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Re: CONSERVATION OF ENERGY



It is with much trepidation that I join this jumbled fray. Anyway, here's
what I'd like to hope my best students would say were I to include Ludwik's
proposed problem on an exam:

*******************************

An iron cube of mass m=1 kg slides horizontally over a long iron plate
which remains at rest. The mass of the plate is 99 kg. The specific heat
capacity of iron is 470 J/(Kg*K), the coefficient of kinetic friction is
0.57. The initial kinetic energy of the cube is 50 J. Friction brings the
cube to rest before its base reaches the plate's boundary. Everything is
in the vacuum.

1) What is the initial speed of the cube? How far will it slide?
(hint: g=9.80 m/s^2)

v_o = sqrt(2*KE_o/m) = 10 m/s
a = F_fric/m = mu*N/m = mu*m*g/m = mu*g
[P.S. Really, Dr. Mallinckrodt, I *do* know
that N is not *always* equal to m*g.
But you did say "horizontally." ;-) ]
d = v_ave*t = (v_o/2)*(v_o/a) = v_o^2/(2*mu*g) = 8.95 m

2) Calculate the change of temperature of iron (both pieces treated as
one object). What assumptions are you making to find the answer?

Assuming that all energy is retained by the two iron pieces (i.e., no EM
radiation, no convection, no sound, etc.), that the initial temperatures
are the same, and that they thermally equilibrate as one object, then
all of the initial kinetic energy of the cube goes into the internal
energy of the single object *just as if* heat had been added in the
amount (m+M)*c*dT = KE_o. Thus,

dT = KE_o/[(m+M)*c] = 1.06 mK

3) How much heat is generated? How much work is done? Explain what happens
to the initial kinetic energy of the iron cube.

Dr. Mallinckrodt!! I can't believe you would ask such an ambiguous,
illposed, and/or misleading question.

"Generated"? Heat can't be generated; it is a thermal *transfer* of
energy. Do you mean, "How much heat transfer is there between the two
objects?" If so, I simply can't answer the question because I don't
have enough details of the process. For instance, if more than 1% of
the dissipated energy of the cube remains in the cube just after its
relative motion ceases, then some will later be transferred (as heat) to
the plate. On the other hand it could go the other way as well.

"Work done?" What *kind* of work? On what? By what? I *can* tell you
that--in the specified frame of reference--the plate does a "pseudowork"
(calculated as if all external forces acted at the center of mass) of
-50 Joules on the cube. However, I can't calculate the "F*dx work"
(calculated with external forces and using the motions of the points of
contact relative to the "lab frame") or the "thermodynamic work"
(calculated with external forces and using the motions of the points of
contact *relative to the CM* of the system or subsystem) on either the
cube or the plate without knowing the details of the process just as
mentioned in the previous paragraph.

For concrete illustration let's suppose that the cube retains 10 J of
its initial 50 J when the sliding stops and that no heat transfer occurs
*during* the sliding process. Eventually the cube will thermally
transfer 9.5 Joules to the plate leaving it with 0.5 Joules so that they
share the 50 Joules in proportion to their masses. Then we would have

F*dx Work (J) Thermodyn W(J) Heat(J) dU(J) dE(J)
cube -40 +10 -9.5 +0.5 -49.5
plate +40 +40 +9.5 +49.5 +49.5
system 0 +50 0 +50 0

Notes: U is the internal energy, E is the total energy. The "F*dx
Work" of -40 J that is done by the plate on the cube serves both to
eliminate its 50 J of bulk kinetic energy *and* to redistribute 10 (of
those) J into internal energy. The system sees its internal energy
increase by 50 J while its total energy remains constant.

What happens to the initial kinetic energy? I think I've already
answered this. Are you looking for something more formal sounding?
Maybe something like, "It is equitably redistributed into the
innumerable microscopic degrees of freedom of the iron pieces."

************************

I'd give it an A!

John
-----------------------------------------------------------------
A. John Mallinckrodt http://www.intranet.csupomona.edu/~ajm
Professor of Physics mailto:ajmallinckro@csupomona.edu
Physics Department voice:909-869-4054
Cal Poly Pomona fax:909-869-5090
Pomona, CA 91768-4031 office:Building 8, Room 223