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Snow sled on a hill

  1. Jan 27, 2016 #1
    1. The problem statement, all variables and given/known data
    Say you push a sled of mass [itex] m [/itex] up a hill that is angled upwards at a certain angle [itex] \theta [/itex] at a constant velocity. The hill has snow on it offering a friction force that is equal to 30% of the sleds weight. If you pushed the sled down the hill with the same amount of force as you did pushing the sled up the hill, how fast does it go down the hill?

    2. Relevant equations

    Newtons Laws

    3. The attempt at a solution

    So, since the sled is moving up the hill at a constant velocity, the net force on the sled must be zero.
    This also means that the force with which you push is equal to the friction force [itex] \mu [/itex] added to [itex] mg sin \theta [/itex].

    So: [itex] F_{push}= \mu + mgsin \theta=.30mg+mg sin \theta= mg(.30+sin \theta) [/itex].

    If I push the sled down the hill with force [itex]F_{push} [/itex] then the sled will accelerate.

    Since the frictional force now points in the opposite direction when I push the sled down the hill, I will have:

    [itex] \mu-F_{push}-mgsin \theta = ma [/itex] which means [itex] ma=.30mg-mg(.30+sin \theta ) -mg sin \theta [/itex]

    So, the sled accelerates down the hill with acceleration [itex] a=-2g sin \theta [/itex].

    Is this correct?
     
  2. jcsd
  3. Jan 27, 2016 #2

    gneill

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    Staff: Mentor

    Looks good. You might want to finish with an expression for the velocity with respect to time, as the problem did ask for how fast it goes.
     
  4. Jan 27, 2016 #3
    Shall I integrate a?
     
  5. Jan 27, 2016 #4

    gneill

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    Staff: Mentor

    You could. Although, since the acceleration is constant you might just use a standard kinematic formula.
     
  6. Jan 27, 2016 #5
    Thanks for the help gneill
     
  7. Jan 27, 2016 #6

    haruspex

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    Yes, if you are taking the positive direction for a as up the slope (which is why you got a negative acceleration at the end).
     
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