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Help finding distance using work and velocity

  1. Oct 16, 2016 #1
    1. The problem statement, all variables and given/known data
    I'm looking to the distance (h) a ski jump should be placed from the top of the hill so that a skier with a mass of 85kg will not exceed 30.0m/s. I am given an initial velocity of 2.0m/s and the total work of frictional forces to 4000J

    2. Relevant equations
    How can I use total work to determine distance with the variables I'm given? I'm not looking for the answer I just need to be pointed in the right direction here.

    3. The attempt at a solution
    I used the work-energy theorem W(total) = Kinetic (final) - Kinetic(initial)
    Kf = 1/2mv^2 = 1/2 * 85kg * (30.0m/s)^2 = 38250J
    Ki = 1/2mv^2 = 1/2 * 85kg * (2m/s)^2 = 170J
    Wtotal = 38250 - 170 = 38080J
    subtract the frictional energy of 4000J
    34080J and this is where I'm stuck. Honestly I don't even know if I'm on the right track.
     
  2. jcsd
  3. Oct 16, 2016 #2

    gneill

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    Staff: Mentor

    Hi SC, Welcome to Physics Forums.

    A question is not a relevant equation. Questions should go in the problem statement. You should list equations you believe may be relevant to the type of problem being solved in the relevant equations section.

    Before tackling the details of a problem like this it's a good idea to list all the places where energy is coming from or going. You've recognized the initial KE, final KE, and the frictional force losses. What other form of energy is in play here? When a skier descends a mountain, why does he gain speed?
     
  4. Oct 16, 2016 #3
    Sorry, I guess I skimmed over that too quickly. I read it as questions.
    The skier accelerates due to the vector of the gravitational force working along the slope of the hill.
    So the potential energy (U) of the skier is transformed to K as he moves down the slope.
    Kf - Ki = Uf - Ui + Efric -> Kf - Ki = mg(Yf - Yi) + Efric Yf-Yi is h
    Am I on the right track now?
     
  5. Oct 16, 2016 #4

    gneill

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    Staff: Mentor

    Yes, that's the idea. As the skier descends he trades gravitational PE for KE, but also loses KE to friction.

    Make sure that you get your signs right to reflect gains or losses. For example, if the skier starts at the top of the hill at height Yi and ends lower down at height Yf, then mg(Yf - Yi) is going to be negative. While it's true that he's losing gravitational PE, his KE should be increased by this amount. Just spend a bit of time making sure that the signs you give the terms reflect what's taking place.
     
  6. Oct 16, 2016 #5
    Thank you for your help, I think I have it.
     
  7. Oct 16, 2016 #6

    gneill

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    Staff: Mentor

    Great! Glad to help.
     
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