Net work required to bring an object with kinetic energy to stop? Am I right in assuming that if an object, let’s say a car, is travelling with kinetic energy of $x$, then the net work required to stop the object would be $ x $ Joules, but in the opposite direction?
 A: Energy does not have a direction. It is a real number, but it is signed. So the work that must be done on the car will be $W=-x$ but that does not imply a directionality for the work.
A: Imagine if the car hit a box and instantly transferred all the energy it had to that box. The car would stop, the box would keep going. The kinetic energy the car had is the same as the work it would do on the box hit by it if that object did not resist:
$x=\frac{1}{2} mv^2$
As Dale pointed out, work is a scalar quantity and does not have a direction, but what I suspect you mean is this: "if the box applied just enough force in the opposite direction to keep from moving, so that the car and box were stopped, how much would that work would that force do?"
Well it would indeed be the same as the kinetic energy, yes.
A: Yes, that is correct.
Work = Force x Distance
Air and Axle friction would eventually stop it, brakes or going uphill really help.  This is why "run away truck" exits on mountain roads are ramps to a higher elevation, using
mg(h2-h1) = 1/2mv$^2$ to bring the truck to a halt.
Soft sandy soil is often used to increase drag and help slow the truck as Drag Force x Distance = Work.
The work against the kinetic energy of the truck can be signed "- Work".  Since the units of kinetic energy are the same as the units of Force x Distance, we can say, at a halt:
Kinetic Energy - (Sum of all Braking Work) = 0
The rate of Work/Second is referred to as Power.
