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!

Homework Help: Derivation of Common Formula

  1. Nov 11, 2003 #1
    Can someone please explain the derivation of the common formula for average kinetic energy: K = 0.5mv^2?
     
  2. jcsd
  3. Nov 11, 2003 #2
    Derivation of KE?

    a = v^2 - v0^2 / 2d

    ΣWork = ΣF * d = mad = 0.5m(v^2 - v0^2) = ke
     
  4. Nov 11, 2003 #3

    HallsofIvy

    User Avatar
    Science Advisor

    First, K = 0.5mv^2 is not "average" kinetic energy- it is kinetic energy- in fact it is the definition of kinetic energy- there is no "derivation". At some time, back at the beginning of "physics" someone notice that, in many dynamics problems that quantity was conserved- so they gave it a name!
     
  5. Nov 16, 2003 #4
    First, K = 0.5mv^2 is not "average" kinetic energy- it is kinetic energy- in fact it is the definition of kinetic energy- there is no "derivation". At some time, back at the beginning of "physics" someone notice that, in many dynamics problems that quantity was conserved- so they gave it a name!

    Quantities, mv^2, 35mv^2, are also conserved!
     
  6. Nov 17, 2003 #5
    So ummmm ... This is relative kinetic energy not average kinetic energy?

    That is, relative to the kinetic energy of the object when v0 = 0

    Interesting derivation. Thanks.
     
  7. Nov 17, 2003 #6
    energy is the ability to do work.

    work is force times distance.

    kinetic energy is that energy a body has due to its motion.

    so lets see:

    [tex]
    W=\int_A^B \mathbf{F}\cdot d\mathbf{s}
    [/tex]

    but by Newton
    [tex]
    \mathbf{F}=m\mathbf{a}
    [/tex]
    so
    [tex]
    W=\int_A^Bm\mathbf{a}\cdot d\mathbf{s} = m\int_{t_{A}}^{t_{B}}\mathbf{a}\cdot\mathbf{v}dt\\
    = \frac{m}{2}\int_{t_{A}}^{t_{B}}\frac{d(v^2)}{dt}\! dt = \frac{1}{2}mv_B^2 - \frac{1}{2}mv_A^2
    [/tex]

    so you see that the work done is equal to the change in kinetic energy between the two points, and this is a derivation of the formula for that kinetic energy.
     
    Last edited: Nov 17, 2003
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook




Loading...