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!

The equivilence of Power and Kinetic Energy

  1. Nov 24, 2009 #1
    This isn't homework, just be going over a few concepts.

    I'm trying to show that the power delivered by a force equals the rate at which the particle is changing.

    Now P = [tex]\vec{F}[/tex] [tex]\bullet[/tex] [tex]\vec{v}[/tex]
    = m[tex]\vec{a}[/tex] [tex]\bullet[/tex] [tex]\vec{v}[/tex]
    = m[tex]\vec{v}[/tex] [tex]\stackrel{\delta}{\delta t}[/tex][tex]V^2[/tex]

    This book is now telling me that the above line = 2[tex]\vec{a}[/tex] [tex]\bullet[/tex] [tex]\vec{v}[/tex]

  2. jcsd
  3. Nov 24, 2009 #2
    I can't really read what you wrote...

    [tex]P = \vec{F} \cdot \vec{v}[/tex]

    [tex]P = m \vec{a} \cdot \vec{v}[/tex]

    [tex]P = m \frac{d \vec{v}}{dt} \cdot \vec{v}[/tex]

    [tex]P = m \frac{d}{dt}\frac{||{v}||^2}{2}[/tex]

    But after that you get that

    [tex]2 \vec{a} \cdot \vec{v} = \frac{d ||{v}||^2}{dt}[/tex]
  4. Nov 24, 2009 #3
    In your 4th line, where does that 2 come from??? I still don't understand the jump from v to a, shouldn't it be only 2a? Not 2a x V?
  5. Nov 24, 2009 #4
    In the fourth line comes from the chain rule, which is also the reason you get 2a dot v.

    [tex]\frac{d}{dt} \frac{v(t)^2}{2} = (2 v(t)) \frac{d}{dt}\frac{v(t)}{2}[/tex]
  6. Nov 24, 2009 #5
    Oh snap, totally forgot about that. Thank you.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook