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Re: ENERGY WITH Q



John Mallinckrodt wrote:

Great! This is a completely acceptable alternate specification of
the problem which will yield the same fundamental physics.

Excellent! We are making rapid progress now.

[snip]

Correct to here.

v = F/m t [7]

No. I get

v = F/2m t [7_corr]

Correction accepted; this was a simple typo.
The correct result appears in my notes, and
the correct result was used in substitutions
below.

We are asked to calculate the CM position (x1+x2)/2
which by equation [4] is just
X = x1 + L/8. [8]

No. I get

X = x1 - L/8 [8_corr]

Correction accepted.

At any rate, using the corrected equations [7_corr] and [8_corr],
I get

X = -L/8
+ m/F v^2
+ 1/2 integral (x1' - x2') dt

Agreed.

... we can easily perform the integral and find that

1/2 integral (x1' - x2') dt = 1/2 (x1 - x2)
= 1/2 (L/4)
= L/8

OK, I see how the math works out in this case.

Now I have some different questions: First, it appears
that the result here (namely the m/F v^2 term) depends
on finding a way to express a second integral in terms
of the square of a first integral.

One cannot in general find such an expression. In
this case it was possible because we were given a
precisely constant force. Very precisely constant.

Now suppose that in contrast, we were considering a
frictional force. Imagine the apparatus in question
sliding along a table. We might know the average
macroscopic force, but we would not in any practical
situation know the instantaneous microscopic force
accurately enough to have any hope of applying the
method suggested here. Note that the spring and
dashpot are under my control; I could choose them
to resonate with some high-frequency component of the
frictional scritching and scratching.

Furthermore, let me point out that this method is
not applicable to ordinary black boxes.
The method requires knowing quite a lot
about the internals of the box, notably the position
and velocity of the contents at the moment of
interest. Typically one has no way of knowing that.

I stand by my assertion that the integral contains
a signifcant oscillatory component. It seems utterly
obvious to me based on a glance at the diagram above,
that if you start shoving on the apparatus at t=0
the contents will shosh for a while.

I hope my explanation has been satisfactory.

I've learned something, but I am still skeptical
that this approach is applicable to real-world
situations.