Does a Constant Force Require Increasing Power (Physics)

General Questions
#1

If an object at rest has, say, one constant force “forward” on it and the object’s mass isn’t changing (and we’re not getting relativistic) then the object will have a constant acceleration. That means its velocity increases in a linear fashion. However, that means its KE increases quadratically. So, it seems to me, the object is gaining more KE every second than it gained last second. (More KE is gained going from 2 to 3 m/s than from 1 to 2m/s.) That means the power from the force is increasing, even though the force is not.

Something I was reading said this makes sense because the object has a greater velocity when travelling fast, and therefore covers a greater displacement each second, so the work done each second is more when you are travelling faster since d is bigger (during each second) and Work = F*d (is true in this simple case with a constant force and since the angle between F and d is 0).

I’m pretty sure about all that, but if you find a flaw fine. But here is my question:

If the force needs an increasing power, does that mean that, say, a car or rocket (ignoring the rocket’s changing mass) needs to burn MORE fuel each second to maintain a constant force when it’s travelling faster than when it’s travelling slower?

I would’ve said before that burning fuel at a constant rate provides a constant force, but now that seems impossible. Burning fuel at a constant rate means a constant rate of energy consumption and thus a constant power (right?), but an INCREASING power is needed to keep increasing the amount of kinetic energy gained each second.

Please help untie the knot in my head about this. I feel like there should be an easy explanation to it but I’ve asked a few sciency people and they couldn’t reconcile the situation for me. As you answer: I’m a high school physics teacher and I know my stuff, though I’m rusty on things like the Hamiltonian and “past” that. I know the whole question breaks down at high speeds, and Newton is wrong, Einstein is (probably) right, but let’s try to pretend this isn’t a question about relativity (assuming that’s possible).

#2

Yup. You have to take into account the kinetic energy of the rocket exhaust, whose velocity will depend on the speed of the rocket at the moment that parcel of exhaust was produced. Then it’ll all work out.

#3

I’m not sure which part the “Yup” refers to. Could you please clarify? Are you saying more fuel needs to be burned to maintain a constant force as you speed up?

Also, you’re saying the fuel changes KE too. Are you saying it was going faster, now it gets “spit out the back” of the rocket and slows down and has less KE? Or are you saying, from the rocket’s perspective the fuel didn’t have KE and now it has some?

#4

Yes, power is equal to force times speed.

#5

It’s an interesting question to try to answer intuitively, and I’ll take a stab at it. You’re substantially on the right track in attributing the key as being the notion of force acting over distance, which in physics is the definition of work, and which is directly reflected in the change of kinetic energy of the object.

The problem with intuition seems to just be implicitly confusing “force” with “work” and “energy”. “Force” is not “energy”. Yes, a constant force produces constant acceleration, but as speed increases the distance over which the force is acting is cumulatively summed, and from the reference frame of an external observer, there is a quadratically increasing amount of work being done and energy expended to maintain a constant acceleration. This is self-evidently true since that’s where the kinetic energy comes from.

And it’s true whether it’s a rocket in space with constant thrust – where the extra energy (from the observer’s reference frame) comes from the propellant mass that’s been accelerated along with the rocket – or whether it’s earthbound devices where motors have to do more work by spinning faster (or, if geared down, spinning with more force). No matter how you do it, as you speed up you always end up with forces acting over cumulatively longer distances, and one way or another, you have to put in the energy to make that happen.

#6

Let me put it in very rough numbers as a practical exercise:

You have a 1kg mass moving a 1m/s. It therefore has a kinetic energy of .5J.

You push this with a force of 1N for 1s. This gives it an acceleration of 1m/s^2, so after one second it has a velocity of 2m/s. It now has a kinetic energy of 2J.

You do the same for another second. Velocity is now 3m/s, energy is 4.5J.

How can 1N for 1s give 1.5J increase in one instance and 2.5J in another?

As pointed out, energy is force times distance. In the first second the distance was 1.5m, and in the second the distance was 2.5m. Everything works out nicely.

So yes, the power from a constant force increases.

#7

I find the question as unintuitive as you do, but I do believe the situation is very different with cars as opposed to rockets.

Cars - yes.

Rockets - no.

With cars, in order to generate the same acceleration force against the road at higher speeds, the engine must run at a higher rpm so the cylinders must be refilled with fuel proportionately more often. To produce the same acceleration force at twice the rpm precisely requires burning twice the fuel.

As for the rocket, maybe the Wiki entry on Oberth effect helps.

