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Why does delta PE = -W?

  1. Oct 10, 2012 #1
    Can someone prove that the change in potential energy is negative work.
    I have a very basic understanding of the concept. I do not understand where it is derived from.
     
  2. jcsd
  3. Oct 10, 2012 #2
  4. Oct 10, 2012 #3
    Unfortunately you can't prove that in the general case. For gravity it's easy. What you can prove is that work done=1/2*m*v^2= kinetic energy, and from conservation of energy (dE/dt=0) you can derive the remaining stuff.
     
  5. Oct 10, 2012 #4
    Thanks that helps.!
     
  6. Oct 10, 2012 #5
    So the proof would be...

    mgy(b)+KE(b)=mgy(a)+KE(a)

    That is: U(b)+K(b)=U(a)+K(a)
    So U(b)-U(a)=K(a)-K(b)=-(K(b)-K(a)
    So U(b)-U(a)=-W ( By work energy theorem )
    Therefore:

    ΔU=-W

    Is this proof correct?
     
  7. Oct 11, 2012 #6
    Yeah it's ok. But another important question is wheather you know where work energy theorem comes from?
     
  8. Oct 11, 2012 #7
    Yeah.
    W= ∫from a to b of Fdx
    =∫from a to b of madx
    =∫from va to vb of mdv/dt vdt. (dx/dt =v so dx =vdt)
    =∫from va to vb of mvdv
    =1/2mv^2 evaluated at va and vb
    = 1/2mvb^2-1/2mva^2
    =KEb-KEa
    Therefore
    W=KEb-KEa
     
    Last edited: Oct 11, 2012
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