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Speed of skiis (someone check my work, please)

  1. Oct 24, 2005 #1
    Law of Conservation of Energy
    (2) A ski starts from rest and slides down a 22 degree incline 75 m long.
    (a) If the coeffiecent of friction is .090, what is the ski's speed at the base of the incline?
    (b) If the snow is level at the foot of the incline and has the same coeffiecent of friction, how far will the ski travel along the level? Use energy methods. correct answer: 2400 m
    ------------------
    My thoughts were to use the equation Wnc = KE + PE

    Seeing as there is no potential energy on the way down I now have Wnc = KE

    Then I had to find the work acting on the skiis... so...

    Work(weight) + Work(friction) = (1/2)mv^2

    Weights force would be (sin22 x m x g)
    frictions would be (.090 x N) = (.090 x cos22 mg)

    I would say the m's cancel out. So then we substitute back into Work(weight) + Work(friction) = (1/2)v^2

    ((sin22 x g) x 75m x cos0) + ((.090 x cos22 x g) x 75m x cos180) = 1/2v^2

    so I get...

    275 + (-61) = .5V^2
    214/.5 = v^2
    sq root of 428 = v
    v = 20.6 = 21.0 Answer in Book: 21 m/s

    Part B

    Would I use a similar equation but instead of the Wnc = KE it would be Worknc = PE
    Work nc = mgh ?
     
  2. jcsd
  3. Oct 24, 2005 #2
    can someone please help me? my exam is friday... :(
     
  4. Oct 24, 2005 #3
    Can't you just plug in the 21 m/s for an initial velocity somewhere?
     
  5. Oct 25, 2005 #4
    Maybe I could use kinematics?

    v2 = vo2 +2a(x-xo) ?

    That doesn't work...

    I was thinking since it says at the bottom of the incline there would be no potential engery so I could use

    PE = KE but that does not work either.

    I cannot figure out an equation in which I am trying to find distance
     
  6. Oct 25, 2005 #5
    also tried first finding a so I could use kinematics.

    a = V^2/75
    21^2/75 = 5.88

    Then I used:

    v2 = vo2 +2a(x-xo)
    0 = 21^2 +2(5.88)(x-75m)
    -441 = 11.76(x-75)
    -36 = x-75
    39... and the correct answer is 2400, what haven't I tried?
     
  7. Oct 25, 2005 #6

    Doc Al

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    Several ways to attack part B:

    Start by finding the friction force.

    Energy methods: Work done by friction = change in Energy

    Kinematics: Use the force to find the acceleration.
     
  8. Oct 25, 2005 #7
    Finding Friction force would be, as previous stated, (but I do not have a mass, so how can I find the friction force?

    How do I find the friction force though?

    I do not have a mass...

    Friction force = .090 x ???
     
  9. Oct 25, 2005 #8
    would it be friction force = .090 x 9.8 m/s = .882 ?

    My problem is writing the questions out...

    (.882 N)(75 m) cos0 = 61.65 J

    61.65 J = change in energy? What does that mean?

    Sorry, Doc Al. I am just getting lost here.
     
  10. Oct 25, 2005 #9

    Doc Al

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    Just find it in terms of the unknown mass. You'll find that you don't need to know the mass to solve the problem. (Sorry if I wasn't clear before.)
    Friction will equal [itex]\mu N[/itex]. What's N?
     
  11. Oct 25, 2005 #10
    N = W = mg.

    f = [itex]\mu mg[/itex]

    So would it just be [itex]\mu g[/itex]
     
  12. Oct 25, 2005 #11

    Doc Al

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    Exactly right.
    Not sure why you dropped the m.

    In any case, now you are ready to solve the problem using energy methods or using kinematics. Do it both ways!

    Energy method: You have the friction force and the distance, so what's the work done by friction? (Don't forget that it's negative.) And what's the change in KE? ([itex]{KE}_f - {KE}_i[/itex])

    Kinematics: You have the friction force, so what's the acceleration?
     
  13. Oct 25, 2005 #12
    Work by f = change

    ([itex]\mu mg[/itex])(d)cos(180) = 1/2mv^2? -1/2mv^2

    ([itex]\mu mg[/itex])(d)cos(180) = -220 - 220 (440)

    That still does not work... am I solving for the D?
     
    Last edited: Oct 25, 2005
  14. Oct 25, 2005 #13
    With Kinematics: F = ma
    (.09 m g) = m (a) ---- m's cancel out
    .09 x 9.8 = a
    .882 m/s

    v2 = vo2 +2a(x-xo)
    21^2 = 0 +2(.882) (x-xo)
    441/1.764 = (x -xo)
    250...

    not getting the answer either way. :(
     
  15. Oct 25, 2005 #14

    Doc Al

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    This is correct. What's the final KE? You found (in part a) the initial speed the skier has entering the level section. I'd rewrite it like this:
    [tex]- \mu m g D = \frac{1}{2}m v_f^2 - \frac{1}{2}m v_i^2[/tex]
     
  16. Oct 25, 2005 #15

    Doc Al

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    This is correct. (Guess what? The book's answer is wrong!) Now do it using energy methods and verify that you get the same answer: D = 250 m.

    (I didn't check the answer to part a; I assume that one's correct.)
     
  17. Oct 25, 2005 #16
    The following code was used to generate this LaTeX image:



    [tex]- \mu m g D = \frac{1}{2}m v_f^2 - \frac{1}{2}m v_i^2[/tex]


    So I substitute with numbers

    -.09 x 9.8 x D = (1/2)(0 cause it will be when skii stops) - (1/2)(21^2)


    -.882 x D = -220.5
    D = 250 m

    And you're saying the book is wrong?
     
  18. Oct 25, 2005 #17

    Doc Al

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    That's what I'm saying.
     
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