Physics

posted by Lala

A 45kg sled is coasting with constant velocity at 3.0m/s over a patch of smooth and level ice. It enters a very rough section of ice that is 11m long in which the force of friction is 15N. With what speed will the sled emerge from the rough ice?

  1. bobpursley

    vf^2=Vi^2 + 2ad where a= force/mass, and it is negative here (it opposes motion). Solve for vf

  2. Lala

    a = F/m = 15N/45kg = 0.33 m/s^2 ,
    vf^2 = vi^2 + 2ad
    vf^2 = 3^2 + 2(0.33)(11)
    vf^2 = 16.3
    vf = 4m/s
    is that correct?

  3. Kyle

    a=F/m=15N/45kg=0.33 m/s^2
    which in this case is Negative
    vf^2=vi^2+2ad
    vf^2=9+2(-.33)(11)
    vf^2=9+(-7.26)
    vf^2=1.74
    vf=sqrt(1.74)
    vf= 1.32 m/s

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