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

[Bob LaMontagne <rlamont@POSTOFFICE.PROVIDENCE.EDU>]

David Rutherford wrote:

> 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). 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.

Agree to here!

> But the
> _total_ work done on both balls, in this case, is twice the work done on
> one of the balls. So the total work done on the two balls, in this case,
> is _exactly _twice_ the work done in the first case.

Ooops! Yes the force in the integral is doubled. BUT, dx in the integral is
not halved. The shorter path of each ball shows up in the limits of
integration!!!! So the factor of 1/2 shows up as 1/4 (1/2 squared) in the
evaluation of the integral. Since there are now two balls, we double the
result, and we had an extra factor of 2 for the force within the integral.
So 1/4x2x2 all cancels - same amount of work is done.

> 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).

Conventional method seems to work just fine.

The fact that I answered this means that I seriously need to get a life ;-)

Bob at PC