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Kinetic and potential energy

  1. Jan 22, 2013 #1
    If a stone sits at the edge of a 155m cliff and falls 43.6m with the velocity at that point of 29.2m/s , what is the kinetic and potential energy? Are they the same or diff?
     
  2. jcsd
  3. Jan 22, 2013 #2
    If I tell you that potential energy is "converted" into kinectic energy during fall. That at the edge of the cliff (155m) kinectic energy equals zero and potential energy is max. At the bottom of the cliff, kinectic energy is max wright before landing and potential is zero also before landing. Will that help solve your problem?

    By the way:
    Kinectic energy = [itex] 1/2 * m * v^2 [/itex]
    Potential energy = [itex]m * g * h[/itex]

    m - mass
    g - gravity constant
    h - hight
     
    Last edited: Jan 22, 2013
  4. Jan 22, 2013 #3
    I forgot to add the mass of the stone is .115kg, but after it falls 43.6m the kinetic vs the potential is different right? U would minus the 43.6 from the 155 to get the height for the formula whereas u would use the velocity to get the kinetic ?
     
  5. Jan 22, 2013 #4

    haruspex

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    That's the correct procedure. The two calculations might or might not produce the same number.
     
  6. Jan 22, 2013 #5
    You are wright about the height. From there you can calculate the potential energy.
    Watch out for velocity since it's different from inicial velocity. As the stone falls it gains velocity.

    I should've add that if you compute the energy at the top you get the total energy of the system. Once the stone starts falling potential energy drops, kinetic energy raises. At the bottom kinetic energy equals potential energy at the top.

    potential energy (top) = kinetic energy (bottom)

    You'll probably need it: [itex]U_{system} = U_{potential} + U_{kinetic}[/itex]
    U - energy
     
    Last edited: Jan 22, 2013
  7. Jan 22, 2013 #6
    Yea, I did the calculations and they are not the same, thanks
     
  8. Jan 22, 2013 #7

    haruspex

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    Quite so. In fact, there's a very easy way to get there. The KE must equal the PE lost in falling 43.6m; the remaining PE is for falling 155m-43.6m. Clearly the two will only be equal when it has fallen half way.
     
  9. Jan 22, 2013 #8
    So ke would be 1/2 0.115kg * 29.2sqrt m/s?
     
  10. Jan 22, 2013 #9

    haruspex

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    Squared, not sqrt.
     
  11. Jan 23, 2013 #10
    Yes. Thank you.
     
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