Confused about kinetic energy

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  • #1
Someperson
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The equation for kinetic energy is:

k = 1/2*m*v2

This confuses me because this indicates it take more and more energy to accelerate...
Lets imagine a rocket in space (assume the mass of the rocket stays the same when accelerating) - the formula suggests it would take more energy (therefore fuel) to go from 50 to 100 km/s than it would from 0 to 50 km/s.

Is it true that it would take more fuel to accelerate from 50 to 100 km/s than it would from 0 to 50 km/s ?????
 

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  • #2
GiuseppeR7
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Kf-Ki=work done (energy inserted)=(1/2)m(Vf-Vi)^2
the energy required to change the speed of the rocket depends on Vf-Vi not on none of the speeds alone.
 
  • #3
Someperson
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I still don't get it:

Imagine 2 rockets, 1 accelerates to 100 m/s and the other accelerated to 50 m/s then to 100 m/s (so both rockets end with the same end velocity)
both have a mass of 1 Kg

Kinetic energy of the first rocket:
1/2*1Kg*(100m/s)2 = 5000 J

Kinetic energy of the second rocket:
1/2*1Kg*(50m/s - 0m/s)2 + 1/2*1Kg*(100m/s - 50m/s)2 = 1Kg*(50m/s)2 = 2500 J

They should end with the same kinetic energy...
 
  • #4
GiuseppeR7
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lol we have both done a SERIOUS mathematical mistake
 
  • #5
Nugatory
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Kf-Ki=work done (energy inserted)=(1/2)m(Vf-Vi)^2
the energy required to change the speed of the rocket depends on Vf-Vi not on none of the speeds alone.

The part in bold is wrong (although by coincidence it gives the right answer if one of Vf or Vi happen to be zero).

The correct formula is ##K_f-K_i = \frac{1}{2}mV_f^2-\frac{1}{2}mV_i^2##.
 
  • #6
GiuseppeR7
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yeah XD
 
  • #7
Nugatory
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Imagine 2 rockets, 1 accelerates to 100 m/s and the other accelerated to 50 m/s then to 100 m/s (so both rockets end with the same end velocity)
both have a mass of 1 Kg

Kinetic energy of the first rocket:
1/2*1Kg*(100m/s)2 = 5000 J

Kinetic energy of the second rocket:
1/2*1Kg*(50m/s - 0m/s)2 + 1/2*1Kg*(100m/s - 50m/s)2 = 1Kg*(50m/s)2 = 2500 J

They should end with the same kinetic energy...

Try calculating the energy required using ##W=Fd## (which is were \frac{1}{2}mv^2## came from in the first place) and you'll see why the part in bold is wrong.
 
  • #8
DrStupid
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more energy (therefore fuel)

That's not correct. In case of rockets more energy does not necessarily mean more fuel. The energetic efficiency of a rocket engine increases with speed.
 
  • #9
I haven't totally read all of the posts related to the topic, but I am familiar with the subject. The force exerted by the rocket is a constant. The time rate of change of the total mass is why the rocket works. The bulk of the payload for a rocket is generally the fuel. Again, as the fuel burns off, the overall mass is reduced and the time rate of change of acceleration ( jerk ) is increased. In this case there are 3 derivatives after position. In order, position, velocity, acceleration and finally jerk.

I hope this helps
 
  • #10
GiuseppeR7
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are you sure? the "energetic efficency" of a rocket depends almost only on the chemistry of it, not on the speed.
 
  • #11
Someperson
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The correct formula is Kf−Ki=12mV2f−12mV2iK_f-K_i = \frac{1}{2}mV_f^2-\frac{1}{2}mV_i^2.
Oh this makes more sense however it still leads me to the same conclusion that it would take more energy (therefore fuel) to go from 50 to 100 km/s than it would from 0 to 50 km/s.

Am I right in saying that burning x amount of fuel will result in a release of kx amount of energy? (where k is an arbitrary constant) assuming the fuel is at the same concentration, heat etc... - I am not worried about the practical side of things in this post (e.g. obviously some fuel will release a different amount energy because of a slightly different condition) I just want to know in theory.

That's not correct. In case of rockets more energy does not necessarily mean more fuel. The energetic efficiency of a rocket engine increases with speed.
Is this the case with only rockets (or other vehicles)? Is it because of some other factor e.g. mass of the rocket changing? If not, I am intrigued to learn more about this.
 
  • #12
Nugatory
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Oh this makes more sense however it still leads me to the same conclusion that it would take more energy (therefore fuel) to go from 50 to 100 km/s than it would from 0 to 50 km/s.
That conclusion is correct.

