1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Find distance before rest given length and angle of incline

  1. Jun 22, 2010 #1
    1. The problem statement, all variables and given/known data
    A skier starts from rest at the top of a hill that is inclined at 9.1° with the horizontal. The hillside is 250 m long, and the coefficient of friction between snow and skis is 0.0750. At the bottom of the hill, the snow is level and the coefficient of friction is unchanged. How far does the skier glide along the horizontal portion of the snow before coming to rest?


    3. The attempt at a solution
    250sin9.1= 39.5 (height of hill)

    v2 = 2gh
    v2 = 2 (9.8 m/s2)(39.5m)
    v= 27.8 m/s

    d= v2 / 2 u g
    d= 27.8 m/s / 2 (0.0750) (9.8 m/s2)
    d= 18.9 m
     
  2. jcsd
  3. Jun 22, 2010 #2
    Do you know the mass of the skier?
     
  4. Jun 22, 2010 #3
    No, i do not know the mass of the skier.
     
  5. Jun 22, 2010 #4
    That is weird because friction force depends on the normal force which depends on the mass of the object.
    By the way, you did not take into account friction here:

    v2 = 2gh
    v2 = 2 (9.8 m/s2)(39.5m)
    v= 27.8 m/s
     

    Attached Files:

  6. Jun 22, 2010 #5
    I used the coefficient of kinetic friction when I solved for the distance
     
  7. Jun 22, 2010 #6
    Yes but there is friction in the slope also. In my opinion this problem should be split into two parts, first find the velocity and the end of the slope then find the distance required to stop. The later one can be solved using work energy theorem, but I still don't see how to do this without the mass. Sorry, maybe someone else can help you more. Good luck.
     
  8. Jun 22, 2010 #7
    all I can give you is theory atm.. .

    the skier gains kinetic energy on the way down. at the bottom of the slope Ek = Eg-W

    W= (coefficient of kinetic friction * mass * gravity)(distance along the slope)

    so then you just take Ek at the bottom and sub in for W=uFn*d

    W/uFn = d
     
  9. Jun 22, 2010 #8
    how do I get the W?
     
  10. Jun 22, 2010 #9
    oh just read the question

    mgh - umg = 1/2mv^2

    9.8*h-0.075*9.8=1/2v^2

    9.725(39.53)=1/2v^2

    384.52*2=v^2

    27.73m/s
     
  11. Jun 22, 2010 #10
    W/uFn = d

    27.73/0.075(mg)=d

    27.73 / (0.735m) = d

    i have no clue how to get beyond this point.

    lol try calculating mass
     
  12. Jun 22, 2010 #11
    this is wrong.. stupid mistake

    27.73 has to be converted to energy .. gl
     
  13. Jun 22, 2010 #12
    okay so I was just playing around with these numbers and I divided 27.73m/s / 9.8m/s^2 and got 2.83m. When I entered this number, the response told me that I was off by a multiple of 10. I only have one submission left before I reach the max, so I want to get this one right. Any ideas?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook