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Kinetic energy and momentum.

  1. Jul 25, 2013 #1
    why is force equal to rate of change in momentum rather than rate of change in kinetic energy.
    If we consider time is 1 sec,and if we consider an object is moving with particular velocity V1 and a force F is applied to change the velocity to V2, then according to newton the force applied is equivalent to change in momentum rather than change in kinetic energy.

    so why is that entire force not converted to energy or kinetic energy but only to momentum.comparing the two forms i.e momentum and KE, where velocity is greater than 1, we observe K.E > Momentum ,so the question becomes why is force reduced to momentum rather than kinetic energy.
     
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  3. Jul 25, 2013 #2

    WannabeNewton

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    Force is defined as the rate of change of momentum; it's a definition.
     
  4. Jul 25, 2013 #3

    hilbert2

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    Momentum is in this case more useful concept than kinetic energy because it is a conserved quantity. Newton's third law states that in an interaction between two bodies, the changes in the momenta of the bodies are equal and opposite, which conserves total momentum. There's no "law of conservation of kinetic energy", so it's not useful to define force as the rate of change of KE.
     
  5. Jul 25, 2013 #4
    Sounds good to me. Furthermore, you might say that what you are talking about, change in kinetic energy is defined as "work". (not a rate of change, but just outright change)
     
  6. Jul 25, 2013 #5

    A.T.

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    If you talk about the change of momentum/kinetic energy then it's better to specify that it the net force/work. Otherwise it is just the transfer of momentum/energy.
     
  7. Jul 25, 2013 #6

    hilbert2

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    That's only true if the potential energy is constant. If you lift an object from the floor and put it on table, you have done work even though the object is at rest in both its initial and final states.

    The time derivative of KE is pretty much a useless quantity, for that reason it hasn't even been named. It's not equal to power unless PE is constant.
     
  8. Jul 25, 2013 #7

    WannabeNewton

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    No, the work energy theorem applies in full generality with no reference to potential energy. The important thing to keep in mind, as A.T. noted, is that the work energy theorem refers to the net work done i.e. net work done = change in kinetic energy. In your example the net work done is zero when integrated along the entire trajectory (work done by gravity + work done by you).
     
  9. Jul 25, 2013 #8

    hilbert2

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    Thanks for clarifying this. For some reason I must have missed the concept of net work in my first year undergraduate mechanics class. Never needed it, though...
     
  10. Jul 26, 2013 #9
    i am still not clear,why is entire force not converted in to energy,why is it getting converted into momentum.I am onto consideration that its the energy that will get converted into force when one apply force to a moving object.so when the force is applied to the object it should be converted into energy rather than momentum.where is that extra v/2 going finally.(considering t=1sec).

    I will be really glad if someone gives solution right onto the point,or if someone can mark mistake in my understanding.
     
  11. Jul 26, 2013 #10

    hilbert2

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    ^ Well, for example, force and momentum are vector quantities and kinetic energy is a scalar quantity. An equation decribing conversion of force into kinetic energy would look messy, complicated and "unnatural". When looking for laws of nature, it's usually assumed a priori that the laws we find will be elegant and simple. This is what we find here, the vector force is equal to the rate of change of vector momentum.
     
  12. Jul 26, 2013 #11
    If you realise that F = ma and that acceleration, a= change in velocity/ time
    = v1-v2/t
    Then it turns out that F = ( mv1 - mv2)/ t
    Which is rate of change of momentum.
     
  13. Jul 26, 2013 #12
    Well, the problem is really a rather meaningless description of the "problem".
    The force is neither "converted" to momentum, nor energy.
    So the question why one and not the other does not have a clear meaning.
    Also, to say "kinetic energy is larger than momentum" is a meaningless statement.
    Like "my age is larger than my height".

    A force acting on a system may result, in general, in changes in both momentum and kinetic energy of the system. There are relationships (equations) to describe the connection between the changes in these two quantities (momentum and KE) and the force.
     
    Last edited: Jul 26, 2013
  14. Jul 27, 2013 #13
    I can understand it might not exactly be the force that is converted into any other form like momentum,but surely there is something which is accepted by the body ("may it not be force") which brings about change in momentum,probably it is not energy either,but when one applies a force to the object he looses that something ,and that something is gained by the object,and it suddenly realize at what velocity it must move.(i considered it to be an energy and it must have been equal to energy lost by the one who applies force). it may sound abstract depending on how well i been able to put that in words.
     
  15. Jul 27, 2013 #14
    Rate of change in kinetic energy with respect to time does not even have units of force. It has units of force times velocity. Rate of change of momentum, on the other hand, does have units of force.
     
