dissipationenergy-conservationfrictionnewtonian-mechanicsperpetual-motion
I am attempting to settle a friendly bet. Would a pendulum swing indefinitely in a hypothetical vacuum (i.e. no air resistance) having a hypothetical frictionless bearing (i.e. no energy lost due to friction) assuming the following
The frictionless vacuum is on Earth (9.8 m/s^2).
The pendulum is already in motion and no other external forces other than gravity act on the pendulum.
You are using 'g' where it shouldn't be used.
Consider two 1 kg objects that fall and hit the ground, one at 10 m/s, the other at 20 m/s. let's say the impact time for both is 0.1 s, that is to say 0.1 seconds passes after the objects first touches the ground until the objects comes to a complete stop.
a = delta v / time
a1 = (10 m/s) / (0.1 s) = 100 m/s/s
a2 = (20 m/s) / (0.1 s) = 200 m/s/s
F = ma
F1 = (1 kg) * (100 m/s/s) = 100 Newtons
F2 = (1 kg) * (200 m/s/s) = 200 Newtons
The ground exerts twice as much force to decelerate the object that was moving at double the speed.
Regardless, no equations will bring you to what you want, there is no 'sweet spot' that'll make the other person go higher.
What you are talking about is a perpetual motion machine.
http://en.wikipedia.org/wiki/Perpetual_motion
There are many clips on you tube showing circus performers and gymnasts on see saws, watch a few and see where their energy input comes from: climbing ladders to gain gravitational potential energy
For a stack of coils, as you say, this is actually a famous result, but not quite in the way that you've phrased it.
If you drop a magnet through the air from some height $h$, it'll generally hit the ground with a speed $v$ that obeys $\frac12 mv^2 = mgh$ --- that its, its gravitational potential energy gets converted into kinetic energy. However if you drop a magnet through a conducting pipe, or through a series of conducting rings as you suggest, it'll strike the ground with a much smaller kinetic energy: instead of being converted entirely into kinetic energy of the magnet, some of the magnet's inital gravitational potential energy is converted to electrical energy in the conducting pipe.
Having done this experiment as a classroom demo (and Youtube doesn't really do this justice --- magnets and pipe are cheap at the hardware store, and you should try to see it on your own) it's pretty clear that what happens is that the falling magnet achieves some terminal velocity. We can try to estimate what that is, but we'll have to correct some errors in your treatment (v32).
You seem to be imagining that your stack of coils is connected like a solenoid, so that all of the current can be extracted from the coils as the magnet falls. That only works if the falling magnet is long compared with the solenoid (as you specify). I'll show the result for both long and short falling magnets.
First, there are a couple of issues with your expression for the induced voltage. You give the induced voltage as a function of the total fall time $\Delta t$. However until we have an equation of motion for the falling magnet, we can't make any assumptions about the total fall time. The voltage produced by a magnet moving into or out of a solenoid depends on the instantaneous velocity $v$ (as my students verify by hooking a solenoid up to an oscilloscope and shoving magnets through them). If the flux produced by putting the magnet through a single coil is $\Phi$, and the stack of coils has $N$ coils and height $h$, the flux through the solenoid increases by $\Phi$ every time your long magnet penetrates another coil. Let's define the coil density $n = N/h$, in which case the height of each coil is $\delta h = h/N = 1/n$. This gives
$$ |V| = \frac{d\Phi_\text{solenoid}}{dt} = \frac{\Phi_\text{coil}}{\delta h / v} = n\Phi_\text{coil} v \tag1 $$
Note that $\Phi_\text{coil}$ depends on both your magnet and your coil. The constant property of a permanent magnet is its magnetic moment, which has units of $\text{current}\cdot\text{area}$. If the magnet mostly fills the coil, you can measure the surface field $B$ near the pole of the magnet, assume that the field within the magnet is pretty uniform, and assign a coil flux $\Phi_\text{coil} = B\cdot A$ proportional to the area $A$ of the magnet. However if there is much empty space between the falling magnet and the sides of the coil, the return field will reduce the flux seen by each coil. I'm going to assume that the magnet mostly fills the coil, so that we don't have to worry about the falling magnet tumbling.
If you connect the two ends of the solenoid across some load resistance $R_\text{ext}$, and the solenoid has internal resistance $R_\text{int}$, the induced voltage (1) will drive a current $I = V/(R_\text{ext} + R_\text{int})$. You get the most current out of the solenoid by shorting the ends together, $R_\text{ext}=0$. We'll call that total circuit resistance $R_\text{ext} + R_\text{int} = R$.
Your statement for the power absorbed by the stack of coils (as of v13) is not correct, because its units are not correct. (Your current version (v32) isn't much better: you have $h=1\,\Omega$ in some places and $h$ a variable height in other places.) The power dissipated by your solenoid will be
$$ P = IV = \frac{V^2}R = \frac{(n\Phi_\text{coil}v)^2}R \tag2 $$
Now, the work done on an object is the scalar product of the force and the displacement, $W = \vec F \cdot \Delta\vec x$, and the power exerted by a constant force is $P = \vec F \cdot \vec v$. The heat deposited in the circuit by (2) must be stolen from the kinetic energy of the relative motion of the solenoid and the falling magnet: the solenoid will, for finite $R$, exert a braking force on the moving magnet. That braking force obeys
\begin{align} P = \vec F_\text{mag} \cdot \vec v &= -\frac{(n\Phi_\text{coil} v)^2}R \\ \vec F_\text{mag} &= - \frac{(n \Phi_\text{coil})^2}R \vec v \tag3 \end{align}
This magnetic force will be present when one end of the magnet is is moving through the solenoid and the other isn't. In the magnet-falling-through-pipe examples above, a short magnet gets a force $\vec F_\text{mag}$ from the current induced where each section of the pipe tries to oppose the increase in flux as the leading edge of the magnet approaches, and a parallel force $\vec F_\text{mag}$ from the opposing current induced where each section the pipe tries to maintain the flux as the trailing edge of the magnet departs. In a solenoid, those opposing currents are connected in series, and so the braking force vanishes to first order if the magnet is moving within the solenoid. (I think there's a second-order braking force that appears if you consider the long solenoid as a set of short solenoids connected in series, but your question is constructed so that we don't have to worry about that.)
