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Work, mass on an incline?

  1. Oct 15, 2004 #1
    Hi all,

    I need help with question

    I need to calculate the work done in pushing a mass up an incline a distance d with slope theta, when the coefficient of kinetic friction is 0.1 and the force is directed 1) parallel to the slope, 2) the force is directed at an angle phi to the slope?

    I know that when ignoring friction the work done is is given by mgh, where h=d*sin(theta). However I'm getting confused on how to determine the problem with the additional forces.

    can anyboby offer some tips on how i could solve this problem.

    Callisto :confused:
     
  2. jcsd
  3. Oct 15, 2004 #2

    quasar987

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    Gold Member

    The work done is the work done by the total force. And the work done by the total force is the sum of the work done by each force individually.

    You are right about the work being done by the "human" force being mgh but more easier would be to find the work of that force using the definition of work, i.e. that

    [tex]W_{human force}=F_{human}dcos\theta[/tex]

    In case a), [itex]\theta = 0[/itex] because the vector d is parallel to the vector F. and so the work is just Fd.

    Remember that the vector frictionnal force is ANTI-parallel to the vector displacement always. This means that [itex]\theta = 180° = \pi[/itex].

    You do the math for [tex]W_{friction}[/tex].
     
    Last edited: Oct 15, 2004
  4. Oct 16, 2004 #3
    I'm really sorry,but i'm still completely stumped!
    do i need to include the work done by the force of gravity as well?
    is the work done by the frictional force given by uk*mgcos(theta)?

    I really dont understand

    Callisto ;/
     
  5. Oct 16, 2004 #4

    Doc Al

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    In each case, calculate what that applied force F must be to have equilibrium (no acceleration). Then the work done by that force is simply its component along the incline times the displacement d.

    Note: the frictional force is u*N, where N is the normal force. It's different in each case.
     
  6. Oct 16, 2004 #5
    So the applied force F must be mg and it's component along the incline is mg*sin(theta). The normal force is mg*cos(theta).
    so mg*sin(theta)-u*mg*cos(theta) gives the net force.
    am i right??
    still a bit stumped
    Callisto`1
     
  7. Oct 16, 2004 #6

    Pyrrhus

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    Actually if the block is going up the slope gravity (the component) acts against it too
     
  8. Oct 16, 2004 #7
    Don't get it
    This is killing me!!

    ARRRGGGGHH!!!
     
  9. Oct 17, 2004 #8

    Doc Al

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    Do it systematically. Start by identifying all the forces acting on the mass. Here's what I see: weight (acting down), normal force, friction (acting down the incline), and the applied force F.

    In case 1, the applied force F is parallel to the incline. So write the conditions for equilibrium and find out what F must be. The work is then just Fd. Try it.
     
  10. Oct 19, 2004 #9
    Alright, for case 1
    i get F=mg*sin(theta)+u*mgcos(theta)

    Case 2 is difficult but...
    N=mg*cos(theta)+F*sin(phi) since F*sin(phi) is the component of the applied force normal to the incline.
    then the component of the applied force parallel to the incline plane is

    F*cos(phi) =mg*sin(theta)+uN

    using these two equations and solving for F i get

    F=(mg*sin(theta)+ u*mgcos(theta))/(cos(phi))-u*sin(phi))

    My brain hurts!
     
  11. Oct 19, 2004 #10

    Doc Al

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    Looks good.

    I assumed that phi is an angle above the incline, so I would say:
    N=mg*cos(theta) - F*sin(phi)
     
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