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Work Problem I think

  1. Nov 16, 2005 #1
    This problem is giving me issues. I don't think I'm using the right equations....

    Problem:
    A 2.0-kg mass is released from rest at the top of a plane inclined at 20 degrees above horizontal. The coefficient of kinetic friction between the mass and the plane is 0.20. What will be the speed of the mass after sliding 4.0 m along the plane.

    A) 2.2 m/s
    B) 3.0 m/s
    C) 3.5 m/s
    D) 5.2 m/s

    Work:
    Here is what I tried.
    Mass= 2.0kg
    Coefficient of Kenetic Friction= 0.20
    Displacement= 4m
    Weight= 9.8*2=19.6N

    W=F*d*cos 0

    W=19.6*4*cos(20)
    W=73.67

    W sub NET= change in KE
    73.67=1/2*2*v(squared)
    v=8.58m/s

    That is not one of my choices :uhh:

    Any help would be fantastic. Thanks in advance.
    Kirk
     
  2. jcsd
  3. Nov 16, 2005 #2
    you need to find the parallel force to the plane, not just the weight,
    Parallel force - friction force or

    mgsin(teta) - u(normal force)
    mg*sin(theta) - umg*cos(theta)

    well, im new here so i dont know how to neter in symbols like theta, coeficient of friction, ....

    u= coeff of friction
     
  4. Nov 16, 2005 #3
    i hope thats right, becuase i only started work-energy theorum today
     
  5. Nov 16, 2005 #4
    looks like a great start for me. I'm gonna give it a shot.

    Thanks,
    Kirk
     
  6. Nov 16, 2005 #5
    Thanks so much! That got me on the right track. I was able to get the right answer, which was C if you wanted to know.

    Kirk:smile:
     
  7. Nov 16, 2005 #6
    Another method that uses "work = change in KE"... if you draw your freebody diagram, 3 forces act on the block, Fgravity, Fnormal, and Ffriction. The definition of Work is W=Fdcos(theta). Quantify the work done by each force, careful when determining theta for each. Fgx4metersxcos(70)... etc. Your total work done = change in KE... you will get an answer matching your one of the choices.
     
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