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Homework Help: Speed and Kinetic Friction

  1. May 30, 2016 #1
    1. The problem statement, all variables and given/known data
    Sam, whose mass is 75 kg, stands at the top of a 12-m-high, 100-m-long snow-covered slope. His skis have a coefficient of kinetic friction on snow of 0.07. If he uses his poles to get started, then glides down, what is his speed at the bottom?

    2. Relevant equations
    v = √(2g(h-μk√(L2-h2)))

    3. The attempt at a solution
    It seems as simple as plugging in the values into the equation, but my result isn't anywhere near what it should be.
    From my understanding:
    g = 9.8
    h = 12m
    L = 100m
    μk = 0.07

    v = √(2(9.8)(12-0.07√(1002-122))) = 152.36
    152.36 m/s? That seems a bit much to me. The back of my book says that it is 9.9 m/s, but how do I get there?
    What did I mess up on?
  2. jcsd
  3. May 30, 2016 #2


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    This is a very specialized formula! (Is this sort of formula supplied for you on exams?)

    When I calculate the left side, I don't get 152.36. Make sure you are doing the calculation correctly using your calculator.
  4. May 30, 2016 #3
    Yeah we have to memorize it. I have no idea where it came from lol

    When I punch it into my calculator I get the same answer. So I tried breaking it down algebraically as follows.
    I still ended up with the same weird number.


    Am I plugging the values into the wrong places?
  5. May 30, 2016 #4


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    See if you can spot the error in going from the 3rd to the 4th line. "Order of operations" is important here.
  6. May 30, 2016 #5
    Rather than multiply, I take the 11.93th root of 9856?


    When I plug it into my calculator I get the correct answer, but I am concerned that when I do it by hand I am going to mess it up. :/
  7. May 30, 2016 #6


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    How would you evaluate ##9 - 2 \cdot 3##

    You have a subtraction and a multiplication. Which operation should you do first?
  8. May 30, 2016 #7
    Oh!!! Duh!!
    Thank you!!!
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