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Frictional Forces-finding velocity and distance

  1. Dec 2, 2016 #1
    1. The problem statement, all variables and given/known data
    A 0.170 kg hockey puck is initially moving at 21.2 m/s [W] along the ice. The coefficient
    of kinetic friction for the puck and the ice is 0.005.

    (a) What is the speed of the puck after traveling 58.5 m?
    (b) After being played on for a while, the ice becomes rougher and the coefficient of
    kinetic friction increases to 0.047. How far will the puck travel if its initial and final
    speeds are the same as before?

    2. Relevant equations
    v2^2=v1^2+2ad
    Fnet=ma
    Ff=μkFn

    3. The attempt at a solution
    a) FN=9.8(0.170)=1.666
    Ff=0.005(1.6660)=0.00833
    Fnet=ma
    0.00833=0.170a =0.049
    v2^2=v1^2+2ad
    v2^2=(21.2)^2+2(0.049)(58.5)
    v2=21.3, but the answer is 21.1

    b)Ff=0.047(1.666)
    Ff=0.078302
    0.078302=0.170a=0.4606
    v2^2=v1^2+2ad
    I'll use the correct v2
    21.1^2=21.2^2+2(0.4606)d
    -4.23=0.9212d
    d=-4.59 but the answer is 6.24
    Not sure where I'm going wrong?
     
  2. jcsd
  3. Dec 2, 2016 #2
    Does friction acting on the puck produce a positive or negative acceleration?
     
  4. Dec 2, 2016 #3
    Negative, so v2^2=(21.2)^2+2(-0.049)(58.5), v2=21.1
    21.1^2=21.2^2+2(-0.4606)d

    -4.23=-0.9212d
    d=4.59....
    That helped fix my mistake in (a) but how about b?
     
  5. Dec 2, 2016 #4
    For part b, you really should use more significant digits than 21.1. I think a more accurate number was 21.06. That should resolve that problem - I think.
     
  6. Dec 2, 2016 #5
    Oh wow that made a difference than you so much!
     
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