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Newron's 2nd Law on a plane

  1. Mar 5, 2009 #1
    1. The problem statement, all variables and given/known data



    A sled moves on the flat surface of a hill. The hill makes an angle θ with the horizontal, and is left orientated.

    The hill: θ = 60°; left oriented; μk = 0.1.

    The sled: mass = 20 kg; starting location = 20 m from the bottom of the hill;
    initial speed = 0 m/s .

    http://img26.imageshack.us/img26/1756/blockoninclineleft.jpg [Broken]

    Solve:
    1. (Fnet)x immediately after motion begins, (N)
    2. (Fnet)x, sliding down the hill, (N)

    2. Relevant equations

    [tex]
    \vec{F}_{net} = \Sigma \vec{F} = m \vec{a}
    [/tex]


    Fn = mg cos θ

    Fn = 98




    3. The attempt at a solution

    I draw an FBD and every time I solve for (Fnet)x I get:

    (Fnet)x = -mg sin θ - μk*Fn

    (Fnet)x = 59.742 - 9.8

    (Fnet)x = 49.9 N

    Every time I input that as my answer it says it's wrong. Also why is it that the kinetic friction force is a positive 9.8. Doesn't the friction go in the opposite direction of the motion?

    Thanks for help
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 6, 2009 #2

    djeitnstine

    User Avatar
    Gold Member

    Welcome to PF.

    If a positive friction indicates that to the right is positive and to the left is negative
     
  4. Mar 6, 2009 #3
    Hi

    Thanks that cleared up the friction.. Do you know what I'm doing wrong with the (Fnet)x ?

    Thanks for the help
     
  5. Mar 6, 2009 #4
    According to my calculations, mg sin(60) (9.8*20*sin(60)) doesn't come out to 59.742...
     
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