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Can anybody explain the formula for Kinetic Energy?

  1. Jun 16, 2007 #1
    I have always plugged numbers into the following forumula:

    KE = 1/2m*v^2 and it works perfectly. :biggrin:

    I don't understand why. :blushing: Can anybody please tell me why it is one half mass, and not all of the mass? And even more perplexing, is why the velocity needs to be squared.
     
  2. jcsd
  3. Jun 16, 2007 #2
    Welcome to PF, fetmar.

    It simply follows from calculating the work done by a force on an object. We call the work done by a force on an object as the change is kinetic energy due to the action of that force.

    [tex]W = \int_{r(t_0)}^{r(t)}{\vec{F}\cdot d\vec{r}}[/tex]

    Write [tex]\vec{F} = m\frac{d\vec{v}}{dt}[/tex] and [tex]d\vec{r} = \vec{v}dt[/tex]
    Now evaluate the integral.

    [edited on request]
     
    Last edited by a moderator: Jun 18, 2007
  4. Jun 16, 2007 #3

    daniel_i_l

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    If you take away the 1/2 and use v instead of v^2 you get mv which is the momentum.
     
  5. Jun 18, 2007 #4
    work is the force x distance.
    the kinematics equation relating acceleration to distance is
    a = vf^2 - vi^2/2d
    F = ma
    so... F = mvf^2 - mvi^2/2d
    multiply this by the distance.. and the change in kinetic energy is
    mvf^2-mvi^2/2

    soo.. 1/2mv^2 = change in kinetic energy.
     
  6. Jun 18, 2007 #5

    russ_watters

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    It isn't half the mass, the 1/2 comes into the equation when evaluating the integral neutrino posted.
     
  7. Jun 18, 2007 #6
    If you understand that Energy is Force * Distance, then it should start to become clear. Imagine you are driving. Keep in mind that the energy required to brake is equal to the kinetic energy of the vehicle while it was in motion... If you are driving at 25 m/s and slow down with a force that causes an acceleration of -5m/s/s, you can do the math and figure out you had a braking distance of 62.5 m. Now, if you were driving at 50 m/s and were to apply the same force, not only would the TIME for you to brake double, your average velocity while slowing down would also double. Therefore the distance you brake for at double the speed, is four times that of the original. In this case, conceptually, the 1/2 is to get the average velocity. If you slow down at an even rate from 50 m/s to a stop, your average velocity was 25 m/s. I really hope this made some kind of sense.
     
  8. Jun 18, 2007 #7
    The deep answer comes from time symmetry via Noether's theorem. You can google that, but it'll only explain why velocity is squared. The factor out the front probably isn't really necessary (we could just change our units and other definitions accordingly) but then we'd still need a new constant in front of all the other energy equations (gravitational potential, electric potential, etc), in other words, it is necessary for consistency.
     
  9. Jun 18, 2007 #8
    f.d=m.a.d=m.(v1^2-v2^2)/2d.d

    =(mv1^2)/2 - (mv2^2)/2=f.d

    newton(force) . distance=delta kinetic energy
     
  10. Jun 18, 2007 #9
    all you people posting that kinematic relationship
    [tex]v_f^2=v_i^2+2ad[/tex]
    don't get that that is the simplified conservation of energy theorem, not that the conservation of energy theorem comes from that.
    [tex]\Delta KE=\Delta U[/tex]

    [tex]KE_f - KE_i = \Delta U[/tex]
    [tex]\frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = mg\Delta h[/tex]
    cancel m and multiply by 2 to clear the fraction
    [tex]v_f^2 - v_i^2 = 2g \Delta h[/tex]
     
    Last edited: Jun 18, 2007
  11. Jun 18, 2007 #10

    Andrew Mason

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    The Kinetic Energy of a body represents the ability of that body to do work due to its motion. To do work is to apply a force through a distance. So ask yourself how much force x distance a body with mass m and velocity v can apply.

    1. Force is the time rate of change of momentum.

    2. Since speed is the time rate of change of distance, distance traveled is the average velocity x time.

    3. So Force x distance is the time rate of change of momentum x average velocity x time, or (since the time cancels out): change of momentum x average velocity.

    Since in going from v to 0, the average velocity is v/2 and the change in momentum is mv. So the work done is v/2 x mv = [itex]\frac{1}{2}mv^2[/itex].

