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Does the normal force do work?

  1. Mar 21, 2014 #1
    1. The problem statement, all variables and given/known data
    Consider the mass m1 on a simple incline plane. It passes from point A downwards with the velocity v0 and stops at point B (still on the plane).

    What is the work done by the normal force on mass m during the complete motion?
    2. Relevant equations



    3. The attempt at a solution
    In my book it says zero but I dont think it is because the mass moves diagonally down which shows movement in x and y axis. Since the normal force acting on the mass isnt perpendicular to the y axis movement. It should do work in the opposite direction. What do you think?
     
  2. jcsd
  3. Mar 21, 2014 #2

    Doc Al

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    Work it out. What are the vertical and horizontal components of the normal force and the displacement? Figure out the work done by each component and add together to get the total.

    Your book is correct. Note that the normal force is perpendicular to the displacement down the incline. But you'll get the same answer when you break it into vertical and horizontal components.
     
  4. Mar 21, 2014 #3

    HallsofIvy

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    You seem to be under the impression that there is a "work in the y-direction" and "work in the x-direction". There isn't. Work is a scalar quantity, not a vector.
     
  5. Mar 21, 2014 #4
    Thank you. My misconception was thinking of it as a vectoral quantity. Now I get it.
     
  6. Mar 21, 2014 #5

    Doc Al

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    You can certainly choose to look at the components of a force separately and consider the work done by each to get the total. (Of course you must add them up!)
     
  7. Mar 21, 2014 #6
    While it may not be a vector there can still be a direction associated with work in the same way that kinetic energy is also a scalar quantity but the velocity still has a direction associated with it. Direction in this sense then doesn't necessarily mean a vector but rather that it is a function of the x and y variables.
     
    Last edited: Mar 21, 2014
  8. Mar 21, 2014 #7

    haruspex

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    To put some algebra around that notion, let the (normal) force vector be expressed as ##\vec F = F_x\vec i + F_y \vec j## and the displacement vector by ##\vec r = r_x\vec i + r_y \vec j##. Then the work done is ##\int \vec F.d\vec r = \int F_x.dx + \int F_y.dy##.
    In the present problem, ##\int F_x.dx ## and ## \int F_y.dy## are equal and opposite, so cancel.
     
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