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Work Done on Inclined Plane

  1. Jul 1, 2015 #1
    1. The problem statement, all variables and given/known data
    Untitled.png


    2. Relevant equations
    W = F d

    3. The attempt at a solution
    W = F d
    30 = F * 2 * cos 0
    30 = 2 F
    F = 15 N [NOT IN CHOICES] !!!
     
  2. jcsd
  3. Jul 1, 2015 #2
    I figured it out finally,
    I have to consider the mgsin 37

    Is that because he said the total work done?
    If he said the work done by force is 30, then the answer is what I previously posted?
     
  4. Jul 1, 2015 #3
    I'm actually somewhat confused by this, and I feel like I shouldn't be.

    First, consider the work-energy theorem.

    ##W = \Delta PE + \Delta KE = \sum(F\cdot d)##

    Second, I would set up a FBD. There are three forces acting on the block: the pulling force, the force due to gravity, and the normal force. The force due to gravity can be decomposed into two forces: one pulling opposite of the pulling force and one pulling the block perpendicularly into the inclined plane. The block's motion is perpendicular to both the normal force and the perpendicular component of the weight pulling it into the inclined plane. Thus, those forces do no work on the block. So, in the direction parallel to the inclined plane, we have ## 30 = -2 mg sin(37^\circ) + 2F##

    The thing is since they are telling us that the work done is 30 J and they are telling us the mass of the block and that the distance traveled is 2 m, and assuming that the inclined plane is frictionless, I'm coming to the conclusion that the work done doesn't merely increase the potential energy of the block, but also the kinetic energy. That's okay, but usually in these kinds of problems, the work done would equal the rise of potential energy (since we are never told anything, nor asked anything about the speed of the block). Nevertheless, it appears to me that this is what we have. Can anyone else give me a sanity check on this?

    Note that my ##F## in my first equation is a generic symbol for forces (and we have four forces in this problem). The ##F## in my second equation is the pulling force that we are looking for.
     
  5. Jul 1, 2015 #4

    haruspex

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    I'm not sure what sign convention you are using for g. You may be taking it as positive up, making its numerical value negative, or, as is more common, positive down. Either way, that is clearly wrong. Suppose |F| were just a little greater than ##|mg sin(37^\circ)|##. Your expression says hardly any work would be done, yet the block would eventually move 2m.
    We need to bear in mind that we are told nothing about acceleration, or even whether it starts from rest.
    The work done in raising the block that height is less than 12J, so presumably there is some net acceleration, and the total work done includes the increase in KE. That part of the work is ##2(F- mg sin(37^\circ))## (g positive down), and the work overcoming gravity is ##2 mg sin(37^\circ)##, so the total work is still 2F.
    Your original answer is correct, and don't let anyone tell you otherwise.
    As a check, suppose the angle were 0. The force F applied over distance d will still do work Fd, not zero.
     
  6. Jul 2, 2015 #5
    We can talk about the total work done by all forces acting on the block, or we can talk about the work done by the force labeled F in the diagram.

    My expression says that the work done by ##F## would almost cancel with the work done by gravity, leaving the kinetic energy of the block at almost zero.

    The work done by ##F## is still ##2F##. The total work done is the work done by ##F## plus the work done by gravity, which gives us the change of kinetic energy.

    If by "total work done" we do not mean to include the work done by gravity, then the total work is the work done by ##F##, which would give us the change of potential energy plus the change of kinetic energy.

    I interpret "total work done" to include the work done by gravity. Therefore, the total work done by all forces acting on the object is the change of its kinetic energy.

    So, taking ##g = 9.8 \frac{m}{s^2}## I stand by ##30 = 2F - 2mgsin(37^\circ)##. But your point is a good one. To be clear, this is the work done by all forces (including gravity). By saying this though, I'm saying that the kinetic energy of the block has increased by 30 J. 30 J is not the increase of potential energy and kinetic energy. 30 J is the increase of kinetic energy. If we say that 30 J is the increase of potential energy and kinetic energy, then the "total work done" would be the total work done by ##F## alone.

    To be fair, the statement of the problem isn't clear whether we are supposed to interpret "total work done" as total work done by all forces including gravity or total work done by ##F##.

    Of course, the work done by gravity is negative. So if the total work done includes the work done by gravity, then the work done by the force ##F## would be ##2F = 30 + 2mgsin(37^\circ)## (where my sign convention gives ##g## a positive value - that is, my ##g## is the magnitude of the acceleration due to gravity). The work done by ##F## equals the change in kinetic and potential energies.
     
    Last edited: Jul 2, 2015
  7. Jul 2, 2015 #6

    haruspex

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    Fair enough - it comes to the question of whether total work done means total work done by F or total work done on the block. I certainly read it as the former.
    But if we take the other view, the work done by gravity is about -23.6N, yes? That makes F about 27N.
     
  8. Jul 2, 2015 #7
    Agreed. That's what I get.
     
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