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Why kinetic energy is proportional to velocity squared

  1. Oct 30, 2007 #1
    Could someone please explain why kinetic energy is proportional to velocity squared and not just velocity.

    A propulsion system that provides a constant force would typically use energy at a constant rate in order to accelerate at a constant rate and thus rate of change of velocity would be proportional to rate of use of energy. So why is rate of gain of energy not also proportional to rate of change of velocity ?

  2. jcsd
  3. Oct 30, 2007 #2
    Consider an object is moving at v. If you exert a force F to change its speed dv in time dt and the distance dS. Then the work needed is :
    A = FdS= ma.dS
    There must be one v in a and another in dS.
  4. Oct 30, 2007 #3


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  5. Oct 30, 2007 #4
    In short, kinetic energy is proportional to velocity squared, rather than velocity, or velocity cubed, or something else, for the following reason. Simply, the conservation of energy imposes this proportionality. The only real way to see this is to see the mathematical origin of the term "kinetic energy". The quantitative value of kinetic energy arises from quantitative (mathematical) consideration of the physical situation.

    Basically, what you're looking for is a certain principle, the so-called "work energy theorem", or principle of conservation of energy. The derivation of this principle is extremely simple, and it will tell you EXACTLY why energy is proportional to velocity *squared*.

    Let's consider the simplest possible case of a rigid body with a constant mass moving in 1-direction. The body starts from rest, at an arbitrary position in space. From Newton's discovery, we know:

    1. F = ma = m [tex]\frac{dv}{dt}[/tex]

    Next, we want to consider what would happen if a force were applied to the rigid body as the rigid body moves from one point to another. So, we multiply each side by an infinitesimal increment of length, and integrate:

    2. int(F ds) = int(m [tex]\frac{dv}{dt}[/tex] ds)

    Next, we realize that we can use a trick from calculus to rewrite the right hand side, considering:

    3. [tex]\frac{dv}{dt}[/tex] = [tex]\frac{dv}{ds}[/tex] [tex]\frac{ds}{dt}[/tex] = [tex]\frac{dv}{ds}[/tex] v

    Putting (3.) into (2.), we get:

    4. int(F ds) = int(m v [tex]\frac{dv}{ds}[/tex] ds) = int(m v dv)

    Integrating the left hand side you get a value which represents force applied over a distance; this is precisely the quantity known as "work". Integrating the right hand side, you get a term which represents .5*m*v^2. This is precisely the quantity which is defined as "kinetic energy". Hence, the proportionality of KE to v^2 comes from the consideration that changes in kinetic energy should be directly proportional to the work done on an object.

    I hope this helps. There's a lot of math and you may not understand everything. It seems like a long-winded answer but really this is the complete answer. I don't really understand the second part of your question; you ask, "why is rate of gain of energy not also proportional to rate of change of velocity ?", and yet, rate of gain of energy IS proportional to rate of change of velocity... it's just not linearly proportional.
    Last edited: Oct 30, 2007
  6. Oct 30, 2007 #5
    Thanks for your responses, but I'm still a bit confused. I understand the math and how kinetic energy is calculated as work done (force x distance). It's just that it doesn't make sense to me when I consider a propulsion system. If you are putting energy in at a rate proportional to time (to create a constant force) how can you get energy out at a rate proportional to time squared (kinetic energy) ?
  7. Oct 30, 2007 #6


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    A constant force on an accelerating object does not deliver energy to the object at a constant rate. The rate at which energy is delivered per unit time is P = F.v, which increases as v increases.

