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Re: Work-energy worries



On Wed, 9 Oct 2002, John S. Denker wrote:

John Mallinckrodt wrote:
... After all, who
among us--other than, perhaps, myself (and I'm not so sure about
that)--is flexible enough to accommodate the notion that the boater
does work on herself.

The boater does work on the boater. In more detail, various parts
of the boater do work on the other parts.

No question, at least about that second thing, but I don't want to get
too far from the *original* question which was, "How much work does
the boater do?" and its context which was the introductory physics
course. I believe that if we restrict our attention to that question,
use the most common definition of work (the one that is presented in
introductory physics--i.e., work done = path integral of the applied
force dotted with the displacement of the point of application), and
answer the question within the context of the introductory course, the
correct answer will be equal to the acquired bulk kinetic energy of
the anchor and the boat.

This hardly seems like a feat of flexibility; it seems
like the only reasonable way to approach the issue.

If you are implying by this that the only reasonable approach is to
say (as I believe you did before) that the work done by the boater is
equal to the acquired bulk kinetic energy of the anchor, the boat, and
the boater, then I disagree for the reasons just stated. If that is
not what you are implying then perhaps we agree.

This is the Nth reincarnation of the "skater" question.
http://www.monmouth.com/~jsd/physics/car-go.htm#sec-skater

I see nothing to take issue with in that document. It supports
everything I have said.

(Note: From here on I am going beyond the context of the original
question. I don't want anyone to think that I would trouble
introductory students with these subtleties.)

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.

If one instead adds up *all* of the works done by the parts of the
boater (note the distinction between this and the sum referred to in
the previous paragraph as this one includes works done on things other
than other parts of the boater like, for instance, the boat and the
anchor), the result will be equal to the acquired bulk kinetic energy
of the anchor and the boat.

John Mallinckrodt
Cal Poly Pomona

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