1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Why is kinetic energy 1/2mv^2 instead of mv^2?

  1. Mar 26, 2009 #1
    The definition of joules, th unit of energy is the 1 J = 1 kg * m2/v2.

    And that 1 joules is the amount of energy needed for the work done by one newton traveling one meter.

    From all this, I got the impression that to get energy, you would have to multiply force times distance. Is work, then just a measurement of energy.

    Assuming my first assumption that energy = F x D, lets sat a 5kg object is accelerating towards earth from 20 feet above at 9.8 m/s2 . To get the force, you would have to do 5 kg x 9m/s2 Then to get energy, that force times distance. So if it traveled 5 meters. The kinetic energy should be F(5kg x 9.8 m/s2) x D(5 meters.)

    Which would end up being 5kg x 49m2/s2.

    Sooo, can someone explain why ke is 1/2 mv2 instead of just mv2.
  2. jcsd
  3. Mar 26, 2009 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Work Energy Theorem
  4. Mar 26, 2009 #3


    User Avatar
    Gold Member

    I hope you'll notice that nowhere did you determine its actual speed after falling the 5 meters. If you use your kinematic equations and solve for the time it took to reach the 5 meters and subsequently found the velocity at that 5meter point, you'd see that the energy is infact, mv^2/2
  5. Mar 26, 2009 #4
    Assuming my first assumption that energy = F x D

    Right, work is the energy added to or taken away from a system when it encounters a force over some distance. The definition of work is actually the integral,
    Where dr is the change in position, the D in your equation.
    The equation for the kinetic energy of an object, K.E.=1/2mv^2 is derived by making a series of substitutions in the above equation and integrating.
    Last edited: Mar 26, 2009
  6. Mar 26, 2009 #5


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    [tex]W=\int_{x_1}^{x_2} F dx=\int_{t_1}^{t_2} m\ddot x \dot x dt=\int_{t_1}^{t_2}\frac{d}{dt}(\frac 1 2 m\dot x^2)dt=\frac 1 2 mv_2^2-\frac 1 2 mv_1^2[/tex]
  7. Mar 26, 2009 #6


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Well, there would be no conceptual problems whatsoever to define a quantity mv^2, and use this instead of "kinetic energy".

    This has already been done; mv^2 is called "vis viva".

    But, the work-"vis viva" theorem is less aesthetic than the equivalent work-energy theorem, due to the explicit inclusion of the factor of 1/2 in the formula.
  8. Mar 26, 2009 #7
    Thx for the replies, everyone. l'll check them all out.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook