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Forces and Motions Question

  1. Aug 12, 2013 #1
    1. The problem statement, all variables and given/known data
    In a tractor-pull competition, a tractor
    applies a force of 1.3 kN to the sled, which
    has mass 1.1 × 104 kg. At that point, the coefficient
    of kinetic friction between the sled
    and the ground has increased to 0.80. What
    is the acceleration of the sled? Explain the
    significance of the sign of the acceleration.


    f=1300n
    m = 1.1 x 10^4 kg
    m =0.80
    a=?
    2. Relevant equations

    F=ma

    m=Ff/Fn

    Fnet= Fx+Fy

    3. The attempt at a solution


    ff=0.80 x (9.8)(1.1x10^4)
    ff= 86240


    86240=(1.1 x 10^4kg) a

    a= 7.85m/s2
     
  2. jcsd
  3. Aug 12, 2013 #2
    There are two forces acting on the sled here. One is by the tractor, the other is due to friction. By Newton's second law of motion, the NET force is equal to the product of mass and acceleration.
    Fnet = ma.

    You've considered only ff in calculating the acceleration.

    Also, make sure if you've taken the weight correctly. An 11 ton sled seems like a lot! :eek:
     
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