Correct me if I am wrong on this, but my line of thinking is the following:
An independent, movable mass on the vehicle is traveling at the same velocity as all of the other components on the vehicle. If a string were to be attached to the axle and this mass and the string were to wrap around the axle and pull the mass, the string would have to exert a tensional force on the mass because the mass would be accelerated from zero in the vehicles reference frame. An equal and opposite force must be exerted at a distance from the center of rotation of the axle, resulting in a moment (torque) being applied opposite of the angular velocity vector of the axle, thus reducing its angular velocity. Since the mass is being moved a given distance within the car's reference frame, work is being done on the mass by the angular motion of the axle.
Interested to hear your thoughts. I think the difference here is you were looking at the vehicle as a single system from an outside reference point (say from an observer).
You are correct that as the vehicle accelerates the mass from the vehicle's frame of reference, it will exert a braking force on the Vehicle. You're forgetting that the same has to happen to decelerate the mass from the vehicle's reference. Unless your mass can continue on a different velocity indefinitely (i.e. detach from the vehicle and continue flying off), then that reduction in speed of your vehicle will only be temporary.
The amount of kinetic energy in your Vehicle remains constant (ignoring any losses). You may have accelerated the mass briefly, but at the end of the run, the mass must be at the same speed as the vehicle. The mass will likely hit some kind of hard-stop, and just transfer that extra energy back to the vehicle.
So yes, if you look at the Vehicle as a closed system, at the end of the run, kinetic energy has to be 0. If you haven't stored any of that kinetic energy in something (like elastic or gravitational potential), then it all must have been dissipated as heat by your brakes.