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Homework Help: 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

    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
    vf=2.68 sec.
    Then I found Kf
    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.
    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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