    Or, if you deliver energy at a constant rate P, then the force must decrease as the velocity increases, and the acceleration decreases accordingly.
  8. Oct 30, 2007 #7
    Jtbell explains very clearly. I just add an example. A falling object (free falll) always has a constant force exerted on : the gravity. But as time elapses, the height reduces faster, that means the potential energy is converted into dynamic energy at higher rate.
  9. Oct 30, 2007 #8
    I don't doubt this is true, I just don't understand it. Why does it take more energy to accelerate from 5 to 10 m/s than from 0 to 5 m/s ? Are these not the same just in a different frame of reference ?
  10. Oct 30, 2007 #9
    By my calculations considering a 1 kg object under gravitational force, the object obtains 12.5 Joules of energy accelerating from 0 to 5 m/s, versus 37.5 Joules to go from 5 to 10 m/s. So, the object actually acquires more energy accelerating from 5 to 10 m/s than from 0 to 5 m/s. The object doesn't "take energy" to move faster; I think you understand this incorrectly. The object takes force to move faster; and it becomes more energetic as a result. I guess you could say it takes more energy to move an object faster, the faster it is already going. However, it seems incorrect to say that it "takes more energy" to move an object faster, because objects do not feel energy directly; they only feel forces.

    Now, as jtbell said, a constant force on an accelerating object does not deliver energy to the object at a constant rate. Why? You tell me! We don't make the rules of nature. We just figure them out. This is the universe man and you can only explain so far.

    Maybe, intuitively, you can understand it by thinking that it takes more energy to make an object move faster, after it is already moving very fast. If you want to kick a soccer ball from rest, you can kick it pretty damn far. But if you want to kick a soccer ball that is already flying very fast, it is hard to kick it so that it moves a lot faster.
    Last edited: Oct 30, 2007
  11. Oct 31, 2007 #10
    Let me try a different wording.

    Say you have a rocket ship at rest with a certain amount of chemically stored energy (fuel). To provide a constant force to constantly accelerate the rocket the chemically stored energy is used up at a constant rate. This energy gets mainly turned into kinetic energy and heat.

    Energy cannot be created or destroyed. The fuel energy is being used at a constant rate and producing heat at a constant rate but producing kinetic energy at an ever increasing rate. It appears that energy is being created out of nowhere !

    I am missing something here, what is it ?
  12. Oct 31, 2007 #11


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    If you burn fuel at a constant rate (kg/sec or liters/sec or whatever) and thereby produce energy at a constant rate (joules/sec), I guarantee that your rate of acceleration will not remain constant, but will decrease as time passes, so that the kinetic energy increases at a constant rate. Specifically, v as a function of time will be proportional to the square root of t.
  13. Oct 31, 2007 #12
    Thanks for guaranteeing that JT, but I'd prefer if you would explain it. Otherwise I'll have to use the Chewbacca Defense.

  14. Oct 31, 2007 #13


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    mordechai9 has given you a detailed derivation of this. In what sense are you still confused? As far as I can tell, you are trying to hold on to an a priori requirement that the change in velocity must be a constant. There's nothing "fundamental" here with respect to this change in velocity. There is something fundamental with conservation of energy, and that should always be your starting point. And the derivation given already in this thread is based on such conservation.

  15. Oct 31, 2007 #14
    The thing to remember is that the energy produced by a gallon of gas will be MEASURED differently in different frames.
  16. Oct 31, 2007 #15
    That sounds promising. Assuming you understand why I am confused that energy is not proportional to v, can frame of reference be used to explain why it is proportional to v squared ?

  17. Oct 31, 2007 #16
    That is why I gave an example where 'it appears to me' that conservation of energy is violated ...
  18. Oct 31, 2007 #17


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    I think the major concept to remember here is that force is not the time derivative of energy... power is the time derivative of energy...

    force is the "distance" derivative of energy...

    Energy is force*distance... not force*time...

    suppose 2 objects have the same mass and acceleration.

    one goes from 0m/s to 5m/s. the other goes from 5m/s to 10m/s.

    both experience the same net force... and this change in velocity occurs in the same time... however the second object travels a further distance... hence it has more work done on it.
  19. Oct 31, 2007 #18


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    But I don't see it.

    All you can say is that the amount of energy being used up by the fuel goes to the increase in KE of the ship. That's it. You are insisting here that this energy goes into the SAME change in velocity, rather than the same change in square of the velocity. There's no rational reason why that needs to be so. If you look carefully, it is [itex]v_f^2 - v_i^2[/itex] that remains constant, not [itex]v_f - v_i[/itex], over the same period of time.

