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Finding Work with friction

  1. Nov 28, 2011 #1
    1. The problem statement, all variables and given/known data
    There is a 2450 kg car being pushed by a group of students. The car is pushed a distance of 12 m in 8.95 seconds. The coefficient of friction is .450.
    a) What is the net work being done?
    b) Work done by street?
    c) Work done by students?


    2. Relevant equations
    Wnet=Kf-K0
    Wf=-Ff*d
    d=[(v0+vf)t]/2
    KE=[mv^2]/2

    3. The attempt at a solution
    For a, Wnet=Kf-K0, and K0=0.
    To find Kf, I started to find Vf, using the displacement formula
    12=[vf*8.95]/2
    vf=2.68 sec.
    Then I found Kf
    Kf=[2450*2.68^2]/2
    Since K0=0, Wnet=8798 J (same value as Kf+0)

    b asks for work done by street so I assumed that meant Wf.
    At this point, I wasn't sure how to find Ff to use in Wf=-Ff*d formula
    I thought Ff=μ*Fn
    Is Fn=m*g? Or do I multiply m*a, after finding out acceleration?
    Since I found Vf in the last step, do I divide that by time to find acceleration?

    c is work done by students, which is Wpush/applied force.
    Wnet=Wpull-Wf
    Do I add Wf to Wnet to find applied force?
    At this point, I don't really know what I'm doing.
    Can someone give me a few tips on the concepts that should be understood about this problem and explain whatever errors I'm making?
     
  2. jcsd
  3. Nov 28, 2011 #2
    Ff is μ*Fn This will give you the force of friction, and you know the distance over which it was acting. And I am not sure what is better, but I personally would have found the acceleration with s = (at^2)/2
    12 = a(8.95^2)/2
    24 = 80.1025a
    a = .2996m/s^2

    Fnet = ma = .2996*2450 = 734.02
    Wnet = Fnet*s = 734.02*12 = 8808J

    Pretty similar answer
     
    Last edited: Nov 28, 2011
  4. Nov 28, 2011 #3
    Thanks for replying. I did get that acceleration and a very close Wnet (8820 J) when I worked it previously, while using the method you demonstrated.

    For Ff, is Fn=m*a? Or m*g? That's what I'm confused about...
     
    Last edited: Nov 28, 2011
  5. Nov 29, 2011 #4
    Fn is the normal force, which is mg :)
     
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