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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?

Abanob Ebrahim
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  • 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. – Charlie Mar 21 '19 at 23:55
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    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. – Charlie Mar 21 '19 at 23:57
  • @Charlie: correct. Put it as an answer. – Gert Mar 22 '19 at 00:03
  • @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? – Abanob Ebrahim Mar 22 '19 at 00:03
  • @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. – Charlie Mar 22 '19 at 00:06
  • @Charlie then what does the 1000000 watts mean in this scenario? There is obviously nothing that is producing 1000000 joules/second in this system. – Abanob Ebrahim Mar 22 '19 at 00:08
  • In other words, the situation you describe about the power input being equal or bigger than the new power of the system only applies if the system is itself losing energy, and thus needs additional input to compensate that lose. If, for example, your car was on a surface with friction, indeed there will be a moment when the engine output will no longer be able to accelerate the car anymore, as the rate energy is lost by friction will be equal to the input of energy by the engine. – Charlie Mar 22 '19 at 00:08
  • @AbanobEbrahim For the frictionless case, those $10^6 watts$ is the net power stored in the moving car. It isn't going anywhere, as there's no way in your scenario for it to lose energy. Only if you extract energy from it somehow, for example, by putting it on a surface with friction or colliding it with some surface, then that power will be transfered to whatever system you want. – Charlie Mar 22 '19 at 00:10
  • Consider the following scenario. Imagine you drop a ball form the Empire State building that has a mass of $0.1\ kg$. For the sake of simplicity, ignore air resistance. You know that the acceleration of gravity is $9.8\ m/s^2$, thus the force (the weight) is $W=mg=0.98\ N$. The ball will continue accelerating with the exact same force until it hits the ground. Now supposed at the mid of its trajectory it has a velocity of $v=100 m/s$, so the power at that instant is $P=Fv=mgv=98 \ Watt$. Does it make sense to say that the Earth is trying to compensate that power? – Charlie Mar 22 '19 at 00:14
  • @Charlie I don't think "power" can be stored. The 10^6 watts are 10^6 joules being stored in the system every second just by applying a force of 2000 N on it while nothing is giving this much energy at this rate. – Abanob Ebrahim Mar 22 '19 at 00:15
  • @AbanobEbrahim Sorry I should've been clearer, by net power I mean the kinetic energy stored in the system each second. At a given time, you have some amount of energy, right? Since the system is accelerating, you're increasing the velocity each second, and thus increasing the total energy each second. If the total energy is, say for example, $10^5 \ Joules$, that doesn't necessarily mean that you inputted the energy in one go, but you inputted slowly over time until it accumulated to that quantity. The same notion is related to the power, it's just energy over each second. – Charlie Mar 22 '19 at 00:18
  • @Charlie. No, but the ball is indeed gaining 98 joules each second, while the system I described above is not gaining 10^6 joules each second. – Abanob Ebrahim Mar 22 '19 at 00:19
  • @Charlie But power is the amount of energy you give in "one go". The one go is a second in this case. The problem is, with this equation it is showing that the system is gaining 10^6 joules every second while it actually it is not. It is gaining energy at a much lower rate because the amount of power can't keep accelerating the car at a constant acceleration. My question is how to reflect this on the equation? Is the force decreasing to show the correct amount of energy given to the system each second or is there something else that needs to chance? – Abanob Ebrahim Mar 22 '19 at 00:23
  • @AbanobEbrahim Indeed, the system you described above is not gaining $10^6\ J$ each second, but it's gaining $149 \ kJ$ each second (assuming 200 hp=149 kW). The $10^6 J$ is the net energy story in the system (as kinetic energy) when it reaches such velocity. – Charlie Mar 22 '19 at 00:27
  • @AbanobEbrahim No, the powet is just the rate of change of energy per unit of time, you don't have to give it in "one go". The engine can accelerate the car until it has such energy, and thus the power you calculate represents the total energy rate if the car were to transfer that power to some other system. If there's no way for it to lose energy, then you don't need to worry about the total power. The input power is always the same (200 hp), and there's nothing stopping the engine from pushing the car at a constant acceleration, since the force doesn't depend on the total power or energy. – Charlie Mar 22 '19 at 00:30
  • Although to be honest, there's actually a restriction if you consider relativistic velocities. In this case, as the car accelerates, it's mass will indeed increase, so the force you need to apply becomes greater each time. For a constant force, there will be a moment when the engine cannot further accelerate the car, and thus it will indeed balance it. In this case, the equations of special relativity offer you description on how the force decreases when the mass of the car increases in the relativistic case. – Charlie Mar 22 '19 at 00:32
  • @Charlie and there's nothing stopping the engine from pushing the car at a constant acceleration Actually there IS. Let's assume the car mass is 1000 kg, and with a force of 2000 N, the car should be accelerating at 2 m/s^2. Now let's say that car reached a velocity of 1000 m/s. Now if the acceleration is constant, the car should be moving at 1002 m/s the next second and therefore the increase in its KE = 0.5 ×(1002^2 - 1000^2) × 1000 which is ~2 MJ. Remember that the engine is supplying only 149 kJ each second, so there is no way the engine is going to accelerate the car at constant a. – Abanob Ebrahim Mar 22 '19 at 00:47
  • @Charlie To keep the system accelerating at a = 2 m/s^2, you have to keep increasing the power output of the engine to match the value calculated from P=Fv. If you are not doing this and keep the power as it is (200 hp), the engine will not be able to supply the system with enough energy to accelerate it as the same rate at higher speeds. My question again is: how is that reflected? Will the now time-dependent acceleration also cause the engine to apply less and less Force to go along with decrease in the rate of acceleration? – Abanob Ebrahim Mar 22 '19 at 00:49
  • @Abanob Ebrahim As I stated, the power input of the engine doesn't need to match the power of the car for the frictionless case, they're independent quantities as Diamini explained in his answer. However, let's assume a more realistic scenario (with friction and chemical reactions) and assume the engine cannot supply enough power at higher speeds. In that case, you would have to assume an expression for the rate of power lost and solve an equation for the force output of the engine in function of the current velocity. It will all depend on the model you adopt. – Charlie Mar 22 '19 at 01:16
  • @Charlie Again, to have constant acceleration, the power input of the engine must match the power of the car calculated from P=Fv.. Use constant power and inevitably the acceleration rate will drop. Check the KE difference example again to prove this. – Abanob Ebrahim Mar 22 '19 at 01:25

