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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]


    whyyyyy?
     
  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.
     
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