I'm a little late to this party, actually to the point where the lights are already off and they're about to lock the doors. But I got to chime in here, because the OP was asking for a conceptual answer to a conceptual question, and what he mostly got in response was math, definitions, and "you're asking the wrong question." Maybe it's because I'm such a newbie myself, but I find that when I'm trying to grasp a concept, I am looking for simple, real-world examples or analogies, not eight lines of calculus followed by a QED. So, bodhi, if you're still out there, here is my answer:
I'm going to start by saying that it really was kind of a wrong question, but let me explain why: Force by itself does not transfer energy; only when it is performed over a distance (work) does energy come into play. I know other people have given that answer on this post, but no one actually provided an example of non-work force, which would have been a lot more helpful, I think. So let's suppose I hang a simple weight from a string. That is force without work - the string is exerting an upward force on the weight, but no energy is changing hands. Assuming no external actors, that weight could hang there until the end of time and have the same energy as when it started.
Therefore, to understand what's happening to the energy, we need to talk about the work that is being done, as the force slows the object, not just the force itself. And that means that when you ask, "why is that entire force not converted to energy or kinetic energy", you really are asking the wrong question, because it is the object's energy that is going into the work of the force, over the distance it takes that force to slow the object down. And yes, all of the kinetic energy lost by the change in velocities does go into that work. (Yes, I know that per relativity, the direction of energy transfer is meaningless, but right now I'm just trying to stay focused on force vs. work.)
Now let's look at the math. Incidentally, morrobay actually did a pretty nice summation up above, but to keep things simple, I'm going to skip the calculus by assuming linear changes, which let's me stick to algebra.
First the momentum, which you've obviously already picked up on; but I'm going to work it out anyhow, in case anyone else comes looking. So when I apply the force, I get:
F = ma = m\left ( \frac{v_{2}-v_{1}}{\Delta t} \right ) = \frac{mv_{2}-mv_{1}}{\Delta t}
That is: F\Delta t = mv_{2} - mv_{1} = p_{2} - p_{1}
So the force over time (called impulse) does equal the change in momentum. You kind of left \Delta t out of your equation, though, by setting it to 1, and that might also have been a source of your confusion. The reason it matters is that the whole system is moving during this \Delta t, which means the force has gone a distance while slowing the object, and that means... work. So let's do the work equation, and see if we can figure out how that energy went to the force:
W = F\Delta x = ma\Delta x = m\frac{\Delta v}{\Delta t} \Delta x = m\left ( \frac{v_{2}-v_{1}}{\Delta t} \right ) \Delta x
We know that the basic equation for distance is \Delta x = v\Delta t. And as I said above, for simplicity we'll assume linear deceleration, so that v just becomes the average of v_{1} and v_{2}. (It still works out the same if you don't assume this, but... calculus.)
That is, \Delta x = \bar {v}\Delta t = \left ( \frac{v_{2} + v_{1}}{2} \right )\Delta t
Now the work equation becomes: W = m\left ( \frac{v_{2}-v_{1}}{\Delta t} \right ) \left ( \frac{v_{2} + v_{1}}{2} \right )\Delta t
The two \Delta t's cancel out, leaving W = \frac {1}{2} m \left ( v_{2}-v_{1} \right ) \left ( v_{2} + v_{1} \right )
I think you can see where I'm going here, but let's finish it by applying the FOIL method (remember that?) to multiply our polynomials:
W = \frac{1}{2}m \left(v_{2}^{2} + v_{2}v_{1} - v_{1}v_{2} - v_{1}^{2} \right) = \frac{1}{2}m \left(v_{2}^{2} - v_{1}^{2} \right) = \frac{1}{2}m v_{2}^{2} - \frac{1}{2}m v_{1}^{2}
or just W = E_{K2} - E_{K1}
So while it's true you can say that "the entire force goes into change of the momentum", the actual work goes into the change in energy (or the energy goes into work, if you prefer).
Hopefully that helps you form a better, conceptual picture.
