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Coefficient of Kinetic Friction on an Incline

  1. Feb 15, 2007 #1
    Here is the problem: A block is released from rest on an inclined plane and moves 3.5 m during the next 4.8 s at an angle of 31 degrees. The acceleration of gravity is 9.8 m/s^2. What is the magnitude of the acceleration of the block? what is the coefficient of Kinetic Friction Muk for the incline?
    - I already identified Xinitial = 0, Vxinitial =0, distance = 3.5 m, time = 4.8 s, theta = 31
    - to get the acceleration I used asubx = gsin(31) which gave me 5.05 m/s^2
    But then again there is this formula Xfinal = Xinitial + Vxinitial*t +0.5at^2 which at the end becomes a=2distance/t^2 giving me a value for a=0.304 m/s^2 and this is the point were I get confused :frown: .

    - For the coefficient of Kinetic friction, I used free body diagrams which gave me a formula of Mk=mgsin(theta)/mgcos(theta) then Mk=tan(theta) I substituded Mk=tan(31) and got an answer of Mk=0.601
    Am I on the right track? Any help will be appreciated :)
     
  2. jcsd
  3. Feb 15, 2007 #2

    Pyrrhus

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    Your acceleration is wrong, because you didn't take in account friction. Start from Newton's 2nd Law.
     
  4. Feb 16, 2007 #3
    Basically I have the wrong formulas then... I found out, by re-reading my book that the Fnet = m*a in Newton's Second Law which means that I have to find first the Net Force? I already have the value for mass = 9Kg and the acting forces are the ones who make the object move? Because according to my problem, it starts moving from rest which mean that only the object's mass affected by gravity makes the force I got confused...
     
  5. Feb 16, 2007 #4
    and the kinetic friction value is not given also, and I don't know if the one I got is right Mk=tan(31) Mk=0.601
     
  6. Feb 16, 2007 #5

    Pyrrhus

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    Are you familiar with the concept of Free Body Diagram and Newton's Laws?, you are just pluggin numbers in formulas...
     
  7. Feb 18, 2007 #6
    yeah I have the hang of free body diagrams but I still get confused... I already found the Mk = 0.56 and I'm missing the acceleration of the object. By using the free body diagrams I got Fy= mgsin(theta)+Mkmgcos(theta) =ma then I got for a= gsin(theta)+Mkgcos(theta)
     
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