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Box sliding down an inclined plane

  1. Oct 9, 2011 #1
    1. The problem statement, all variables and given/known data
    A 2kg box is at the top of an inclined plane that is 10m long. The inclined plane makes an angle of 30° with the ground. The μk of the ramp is 0.36 and the μk of the flat ground at the end of the ramp is 0.50. How far does the box travel along the flat ground until it stops?

    I'm not quite sure what to do after finding the acceleration, but I think you need to find the final velocity of the box when it finishes going down the ramp? I can't seem to figure out what to do past this point, so any help would be greatly appreciated. Thanks for your time.

    2. Relevant equations



    3. The attempt at a solution
    1. Fg=mg
    Fg=2kg(10m/s^2)
    Fg=20N

    2.Fgx=mgsinθ
    Fgx=2kg(10m/s^2)sin30°
    Fgx=10 N

    3. Fgy=mgcosθ
    Fgy=2kg(10m/s^2)cos30°
    Fgy=17.32N

    4.a=g(sinθ-μkcosθ)
    a=10m/s^2(sin30°-0.36cos30°)
    a=1.88m/s^2

    5.vf^2=vi^2 +2ad
    Vf^2=(0m/s)^2 +2(1.88m/s^2)(10m)
    Vf=6.13m/s
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 9, 2011 #2

    Doc Al

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    Staff: Mentor

    Sounds good.
    Figure out the new acceleration of the box, once it starts sliding on the flat ground.
     
  4. Oct 9, 2011 #3
    Figure out the new acceleration of the box, once it starts sliding on the flat ground.[/QUOTE]

    How would you find this? Would you use a=F(net)m? If yes, how would you find the applied force?
     
  5. Oct 9, 2011 #4

    Doc Al

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    Staff: Mentor

    I assume you mean a = F(net)/m. But yes.
    What force acts on the box as it slides across the flat ground?
     
  6. Oct 9, 2011 #5
    Friction?
    So on the flat surface you find the force of friction?
    FFk=μk*FN
    FFk=0.50*20N
    FFk=10N
    How would you find the applied force using this though?
    With this doesn't the formula a=F(net)/m still leave two variables?
     
  7. Oct 10, 2011 #6

    Doc Al

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    Staff: Mentor

    Right.
    There's no 'applied force'. Friction is the only (horizontal) force acting.
    No. You know the mass and the net force.
     
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