Why is the net work of a falling object impacting the ground zero? In my physics textbook, I found the following problem:
"A 2,100 kg pile driver is used to drive a steel I-Beam into the ground. The pile driver falls 5.00m before coming into contact with the top of the beam, and it drives the beam 0.12m farther into the ground before coming to rest. Using energy considerations, calculate the average force the beam exerts on the pile driver while the pile driver is brought to rest."
My question doesn't specifically pertain to the question of the problem but rather to the conceptual understanding of the problem. 
I follow that the change in kinetic energy is zero since the pile driver started with vi = 0 and vf = 0. However, I do not understand why this would result in the Wnet being equal to zero. In my understanding, work is the transfer of energy. And since the pile driver started with potential energy and ended with no energy, I do not follow how no work was done on the system. This problem requires you to say that Wnet = 0 in order to solve it, so I'm just curious how that is possible if there was a transfer of energy within the system? 
Thanks for any help, this has really been breaking my brain since we only recently learned about work in our physics class. 
 A: Let's define "initial" as before the pile driver falls and "final" as after it comes to rest on the beam, as it seems you are doing in your question. Now, you are right to say that since the change in kinetic energy is $0$ over this time interval, it must be that the net work done on the pile driver is $0$.

And since the beam started with potential energy and ended with no energy, I do not follow how no work was done on the system.

This statement seems to have some confusion. I think you mean "pile driver" instead of "beam" and "system". But let's say we have made those changes. You are right to say that the pile driver has potential energy initially and then has no potential energy after coming to rest. But you have to keep in mind that the change in potential energy is just the opposite of the work done by gravity ($W_{grav}=-\Delta U$). So gravity does positive work on the pile driver.
But what other forces do work on the pile driver? Well there is the force between the pile driver and the beam when they come into contact. Since the pile driver is moving down and this force pushes up on the pile driver, this force does negative work. In fact, it does enough work so that the total work done on the pile driver is $0$.
So really what we have is 
$$W_{net}=W_{grav}+W_{beam}=-\Delta U+W_{beam}=\Delta K=0$$
So as we see, saying the net work is $0$ does not mean that no forces do work the entire time. It just means that over the time interval in question, the sum of all of the work done on the pile driver is $0$. If, for example, we said that "after" is instead when the pile driver and the beam come into contact, then the pile driver is still moving, so the net work is no longer $0$ (in fact in that case the net work is just the work done by gravity).

In my understanding, work is the transfer of energy...so I'm just curious how that is possible if there was a transfer of energy within the system?

You are right about this, but if over some time period your kinetic energy has ended up at where it started, then an equal amount of energy has been transferred to and from the object, resulting in no net work.
A: 
Net work done in a cyclic process under the influence of a conservative force is 0

When at top, the work done is
 Wg1 = mgh cos180°= -mg
And  work done by gravity to bring it down is Wg2= mgh cos0 = mgh.
Now since gravitational force is a conservative force, so W total = Wg1 + Wg2 = 0
Assuming that acceleration due to gravity is same for small h.
