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Girl slides down a slide. What is the friction?

  1. Oct 20, 2008 #1
    1. The problem statement, all variables and given/known data
    A 17 kg girl slides down a playground slide that is 2.7 m high. When she reaches the bottom of the slide, her speed is 1.3 m/s. What is the cohefficient of friction?


    2. Relevant equations
    F=ma
    Vf^2 = Vi^2 + 2ad

    3. The attempt at a solution

    I calculated the normal force to be 166.77 N (sin(theta)*m*g = sin(20)*17 kg*9.81 m/s^2 = 166.77 N)

    I found distance of the slide by (Hyp = Opp/Sin(theta) = 2.7m /sin (20) = 7.89 m)

    I then filled in the equation Vf^2 = Vi^2 + 2ad ( (1.3^2 = 0^2 + 2a *7.89m) = (1.69 = 15.788 a) = (a = .107m/s^2 )

    I then used F=ma ((F = 17kg *.107m/s^2)= (F =1.82 N))

    Then i used mew = 1.82N /166.77 N = .0109

    .0109 is wrong. Ehhh how do i do this??!1qWsdf
     
  2. jcsd
  3. Oct 20, 2008 #2

    LowlyPion

    User Avatar
    Homework Helper

    How did you know it was 20 degrees?

    Anyway figure how much potential energy she had at the top.

    PE = m*g*h

    Then how much KE does she have at the bottom? mV2/2

    The difference is the work that went to friction. = u*m*g*cosθ*d
     
  4. Oct 20, 2008 #3
    Thanks that helped a lot. And it was in the picture. Sorry bout that
     
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