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Sliding down a hill with friction

  1. Oct 6, 2009 #1
    1. The problem statement, all variables and given/known data

    Johnny jumps off a swing, lands sitting down on a grassy 20 degree slope, and slides 3.5m down the slope before stopping. The coefficient of kinetic friction between grass and the seat of Johnny's pants is 0.5

    2. Relevant equations

    F=ma
    Frictional force = [tex]\mu[/tex]N
    V[tex]^{2}[/tex]=V[tex]_{2}[/tex]+2a(x)
    V final is the first V and V intial is the second V
    Sorry about the equations, I'm still inexperienced with Latex.
    Also, you may use any kinematic equation or Newtons equations. I listed these because this is how I approached the problem.
    3. The attempt at a solution

    Ok, I thought/think this was/is a very easy problem. However, masteringphysics.com does not agree with my answer. Hopefully someone can point out my errors.

    I'm not sure exactly where to start. Breaking the forces into vectors gives me:

    mg-----------(0,-mg)
    N------------(0,mgcos(20))
    Friction------(-.5mgcos(20),0)
    Force Johnny (mgcos(20),0)

    Is this the correct breakdown?

    I know that the net force is 0 because Johnny eventually stops. I know that Frictional force is heading up parallel to the slope of the grass. I'm pretty sure the Frictional force is .5m9.8cos(20) where m is the mass of Johnny. I evaluated that to get that the force of friction is 4.6m. Here is where I run into all of my problems. I know that I want to find the acceleration in order to solve for the initial velocity. I think there is something wrong with my vectors. Could someone give me a push in the right direction in order to solve for acceleration?
     
  2. jcsd
  3. Oct 6, 2009 #2
    You cannot say that Fnet = 0 because you are trying to find the acceleration while he's sliding down the hill. When he's stopped you know the net force is 0, but he is also not accelerating, so you simply cannot do this.

    Below is what your given and what is required for the question...

    Mu = 0.5
    d = 3.5 m
    Vf = 0 m/s
    a = ? m/s

    If you draw an FBD, which I assume you did, there are only 3 forces acting on the person sliding down the hill: Fn, Fg and Ff.

    http://img3.imageshack.us/img3/7661/47955373.th.jpg [Broken]

    You know the y component of Fg will equal Fn, hence: Fn = cos20mg

    Next, you know Fgy and Fn will equal 0 because it is not accelerating up, nor down.

    Fgx is the only force that is causing the person to slide down the hill and friction is the only force that is opposing Fgx.
    Fnet = Fgx - Ff = ma

    Fgx also equals: sin20mg

    When you solve for Ff you get:
    Mu = Ff/Fn
    0.5 = Ff/cos20mg
    Ff = 0.5(cos20)(9.8 m/s^2)(m)

    You can now substitute Fgx and Ff back into the equation: Fgx - Ff = ma
    sin20mg - 0.5(cos20)(9.8 m/s^2)(m) = ma

    You notice that in both (sin20mg) and (0.5(cos20)(9.8 m/s^2)(m)) there is an m, so you can factor it out:
    m[(sin20)(9.8 m/s^2) - 0.5(cos20)(9.8 m/s^2)] = ma

    Now you can divide by m on both sides because they will cancel out. You get:
    a = sin20(9.8 m/s^2) -0.5(cos20)(9.8m/s^2)
    a = -1.252696 m/s^2

    Now you know the acceleration and you already know 2 other pieces of information. You can now use your regular motion equations to solve for Vi...

    Hope this helps...
     
    Last edited by a moderator: May 4, 2017
  4. Oct 7, 2009 #3
    That helped a lot. Thank you very much. I see now that my post was extremely confusing. I'm sorry about that. I reviewed my FBD drawing skills and I understand exactly where I went wrong. I really appreciate your time.

    Does anyone know how to mark this as solved? I looked under thread tools, but I didn't see a "mark as solved" option.
     
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