  16. Jul 27, 2013 #15

    morrobay

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    Another way to look at it :

    ∫ F dt = ΔP

    ∫ F dx = ΔKE
     
    Last edited: Jul 28, 2013
  17. Jul 28, 2013 #16

    morrobay

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    With a = v dv/dx

    W = x∫x0 F dx = xx0 mv dv/dx dx = vv0 = mvdv = 1/2 mv2 - 1/2mv20
     
  18. Jul 28, 2013 #17
    Thanks to all for replying on my post,i was seriously looking for that 1 common thing to how can i relate force,momentum,kinetic energy,displacement,velocity and time with drilling deep into the idea of classical mechanics to understand the very basic idea. As for now i have solved my mystery,i am thankful to each one of you to provide me with certain hints on classical mechanics.

    Thanks n cheers!!
     
  19. Aug 26, 2015 #18
    I'm a little late to this party, actually to the point where the lights are already off and they're about to lock the doors. But I gotta chime in here, because the OP was asking for a conceptual answer to a conceptual question, and what he mostly got in response was math, definitions, and "you're asking the wrong question." Maybe it's because I'm such a newbie myself, but I find that when I'm trying to grasp a concept, I am looking for simple, real-world examples or analogies, not eight lines of calculus followed by a QED. So, bodhi, if you're still out there, here is my answer:

    I'm going to start by saying that it really was kind of a wrong question, but let me explain why: Force by itself does not transfer energy; only when it is performed over a distance (work) does energy come into play. I know other people have given that answer on this post, but no one actually provided an example of non-work force, which would have been a lot more helpful, I think. So let's suppose I hang a simple weight from a string. That is force without work - the string is exerting an upward force on the weight, but no energy is changing hands. Assuming no external actors, that weight could hang there until the end of time and have the same energy as when it started.

    Therefore, to understand what's happening to the energy, we need to talk about the work that is being done, as the force slows the object, not just the force itself. And that means that when you ask, "why is that entire force not converted to energy or kinetic energy", you really are asking the wrong question, because it is the object's energy that is going into the work of the force, over the distance it takes that force to slow the object down. And yes, all of the kinetic energy lost by the change in velocities does go into that work. (Yes, I know that per relativity, the direction of energy transfer is meaningless, but right now I'm just trying to stay focused on force vs. work.)

    Now let's look at the math. Incidentally, morrobay actually did a pretty nice summation up above, but to keep things simple, I'm going to skip the calculus by assuming linear changes, which lets me stick to algebra.

    First the momentum, which you've obviously already picked up on; but I'm going to work it out anyhow, in case anyone else comes looking. So when I apply the force, I get:

    [itex]F = ma = m\left ( \frac{v_{2}-v_{1}}{\Delta t} \right ) = \frac{mv_{2}-mv_{1}}{\Delta t} [/itex]

    That is: [itex]F\Delta t = mv_{2} - mv_{1} = p_{2} - p_{1}[/itex]

    So the force over time (called impulse) does equal the change in momentum. You kind of left [itex]\Delta t[/itex] out of your equation, though, by setting it to 1, and that might also have been a source of your confusion. The reason it matters is that the whole system is moving during this [itex]\Delta t[/itex], which means the force has gone a distance while slowing the object, and that means... work. So let's do the work equation, and see if we can figure out how that energy went to the force:

    [itex]W = F\Delta x = ma\Delta x = m\frac{\Delta v}{\Delta t} \Delta x = m\left ( \frac{v_{2}-v_{1}}{\Delta t} \right ) \Delta x[/itex]

    We know that the basic equation for distance is [itex]\Delta x = v\Delta t[/itex]. And as I said above, for simplicity we'll assume linear deceleration, so that [itex]v[/itex] just becomes the average of [itex]v_{1}[/itex] and [itex]v_{2}.[/itex] (It still works out the same if you don't assume this, but... calculus.)

    That is, [itex]\Delta x = \bar {v}\Delta t = \left ( \frac{v_{2} + v_{1}}{2} \right )\Delta t[/itex]

    Now the work equation becomes: [itex]W = m\left ( \frac{v_{2}-v_{1}}{\Delta t} \right ) \left ( \frac{v_{2} + v_{1}}{2} \right )\Delta t[/itex]

    The two [itex]\Delta t[/itex]'s cancel out, leaving [itex]W = \frac {1}{2} m \left ( v_{2}-v_{1} \right ) \left ( v_{2} + v_{1} \right )[/itex]

    I think you can see where I'm going here, but let's finish it by applying the FOIL method (remember that?) to multiply our polynomials:

    [itex]W = \frac{1}{2}m \left(v_{2}^{2} + v_{2}v_{1} - v_{1}v_{2} - v_{1}^{2} \right) = \frac{1}{2}m \left(v_{2}^{2} - v_{1}^{2} \right) = \frac{1}{2}m v_{2}^{2} - \frac{1}{2}m v_{1}^{2}[/itex]

    or just [itex]W = E_{K2} - E_{K1}[/itex]

    So while it's true you can say that "the entire force goes into change of the momentum", the actual work goes into the change in energy (or the energy goes into work, if you prefer).

    Hopefully that helps you form a better, conceptual picture.
     
    Last edited: Aug 26, 2015
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