Re: a work-energy situation

[LUDWIK KOWALSKI <kowalskil@alpha.montclair.edu>]

On Fri, 11 Jul 1997 W. Barlow Newbolt <wnewbolt@liberty.uc.wlu.edu> wrote:

> I would suppose that if the whole process is adiabatic (dQ = 0) and no 
> work is done on the surroundings of this apparatus (dW = 0); then the 
> change in internal energy for the system is zero (dU = 0). (It is OK if 
> one part of the apparatus does work on another.) My guess would be (3) 
> since the kinetic energy of the sliding mass is already part of the 
> internal energy of the system.  So, I say dU = 0 -- no change in internal 
> energy.

He was responding to the problem below:

> >  An iron cube whose mass is m slides horizontally over a long iron 
> >  plate. The mass of the plate M is much larger than m. The plate is 
> >  at rest while the cube has an initial kinetic energy K. Friction 
> >  brings the cube to rest before its base reaches the plate's boundary. 
> >  Everything is in the vacuum and the internal energy, dU, added to 
> >  both pieces of iron is not radiated in any substantial amount. 
> >  [In principle, dU can be calculated from the observed change of 
> >  temperature but no thermometer is available.] Which of the following 
> >  is correct and why? 

> >  1. The internal energy dU is equal to K.
> >  2. Only part of K goes into dU, the rest is used to do work W=R*x,
> >     where R is the force of kinetic friction and x is the distance 
> >     the cube slides before comming to rest.
> >  3. Neither of the above is correct.
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Yes, the process is adiabatic but both metallic objects become warmer.
How would you answer, Barlow, if the terminology was changed and the
adjective "thermal" was used instead of "internal"? That is what I
really had in mind. Sorry for not making this clear.
                                                       Ludwik Kowalski