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Homework Help: What is The New velocity of the box?

  1. Sep 29, 2010 #1
    1. The problem statement, all variables and given/known data
    A 24 kg box is being pushed across the floor by a constant force ‹ 107, 0, 0 › N. The coefficient of kinetic friction for the table and box is 0.18. At t = 8.0 s the box is at location ‹ 11, 2, −4 › m, traveling with velocity ‹ 5, 0, 0 › m/s. What is its position and velocity at t = 9.7 s?



    2. Relevant equations



    3. The attempt at a solution
    i got <12.11,0,0> and <7.58,0,0> but they were both wrong
     
  2. jcsd
  3. Sep 29, 2010 #2

    rock.freak667

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    Since the force is in the x-direction, you need to get the frictional force which would be Frictional force = <μmg,0,0>, then find the resultant force,F. (and F=ma :wink: )
     
  4. Sep 29, 2010 #3
    yea but i still dont get how to find v final at 9.7s
     
  5. Sep 29, 2010 #4

    rock.freak667

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    Using my suggestion, once you get the resultant force vector and apply F=ma, you can get the vector a.

    Then you can use v(t) = ∫a(t) dt
     
  6. Sep 29, 2010 #5
    can you show me how to get the answer?
     
  7. Sep 29, 2010 #6

    rock.freak667

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    I can't pose the solution for you, but I will start you off.

    Resultant force, ma(t) = Constant force - frictional force
     
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