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Is the classical formula for kinetic energy wrong?

  1. Nov 28, 2013 #1
    Recently I have been troubled by several things which have been stated in my school physics classes, mainly concerning the classical kinetic energy formula and other formulae derived from it (as well as the 'Work Energy Theorem').

    The classical formula for kinetic energy '(mv^2)/2' was first found by two lone french scientists in the early 18th century, with neither proof nor definitive experimental evidence for the key to this formula - the v squared coefficient. If one looks up the proof for this formula, there are several proofs to choose from. The problem with them all is that they utilise the 'Work Energy Theorem', which in itself uses the Kinetic Energy formula in its proof - it doesn't take much to realise something is wrong.

    Consider this. In a friction-less environment, away from any gravitational influence, there is a rocket. It has an engine with an unlimited fuel supply (to keep mass the same) that is burning at a constant rate. Its rate of fuel consumption is constant (the chemical energy contained within it is being used at a constant rate), and therefore the engine is providing a constant force. This is where I am most likely to wrong in my deduction (if I am).

    Therefore the energy transformed from chemical energy to kinetic energy can be said to be proportional to the force provided by the engine multiplied by the time it is running. Most importantly, neither the rate of fuel nor the force provided by the engine will change depending on how fast the rocket is moving. From this I get the formula:

    E (transformed)=f*t

    Two other formulae which are well known and I assume are accurate:

    F=ma
    v=at

    Simple algebraic manipulation will find that E=mat, and therefore E=mv

    Mass*velocity, as we all know, is the formula for momentum, NOT kinetic energy. But what exactly is the difference? Why should kinetic energy be proportional to velocity squared and not just velocity. It just doesn't make any sense to me. Isaac newton thought the same, that the 'quantity of motion ' should be mv and not (mv^2)/2, and after all, momentum is conserved in all collisions, not kinetic energy, so surely the definitions should be reversed at least.

    Anyway, I would like someone to tell me exactly where I have gone wrong in these deductions (if I have that is). If I have gone wrong, then I will at least understand why, and if I haven't, then why exactly has no one else noticed this?
     
  2. jcsd
  3. Nov 28, 2013 #2

    WannabeNewton

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    Energy is not force times time. That doesn't even make sense dimensionally. ##[E] = \text{Joules} = \frac{kg\cdot m^2}{s^2}## but ##[F] = \text{Newtons} = \frac{kg\cdot m}{s^2}## so ##[Ft] = \frac{kg\cdot m}{s} = [p]## where ##p## is momentum. In fact ##Ft## is a quantity called impulse and it is equal to the change in momentum. Note however that ##[\frac{E}{d}] = \frac{kg\cdot m}{s^2} = [F]##.
     
  4. Nov 28, 2013 #3
    Yes, I understand that that is what is taught and accepted, but surely impulse should be equal to the change in energy instead, as the increase in kinetic energy should be proportional to the decrease in the chemical energy in this case. If the kinetic energy increases in proportion to velocity squared, the rocket would use more energy to provide the same acceleration at higher velocities, which defies common sense, as the same amount of fuel is being used.
     
  5. Nov 28, 2013 #4

    Integral

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    Please make an effort to learn and understand Physics before attempting to re-do it.

    Note that we have rules against speculation and personal theories here. You are bordering on this which will get you infractions leading up to a ban if you persist.

    Please change your though processes from, I do not understand, therefore it is wrong. to I do not understand, so let me read more, think more, and make an effort to understand.
     
  6. Nov 28, 2013 #5
    What I am really looking for is someone to tell me where in what I have said i have been incorrect. I do not believe the kinetic energy formula to be wrong. It would be unreasonable for me to think so. I would just like to know why what I have said isn't valid.
     
  7. Nov 28, 2013 #6

    BruceW

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    to me, it looks like you have a problem with the fact that at higher velocities, it takes more energy to make the same change in velocity. yeah, maybe this is counter-intuitive. but it's true :)
     
  8. Nov 28, 2013 #7
    seanhenley,
    I think that is a good attempt at a derivation. It sounds logical, but you left out an important feature. When you do these sort of things you have to define your system and look at everything that is going on.

    You have examined only half the process and seem to have overlooked the fact in that if the chemical conversion is providing the thrust to the rocket, it is also expelling something out the back of the rocket to do so. That something also has a velocity, momentum and kinetic energy.

    If you re-analyze you should be able to reconcil your equations.

    For example, at startup, the rocket has zero velocity. The chemical conversion is providing no increase in kinetic energy to the rocket as the stuff coming out the back has it all. As the rocket increases in speed...
     
  9. Nov 29, 2013 #8

    sophiecentaur

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    You have been told where you are incorrect. You say Energy is Force times time. It is not. Work done is Force times Distance.
    Force times time is Impulse, which is equal to the change in Momentum.
    If you do the right sums then you will get the right answer. Any other way and you are likely to get the wrong answer. You will not find any loopholes in Physics at this level.
     
  10. Nov 29, 2013 #9
    If it takes more enrgy to provide the same acceleration, and f=ma is true, does that mean that it would take the more energy to provide the same force, or is the acceleration less at higher velocities irrespective of the force?
     
  11. Nov 29, 2013 #10

    Philip Wood

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    A good starting definition of energy is 'the amount of work a body or system can do'.

    So the kinetic energy of a moving body is how much work it can do as it comes to rest.

