Re: Work-energy worries

["John S. Denker" <jsd@MONMOUTH.COM>]

John Mallinckrodt wrote:
> ... 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),

Yes, that's the only definition of force that makes sense
to me.

> 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.

To the extent that that excludes the boater, I disagree.

> 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.

No.

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.

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.

This posting is the position of the writer, not that of Olga, Irina, or
Masha.

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