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



David,

Your statement that the force in the second case is twice the force in the
second case is not correct. If force is a function of the separation
between the balls, and the starting and stopping separation is the same in
both cases, then the forces are the same, as a function of separation. The
tricky part is that the forces differ in their dependence on x; see below.

As far as the integral, you must do this carefully. The integration
variable is x. In the first case, you can safely put the origin at the
stationary ball, so that x ranges from the initial separation to the final
separation. In the second case, the safest origin is midway between the
balls. Then, x ranges from half of the initial separation to half of the
final separation. Also, in the second case, the force is -2kx, since the
separation is equal in magnitude to twice x. When I do the integral in the
second case, for one ball, I get (1/4) k ((final separation)^2 - (initial
separation)^2). Double for two balls, and I get the same answer as for the
first case.

Michael Burns-Kaurin
Spelman College





David Rutherford
<drutherford@SOFT To: PHYS-L@lists.nau.edu
COM.NET> cc:
Sent by: Forum Subject: Re: There's work, and then there's work
for Physics
Educators
<PHYS-L@lists.nau
.edu>


01/30/2003 01:59
PM
Please respond to
Forum for Physics
Educators






Robert Cohen wrote:

I may be totally misinterpreting DR's question but let me rephrase it=
in a way that I think my student's would ask it. If we have two bal=
ls connected by a spring and we compress the spring by holding one ba=
ll stationary while moving the other ball toward it, haven't we done =
work on*both* balls?

After all, our muscles are working to keep the stationary ball steady=
. Where does that energy go if not into the stationary ball or at le=
ast the two-ball configuration?

Note: To address this, I find that JG's terminology is helpful. I al=
so find it helpful to compare two situations. In one situation, one =
ball is held stationary while the other is brought toward it. In the=
other, both balls are brought toward each other. Compare the stored=
energy in both cases.

Recess is over. Back to "work" class.

The work done _on the balls_ must be the same in both cases. We don't
consider the work that the balls do on the person bringing the balls
together, since we only want to know the energy of the _balls_
after they are brought together.

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.

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. 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.

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).

--
Dave Rutherford
"New Transformation Equations and the Electric Field Four-vector"
http://www.softcom.net/users/der555/newtransform.pdf

Applications:
"4/3 Problem Resolution"
http://www.softcom.net/users/der555/elecmass.pdf
"Action-reaction Paradox Resolution"
http://www.softcom.net/users/der555/actreact.pdf
"Energy Density Correction"
http://www.softcom.net/users/der555/enerdens.pdf
"Proposed Quantum Mechanical Connection"
http://www.softcom.net/users/der555/quantum.pdf