quantum-tunneling
According to a classical mechanics a particle that do not have enough energy will not be able to surmount a potential barrier.
But what if we have a lot of particles(e.g. electrons)? They can interact with each other(Coulomb's law) and transfer enough energy for some of them to surmount a barrier.
Does it explain a tunneling phenomenon without quantum theory?
Not a good analogy. Here's why.
Your cork wall system is very complicated. That wall isn't just a potential with finite height. Rather, it contains holes where the potential is zero (or near zero), so that the particle can just propagate further and it contains full woody parts where the potential is infinite and the particle bounces back. So it is a very complicated many particle system which you can reduce to one parameter (like penetration depth, possibly depending on energy). But the resulting system will be probabilistic and very different from just a single particle with finite height potential. Because in this simple system the particle will bounce back with 100% probability if its energy is below the potential and pass with 100% probability if its energy is bigger.
Now, if you take this simple system (with just one degree of freedom) and investigate it from the point of view of quantum mechanics you'll discover that there's non-zero probability that particle will pass through the potential even if it doesn't have sufficient energy to pass it.
Why is this strange? Well, the particle has a total energy $E = T + V$. This means that its kinetic energy is ${1\over 2}mv^2 = E - V$. As you can see, this equation makes no sense classically if the quantity on the r.h.s. is negative because you'll obtain imaginary velocity. But in quantum mechanics this still makes sense because the particle is also a wave and that imaginary factor means that this wave will exponentially decay, rather than just propagate freely.
This model is something that I've seen a lot of students (including myself) struggle with, and a big part of it is that calling it the "Coulomb barrier" is quite over-simplified.
Going back to fundamental physics principles, the Coulomb force is repulsive as you say, so I'll write this as $F\propto 1/r^2$, where $F$ is the force and $r$ is the distance from the center of the nucleus and the positive sign indicates that the force is in the positive $r$ direction, away from the nucleus. The potential, which is what's shown on the graph, is the negative of the integral of the field (which is proportional to the force), this results in $E\propto 1/r$. For an analog, imagine lifting a weight where "up" is the positive z-axis. The path involves an increase in z-coordinate and the force from the field (gravitational) is negative. The negative integral of the field over distance is the potential.
The missing part is why they "chop off" this $1/r$ function at some radius, presumably the radius of the nucleus. Well, this allows us to infer some things about the other force present, which is the nuclear force. It would appear that the force acts weakly beyond the radius of the nucleus, since the $1/r$ form is unaltered beyond that point. It also absolutely must have a potential that increases faster than $1/r$ as $r \rightarrow 0$. Both of these would be loosely satisfied if the nuclear potential was $-1/r^n$ (note the negative) where $n>1$. The alpha particle, while in the nucleus, by the way, has kinetic energy which is the reason its energy is higher than the hypothetically lowest energy level possible for it. This has to do with the fact that quantum mechanics only allows certain (quantized) energy levels.
Just For an example of how this is possible, I'll say $E_{nuclear}=-1/r^4$ (I am not saying this is how the force acts, it's just for utility).
$$E(r) = E_{nuclear}(r) + E_{Coulomb}(r)$$ $$E(r) = 1/r - 1/r^4$$
$$F(r) = -\frac{d}{dr} \left( E_{nuclear}(r) + E_{Coulomb}(r) \right) = F_{nuclear}(r) + F_{Coulomb}(r) $$ $$F(r) = 1/r^2 - 1/r^5$$
My intent is that this answer allows you to explicitly answer to yourself what the potential and force contributions are, and give some very rudimentary example for how the "edge" can appear. Beyond that I hope it's clear how someone may take the shape of the above curve and lump it into a "wall".
Best Answer
This will help:
Please note that the energy is the same , inside and outside the barrier. Also it works on one particle. Your "model" would give a statisticacl gaussian distribution for the probability of finding the tunneling particle whereas alpha decays of a nucleus, for example, have an exponential fall off distribution.
No, tunneling is a quantum mechanical phenomenon.