What happens to $P = F × v$ for a car moving on a frictionless surface and in vacuum? I'm considering a scenario in which a car with an engine capable of producing maximum power of 200 hp is moving on a frictionless surface and in vacuum. Since no work is lost due to friction or air drag, the car should accelerate indefinitely. For this scenario, let's assume that the 200 hp engine is working at full power and producing 2000 N at the wheels.
Now since the car will be accelerating indefinitely, there will be a point at which the velocity of the car multiplied by the force will result in a power requirement more than what the engine of the car can produce.
To illustrate this, let's assume that the car has accelerated to a velocity of 500 m/s, the power then will be: P = F × v = 2000 N × 500 m/s which is equal to $10^6$ watts or about 1300 hp.
My question simply is: since the engine is incapable of producing any thing above 200 hp, then how will this situation alter the power equation to reflect this?
 A: As cars work due to friction, I'm going to assume that you mean a system without any drag rather than no friction.  So that 100% of the power of the engine is developed into increasing the KE of the car.

For this scenario, let's assume that the 200 hp engine is working at full power and producing 2000 N at the wheels.

Unfortunately, we can't do that with a real engine.  For any real engine, the ability to develop force/torque decreases as the speed gets higher.  In fact, you can use the speed and the power to find the max force at that speed. 
At high speed the engine will still be able to accelerate the vehicle, but with ever decreasing amounts of force/torque.

But my question here is about the physical quantities rather than the true capability of an ICE. In other words and to make things simpler, let's use a 200 hp rocket producing 2000 N

This isn't a limitation of an internal combustion engine (or any engine).  It's a limitation of how the force is produced.  You only have two choices for producing the force:


*

*You're pushing against some external mass (like the earth)

*You're pushing against some mass you have with you (you're a rocket)


My answer above is limited to the first case.  As your speed increases relative to the reaction mass, your ability to produce torque decreases.  This doesn't matter if it's an ICE, an electric motor, a spring, or anything.  
If you bring the reaction mass with you, then you are producing constant force, not constant power.  But at the beginning, your system is horribly inefficient from an energy point of view.  Whereas in the first case all of the energy of the engine can go into the KE of the car, in case of the rocket most of the energy is going into the KE of the exhaust.   
At high speeds (when the rocket is going at speeds approaching the exhaust velocity), additional power comes from the fact that the KE of the now-accelerated fuel is reduced as it leaves the rocket.
A rocket can produce constant thrust, but not constant power.  The power will change as it accelerates.
Here's one last way to think about it:  The transmission from your power unit (engine) to your reaction mass (the ground) can be considered to be a moveable lever.
You have a choice with a lever, you can shorten the lever so that it produces high speed but reduces the force you apply, or you can lengthen the lever so that it produces lower speed, but increases the force you apply.
As your speed relative to the reaction mass increases, you have to bias your lever more to the "speed" side, which will reduce your applied force.  In a car this happens through the gears in the transmission, but is true regardless of the method applied.
A: 
For this scenario, let's assume that the 200 hp engine is working at full power and producing 2000 N at the wheels.

200 hp is approximately 150 kW so I am just going to use that for this answer. 
Because $P=F\cdot v$ if you specify both $P$ and $F$ then there is only one possible $v$. In this case $P=150\text{ kW}$ and $F=2\text{ kN}$ implies $v=75\text{ m/s}$. No other velocity, either higher or lower, is possible to meet that combination of power and force. 
If the car continues accelerating at peak power then the force will necessarily decrease as the velocity increases. Under the idealized conditions you listed you can continue accelerating indefinitely, but at progressively lower force and lower accelerations. This is directly implied by $P=F\cdot v$
A: I note your question says the car is

moving on a frictionless surface and in vacuum

Let's go back to work for a moment: the work formulation is: $W = F\cdot{d}$
whereby in your situation it means "the additional distance covered by the car due to the engine power input during a certain time interval" 
Following from that definition, power $P$ is "The additional velocity given to the car by force $F$", i.e.
$P = {F\cdot d \over{t}}$
Therefore, the expression $P = F \cdot v$ means the power $P$ required to $\underline{increase}$ the velocity of the car using force $F$, is that force $F$ multiplied by the $\underline{additional}$ velocity given to the car during the applicable time interval.
Now, if we simplify the situation and say that friction in the engine does not increase, even then, the engine will not be able to accelerate the car beyond a certain speed because of chemical limitations: there is a minimum time required to combust the fuel-air mixture in the engine cylinders.
Note that in the real world, the force would be required to maintain a velocity because air resistance, friction and so on are trying to slow down the car with a 'power' output equal to the power input of the force being used to maintain that speed.
