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Simple energy problem

  1. Jun 9, 2008 #1
    1. The problem statement, all variables and given/known data

    The problem states that a skier is on a 20m hill where the bottom makes a 20 degree angle with the horizontal. if the hill has a coefficient of friction of .210, and then the ground levels and the flatland has the same coefficient (.210) what is the horizontal distance the skier travels before coming to a stop. (horizontal after the hill)

    2. Relevant equations
    The work of the non-constant force on the hill is just the negative force of friction times the
    change of x. This equals (KE(f) - KE(i)) + (PE(f) -PE(i)). Since KE(i) and PE(f) are zero the whole equation becomes 0.210 * the normal force * the change of x = .5v^2 - (20 * 9.8).
    (all the masses cancel out)


    3. The attempt at a solution
    Now to solve for the normal I used 9.8cos(20). The normal being mg on a 20 degree hill.
    After the hill the normal became 9.8 (all mass cancels out). After solving for v then plugging into v^2 = Vo^2 +2ax I got 40.6. (a = 2.058) The answer in the book is 40.3m.
    I redid the problem using the normal as just 9.8 for the friction force on the hill. This I got 36.8m.
    The first two examples I was using the change of x to be the hypotenuse of the triangle.
    (20/sin(20)) When I used the change of x to be just the x direction of the hill and the normal to be just 9.8, I got 40.3m which is the answer in the book. I was wondering why the non-constant force on the slope would just be .210m(9.8) times 54.9m instead of .210m(9.8)cos(20) times 58.5m.
    I'm sorry all this reads confusing but if you write it out it makes it a lot easier.
     
  2. jcsd
  3. Jun 9, 2008 #2

    tiny-tim

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    Welcome to PF!

    Hi format22! Welcome to PF! :smile:
    It's a little difficult to tell, since you should have shown your full working, but I think you may have used the wrong initial equation.

    KE = .5v^2 = PE gained minus work lost to friction. :smile:
     
  4. Jun 9, 2008 #3
    I think I did type it in wrong. For the initial I used.. work(non constant) = KE(f) - PE(i).
    From this I solved for v in the KE(f). As far as the problem goes I think I am given all the variables except for v but I think my problem lies in the numbers I come up with for the work of the non constant force. I am using -0.210 * m(9.8)cos(20) * the hypotenuse on the triangle (-Friction force times change of x). Is this wrong because I can't see where else my problem would be?
     
  5. Jun 10, 2008 #4

    tiny-tim

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    Hi format22! :smile:

    Sorry, but if you don't show us your detailed answer, how can we find out where you've gone wrong? :wink:
     
  6. Jun 10, 2008 #5
    Sorry for the confusion Tiny Tim. It sucks when you spend hours on a problem just to find you have made a simple rounding error. But thanks anyway for the help.
     
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