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Velocity of skier slides off a frictionless hill

  1. Nov 12, 2007 #1
    a Skier of mass 60 kg pushes off the top of a frictionless hill with an initial speed of 5.0 m/s. How fast will she be moving after dropping 20m in elevation?

    i dont know how to compute this without the angle
     
  2. jcsd
  3. Nov 12, 2007 #2

    Doc Al

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    Staff: Mentor

    What's conserved?
     
  4. Nov 12, 2007 #3
    Use potential energy formula :

    E = m*g*(change in height)

    E = 60*9.81*20 = 11772 J

    So skier loses 11772 J of potentiaql energy on descent, this is all converted into kinetic energy. So using KE formula :

    KE = 0.5*m*(v^2)

    V = sqrt(KE/0.5*m) = 19.81 m/s

    So add the initial velocity to this to get :

    5 + 19.81 = 24.81 m/s
     
  5. Nov 12, 2007 #4

    Doc Al

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    (1) Please reread the forum rules about posting complete solutions.
    (2) This solution is not correct.
     
  6. Nov 12, 2007 #5
    The books says the answer is 20m/s?
     
  7. Nov 12, 2007 #6

    Doc Al

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    Use conservation of energy, but be sure to apply it correctly.
     
  8. Nov 12, 2007 #7
    E = 60*9.81*20 = 11772 J
    KE = 0.5*m*(v^2)

    I'm not exactly sure how to manipulate the KE formula to where it will come up with a reasonable answer
     
  9. Nov 12, 2007 #8

    Doc Al

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    That's the increase in KE as the skier goes down the hill. What's the initial KE? Final KE? Final speed?
     
  10. Nov 12, 2007 #9
    initial = 750J
    final = 12522J

    so v^2 =12522J/(.5*m) = sqrt 417.4 = 20.4

    And do you not add the initial push off speed since you used it to get the initial KE?
     
  11. Nov 12, 2007 #10
    no you have already accounted for it since you used KE+PE at the beginning and got total KE at the bottom. basically because you had the KEi in the equation it's accounted for.
     
  12. Nov 12, 2007 #11
    sweet, thanks
     
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