    AM
     
  12. Jun 18, 2007 #11

    rbj

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    Fetmar,

    can you deal with regular kinematics? like do you know why the distance traversed by an accelerating particle is:

    [tex] d = \frac{1}{2} a t^2 \ + \ v_0 t [/tex]
    ?

    if you know why that is true, then i can explain kinetic energy, in terms of work, to you pretty succintly.
     
  13. Jun 19, 2007 #12

    robphy

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    Most likely, this implicitly assumes a constant acceleration,
    as do the other responses with an acceleration "a".

    neutrino's reply is more general since it doesn't require a constant acceleration.
     
  14. Jun 23, 2007 #13
    To put it simply, it's the integral of momentum.


    On a different note, does the integral of (F=) ma have any physical significance?
     
  15. Jun 23, 2007 #14

    jtbell

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    The integral of F with respect to position is the work done on an object by F, which equals the change in the object's kinetic energy:

    [tex]\int {\vec F \cdot d \vec r} = \Delta K[/tex]

    The integral of F with respect to time is the impulse delivered to an object by F, which equals the change in the object's momentum:

    [tex]\int {\vec F dt} = \Delta \vec p[/tex]
     
  16. Jun 24, 2007 #15

    Andrew Mason

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    There is no need to assume a constant acceleration (ie. constant force). All you need to know is that change in momentum is always average force x time interval. This is because the definition of average force is:

    [tex]\bar F = \frac{\Delta p}{\Delta t}[/tex]

    Hence:

    [tex]\bar F \Delta d = \bar F \Delta t \frac{\Delta d}{\Delta t} = \Delta p \bar v[/tex]

    AM
     
  17. Jun 24, 2007 #16

    robphy

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    The key point I am trying highlight is that
    is not generally true.
     
  18. Jun 24, 2007 #17

    Andrew Mason

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    It is if there is only deceleration. If an object of mass m travelling at speed v experiences negative acceleration of any kind (constant, not constant or arbitrary) such that it ends up at rest (speed = 0) and stays stopped, its average [of the beginning and end speed] over the entire period of deceleration will be v/2. The average is always v/2.

    [Edit][PS: see below]

    Andrew Mason
     
    Last edited: Jun 25, 2007
  19. Jun 24, 2007 #18

    nrqed

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    Just out of curiosity: what definition of average do you use? If one uses a time average [itex] = \frac{1}{T} \int_0^T v_x dt [/itex] it does nto seem to me that the average will always be v/2.
     
  20. Jun 25, 2007 #19

    Andrew Mason

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    Well, you are right, generally.

    But consider this when the mass is subject to only deceleration. For a reduction in speed from v0 to v1 over an arbitrarily short time interval the change in momentum is m(v1-v0) and the average speed is arbitrarily close to (v1+v0)/2. If you combine n of these intervals:

    [tex]E = \sum_{j=0}^n m(v_{j+1} - v_j)(v_{j+1} + v_j)/2 = \frac{m}{2}\sum_{j=0}^n (v_j^2 - v_{j+1}^2) = = \frac{m}{2}\left((v_0^2 - v_{1}^2) + (v_1^2 - v_2^2).... + (v_{n-2}^2 - v_{n-1}^2) + (v_{n-1}^2 - v_n^2)\right)[/tex]

    But the intermediate terms all cancel out and you are left with:

    [tex]E = m(v_0^2-v_{n}^2)/2 = m(v_n - v_0)(v_n + v0)/2 = \Delta p(v_0+v_n)/2[/tex]

    If v_n = 0 then:

    [tex]W = m(v_0)(v_0/2) = \Delta p(v_0)/2[/tex]

    So, to be perfectly correct, one should say that the work done is the change in momentum multiplied by the average of the beginning and end speeds ([itex]W = \Delta p (v_i + v_f)/2[/itex]).

    AM
     
    Last edited: Jun 25, 2007
  21. Jun 26, 2007 #20
    Scientists came up with the idea of work after noticing that the quantity Fd (force times distance) is conserved. This led them to call that quantity work, W.
    They then sought to manipulate the equation to see what they can extract from it and they did this: W=ΔKE=Fd=mad=
    m* (Vf-V0)/t * (Vf+V0)/2 * t=(m/2) (Vf^2-V0^2)=KE(final)-KE(initial)
     
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