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Homework Help: Plane facts, I

  1. Nov 23, 2007 #1
    [SOLVED] Plane facts, I

    Is this right?

    A ramp has an incline angle of 25 degress. How much work must you do if you are to drag a 50 kg mass a distance of 100m upward along the ramp? Assume that the coefficient of kinetic friction is uk = .35 and that you apply the force parallel to the plane.

    uk mg cos(25) = (.35)(50)(9.8)cos(25) = 155N
    Wfk = (155)(100)cos(180) = -15.5*10^3J

    W = (100)(100)(1) = 1*10^4J

    Wtot = 1*10^4 + 15.5*10^3 = 25.5*10^3J
  2. jcsd
  3. Nov 23, 2007 #2


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    Not sure what the second equation (100)(100)(1) means
    but you should have (friction force * linear distance) + (weight * vertical displacement)
  4. Nov 23, 2007 #3

    Assuming the coordinate system is established with the x axis parallel with the ramp you would use the following:

    W = F*d where

    F = Fgravityx + Ff where

    Fgravityx = mgsin(25)

    Ff = u*N and

    u = coefficient of friction
    N = mgcos(25)

    W = d(Fgravityx + Ff)

    Hope this helps
  5. Nov 23, 2007 #4
    Thank you for replying me.

    The second statement is (Work = friction force * distance * cos 25 degrees)

    When I get the answer, I found the total work done by adding the work done by friction and the work done by the force to get the total work done.

    Is that right?
  6. Nov 23, 2007 #5


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    Work is always force * distance.
    The easiest way is to consider both the force to overcome friction * the 100m distance
    Plus the vertical force (weight) * the vertical displacement ( 100m * sin(25) )
  7. Nov 23, 2007 #6
    After a few rotations, the dragon has the same radius of rotation, but a shorter tail (period). Explain what effect this would have on the horizontal force acting on Jam:
  8. Nov 23, 2007 #7
    Thank you all so much.
    Correct me if I am wrong.

    Isn't Wgravity = mgh
    and h = d sin(25)?

    That should make Wnet = Wgravity + Wf + Wfk
    Which should give you negative for Wnet, but work cannot be negative?
  9. Nov 23, 2007 #8
    The work is due to the Force of gravity plus the Force of friction. Not sure what your third force is in your equation. Your equation is correct for the force due to gravity.
  10. Nov 23, 2007 #9
    Thank you for replying back so quick.
    I am a little confused, the question asked to assume that you apply the force parallel to the plane. So I made the constant force applied equal to the distance (100N).
    So, that third force was the work done by the force applied. Am I reading the question right, or should I remove the third force?
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