In article , Gary Coffman says...

Well, Feynman disagrees (at least partly). He gives examples in his
"Lectures on Physics" of cases where work must be negative.

But 'negative work' is trivial, the example that I
gave from H&R showed that the value calculated is
really the difference between an initial velocity
and a final velocity of the body. If the final
velocity is less than the initial, the "work" is
negative, ie the body has supplied some energy
to the thing that deaccelerated it.

Basically the limits of integration can be from
a small number to a large number, or from a large
to a small. For the second case the integral
winds up being negative in sign. This isn't
anything fancy, it just tells you what way the
work 'went.'

This is *why* bodies in orbit, or frictionless pendulums,
show zero net work.

Not really. The real reason is that the dot product
of the force and the displacement has cos(theta) = 0
because the force is radial, and the displacement is
at 90 degrees to that.

Jim

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