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Energy Question- Major help needed

  1. Oct 19, 2006 #1
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    Problem A skier starts from rest at the top of a frictionless incline of height 20.0 m, as in Figure 5.19. As the bottom of the incline, the skier encounters a horizontal surface where the coefficient of kinetic friction between skis and snow is 0.225.

    Find the horizontal distance the skier travels before coming to rest if the incline also has a coefficient of kinetic friction equal to 0.225.

    I tried setting it up like this. I found the height(y) of the incline by doing 20/sin20= 58.476

    -0.225mgcos20(58.476)- 0.225mgd= 0-mgy

    Now I dont know what to do, help. How do I get rid of the m's, cause there are 3 this time. Am I doing something wrong.
     
  2. jcsd
  3. Oct 19, 2006 #2
    All of the terms in your equation have an m in them. Just divide it out.
     
  4. Oct 19, 2006 #3
    I cant get it right, please tell me if I am doing something wrong. And how do I divide the m's out, there are 3 of them.
     
  5. Oct 19, 2006 #4
    can someone help me please
     
  6. Oct 19, 2006 #5
    First off they give you the height in the diagram to be 20.0 m. Second off, solving this problem in symbols first makes it much easier.

    [tex]\Delta KE = \Delta U_g[/tex]
    [tex]\frac {1} {2} mv_f^2-\frac {1} {2}mv_i^2 = mgy_i - mgy_f[/tex]
    the initial kinetic is 0, the final gravitational potential energy is 0.
    [tex]\frac {1} {2} mv_f^2 = mgy_i[/tex]
    you can cancel the masses by dividing both sides by m...
    [tex]v_f= \sqrt{2gy_i}[/tex]

    You got the final velocity at the bottom of the hill. From there you calculate the acceleration he has from friction. Then it is a simple motion problem.
     
  7. Oct 19, 2006 #6
    Bishopuser I dont understand that method

    Here is how I attempted it

    for the coefficient of friction I will use the symbol (uk) ok and for the height/distance of the incline as D ok


    ok so in symbolic format, this is how I know how to do it

    -ukND-ukmgd= 0 + mgy1

    I found the height(y) of the incline by doing 20/sin20= 58.476
    N= normal force= mgcos20

    so its

    -uk(mgcos20)58.476-ukmgd= mgy1

    Is this right so far? Now I have trouble with the masses and I am confused what to do here.
     
  8. Oct 19, 2006 #7
    look at your first equation this way:
    m[-0.225gcos20(58.476)- 0.225gd]= mgy
    now does dividing the mass out make sense?

    You may need a brush-up on your algebra skills a bit.
     
  9. Oct 19, 2006 #8

    PhanthomJay

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    Is the incline frictionless or with friction? The problem reads both ways. Also, you have calculated the length of ther incline, not the height of the incline. The height is given as y = 20.
     
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