Am I right in saying that burning x amount of fuel will result in a release of kx amount of energy? (where k is an arbitrary constant) assuming the fuel is at the same concentration, heat etc... - I am not worried about the practical side of things in this post (e.g. obviously some fuel will release a different amount energy because of a slightly different condition) I just want to know in theory.
Yes, you're right about that too.

Is this the case with only rockets (or other vehicles)? Is it because of some other factor e.g. mass of the rocket changing? If not, I am intrigued to learn more about this.
It is the case for everything; as I suggested above you could try starting with the equation ##W=Fd##, and it may make more sense. You can even derive ##E_k=\frac{1}{2}mV^2## from ##W=Fd## (but simplify the problem by keeping ##F## and ##m## constant or you'll need some calculus).
 
  • #13
GiuseppeR7
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if you have a rocket, in general the final speed is given by this: Vf=g ISP Ln(Final Mass/Initial Mass).
Where the ISP is defined as the Thrust/Wheight of fuel flow.
 
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  • #14
Someperson
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So I see that it does take more fuel to accelerate at a constant rate the higher the velocity? this seems very un-intuitive to me. In every physics simulator I've played (e.g. KSP) fuel always seems to be consumed at the same rate - is this just an illusion or are the physics wrong in these games?

This seems like it leads to a paradox:
Imagine 2 reference frames one moving at 0 m/s and the other at 1 m/s
A rocket it also moving at 1 m/s - it then accelerates to 2 m/s.
The rocket weighs 1 Kg.

The 2 reference frames must agree on the amount of fuel being used (right?)
The first reference frame (the one moving at 0 m/s) thinks the rocket gained 1.5 J of kinetic energy
however the second reference frame thinks the rocket gained 0.5 J of kinetic energy

If burning x amount of fuel release kx amount of energy and both reference frames agree that x amount of fuel was burned then both must also agree on k as well however as shown above they don't.

what is the mistake I am making?
 
  • #15
jbriggs444
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are you sure? the "energetic efficency" of a rocket depends almost only on the chemistry of it, not on the speed.
Yes, it is a fact -- if you compute "energetic efficiency" as the energy added to the payload by a fixed quantity of fuel. The fuel delivers a fixed delta-v. For a rocket moving at greater speed, a fixed delta-v corresponds to a greater increase in payload energy. This is the Oberth effect.

It seems like a violation of conservation of energy until you consider the energy in the exhaust stream.
 
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  • #16
GiuseppeR7
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Intuitively: the rocket is creating a thrust, the rocket is moving so work is done, energy from the chemistry is released. But at higher speed (a being constant because T is) in a unit time a change in speed is a*t but in that time much more space is done by the rocket and so the work is bigger since it is W=T*space and so the energy transfered.
The Kinetic Energy is of course dependent on the frame of reference!
jbriggs444: yeah, now i get to what you where referring...
 
  • #17
PeroK
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So I see that it does take more fuel to accelerate at a constant rate the higher the velocity? this seems very un-intuitive to me. In every physics simulator I've played (e.g. KSP) fuel always seems to be consumed at the same rate - is this just an illusion or are the physics wrong in these games?

This seems like it leads to a paradox:
Imagine 2 reference frames one moving at 0 m/s and the other at 1 m/s
A rocket it also moving at 1 m/s - it then accelerates to 2 m/s.
The rocket weighs 1 Kg.

The 2 reference frames must agree on the amount of fuel being used (right?)
The first reference frame (the one moving at 0 m/s) thinks the rocket gained 1.5 J of kinetic energy
however the second reference frame thinks the rocket gained 0.5 J of kinetic energy

If burning x amount of fuel release kx amount of energy and both reference frames agree that x amount of fuel was burned then both must also agree on k as well however as shown above they don't.

what is the mistake I am making?

I don't know if you ride a bicycle. But, it's easy to get up to about 30 kmph. Especially if the wind is behind you. But, at about this speed it gets very difficult to accelerate further. You can tell it's not just the air or the mechnical inefficiency. At that speed, all your power is used to generate a small force that overcomes the various small retarding forces.

You may be getting confused with a constant force creating constant accleration. But, the faster you go, the more Power is required to maintain a constant force. That's what you feel when you reach your maximum cycling speed. You have no more power to generate an accelerating force.

To deal with your paradox. Suppose two cars are moving alongside each other. The cars are not moving relative to each other, but both cars know they are moving relative to the road. And, as that is where the accelerating force must act, they will understand fuel consumption etc.

Another example would be if you are on a moving walkway, going at, say, 10kpm. And someone is running alongside it. It's much easier for you to accelerate relative to the walkway than it is for the runner to accelerate.

If you move to space where there is no medium to get a grip on, then you need other means of propulsion and this complicates the issue. But, the same principle applies: the fuel consumption would depend on the mechanism and that mechanism would be observed and the amount of fuel consumption understood by another observer regardless of the relative motion.
 
