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Calculate the coefficient of friction of a box sliding down a hill?

  1. Apr 21, 2012 #1
    1. The problem statement, all variables and given/known data

    A block slides down a hill. The hill has a gentle 20° slope with respect to
    the horizontal. The block has an initial speed of 6 m/s. If the block comes to a stop
    after traveling a total distance of 11 m along the slope, what is the coefficient of kinetic
    friction between the block and the hill?


    2. Relevant equations

    F=ma
    Fkk*n

    3. The attempt at a solution

    The first thing I did was draw a FBD. The box is going forward in the positive direction so the kinetic friction is acting in the opposite direction. There is force of gravity acting on the box and normal force, these forces are in the opposite direction on y axis. My x axis with was with respect to the slop, so kinetic friction is in the negative direction and the box is in position direction, on the x axis.
    Now I don't know what to do next...help?
     
  2. jcsd
  3. Apr 21, 2012 #2
    Use conservation of energy.
     
  4. Apr 21, 2012 #3
    Okay so use
    1/2*m*v2=m*g*h
    and I solve for m, this way then I can find gravity and find the two components of gravity. The y component of gravity will give the normal force. But then how do i find the kinetic friction?
     
  5. Apr 21, 2012 #4
    What is initial energy and what is the final energy of the block. The difference is work done, which is friction.
     
  6. Apr 21, 2012 #5
    Ok wouldn't the m cancel out with the conservation of energy?
     
  7. Apr 21, 2012 #6
    Yes, All have m as factor.
    [itex]KE=\frac {1}{2}mv^2[/itex]
    [itex]PE=mgh[/itex]
    [itex]friction=μN=μmgCosθ[/itex]
     
  8. Apr 21, 2012 #7
    Oh I see now
    So I can set up the equation like this:

    1/2mv2-mgh=μmgCosθ

    Isolating for μ:

    μ=(v2/2-gh)/gcosθ


    μ=(6m/s2/2-(9.80 m/s2)(11m))/(9.80 m/s2)cos 20°

    μ=-9.75

    Is this correct?
     
  9. Apr 21, 2012 #8
    Your equation should be both sides with equal units, joules.
    Your initial energy should be PE+KE.
     
  10. Apr 21, 2012 #9
    How come the initial energy is PE+KE? I thought originally the box was sliding down so all energy is KE and then it comes to a stop so all energy is PE
     
  11. Apr 21, 2012 #10

    tms

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    Mechanical energy is not conserved in this problem; there is friction.
     
  12. Apr 21, 2012 #11

    tms

    User Avatar

    There is always both kinetic and potential energy, even though at some points one or the other may be zero. Of course, by `always' I mean always when there is something around to give rise to potential energy.
     
  13. Apr 21, 2012 #12
    Work-Energy Principle.
     
  14. Apr 23, 2012 #13

    tms

    User Avatar

    Never mind.
     
    Last edited: Apr 23, 2012
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