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Confused about kinetic energy

  1. Jan 31, 2015 #1
    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 ?????
  2. jcsd
  3. Jan 31, 2015 #2
    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.
  4. Jan 31, 2015 #3
    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...
  5. Jan 31, 2015 #4
    lol we have both done a SERIOUS mathematical mistake
  6. Jan 31, 2015 #5


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    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##.
  7. Jan 31, 2015 #6
    yeah XD
  8. Jan 31, 2015 #7


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    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.
  9. Jan 31, 2015 #8
    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.
  10. Jan 31, 2015 #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
  11. Jan 31, 2015 #10
    are you sure? the "energetic efficency" of a rocket depends almost only on the chemistry of it, not on the speed.
  12. Jan 31, 2015 #11
    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.

    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.
  13. Jan 31, 2015 #12


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    That conclusion is correct.

    Yes, you're right about that too.

    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).
  14. Jan 31, 2015 #13
    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.
    Last edited: Jan 31, 2015
  15. Jan 31, 2015 #14
    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?
  16. Jan 31, 2015 #15


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    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.
  17. Jan 31, 2015 #16
    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...
  18. Jan 31, 2015 #17


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    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.
    Last edited: Jan 31, 2015
  19. Jan 31, 2015 #18


    Staff: Mentor

    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.
  20. Jan 31, 2015 #19


    Staff: Mentor

    The mistake is neglecting the KE that goes into the exhaust. That is why I always recommend starting with conservation of momentum for rockets.
  21. Feb 1, 2015 #20


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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?
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