In that case, a free-body diagram for the falling magnet gives a net force of
\begin{align} \vec F_\text{net} &= \vec F_\text{grav} + \vec F_\text{mag} \\ m \vec a &= m \vec g - \frac{(n \Phi_\text{coil})^2}R \vec v \\ \frac{d\vec v}{dt} &= \vec g - \frac1\tau \vec v & \text{with } \tau &= \frac{mR}{(n\Phi_\text{coil})^2} \tag4 \end{align}
where the constant $\tau$ has units of time. The equation of motion which obeys the Newtonian constraint (4) is
$$ \vec v(t) = \tau \vec g + \left( \vec v_0 - \tau \vec g \right) e^{-t/\tau} \tag5 $$
where $\vec v_0$ is the velocity of the falling magnet at $t=0$, and the falling magnet exponentially approaches a terminal velocity $\tau \vec g = \vec gmR/(n\Phi_\text{coil})^2$ with time constant $\tau$. This does not give a constant acceleration $\frac{d\vec v}{dt} = \vec g + \vec f$, as you assume in your question.
Let's consider how energy conservation works in some limiting cases. The only parameter in our equation of motion is the time constant $\tau$, so the interesting limits are where $\tau$ is very brief or very long compared to the time $\Delta t$ that's required for the magnet to complete its fall through the tube.
First the case where the time constant is very brief, $\tau \ll \Delta t$. That is the limit when the flux from the falling magnet $\Phi_\text{coil}$ is very strong, when the mass $m$ of the falling magnet is very small, when the coil density $n$ of the solenoid is very large, or when the load resistance $R$ is very small. In that case, no matter what the falling magnet's initial velocity, it rapidly approaches a constant velocity $\tau\vec g$ and maintains that speed for the duration $\Delta t \approx h/\tau g$ of its descent through the tube. At the top of the tube the magnet has only potential energy $U_\text{top} = mgh$; at the bottom it has kinetic energy
$$K_\text{bottom} = \frac12 m (\tau g)^2 = \frac12 m (\tau g) \left( \frac h{\Delta t} \right) = U_\text{top} \frac \tau{2\Delta t} \ll U_\text{top} \tag6 $$
If the fall speed $v = \tau g$ is constant, the current generated will be constant, and the energy deposited in the resistor $R$ will be
$$ E = P\Delta t = \frac{(n\Phi_\text{coil}v)^2}R \cdot \frac hv = \frac{(n\Phi_\text{coil})^2h}R \frac{mRg}{(n\Phi_\text{coil})^2} = mgh = U_\text{top} \tag7 $$
That is to say, if the speed all the way down is constant, the kinetic energy doesn't change, and all of the gravitational potential energy gets spent as electrical energy in the resistor. There's no perpetual motion here; it is totally ordinary Newtonian physics.
Now let's consider the other limit, where the travel time $\Delta t$ is much briefer than the time constant $\tau$. I'll point out in particular that this covers the case from your question where the coils are expanded to the size of a football field while the magnet remains small, because that change in the relative sizes means the coil flux $\Phi_\text{coil}$ includes both the strong field inside of the magnet and the oppositely-directed returning dipole field that surrounds it. In this $\Delta t \ll \tau$ limit, the exponential in the equation of motion (5) can be approximated at all times by
$$ e^{-t/\tau} \approx 1 - \frac t\tau + \cdots $$
and so the equation of motion (5) itself becomes
\begin{align} v &= \tau g + (v_0 - \tau g)\left(1 - \frac t\tau\right) \\ &= \tau g + \left(v_0 - \frac{v_0 t}\tau - \tau g + g t\right) \\ &= v_0 + gt + v_0 \frac t\tau \tag8 \\&\approx v_0 + gt \end{align}
which is, as you said, pretty close to free-fall.
If you wanted to build an oscillatory system ("a system that powers its own motion"), an interesting thing to explore would be an ideal solenoid (self-inductance $L$, internal resistance $R_\text{int}=0$) connected so that the induced current drives a capacitance $C$, chosen so that the oscillatory time constant $1/\sqrt{LC}$ is comparable to the crossing time $\Delta t$. (Many years ago I had a desk toy that worked something like this: a pendulum with a magnetic bottom that swung near a solenoid; that solenoid was powered by a battery that lasted for weeks if the pendulum was oscillating, but died quite rapidly if the pendulum got stopped.) That'd be the subject of another question, though.
Best Answer
The real world is full of small effects that only matter when you've eliminated everything else. For example, if the pendulum has a non-zero conductivity its motion through the Earth's magnetic field would cause eddy currents and dissipate energy. This would be a tiny effect, but it would mean the pendulum wouldn't oscillate for ever. I imagine the more creative minds hereabouts could come up with a number of vanishingly small effects that would eventually damp the pendulum.
If you manage to eliminate all these effects then yes, the pendulum will oscillate forever. However you're just asking whether if all sources of energy dissipation are removed will any energy be dissipated, and the answer is obviously no.