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Mgh = 1/2mv^2

  1. Sep 29, 2006 #1
    I just wanna know in what conditions does mgh = 1/2mv^2 when a guy goes down a slope and in what conditions does mgh =/= 1/2mv^2. Thanks in advance.
     
  2. jcsd
  3. Sep 29, 2006 #2
    according to COE, (initial)PE + KE = PE + KE(final) so i think that its better to use this instead of your equation.For example,if the object were to be released with some initial velocity,the KE(initial) would not be 0 so total initial energy is mgh+1/2mv^2 instead of mgh.Btw,it is more precise to write change of PE = -change of KE as this shows that energy is conserved.
     
  4. Sep 29, 2006 #3

    HallsofIvy

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    And energy is conserved as long as the are no "non-conservative" forces- i.e. as long as there is no friction.
     
  5. Sep 30, 2006 #4
    o_O
    So when there is no friction P.E = K.E?
     
  6. Sep 30, 2006 #5
    Yes, decrease in PE= increase in KE and vice versa. If there's friction, decrease in PE= increase in KE + energy 'lost' to surroundings, so increase in KE<decrease in PE.
     
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