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  • #18
Dale
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are you sure? the "energetic efficency" of a rocket depends almost only on the chemistry of it, not on the speed.
Dr Stupid is definitely correct on this. The efficiency of a rocket engine most definitely does depend on the speed.

It is best to analyze a rocket using conservation of momentum first, since that forces you to consider the exhaust. Then, when you apply conservation of energy it is easier to remember that the exhaust has kinetic energy too. Most of the fuel's energy goes into accelerating the exhaust, not the rocket.
 
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  • #19
Dale
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If burning x amount of fuel release kx amount of energy and both reference frames agree that x amount of fuel was burned then both must also agree on k as well however as shown above they don't.

what is the mistake I am making?
The mistake is neglecting the KE that goes into the exhaust. That is why I always recommend starting with conservation of momentum for rockets.
 
  • #20
Drakkith
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The mistake is neglecting the KE that goes into the exhaust. That is why I always recommend starting with conservation of momentum for rockets.

Does the fact that you have to accelerate the fuel as well as the rocket have anything to do with this?
 
  • #21
GiuseppeR7
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Dr Stupid is definitely correct on this. The efficiency of a rocket engine most definitely does depend on the speed.

It is best to analyze a rocket using conservation of momentum first, since that forces you to consider the exhaust. Then, when you apply conservation of energy it is easier to remember that the exhaust has kinetic energy too. Most of the fuel's energy goes into accelerating the exhaust, not the rocket.

yeah, i was thinking about another "efficiency" definition.
 
  • #22
jbriggs444
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Does the fact that you have to accelerate the fuel as well as the rocket have anything to do with this?

The Oberth effect (increasing efficiency with increasing speed) and the Tsiolkovsky rocket equation (increasing speed as the log of the fuel to payload ratio rather than in direct proportion to fuel consumption) are independent effects in the sense I think you mean.
 
  • #23
Dale
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Does the fact that you have to accelerate the fuel as well as the rocket have anything to do with this?
Both facts are certainly important in a real rocket, but I think that the thing which causes the confusion of the OP is mostly forgetting the KE of the exhaust rather than the fuel-as-payload effect. You can contrive scenarios to neglect the fuel-as-payload effect, but the KE of the exhaust effect is always there if you look at different frames.
 
  • #24
are you sure? the "energetic efficency" of a rocket depends almost only on the chemistry of it, not on the speed.
I did not want to delve into the efficiency of the rocket. I was just giving the general physical reason that a rocket works I.e why it has a time rate of change in acceleration. Since there is a time rate of change in acceleration there must be a time rate of change in velocity. Since there is jerk for rockets the time rate of change in velocity is non linear. There are even terms for the time rate of change of jerk but generally these are stated as the number if derivatives from position. The following link will take you to an explanation of the preceding.

https://en.m.wikipedia.org/wiki/Jerk_(physics)
 
  • #25
GiuseppeR7
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In physics there are no explanations to "why" stuff work. There are only descriptions about it. One of the explanations for the rocket can be: the rocket create a thrust because of the pressure distribuition inside the nozzle, the gas accelereted trough the nozzle is an effect of the pressure distribuition. Honestly i can not understand anything of what you are saying.
 
  • #26
Drakkith
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Both facts are certainly important in a real rocket, but I think that the thing which causes the confusion of the OP is mostly forgetting the KE of the exhaust rather than the fuel-as-payload effect. You can contrive scenarios to neglect the fuel-as-payload effect, but the KE of the exhaust effect is always there if you look at different frames.

Hmmm. Just looking at it from a simple W=FD perspective, it seems pretty obvious that it takes more work to accelerate to higher velocities since you're applying a force over a larger distance than before. However, I have no idea how to include the fuel exhaust into this.
 
  • #27
Dale
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Hmmm. Just looking at it from a simple W=FD perspective, it seems pretty obvious that it takes more work to accelerate to higher velocities since you're applying a force over a larger distance than before. However, I have no idea how to include the fuel exhaust into this.
Here is an extremely simplified example:

Suppose that we have a spaceship of mass 1010 kg (including cargo) at rest in space. Now, to make things simple, instead of burning a steady stream of fuel, let's suppose that we throw a 10 kg moon rock out the back at a speed of -1000 m/s. Now, the momentum of the rock is -10,000 kg*m/s, so the momentum of the ship is +10,000 kg*m/s which corresponds to a speed of 10 m/s. If you work it out, the KE has changed from 0 to 5.05 MJ, with 5 MJ in the rock and 0.05 MJ in the ship.

