Re: There's work, and then there's work

[John Mallinckrodt <ajm@CSUPOMONA.EDU>]

> In the case of one ball being held stationary, the work done on the
> balls (conventionally) is entirely due to the work done on the ball
> that's moved. Say that the moved ball experiences a force F(x) at each
> point along its path. To get the total work done on the balls
> (conventionally), you need to integrate F.dx from the initial to the
> final position for the moved ball. Say you get a value W for the work
> done on the moved ball, in this situation. But conventionally, since no
> work is done on the stationary ball, W is the total work done on the two
> balls.

Check!

> In the case of both balls being moved toward each other, say that each
> ball experiences equal, but opposite, forces F(x) (not necessarily the
> same F(x) as in the first case) at each point along its path. To get the
> total work done on the balls (conventionally), you need to integrate
> F.dx from the initial to the final position for each ball. Or you could
> just double the work done on one ball, in this case. Calculating the
> work done on the same ball as before, the distance it moves is half the
> distance it moved in the first case, _but_ the force it experiences at
> each point along its path is exactly _twice_ the force that it
> experienced at that point, in the first case (look at the force as a
> function of the distance between the balls).

Check!  (This is looking hopeful!)

> So the work done on that ball is the same as the work done on it, in
> the first case, since it moves half the distance, but experiences
> twice the force. ... So the total work done on the two balls, in
> this case, is _exactly _twice_ the work done in the first case.

Doh! Too bad; wrong again.

Take a look at http://www.csupomona.edu/~ajm/special/workonballs.pdf

Note that when both balls move, each ball moves half as far (just as
you said) AND that the force on the ball at any specific point along
that path is twice what it was in the case where only one ball moved
(just as you said.)  Now we simply need to look at the resulting area
under that curve to see that it is half the area that one gets when
only one ball moves--NOT the same area.  So, double it (just as you
said) and you get the SAME result as when only one ball moved.

> Conventionally, this discrepancy can't be explained. It's only by
> assuming that work is done on the stationary ball, that you can account
> for the differences in work (energy).

Since there is no discrepancy, there is no need to throw out
Newtonian mechanics and replace it with your own.

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