#8

It’s fairly intuitively obvious for ground-based vehicles like a car that more work is being done when accelerating at higher speeds – one can show that the engine’s pistons have to move further per unit time, so consume more fuel per unit time, or an electric motor has to spin faster, as noted before. Actually, the same principle applies to rocket ships in a vacuum, though it’s less intuitively obvious. Consider this simple thought experiment. Suppose you’re on a space station, which we will consider to be our stationary frame of reference. A spaceship that is stationary with respect to the station fires its thrusters for one second and moves away at a speed of 1 m/s.

Now imagine a second identical spaceship, with exactly the same mass and fuel and engines, that goes past at 5 m/s. It fires the identical thrusters for one second and now increases its speed from 5 to 6 m/s. The paradox appears to be that, with exactly the same thrust and consuming the same energy, according to the KE calculations that give us the quadratic progression, spaceship #2 has somehow gained more than ten times the kinetic energy increase of spaceship #1 from the same one-second thrust.

Probably the best way to resolve this paradox is to imagine yourself in the frame of reference of spaceship #2, moving alongside it at 5 m/s. Then you would agree with the observer of spaceship #1 that a 1 m/s gain in velocity produces exactly the same gain in KE that he observed, and it would be true for you in every practical sense, because if the ship smashed into you, the damage and energy dissipation would correspond to a velocity of only 1 m/s.

So we can conclude that there’s no actual paradox. In familiar contexts more work has to be done per unit time to push things to go faster, but force remains the same for any given acceleration, ignoring friction and other practicalities. For rockets in space, we note that even in classical physics we need to account for different energy views from different frames of reference.

#9

Yup, power is force times speed, like Chronos answered.

And, there are various components here that have kinetic energy: the rocket, the unburnt fuel, the burnt fuel (exhaust). From any reference frame the sum of their kinetic plus chemical plus radiated energy will stay constant throughout the experiment.

I pictured a similar brain teaser about 40 years ago. Two rocket ships are attached to winches on some asteroid that is big enough they’re not going to move it enough to matter, but small enough that their gravity can be ignored. Each winch turns an electric generator as it unwinds, freewheeling at first because we don’t have a load on the generators. Both rockets start off at the same time, and after a minute we attach an electrical load to one winch, and do likewise for the other winch at two minutes. The second rocket will be going twice as fast, having accelerated similarly for twice as long (I’m assuming the rockets have enough fuel that their mass stays pretty much the same for the duration). Both electrical loads create a drag equal to the rocket thrust so the rockets now travel at constant speeds, and the second system will be cranking out twice as much electrical power as the first.

#10

Thanks everyone. I think I’m getting close to wrapping my mind around it. Putting your replies together it seems like you all are saying that the force becomes more efficient in producing KE for the object as the speed of the object increases. Q1: Is that correct?

But if the force remains constant eventually the increase in KE in 1s will get larger and larger, as Naita showed. So an identical rocket going by at 100m/s would get a very large increase in KE with the same amount of fuel (right?). Still trying to stay non-relativistic, Q2: I still don’t see how the same fuel burn could provide so much (eventually infinite) of an increase in KE. And it seems to me I run into the same question in the Oberth Effect article posted by Frankenstein Monster,
https://en.wikipedia.org/wiki/Oberth_effect#cite_note-ways-3 it says " The Oberth effect also can be used to understand the behavior of multi-stage rockets: the upper stage can generate much more usable kinetic energy than the total chemical energy of the propellants it carries." I understand that the upper stage could be more efficiently used when the rocket is already going fast (well, I sort of understand why), but I don’t get how it could produce more usable KE than the chemical energy of the propellants. It sounds like energy from nowhere, like energy isn’t conserved.

Wolpup, I think you were getting at this, but I don’t understand the explanation of the paradox.

Maybe I can’t generalize things so much, but all this makes me wonder if we can say that practical things on earth have trouble continuing to accelerate not just because of an increase in air resistance but because they can’t provide the power to continue providing that same force while now travelling at a higher speed. So when I run fast the reason I can’t speed up isn’t just because of the biomechanics of running but a real physics issue of needing to provide MORE power now that I’m travelling faster? Q4: Or is that all wrong and it’s about the efficiency of the force in providing KE?

#11

Picture the rocket coasting past you at high speed, when all of a sudden it throws a bunch of mass out the back. That mass was part of the rocket, and so already had a high kinetic energy. Now that mass is moving more slowly, and so has a lower kinetic energy. That kinetic energy the propellant lost went somewhere. Specifically, it went into the rocket.