  20. Oct 31, 2007 #19


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    Assume the same constant force in both cases, therefore the same acceleration. Therefore both cases take the same amount of time. In the first case, the average speed is higher, so the object travels further than in the second case. Therefore, W = f.d is larger in the first case.
  21. Oct 31, 2007 #20

    The kinetic energy has squared v in its calculation as explained above, so as a matter of course, energy to accelerate from 0m/s to 5m/s must be different from 5m/s to 10m/s.
  22. Oct 31, 2007 #21
    The way i have learned to think about this kind of problem goes back to Clerk Maxwell's dictum to always change all energies to a moving mass before doing any sort of comparison. I find that gets me out of this whole confusion of why does a gallon of gas in one frame not produce the same energy as a gallon of gas in another frame. So, instead of supplying fuel to some engine, I bring in energy in the form of a moving mass from an outside frame (that way, frame recoil momentum goes away) and add it as needed in the form of an elastic collision. Then, from what ever frame I choose, I can measure that added energy, and it will be different for each frame.
  23. Oct 31, 2007 #22
    But the reason I say that it goes into the same change in velocity is because it is constant acceleration caused by a constant force caused by a constant supply of energy.

    I am burning fuel (releasing energy) at a constant rate in time to create a constant force and yet my rocket is not gaining energy at a constant rate in time. So what is happening in terms of conservation of energy ?

    Are we agreed that a constant supply of energy can / will generate constant thrust / force ?

  24. Oct 31, 2007 #23
    I'm glad at least one person doesn't think I'm mad / stupid, even if I didn't fully understand your explanation. :rolleyes:
  25. Oct 31, 2007 #24


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    Why do you say this? Where is this "increase" in energy?

    Let's say the fuel burns an amount of E per unit time.

    Now let's do this 2 ways, via a simple conservation of energy and then via the amount of work done by the force being applied. Already we can see that

    [tex]E = \Delta(KE)[/tex]

    which means that

    [tex]E = 1/2 m(v_f^2 - v_i^2)[/tex]

    This E is a constant throughout, and so is the difference in the square of the velocity, NOT the difference in velocity. That's why you were making a mistake when you comparing the change in going from 0-5 m/s versus 10-15 m/s. This is NOT the fundamental quantity that you should be dealing.

    Now, let's consider this via the force applied, since you were worried about the acceleration. Rewriting the first equation using work done

    [tex]E = \int{F ds}[/tex]

    where ds is the differential displacement. Now, it has already been pointed out that we can write F as

    [tex]F = ma = m dv/dt = m \frac{dv}{ds} \frac{ds}{dt} = m \frac{dv}{ds} v[/tex]

    This means that

    [tex]E = m\int{v dv}[/tex]

    Guess what? When you do the integral, you end up with the exact same expression as the one I got using conservation of energy.

    Again, there's gain or change in energy over the unit time period in which E is burned from the fuel. What you are confusing with here is the insistance that [itex]\Delta(v)[/itex] be constant, where as in actuality, it is [itex]\Delta(v^2)[/itex], which is a direct result from the conservation of energy.

    So there are no energy being created nor loss here.

  26. Oct 31, 2007 #25
    No, I think you're looking hard for a way to understand this. And, you should know that many people, many of them with good technical educations have trouble with this same point. The tricky thing to get hold of is that momentum (kg-m/s) is conserved, and energy (kg-m^2/s^2) is conserved, but velocity (m/s) is not. And, if you think about it a little bit, you might see that, in order to conserve momentum for example, you have to let go of trying to conserve velocity.

    But, the main thing is that, in all external frames, the energy of the fuel you burn will be measured differently depending on whether you're going from 0 to 5 m/s as opposed to going from 15 to 20 m/s. And the important word there is measured.

    Do you know about Galilean transformations? Here's a good link that has some explanations. Some of the other posters have been trying to get you to see that, if you are already moving when you start the acceleration, you will travel farther during the time you accelerate. You can see that in the hyperphysics link.
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