Now, suppose we have the same scenario, but this time the ship is already moving at 100 m/s. This time the final speed of the rock is -900 m/s and the final speed of the ship is 110 m/s. If you work it out, this time the KE started at 5.05 MJ (5 MJ ship and 0.05 MJ rock), and changed to 10.10 MJ, with 4.05 MJ in the rock and 6.05 MJ in the ship.

So, the total change in energy is the same 5.05 MJ in both cases. In the first case 5 MJ went into the KE of the rock and only 0.05 MJ went into the ship, but in the second case 4 MJ went into the rock and 1.05 MJ went into the ship.

So the efficiency increased as less energy was being thrown away in the KE of the exhaust. The actual rocket equation is more complicated because it is continuous, but that is the basic principle.
 
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  • #28
zanick
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I think you have to remove all the variables from the set up of the question. call the rocket using a fuel that doesn't burn off. say it uses the photons from the sun. And with your imagination, say that the Force is equal to x..... as was said the force is constant, so the acceleration is constant, and the power goes up proportionally with the speed. the fuel use rate is exactly the same from 0-50m/s as it is from 50 to 100m/s. (in space) now, on land we have aero loses that go up with the square of velocity.
KE goes up with the square of speed..... so, for every second of change of speed, here is a HP-second associated with it. if the rate of change is the same, the hp-second (or HP x 746 =Joules) will be the same. as you go faster and faster, it all accumulates for higher and higher KE.

on earth, the aero dynamic forces go up with square, and power is force x speed, so the power required and used, goes up with the cube of speed. the KE only goes up with the square of speed. there is the difference.
 
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  • #29
Joel A. Levitt
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The equation for kinetic energy is:

k = 1/2*m*v2

This confuses me because this indicates it take more and more energy to accelerate...
Lets imagine a rocket in space (assume the mass of the rocket stays the same when accelerating) - the formula suggests it would take more energy (therefore fuel) to go from 50 to 100 km/s than it would from 0 to 50 km/s.

Is it true that it would take more fuel to accelerate from 50 to 100 km/s than it would from 0 to 50 km/s ?????

My guess is that you are a student in a first course in mechanics. Is my guess is correct? If my guess is correct, are you truly confused about kinetic energy or are you just having fun at the expense of the responders? If you are truly confused, I will try to explain work and energy.
 
  • #30
zanick
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My guess is that you are a student in a first course in mechanics. Is my guess is correct? If my guess is correct, are you truly confused about kinetic energy or are you just having fun at the expense of the responders? If you are truly confused, I will try to explain work and energy.
from the very first post, why didn't anyone just break out the math. KE change is the same from 0-50 as it is from 50 to 100. wouldn't that be the answer to the question right there?
 
  • #31
Drakkith
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from the very first post, why didn't anyone just break out the math. KE change is the same from 0-50 as it is from 50 to 100. wouldn't that be the answer to the question right there?

I don't think that's correct. KE for a 1kg object at 50 m/s: 1250 joules. KE for a 1 kg object at 100 m/s: 5000 joules. Change from 0 to 50 is 1250 joules. Change from 50 to 100 is 3750 joules. That's 3 times as much change in KE.
 
  • #32
zanick
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I don't think that's correct. KE for a 1kg object at 50 m/s: 1250 joules. KE for a 1 kg object at 100 m/s: 5000 joules. Change from 0 to 50 is 1250 joules. Change from 50 to 100 is 3750 joules. That's 3 times as much change in KE.
that's right, at a constant HP (or fuel flow) acceleration goes down proportional to velocity... (which means so does thrust and force). So, for constant thrust, fuel consumption has to go up with speed.
Now Im confused!
 
  • #33
Drakkith
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that's right, at a constant HP (or fuel flow) acceleration goes down proportional to velocity... (which means so does thrust and force). So, for constant thrust, fuel consumption has to go up with speed.
Now Im confused!

I don't think that's true either. Acceleration should remain constant, but the faster you go, the large the distance is that you apply the force over, meaning that work goes up.
 
  • #34
zanick
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actually, that is true.... acceleratrion is proportional to power, given by the Newtonian identity, acceleration = power/(mass x velocity)
so yes, as speed goes up, with constant power, acceleration will go down proportionately. for constant acceleration (or force) , power will have to go up with velocity.
Power = Force x Velocity
 
  • #35
Drakkith
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actually, that is true.... acceleratrion is proportional to power, given by the Newtonian identity, acceleration = power/(mass x velocity)
so yes, as speed goes up, with constant power, acceleration will go down proportionately. for constant acceleration (or force) , power will have to go up with velocity.
Power = Force x Velocity

I don't think that applies to rockets, which is what we're talking about here. A rocket actually has to reduce power (and thrust) to maintain a steady acceleration.
 

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