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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.

BowlOfRed
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  • 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. – Abanob Ebrahim Mar 22 '19 at 01:57
  • My answer is not specific to ICE. Amended. – BowlOfRed Mar 22 '19 at 03:00
  • @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. – BowlOfRed Mar 22 '19 at 03:15
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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 $P=F\cdot v$

Dale
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  • 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? – Abanob Ebrahim Mar 22 '19 at 02:04
  • Yes, that is exactly correct – Dale Mar 22 '19 at 02:30
  • 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? – Abanob Ebrahim Mar 22 '19 at 02:35
  • “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? – Dale Mar 22 '19 at 02:56
  • 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. – Abanob Ebrahim Mar 22 '19 at 03:06
  • Rockets are very much not a simplification and if you do consider a rocket you cannot neglect the KE nor the momentum of the exhaust. With a rocket the engine produces a constant thrust, but the power delivered to the vehicle increases as the kinetic energy removed from the exhaust increases without bound. An engine that produces constant force at all speeds has no maximum power. – Dale Mar 22 '19 at 03:15
  • See my answer here: https://physics.stackexchange.com/questions/428952/why-is-the-work-done-by-a-rocket-engine-greater-at-higher-speeds – Dale Mar 22 '19 at 03:16
  • Great answer. I have question related to your answer on the rocket question and I hope you will consider answering it. Consider a scenario of a turbojet aircraft (say F-16) flying at max thrust (afterburner 127 KN) and max speed (2,120 km/h). The F100 engine is consuming 6 kg/s of fuel to achieve this thrust. Now if I understand your answer correctly, this represents the 'low speed' in rockets in which the exhaust velocity is higher than the system velocity. My question: is the 6 kg/s of fuel contributing any of its gained KE to the aircraft or is it powered solely by its fuel power? – Abanob Ebrahim Mar 23 '19 at 22:08
  • You are asking if a jet engine behaves at least partially like a rocket? – Dale Mar 23 '19 at 23:04
  • Yes, but only at afterburner since the engine requires much more fuel at lower efficiency. So if I understand your answer correctly, and even if a jet engine behaved partially as a rocket while using the afterburner, the exhaust would not be contributing any of its gained KE to the aircraft, similar to example you gave when the system was at rest or moving at a velocity < 5 m/s. And then the fuel chemical energy is the only energy doing work here. So do I understand that correctly? – Abanob Ebrahim Mar 24 '19 at 07:13
  • I did a little bit of reading about afterburners. I am not very familiar with them, but from what I read I would say yes, they are at least somewhat like a low-speed rocket. You might consider asking a new question specifically about that to get opinions from people with more knowledge of afterburners – Dale Mar 24 '19 at 11:10
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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.

Dlamini
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  • 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 – Dale Mar 22 '19 at 01:53
  • @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. – Dlamini Mar 22 '19 at 02:25
  • 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 – Dale Mar 22 '19 at 03:28