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!

Work done by friction on a skier and resulting distance the skier travels

  1. Feb 29, 2012 #1
    1. The problem statement, all variables and given/known data
    A skier starts from rest on a 20m high, 20° slope. μk=0.210
    Find the horizontal distance traveled by the skier.
    From this, for the equations below we know that:
    yf = 0
    vi = 0
    vf = 0

    2. Relevant equations
    Wnet = Wnc + Wg = ΔKE
    Wnet = -fkd
    Wnc = ΔKE + ΔPE
    Wnc = ΔKE + mg(yf - yi)
    KE = 1/2mv2
    PE = mgy
    fk = μkmg

    3. The attempt at a solution
    So I went with the work of a non-conservative force
    Wnc = (KEf - KEi) + (PEf - PEi)
    Wnc = (1/2mvf2 - 1/2mvi2) + (mgyf - mgyi)
    From given, I eliminated all 0 quantities, leaving me with
    Wnc = -mgyi
    Then plugged in Wnet = -fkd = -μkmgd so,
    kmgd = -mgyi
    eliminated like terms (m, g):
    kd = -yi
    and solved for d
    d = yik
    and plugged in the knowns
    d = 20m/0.210
    d = 95.2m
    I realize this is the distance traveled from the top of the hill to the end of motion, but all they want is the horizontal distance, so now I have to solve for the distance traveled from the top of the hill to the bottom of the hill. The only thing I think that changes between the above work and the distance from the top to bottom is the final velocity which will be nonzero.
    So my question is, how do I find the distance traveled from the top of the hill to the bottom?
    Or am I going about this wrong? Is there a more direct way to find just the horizontal distance traveled?

    Oh I should add, the answer given by the book is 40.3m
     
  2. jcsd
  3. Feb 29, 2012 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Is this the full statement of the problem?

    I'm guessing that after the skier gets to the bottom of the slope, she skis over a horizontal stretch of ground? Perhaps it's that horizontal distance that they want.
     
  4. Feb 29, 2012 #3
    AH! I figured it out...

    I had to find the length of the bottom of the triangle formed by the horizontal and the slope of the hill. sin20° = 20/x (where x is the hypotenuse, or the length of the slope of the hill)
    from that I got x ≈ 58.5m
    Then,
    cos20° = x/58.5m (where x is the length of the bottom of the triangle)
    x ≈ 54.9m
    then subtract that from the total length traveled by the skier, 95.2m - 54.9m = 40.3m which is the answer given by the book.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Similar Discussions: Work done by friction on a skier and resulting distance the skier travels
  1. Friction and a skier (Replies: 3)

Loading...