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Toboggan down hill, FIND ANGLE?

  1. Dec 15, 2007 #1
    1. The problem statement, all variables and given/known data

    A girl starting from rest, toboggans down a hill of height 15m. The mass of the girl and toboggan is 42kg. The speed of the girl and toboggan at bottom of hill is 2.9m/s. If force of friction is 112 N, what is the slope of the hill? Ignore air resistance.

    2. Relevant equations

    KE=PE, 15sin(theta), F=mg

    3. The attempt at a solution

    I started by equating initial and final energies by saying KE final is equal to PE initial plus the force of friction. (.5mv)squared = mgh - (mu)Fn
    I ended up with 0.1429=sin(theta) which equals 8.2 degrees. The answer was 16 degrees, where did I go wrong?
  2. jcsd
  3. Dec 15, 2007 #2
    What is 15sin([tex]\vartheta[/tex]) meant to be?
  4. Dec 15, 2007 #3
    I am trying to figure out the distance down the slope that she travels so I can use that along with the height (15m) to find the angle.
  5. Dec 15, 2007 #4
    15sin(theta) will not give you the distance travelled down the slope.
  6. Dec 15, 2007 #5
    Well I suppose I am stuck then.
  7. Dec 15, 2007 #6
    You know what to do, you just have your trig relations mixed up a bit. If sintheta = height/hypotenuse, then 15sintheta = height x height/hypotenuse...
  8. Dec 15, 2007 #7
    I should know this, but that still leaves me with unknowns for theta and hypotenuse.
    I am drawing a blank on how to find the hypotenuse with what is given.
    Thanks for the replies by the way.
  9. Dec 15, 2007 #8
    Sintheta = height/hypotenuse where hypotenuse = the distance travelled down the slope. Since we know the height, we can state the hypotenuse in terms of the height and theta, that is hypotenuse = height/sintheta. This way you can use 15m/sintheta in the expression for the work done by friction, and then solve for theta, which was your idea to begin with. You just misused the sin ratio!
  10. Dec 15, 2007 #9


  11. Dec 15, 2007 #10
    Beautiful. Thanks much!
  12. Dec 16, 2007 #11
    So how did you determine W was 5997.4?
  13. Dec 16, 2007 #12


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    Staff Emeritus
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    Gold Member

    I have to disagree with Squeezebox, the work done is not as quoted. To calculate the work done, you can apply conservation of energy.
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