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Kinematics skier problem

  1. Oct 23, 2009 #1
    1. The problem statement, all variables and given/known data

    Part A. A 64.0-kg skier starts from rest at the top of a ski slope of height 70.0 m.
    If frictional forces do −1.00×104 J of work on her as she descends, how fast is she going at the bottom of the slope?
    I already got the right answer for this one.. I have problems in this second part.

    Part B. Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is 0.19. If the patch is of width 69.0 m and the average force of air resistance on the skier is 140 N, how fast is she going after crossing the patch?

    2. Relevant equations

    Vinitial= 32.5 (got it from part A)
    Vfinal= ???
    Distance = 69.0 m
    Fnormal= mg = 627.84N
    coefficient of friction = 0.19
    m= 64.0 kg

    Ffriction= Fnormal * coefficient of friction

    Wfriction= -Ffriction*Distance
    Wairresistance= -Fairresistance*Distance (is this right?)
    Change in kinetic energy= (1/2)mvfinal2-(1/2)mvinitial2

    3. The attempt at a solution

    So I got the equation..
    Wfriction+Wairresistance=(1/2)mvfinal2-(1/2)mviinitial2

    and i plugged in..

    -8222.59 + -9660 = (1/2)(64)vfinal2-(1/2)(64)(32.5)


    I know I'm doing something wrong somewhere because I can't take the square root of a negative number, but i don't know where.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 23, 2009 #2
    just noticed i put this in the wrong section, sorry!
     
  4. Oct 23, 2009 #3

    berkeman

    User Avatar

    Staff: Mentor

    No, I moved it here from Advanced Physics. This belongs here in Intro Physics. Check your PMs.
     
  5. Oct 24, 2009 #4

    alphysicist

    User Avatar
    Homework Helper

    Hi Fera09,


    In your last line, did you forget to square the 32.5 (or is that just a typing error)?
     
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