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Inclined plane question?

  1. Mar 3, 2012 #1
    1. The problem statement, all variables and given/known data

    Boom-Boom Slapshot, Canadian hockey star, slides down a 50m long ice covered hill on his skates. The frictionless hill is inclined at 35° to the horizontal. Once he reaches the bottom of the hill, the ice is covered with deep snow that has a coefficient of kinetic friction of .50
    How far into the snow will BoomBoom go before coming to rest?

    3. The attempt at a solution

    I've figured out acceleration on the inclined plane which is 5.62m/s^2. But, I was having trouble with including friction into the other part as well as normal force.

    a = sin35 * 9.8 = 5.62m/s^2

    Using the formula Vf^2 = Vi^2 + 2(a)(d)

    I found the velocity which is 23.71m/s

    I got an answer of 175m but the text book says the answer is 57m. The textbook has been wrong few times, I just want to confirm this, thank you. :)
  2. jcsd
  3. Mar 3, 2012 #2
    How did you use that velocity to get 175m?
    When I calculated it, I got 57m.
  4. Mar 3, 2012 #3
    a1 = g*sin35° = 5.62 m/s²

    Vmax = √[2*a1*s] = √[2*5.62*50] = 23.7 m/s

    a2 = -g[sin35° - µ*cos35°] = -1.607 m/s²

    d2 = -Vmax²/(2*a2) = 174.76 m

    Did I do something wrong?
  5. Mar 3, 2012 #4
    Ah, the problem is the interpretation of the question.
    I think the problem was intended to mean that he starts travelling horizontally (at the bottom of the hill) when he enters the snow.
  6. Mar 3, 2012 #5
    Can you expand on that please? I'm really confused after trying 4 times.

    Thank you. Just pointing me in the right direction will do. :)
  7. Mar 3, 2012 #6
    After he reaches Vmax, the incline becomes 0°
  8. Mar 4, 2012 #7
    I need more help please, I'm kind of lost. =\
  9. Mar 4, 2012 #8


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    Staff Emeritus
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    What is it that you don't understand regarding what Browne said in his last two posts?

    It would be difficult to restate what he said in a manner that's any clearer !
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