Is stopping something work? If somebody pushes against a mass moving with $3 \frac{m}{s}$ to slow it down to $2 \frac{m}{s}$, he will drain the moving system of kinetic energy. Does he do work then or does he consume work?
My homework problem is that somebody is on a train, splits the cars and pushes one away making all other ones slower. The total energy of the system is lower, but he did some work obviously since he pushed it.
http://wstaw.org/m/2011/11/19/m12.png
So do I add the two changes in kinetic energy (which yields a negative number) or do I add their absolute values?
 A: If you actually gave us your complete homework problem, then your homework problem is wrong; it describes an impossible occurrence. 
Assuming all the cars are the same weight, if you push one car away from the other three, in order to maintain conservation of momentum, the change in velocity of the lead car should be three times that of the change in velocity of the others. That is, the forward car could now be going at 6 m/s and the other three at 2 m/s. The kinetic energy before the push is then proportional to $$\frac{1}{2} \cdot 4 \cdot 3^2 = 18.$$ The kinetic energy after the push is now proportional to $$ \frac{1}{2}\left( 3 \cdot 2^2 + 1 \cdot 6^2 \right) = 24, $$ an increase. 
Your intuition that you need to do work to push the car away is completely correct.
A: Work is signed.  When you apply a force that slows something down, you do negative work on it.
With the train example, you would not use the absolute values.  The work done would be positive on cars that get faster and negative on cars that get slower.
To tell whether the work will be positive or negative, use the formula
$$\mathrm{d}W = \vec{F}\cdot \mathrm{d}\vec{x}$$
If the angle between the force and displacement is acute, the work is positive.  If it is right, the work is zero.  If it's obtuse, the work is negative.