    Let's apply this to a block of ice of mass m and speed u gliding without friction (!) on a flat surface. A bystander throws a rope round it and brings it to rest in a distance s by exerting a constant force on it. So, by the definition above,

    Initial KE of ice-block = Fs .

    Here, F is the force exerted by the block on the rope. So the force exerted by the rope on the block is -F, and the block's acceleration, a, is [itex]\frac{-F}{m}[/itex].

    But the initial and final speeds, u and 0, of the block are related by [itex]v^2 = u^2 + 2as[/itex], so [itex]0 = u^2 + 2 (\frac{-F}{m})s[/itex].

    Making Fs the subject, we arrive at [itex]KE = \frac{1}{2}mu^2[/itex].

    Note that there is no need to insist that the retarding force is constant; you get the same result for KE if you let the body do work against a changing force, but you have to ditch [itex]v^2 = u^2 + 2as[/itex], and use a bit of simple calculus instead.

    All the above is clearcut and logical - or if it isn't, it's my fault - and not circular. This is because I started with a useful and workable definition of energy. This definition needs modification when you apply it to in thermal Physics, but these modifications are easily taken in one's stride. If energy is defined in this simplistic way, the so-called work-energy theorem is redundant or a mere tautology.
     
    Last edited: Nov 29, 2013
  12. Nov 29, 2013 #11
    I know that it isn't, but what I have said in the beginning would seem to indicate that it is. What is wrong with that reasoning?

    Maybe my rocket example wasn't appropriate, as I hadn't included the propellant. I assumed it could be ignored. Instead you could think about some kind of train (still in a frictionless environment), that is powered by constanly running electic motors (it still works).
     
  13. Nov 29, 2013 #12
    If the fuel is burning at a constant rate, then the energy is being spent at a constant rate. The rate of energy is power, so the rocket generates constant power.

    But that does not mean that constant force is being generated. Power is force times velocity.
     
  14. Nov 29, 2013 #13
    It is reasonable to assume as F*t increases the kinetic energy increases but a big mistake is made when it is assumed,without justification, that KE is proportional to Ft. Even if they were proportional a constant of proportionality would be needed this needing the right units to make the equation dimensionally balanced. It just doesn't work.
     
  15. Nov 29, 2013 #14

    Philip Wood

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    I don't want to sound patronising, but technical terms in Physics have strict definitions, and can't sensibly be used to mean what a person thinks they ought to mean. I'd strongly advise you to stick at first to the definition of energy as amount of work a system can do. But I'm assuming the question was asked out of a desire to understand Physics.
     
    Last edited: Nov 29, 2013
  16. Nov 29, 2013 #15

    sophiecentaur

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    There is a massive confusion between Kinetic Energy and Momentum - and not for the first time on these forums. This explains where the OP is struggling, I think.
     
  17. Nov 29, 2013 #16
    The massive confusion between Kinetic Energy and Momentum is the history of physics, not just of these forums.
     
  18. Nov 29, 2013 #17

    sophiecentaur

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    haha!
     
  19. Nov 29, 2013 #18

    BruceW

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    To be precise, it takes more energy to provide the same change in velocity. So this means it takes more energy to provide the same average acceleration, over some time interval. And yes, f=ma so this means it takes more energy to provide the same average force, over some time interval.

    edit: and thinking about it the other way around, if you are exerting a constant force on some object, then initially you are hardly using any energy, but when the object gets faster and faster, you are using up more and more energy per unit time, even though you are still exerting the same constant force on the object. Yes, this is counter-intuitive, but that's how it works. I think maybe the reason it is counter-intuitive is because we live in a world where friction is very important. And because our muscles work in antagonistic pairs, so even though we might be doing no work on an object, our muscles could still be using a lot of energy. For example, if we are holding a heavy book steady in the air, we are doing no work to change the GPE of the book. But our muscles can't keep the book exactly steady, so they are doing work by moving the book up and down very slightly, and with a large force, since the book is heavy.

    Another example, if a big boulder is sitting on the ground, then initially it has a large contact area with the ground, since the ground has been squashed around the boulder. So the friction with the ground is very great to begin with. So when you start pushing the boulder, there is a lot of initial friction, but then once the boulder starts moving, there is less friction. In this case, most of your energy is being lost as friction, not being converted to KE of the boulder. When the boulder gets to higher velocity, you need to use more energy to provide the same change in velocity. But you don't notice this. What you notice is that you use a lot of energy as friction initially, and less as the boulder moves faster.
     
    Last edited: Nov 29, 2013
  20. Nov 29, 2013 #19

    Dale

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    Yes. For a constant force the same change in velocity requires the same amount of time, and over that same amount of time at a higher velocity you have travelled a further distance. Since work is f.d that implies that the same change in velocity required a greater amount of work since the same force was applied over a larger distance.
     
  21. Nov 29, 2013 #20

    Dale

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    I don't know what your reasoning is now that you recognize that the rocket is wrong. What reasoning remains that you still think is correct?

    Personally, I think that your whole approach is off. There are two quantities: mv and 1/2 mv². Each quantity is useful: mv is useful because of its relationship with f*t, and 1/2 mv² is useful because of its relationship with f.d, and both are useful because in the right circumstances they are conserved.

    So since they are both useful it is also useful to have names for both. The name for the first is momenum and the name for the second is energy. There is no reason for the names, but it is silly to try to attach one name to the opposite definition. You would then need to find a new name for the other because regardless of the name change they are both useful quantities.
     
    Last edited: Nov 29, 2013
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