Ronald Raygun wrote:

OK, then. Let's add deliberate energy-wasting drag.

As before, the pleb is accelerated backwards for 1 second and, relative
to the ground, is brought to a stop in a distance of 0.5m.

As before, the pleb loses 30 J of kinetic energy, but the hero's
kinetic energy does not change.

This time, the hero's arms are exerting 60 N of force over a distance
of only 1.5 m, so they are doing 90 J of work.

The energy balances: 90 J stored energy plus 30 J kinetic energy from
the pleb give us the 120 J for the hero's lightbulbs.

Assuming Rick approves this, the next step is trivial. We replace the
hero's arm with an electric robot arm programmed to push the pleb.
We plug this robot arm into the generator instead of one of the two
60W lightbulbs, and replace the other with a 30W bulb. That way the
arm gets the same 90W as the hero's arm had.


Now, how does this all work if we want to look at it, as dennis
seems to, entirely from a frame of reference moving with the hero.

In the hero's frame, obviously Vh and Sh and Eh are zero.

Ap = -A
Vp = -A t + Vp0
Sp = -1/2 A t^2 + Vp0*t
Ep = 1/2 M Vp^2

Initial condition:
Vp0 = -1 m/s

Reminder:
A = 1 m/s^2
M = 60kg

Calculate:
Vp1 = -2 m/s
Sp1 = -1.5 m
Ep0 = 30 J
Ep1 = 120 J
Arm work = 90 J

Now we have a problem getting the energies to balance. The hero has done
90 J work, but now instead of gaining 30 J from the pleb, we're losing
90 J to him. Dear oh dear oh dear, our energy budget has been completely
depleted to zero and there's nothing left for the lightbulb.

There's something wrong!

No there isn't.

It's easy to forget what we took for granted when our reference frame was
the ground. Notice how once we moved the frame to the hero, his own
acceleration, speed, distance, and kinetic energy are all zero, as were
those of the ground before we moved. Well, now that the frame is moving
with the hero, the ground's motions are of interest.

The wheel drag is a force which the ground is exerting on the hero's
wheels, but of course the same force is pushing back against the ground,
moving it forward, i.e. tending to reduce the relative speed between the
two. The ground is hugely massive, of course, but we do slow it down
ever so slightly, so we need to take into account the way its kinetic
energy changes.

The Earth is initially moving at speed Ve0 (equal to -2 m/s which we'll
denote by unsuffixed V) in the hero's frame. We put a nominal figure on
the earth's mass (which for convenience we model as flat and thin so we
don't need to worry about rotational effects). Let's say its mass is H
(for huge). Its kinetic energy is therefore

Ee = 1/2 H Ve^2

We are slowing down the Earth using a puny 60 N of force, giving an
acceleration of -60N/H. Another way of expressing this is (M/H)*A.

Ve = M/H A t + V

We calculate:
Ve1 = (M/H m/s + V) m/s
Ee0 = 1/2 H V^2
Ee1 = 1/2 H (M/H m/s + V)^2

The energy difference Ee1-Ee0
= 1/2 H ((M/H m/s)^2 + 2 M/H m/s V + V^2 - V^2)
= 1/2 M (M/H m/s + 2 V m/s)

Because H is huge compared to M, we can ignore the M/H term since it'll be
vanishingly small compared to 2V. The earth is losing M V m/s = 120 J of
kinetic energy.

Now it balances again: The hero puts in 90 J of work, the Earth contributes
120 J, the pleb takes away 90 J, and so we have 120 J left over for the
lightbulb.