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Energy Conservation Spring Pulling Mass Up Incline Plane

  1. Oct 8, 2013 #1
    1. The problem statement, all variables and given/known data
    http://imgur.com/xIckJqW
    A block of mass m rests on a plane inclined at an angle θ with the horizontal. A spring with force constant k is attached to the block. The coefficient of static friction between the block and plane is μs. The spring is pulled upward along the plane very slowly.

    (a) What is the extension of the spring the instant the block begins to move?

    (b) The block stops moving just as the extension of the contracting spring reaches zero. Express μk (the coefficient of kinetic friction) in terms of μs and θ.


    2. Relevant equations

    F = -kx

    Esys = Emech + Etherm + Echem + Eother

    G.P.E.: U = U0 + mgh

    Spring Potential Energy: U = (1/2)kx2

    Kinetic Energy: K = (1/2)mv2

    Fn = mgcosθ

    3. The attempt at a solution

    Solution For Part (a):

    X forces:
    (-kx) , mgsinθ , Fnμs
    Y forces:
    Fn, Fgcosθ

    ƩFx = -Fnμs - mgsinθ + (-kx) = 0
    Solve for x:
    x = (-mg/k)(μscosθ+sinθ)

    Attempt at a solution for part (b):

    E1 = (1/2)mv02 + (1/2)kx2
    E2 = mgh + fkx
    fkx = Fnμkx = mgμkxcosθ

    I'm just not really sure how to go about this question at all. I'm unsure about assuming that there is an initial velocity, but i'm basing that on the previous problem.
     
  2. jcsd
  3. Oct 8, 2013 #2

    Doc Al

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    Staff: Mentor

    Looks good.

    The mass is initially at rest, until the spring force is great enough to overcome static friction. So you are starting with a mass at the end of a stretched spring. What's the energy of the system at that moment?
     
  4. Oct 8, 2013 #3

    gneill

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    Staff: Mentor

    For part (b) you can take the starting condition being the instant just before the block begins to move (so the spring is at maximum extension), and ends when the block has just come to rest. So the block is at rest at both ends, making life simpler :smile:

    EDIT: Doc Al got there before me!
     
  5. Oct 8, 2013 #4
    So the Energy at the initial point would just be the potential energy of the spring?
     
  6. Oct 8, 2013 #5
    Ok so starting now from
    (1/2)kx2 = mgx(sinθ+μkcosθ)
    Sub in x from previous solution;
    (-1/2)mg(μscosθ+sinθ) = mg(sinθ+μkcosθ)
    With a small amount of arithmetic becomes;
    μk= (-1/2)(μs+3tanθ)

    Does that final solution check out?
     
  7. Oct 8, 2013 #6

    Doc Al

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    Staff: Mentor

    Good.

    Get rid of that minus sign. (I should have mentioned that earlier--you had the sign of the spring force wrong.)

    Redo it, after the above correction.
     
  8. Oct 8, 2013 #7
    After the above corrections I came to a final answer of:

    μk = (1/2)(μs - tanθ)
     
  9. Oct 8, 2013 #8

    Doc Al

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    Staff: Mentor

    Looks good to me.
     
  10. Oct 8, 2013 #9
    Thanks guys!
     
    Last edited: Oct 8, 2013
  11. Apr 3, 2016 #10
    How did you get that equation?
     
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