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Terminal Speed for Skier Going Down Slope

  1. Sep 27, 2013 #1
    1. The problem statement, all variables and given/known data

    What is the terminal speed for an 75.0kg skier going down a 35.0∘ snow-covered slope on wooden skis μk= 0.060?

    Assume that the skier is 1.60m tall and 0.300m wide.

    Express your answer using two significant figures with the appropriate units.

    Area skier = A = (1.60m x 0.300m) = .48 m^2

    2. Relevant equations

    D = (1/4)A*v^2
    Fnet = D

    3. The attempt at a solution

    (mgsinθ) - (μk*mgcosθ) = (1/4)A*v^2

    => vterminal = sqrt((4((mgsinθ) - (μk*mgcosθ))) /A )

    vterminal = sqrt((4((75.0 * 9.80 * sin(35)) - (.060 * 75.0 * 9.80 * cos(35))))/.48)

    vterminal = 56.68 m/s = 57 m/s (rounded to 2 sig figs)

    Mastering Physics says that this answer is wrong but I cannot find my error. I have checked over the problem and my solution about a dozen times. I figure more experienced eyes may help find my ruinous mistake. Thank you! :)
     
  2. jcsd
  3. Sep 27, 2013 #2

    TSny

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    Your work looks good to me. The data in the problem is given to 3 significant figures. Maybe you are supposed to keep 3 sig figs in the answer. Don't know.
     
  4. Sep 27, 2013 #3
    Thanks for double checking for me, I really appreciate it. :)

    I actually tried submitting it with 3 sig figs as well but it said it was wrong. I am going to email my professor and see if they can take a look.
     
  5. Sep 27, 2013 #4

    PhanthomJay

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    I believe this equation for drag force assumes a spherical shape for the drag coefficient (C_drag = about 0.5) . If the skier is standing more or less upright rather tham curled up like a ball, the drag coefficient is higher, and thus the terminal speed will be less. The quadratic drag force is actually 1/2(C_drag)ρ(A)v^2, where ρ is the air density(about 1.2 kg/m^3), and C_drag for a flat exposed surface might be more like 1.0 or so.
    When I last went snow tubing, I knew my speed was much faster when lying back instead of sitting up. And faster than that when going with my grandkids 2 or more per tube and faster than that with several tube loads of kids hooked together in a lying down position! What a rush!
     
  6. Sep 28, 2013 #5
    It looks like you were right, PhantomJay. I should have seen this before but I didn't check my units. The answer is as follows:

    sqrt((2mg/ACp)*(sinθ-μkcosθ))

    If you do it this way then you get all of the correct units (m/s) in the end after taking the square root. I used the following:

    p (rho) = 1.2 kg/m^3
    m = 75.0 kg
    g = 9.80 m/s^2
    θ = 35 degrees
    μk = 0.060
    C = 0.8

    When you plug all of these into the equation you get a terminal velocity of v = 41 m/s (to 2 significant figures), which is the correct answer.

    The only thing I am still unclear on is why we use a drag co-efficient of C = 0.8, which closely resembles a cylinder falling end-down, instead of a drag co-efficient of C = 1.1, which closely resembles a cyclinder falling side-down. I feel like a skier standing upright is affected by drag more similarly to the side of a cylinder than the bottom of one. Perhaps it has to do with the fact that they are falling at an angle? I don't know.

    Anyways, thank you all for your help! :)
     
  7. Mar 14, 2016 #6
    Thank you So much Sir/Madam, i tried a thousand times, spent an hour and 15 mins trying, but failing. Thank you again!
     
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