Re: a work-energy situation

["W. Barlow Newbolt" <wnewbolt@liberty.uc.wlu.edu>]

On Thu, 10 Jul 1997, LUDWIK KOWALSKI wrote:

> An iron cube whose mass is M slides horizontally over a long iron plate
> of identical mass. The plate is a 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 lost. [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.
>
> I am really confused in this situation. My frame of reference (for K and x)
> is attached to the plate.
>                                         Make sure I understand you.
>                                               Ludwik Kowalski

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.

W. Barlow Newbolt              540-463-8881  (telephone)
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Washington and Lee University  newbolt.w@fs.science.wlu.edu	
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