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Work done by friction force problem

  1. Apr 9, 2007 #1

    r16

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    1. The problem statement, all variables and given/known data
    A 50 kg trunk is pulled 6.0 meters up a 30 degree incline at a constant velocity. The coefficient of kenetic friction is .2. What is a) the work done by the applied force, b) the work done by gravity and c) the work done by the frictional force

    I set up my x axis in the direction of the incline and my y axis normal to the incline.

    [itex]F_N[/itex] normal force
    [itex]F_g[/itex] gravity
    [itex]F_A[/itex] applied force
    [itex]F_f[/itex] frictional force

    2. Relevant equations
    [tex]\sum F = 0[/tex]
    [tex]\sum F_x = 0 = F_A - F_f - F_{gx}[/tex]
    [tex]\sum F_y = 0 = F_N - F_{gy}[/tex]
    [tex] F_N = F_{gy} = F_g \cos 30 = (9.8)(50)(\cos 30) = 420N [/tex]
    [tex] F_f = \mu_k F_N = (420)(.2) = 85N[/tex]
    [tex] F_A = F_f + F_{gy} = \mu_k F_N + \sin 30 F_g = 85 + (.5)(9.8)(50) = 330N [/tex]

    [tex]W_f = f_k \cdot \delta d[/tex]
    [tex]W_g = mgh = (9.8)(50)(6 \sin 30) = 1500J[/tex]
    [tex]W_f = F \cdot d = (85)(6.0)(\cos 0) = 510J[/tex]
    [tex]W_a = F \cdot d = (330)(6.0)(\cos 0) = 2000J[/tex]

    However, the answers in the book says that a) 2200J b) 1500J c) 700
    I can't seem to find my error. Any suggestions?


    3. The attempt at a solution
     
    Last edited: Apr 9, 2007
  2. jcsd
  3. Apr 9, 2007 #2
    I must be missing something as well or book is in error. Maybe others will see something wrong. By the way, very nice presentation of problem and work.
     
  4. Apr 9, 2007 #3
    The work done by the applied force is mgh + W(friction).
    W(friction)=mg(6cos30)(cos30). (The work done due to frictional force is mgx where x is the horizontal distance moved.
     
  5. Apr 9, 2007 #4
    chaos,
    For the frictional force, I dont follow. You say one thing and express another, ie from your verbal comments, the eqn should be mg*mu*cos(30)*6. But you have two cosines in the math expression. I agree with the worded part. In any event the frictional work just got smaller if we multiply again by
    1/2sqrt(3). The book result is bigger than the OP has calculated.
     
  6. Apr 9, 2007 #5

    r16

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    I believe that this is incorrect because it assumes that the [itex]F_f[/itex] is conservative, however it is non-conservative. I believe what confused chaos is the use of [itex]x[/itex] in the definition for [itex]W_f[/itex] where they really meant [itex]s[/itex]. If we change the definition of the work done by a non-conservative force to a more generic:

    [tex] W = \int_S F \cdot ds [/tex]

    Where work is a line integral of the dot product of the [itex]F_f[/itex] and [itex]ds[/itex] over the curve. If we view the hypotenuse as a curve (which it is, albeit a simplistic one) the work done cannot be determined by the dispacement ([itex](6 \cos 30)(\cos 30) = [/itex]displacement vector) but you need to know the path it took to get to there.
     
    Last edited: Apr 9, 2007
  7. Apr 9, 2007 #6
    totally agree, its a line integral, one could devise a bunch of humps and valleys where the net horizontal displacement had little to do with the amt of f frictional work. Inthe extreme case, one could go nowhere and burn a lot of energy.
     
  8. Apr 9, 2007 #7
    Your work looks correct to me the in your first post.

    I dont buy this comment:

    Work due to friction is the force dotted with the displacement. In this case, its 6m up the ramp.

    Good luck, hooah.
     
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