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Skier/friction problem

  1. Oct 23, 2013 #1
    Well, I've got the right answer on this problem, but I don't know if/why the math makes sense. I feel like I got the right answer by coincidence.

    I'm rephrasing the question because it's part of a multipart question.

    A skier with mass of 100kg starts at point A with a velocity of 31.3m/s. Find the skier's velocity at point B if the distance between A and B is 20m and a constant frictional force of 50N opposes his motion.



    K=1/2mg2

    v=√(v02-2μκmgx

    Those are the equations I thought relevant.

    The first equation, K=1/2mg2 is what I used to get the skier's velocity at A (previous part of the problem), which is 31.3m/s. 31.3m/s is correct; I have the answers.

    Also, from the second equation, √(31.32-(2x50Nx20m)/100kg))=30.98m/s ≈ 31m/s

    I also know that 31 m/s is correct, because again, I have the answers. But I am particularly confused about why I divided by 100kg. I was trying to find a way to get rid of the "kg" in the Newton unit so I would have like units in both terms. Dividing by 100kg works to make the units match, but I don't understand if/why it works to get the right answer.

    I appreciate any guidance. Thanks!
     
  2. jcsd
  3. Oct 23, 2013 #2
    Re do the math - and use the "units of a Newton" as appropriate. On any mathematical problem where units become confusing - make sure you are using the most fundamental units available.
     
  4. Oct 23, 2013 #3
    That second equation comes from conservation of energy: The final kinetic energy of the skier is equal to the initial kinetic energy added to the work done on the skier/
    The work done is force times displacement. It's negative since the force is inthe opposite direction as the displacement. Since the force is simply given, and you don't have to compute it from the mass of the skier and a friction coefficient, it doesn't involve mass, and you should replace [itex] \mu_k g m [/itex] in your second equation with F (wich is 50 newton)

    Since kinetic energy is (1/2)mv^2, conservation of energy means:

    [tex] \frac{1}{2}m v_i^2 - F x = \frac{1}{2}m v_f^2 [/tex]

    divide this by (1/2)m, and you see why you end up with 2 F x/m.
     
  5. Oct 23, 2013 #4

    CAF123

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    Gold Member

    Check the dimensions of your eqn v = √vo2-2μkmgx.
     
  6. Oct 23, 2013 #5
    Thank you very much all of you! I get the concept now. My mind was just stuck in a rut, I think. Once confused, I am not easily unconfused.
     
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