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F = m dv/dt => F dot dr = m v dot dv
Integrate both sides to get desired result W = delta(K). Add
subscript "com" to W and subscript "TR" to K if you prefer (but you
said this wasn't an argument over nomenclature).
Hint: I've got a pocketful of counterexamples, so
any such derivation is going to be quite a rare bird.
Fire away.
VIEWPOINT 1: VIEW THE ENTIRE SYSTEM (two blocks, spring, and table)
AS ONE BULK WHOLE.
What I'd say is that the interaction with the table has converted
*one form* of internal energy (namely macroscopic rotational and
vibrational) to *another form* of internal energy (namely microscopic
kinetic and potential).
Bulk energy refers purely to translational KE
of the system's com BECAUSE OF OUR VIEWPOINT. This is zero both
before and after the interaction, so W-K correctly predicts 0=0.
Also, there are no external forces on the system, so FLT correctly
predicts 0=0. This view is thus not very helpful. Let's go on.
VIEWPOINT 2: BREAK THE SYSTEM INTO MACROSCOPIC PARTS.
Suppose we're not happy with calling the bulk rotational and
vibrational "internal energy". Fine, I can break the system into
three parts: the two masses and the table. If you will permit me, I
will take the spring to be massless (ideal) so it doesn't count as a
part.
In that case, we have mechanical energy = sum of K and U for
the parts = mv^2 + kx^2/2 where I assumed for simplicity that each
block had the same mass m and speed v and the "centrifugal" stretch
of the spring is x.
And we have initial internal energy = E1 (which
reflects the temperatures and phases of the blocks and table). W_ext
= delta (E_mech) + delta (E_int) now predicts 0 = -mv^2 - kx^2/2 + E2
- E1 where E2 is the final internal energy and where I assumed spring
relaxes fully.