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Homework Help: A skier of mass 59.0kg starts from rest at the top of a ski slope of height 70.0m

  1. Feb 18, 2008 #1
    1. The problem statement, all variables and given/known data

    A skier of mass 59.0kg starts from rest at the top of a ski slope of height 70.0m .

    2. Relevant equations

    3. The attempt at a solution

    If frictional forces do −1.08×10^4J of work on her as she descends, how fast is she going at the bottom of the slope?

    (1/2)mV^2=mgh + (1/2)mV^2 + -1.08 x10^4

    Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is = 0.180. If the patch is of width 65.0m and the average force of air resistance on the skier is 170N , how fast is she going after crossing the patch?


    Where an I going wrong with this?

    C. After crossing the patch of soft snow, the skier hits a snowdrift and penetrates a distance 3.00m into it before coming to a stop. What is the average force exerted on her by the snowdrift as it stops her?
  2. jcsd
  3. Feb 18, 2008 #2
    first of all, please don't put in numbers. Use variables. The number one source of error is using numbers before having derived.

    So in your first part, you have 1/2mvi^2 = 1/2mvf^2 + mgh - Et (friction). There is a flaw in your equation there. The skier didn't gain potential energy coming down the hill, and the thermal heat wasn't gained, it was lost.

    Try to fix that.

    Then in your second part, try doing an energy balance equation again. Try to do the first part first.
  4. Feb 18, 2008 #3
    Well first off, I notice that you used the correct value of 31.7 m/s in the second part of the question, but for the first part you wrote it as 31.4 m/s. It might be a typo, but if not it'll need correcting.

    For the second part of the question, start by calculating the frictional force acting on the skier using the formula:
    [tex]F = \mu R[/tex]

    Once you have that force, you can add it to the force due to air resistance, then you can calculate the work done on the skier with:
    [tex]W = Fd[/tex]

    Then you can subtract that from the skiers initial kinetic energy, to find the KE after the patch of soft snow and therefore also their velocity.

    For the final part of the question, you know their initial velocity (the answer from part b), their final velocity (0) and the distance it takes to stop. So you'd need to calculate their deceleration with:
    [tex]v^{2} = u^{2} + 2as[/tex]

    And then plug that into:
    [tex]F = ma[/tex]
  5. Feb 18, 2008 #4
    yea, but lavalamp. he did the first part wrong.
  6. Feb 18, 2008 #5
    While I admit that the working out could be ... improved upon. The poster did get the correct answer.

    It is possible that for the work that will be handed in it was laid out correctly, since I find that when typing lots of people tend to get lazy and write things that are incorrect but where the meaning is (usually) clear.
  7. Feb 18, 2008 #6
    What does U^2 and R stand for in those equations
  8. Feb 18, 2008 #7
    u is initial velocity, and R is the normal force.
  9. Feb 18, 2008 #8
    Thanks for all the help, I figured it out
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