#12

Thanks Chronos. That exact thought occurred to me this morning as I was mulling the posts of everyone here. The burned chemicals not only lose chemical potential energy but also lose KE.

If one were to pretend that the burned chemicals have a constant force on them (for a short time) then their loss in KE would be quadratic as well. So the faster the rocket (and fuel) is going the greater the change in KE for the burned fuel when it loses, say, 1 m/s of speed.

So, with this effect, does that mean the rocket can burn the same amount of fuel each second as it accelerates (non-relativistically), or would it still need to burn more fuel each second to keep a constant force?

#13

Yup, a constant fuel burn rate will result in a constant force (and thus, an increasing acceleration, as the mass decreases). This is easiest to see from the point of view of someone on the spaceship: In their frame, they’re at rest, no matter what speed anyone else says they’re going at, and so burning the same fuel will always have the same result on them.

#14

Right, I actually imagined myself watching the fuel burn and leave my ship. No matter how fast I’m going it’d always leave with the same speed compared to me. So a constant fuel-burning rate would mean the fuel burned each second would have the same loss in momentum, which means my rocket gets the same gain in momentum (ignoring the fact that the rocket is losing mass). But that means a constant dp/dt, which means a constant force. So a constant burn rate makes a constant force.

The conundrum was that an observer would also agree that the burn rate was a constant J/s and that it was providing a constant force, but the observer would say that the force was providing an ever-INCREASING power (increasing J/s). So it seemed problematic that a constant J/s was causing an increasing J/s. The piece I was missing was that as the rocket speeds up the unburned fuel’s (quadratic) gain in KE each second means that when it is burned it has a (“quadratic”) loss of KE of as it becomes the burnt fuel being shot out the back. (The fuel burned in each successive second loses more KE than the fuel burned in the previous second, and that energy has to go somewhere.)

I think I’m still going to have to think a bit about the specifics of how that works out (i.e. how the rocket “knows” to gain the KE lost by the burned fuel) but the major knot has been untied.

Thanks!

#15

About the rocket “knowing” about the burned fuel? That’s a conservation of momentum thing, iirc.

#16

It seems that you consider the KE to be an absolute physical value, which is not. It’s relative, just as PE.
Kinetic energy is defined as an amount of work needed to accelerate an object from V0 to V1 (or V1 to V2, it doesn’t matter), so EK=m(V1-V0)squared/2. For the sake of simplification V0 is heedlessly assumed to be always zero. That’s where all hell breaks loose if one doesn’t have a good enough conceptual understanding of the subject. So there’s no conundrum or paradox, the amount of KE gained on equal time intervals is the same, assuming the acceleration is constant, because every other second V1-V0 = V2-V1= V3-V2 and so on.

#17

Correct me if I’m wrong, but isn’t the kinetic loss of speed of the spent fuel actually what drives the rocket forwards? The point of burning fuel is to release trhe energy bound in the fuel but what actually accelerates the ship is the physical mass of the spent fuel being tosseed back – the reason we burn rocket fuel rather than say tossing crap off the back of the ship is that it contains a lot of energy for its mass, meaning each particle gets accelerated away from the ship at a very high velocity, meaning a high force for Newton’s laws to act upon the ship in turn.

#18

I believe I agree with you on the first point. KE is relative to the reference frame.

However, the change in KE (delta KE) is = 1/2 m V2^2 - 1/2 m V1^2, it is not = 1/2 m (V2 - V1)^2.

#19

You are probably right. The only way the rocket gains speed is by gaining momentum, and the only thing that loses momentum is the burnt fuel. The energy of the burnt fuel not only causes the force to speed up the rocket and slow down the burn fuel (from an observer’s reference frame), but it also generates a whole heap of “heat” that is not converted into increasing the rocket’s KE.

(But the “heat” does cause the burnt fuel’s particles to be moving fast and cause a pressure on the bottom of the rocket, pushing it forward, which is the real cause of the force on the rocket on the molecular level, as I understand it. All this business of momentum and energy are mathematical constructs man has created to simplify our lives and help us solve problems faster and more easily.)

#20

Momentum is the conserved quantity here, the heat of the gasses does allow the ability to expel mass at speed, but it doesn’t add to the velocity in itself.

Here it the ideal rocket equation, to avoid tensors.

Δv = v[sub]e[/sub]ln(m[sub]0[/sub]/m[sub]f[/sub])

Where:
Δv = max change in velocity
m[sub]0[/sub] = the initial mass
m[sub]f[/sub] = the final mass
v[sub]e[/sub] = the exhaust velocity

You can see that it is the change in mass* (or the natural log of it)* that is important.