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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?

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  • $\begingroup$ I think you're confusing the power provided by the engine with the actual power of the system. The part that accelerates the system, in this case the engine, can indeed have a smaller power than the total power that the system can achieve, as you're moving on a frictionless surface and there's nothing from stopping you from adding more power. $\endgroup$ – Charlie Mar 21 '19 at 23:55
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    $\begingroup$ To exemplify this, consider a rocket which moves in the vacuum (assume it has infinite, massless fuel). It will continue pushing the rocket even to relativistic speeds if there's nothing to stop it, and ultimately the net power of the rocket $P=F\cdot v$ will be obviously way bigger than the power that the engine is able to provide, yet it will continue accelerating assymptotically towards the speed of light (it will never each it, though, as that would need an infinite amount of energy). So I think the same idea applies to your example. $\endgroup$ – Charlie Mar 21 '19 at 23:57
  • $\begingroup$ @Charlie: correct. Put it as an answer. $\endgroup$ – Gert Mar 22 '19 at 0:03
  • $\begingroup$ @Charlie. I understand what you mean. But now if we consider that F= ma, then it is obvious that the car should have a constant acceleration, but at the certain point, the engine will be unable to accelerate the car at the same rate since the system is increasingly requiring more power to accelerate at the same rate at higher velocities. So what will happen, will the force drop to compensate for this? $\endgroup$ – Abanob Ebrahim Mar 22 '19 at 0:03
  • $\begingroup$ @AbanobEbrahim not exactly. As you can see from $F=ma$, the acceleration will indeed be constant if the force is constant, so there's nothing stopping a small force from accelerating an object to great velocities. The power you provide doesn't have to be more or equal to the net power of the system, as there's no reason to compensate since you aren't losing energy (as you're on a frictionless surface). You will need to compensate only if there's a mechanism that causes the system to lose energy, like friction or radiation. $\endgroup$ – Charlie Mar 22 '19 at 0:06
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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.

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  • $\begingroup$ I agree with you from an ICE point of view. 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. $\endgroup$ – Abanob Ebrahim Mar 22 '19 at 1:57
  • $\begingroup$ My answer is not specific to ICE. Amended. $\endgroup$ – BowlOfRed Mar 22 '19 at 3:00
  • $\begingroup$ @AbanobEbrahim, added an analogy to a lever. That may be a bit more what you're looking for. Think of moving a lever's fulcrum closer to you. It allows you to keep up with a fast moving load, but it decreases the force you can apply. $\endgroup$ – BowlOfRed Mar 22 '19 at 3:15
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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 $P=F\cdot v$

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  • $\begingroup$ Thank you. I think your answer makes the most sense to me. If I understand you correctly, and in order to keep the force and acceleration constant, we would alternatively need to progressively increase the power of the engine to match the value calculated from P=Fv at constant force and increasing speed. Is that correct? $\endgroup$ – Abanob Ebrahim Mar 22 '19 at 2:04
  • $\begingroup$ Yes, that is exactly correct $\endgroup$ – Dale Mar 22 '19 at 2:30
  • $\begingroup$ Great. That makes perfect sense mathematically and from the equations. But practically in case we want to keep the power constant, how can I conceive the idea that the force will inevitably have to decrease while knowing that the engine should still produce 2000 N as this is independent on the system? $\endgroup$ – Abanob Ebrahim Mar 22 '19 at 2:35
  • $\begingroup$ “knowing that the engine should still produce 2000 N as this is independent on the system”. I don’t know where this idea came from, but it is not correct. The force produced by a drive train is not independent of the speed. Sometimes a force is listed as the force at v=0. Could you be misinterpreting the v=0 max force as a max force at all v? $\endgroup$ – Dale Mar 22 '19 at 2:56
  • $\begingroup$ No I understand this. But can't we for simplicity in this scenario just picture the engine as a rocket with massless fuel while neglecting the KE of the exhaust? By doing this we will skip the problems arising specifically from internal combustion engines since this scenario just requires a source of constant force of 2000 N. $\endgroup$ – Abanob Ebrahim Mar 22 '19 at 3:06
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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.

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  • $\begingroup$ This answer is not correct. P=F.v is valid at every instant. There is no interval involved. At every instant the instantaneous power is equal to the force times the instantaneous velocity. There is no need to determine an interval nor what is additional velocity $\endgroup$ – Dale Mar 22 '19 at 1:53
  • $\begingroup$ @Dale It is useful and perhaps necessary to consider time since the question states no resistance, so no power would be required to maintain any velocity. In the real world, yes, 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. Consideration of time is valid. You may wish to simplify it for your own understanding, that's fine. $\endgroup$ – Dlamini Mar 22 '19 at 2:25
  • $\begingroup$ It has nothing to do with simplification for my understanding. It is just about the meaning of the terms in the equation $P=F\cdot v$. P is instantaneous power and v is instantaneous velocity and F is the force whose power we wish to calculate. There is no “additional velocity” nor any “applicable time interval” involved. Those concepts have no part in the equation. Bringing them in is simply wrong. Consideration of a time interval is not valid for a formula which uses only instantaneous quantities $\endgroup$ – Dale Mar 22 '19 at 3:28

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