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Kinetic Friction of shoved box of books

  1. Sep 24, 2007 #1
    1. The problem statement, all variables and given/known data

    A box of books weighing 300 N is shoved across the floor of an apartment by a force of 400 N exerted downward at an angle of 35.2 degrees below the horizontal. If the coefficient of kinetic friction between the box and the floor is .570, how long does it take the box to move 4.00 M, starting from rest.


    2. Relevant equations

    E Fx = 400 cos (35.2) - mK Fn = 300 (a)

    E Fy = Fn - Fmg - 400 sin (35.2)



    3. The attempt at a solution

    400 cos (35.2) -(.570) (300) (9.8) + 400 Sin (30) = 300 A

    A= -3.83 m/s2

    4m = (1/2) (-3.83) (t^2)

    t= 1.44s







    This is what I have so far... I don't know if its correct!?!? Can someone help me out and check it??? Thanks!!!
     
  2. jcsd
  3. Sep 24, 2007 #2

    400 cos (35.2) -(.570) (300) (9.8) + 400 Sin (30) = 300 A

    This equation looks funny to me.. can you elaborate on how you got this?

    [Edit]

    I forgot to mention this, but on these type of problems, it is essential that you draw a free body diagram. Scan and attach one if you drew it.
     
    Last edited: Sep 24, 2007
  4. Sep 24, 2007 #3
    I substituted arranged the E Fy to solve for FN and then substituted that in for the FN in my Efx equation that I have listed up top.

    Thanks for the help! I don't have a scanner to put my sketch on but I have the applied force coming in from the left side from above the block at an angle of 30 degrees below horizontal, the normal force going up from the box and the Fgrav headed down. The last force I ID'd was the F kf going to the left from the box.
     
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