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Re: Work-energy worries



Sorry! A few small details in my last message were not clear. Please
discard it and replace it with this version:

And there is a *genuine* point of departure. I stand by everything I
have written in this thread and am content now to leave it to others
to judge the merits of the arguments that have been put forward.

John Mallinckrodt

Oh dear, I guess I'm an other.

> JD has suggested (correctly that the various parts of the boater do
work *on other parts of the boater*. That is, of course, correct.
But if one adds up all of *those* works (using the definition of work
given above) to get the "total work done by the boater", one will find
that the result is zero. This is because the work that any part a
does on any part b is necessarily equal and opposite to the work done
> by part b on part a.

There's a law that says that the forces are equal and
opposite. But work is not equal to force. There's a dx
involved, and the dx is commonly very different.

I have to return to something like my previous example. I stretch and
then release a mass on a horizontal spring (on the surface of a
heavy, frictionless table with the other end of the spring attached
to a post on the table). System A = mass. System B = spring. System B
does work W_A on the mass as the spring relaxes toward equilibrium.
Consequently the mass gains KE.

Now for the key question: How much work W_B does the mass do on
system B? Justify your answer *in terms of integrals over appropriate
dx values*.

I believe this is a reasonable first crack at a model for the
boater's hand and the muscles in the boater's arm. The "total work"
on the boater will be W_A + W_B.

Example: Suppose I start at rest and then start waving my
hand around in a circle. Suddenly I've got KE that I
didn't have before. By the work/KE theorem, the total
work (summed over the various parts) is nonzero.

The standard version of the work-energy theorem refers to bulk
translational KE. However, if you insist on referring to rotational
(i.e. internal) KE, I agree with your analysis. Clearly if the hand
gained energy, the rest of the body lost energy. So yes, the rest of
the body did work torque*(angular displacement) on the hand. I would
*not* call this an application of the work/KE theorem however. It is
an application of the first law of thermodynamics.

I agree with JohnD. Yes indeed JohnM's approach of using the idea of
pseudowork seems to me to be quite unnecessary. Each time the ice skater
problem comes up, I get out JohnM's work paper and try to get some help out
of it. I am still trying. But I keep the reprint. <g>

Especially for intro students, W=dKE ought to be enough, if the various
systems are broken down appropriately.

Jim Green

To fully analyze the ice-skater problem, I personally find it helpful
to apply *both* the work-energy theorem and the first law:

Neglect the friction between the skates and the ice and assume that a
person of mass m pushes off a rigid, stationary wall starting from
rest by extending his arms a distance d, to end up with a final speed
u. (a) What average force did he exert on the wall? (b) What minimum
amount of food energy was burned in this process?

(a) Take the system to be the person. Apply work-energy to get
Nd = 0.5mv^2
where N is the average normal force of the wall on the skater's
hands. By Newton's third law, this is equal and opposite to his
pushing force.

(b) Apply the first law to find
0 = 0.5mv^2 + delta(E_int)
since the hand on the wall does not move. The internal energy came
from the chemical energy of the food he ate.

I agree with you that (a) is standard introductory mechanics.

But be very careful Jim: it is the pseudowork (Nd) on the left-hand
side of (a) and it is the thermodynamic work (0) on the left-hand
side of (b)! Please think about this. If the wall did thermodynamic
work on the skater, we would be saying that the wall lost energy,
which I hope you agree would be a very strange thing to say.
--
Carl E. Mungan, Asst. Prof. of Physics 410-293-6680 (O) -3729 (F)
U.S. Naval Academy, Stop 9C, Annapolis, MD 21402-5026
mungan@usna.edu http://physics.usna.edu/physics/faculty/mungan/

This posting is the position of the writer, not that of SUNY-BSC, NAU or the AAPT.