing, then the _system_ is not doing work.  But A might
still be doing work on B.

2) To say that somehow "external kinetic energy" is being "transferred" to
internal energy (mistakenly called "thermal energy" above) leads us right
back to the terrible idea that energy is a fluid being moved somehow.

No! If the energy of _any_ system is changed, _work_ in some form must be
done on that system.

If Block B gets warmer, then (net) _work_ in some form is being done on it.

Just me over here in my little corner.

Jim Green
mailto:JMGreen@sisna.com
http://users.sisna.com/jmgreen