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Work energy theorem involving a pulley

  1. Oct 21, 2009 #1
    So, yeah, my midterm is on friday, and i can't even figure out how to do questions like these because i just don't understand the way my prof teaches and the textbook has nothing relevant.

    So far, i've used work energy theorem to find the work done to stop the block. then i wrote out a net work equation stating the Wnet=Wfriction+Wtension, and from there went on to solve u, where Wnet and Wtension were -, and Wfriction was positive. my final answer was u=0.72. i am not sure if this is the right steps and answer.

    once this image is approved, can someone please just give me a basic step by step breakdown that doesn't complicate things too much? thank you.
     

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    Last edited: Oct 21, 2009
  2. jcsd
  3. Oct 22, 2009 #2

    Doc Al

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    The work done by friction will be negative, since it acts opposite to the direction of the block's motion.
    Show exactly what you did.
     
  4. Oct 22, 2009 #3
    My approach:

    Use Work-Energy Theorem,
    Loss in K.E. = Work done against friction

    1/2 (m1 + m2) v^2 - 0 = (mu)Nx

    where (mu) is the coefficient of kinetic friction;
    N is the Normal reaction by the table to the 8 kg block.. and N = 8g
    and x is the distance travelled by the block before it comes to rest.
     
    Last edited: Oct 22, 2009
  5. Oct 22, 2009 #4

    Doc Al

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    (1) This is incorrect.
    (2) Please let the OP solve the problem.
     
  6. Oct 22, 2009 #5
    35ak7jn.jpg
    That's my work. I don't feel familiar enough with the math text on this site yet, so scanned image will do.

    so, that's my work, but i'm not sure if it's right, because my friend got 0.072=mu
     
  7. Oct 22, 2009 #6

    Doc Al

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    Comments:
    (1) Your calculation of the net work is correct.
    (2) When you set that net work equal to various individual work terms, those terms are incorrect:
    (2a) The sign of the work done by friction is incorrect--it should be negative.
    (2b) What you call Wt8 and Wg6 end up being force terms, not work terms. And they have the wrong sign.

    Hint: Only work done by external forces count. (The tension is an internal force to the system; the work it does on the 8kg block is canceled by the work it does on the 6kg block.)

    Note that the tension does not equal mg = (6)(9.8).
     
  8. Oct 22, 2009 #7
    Sorry, but I'm not sure if i understand the whole tension thing. What I understand is when the 8kg block pulls left on the rope and the 6kg block pulls down on the rope the tension is the same throughout the rope, and therefore as it pulls both blocks it cancels out? I just don't see how that works considering they're different masses.

    Okay, so tension is not involved. Also, Work done by gravity is no longer involved because it's a "force term". so I'll have Work(net)=-Work(friction) ?

    May you please explain the difference between a work term and a force term to me?
     
  9. Oct 22, 2009 #8

    Doc Al

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    The fact that the rope and pulley are massless and that the pulley is frictionless allows you to say that the tension is the same throughout the rope. That tension does positive work on the 8 kg mass and negative work on the 6 kg mass. So it cancels.
    Right.
    No, work done by gravity is definitely involved.
    When I said that your terms were "force terms" I meant that you were mixing up force (like mg) with work (like mgd). Those are different things with different units--you can't add work and force terms. If you have an equation for work, every term must be a work term.

    Make sense?
     
  10. Oct 22, 2009 #9
    Oh! Okay I gotcha now. I just didn't include the distance that gravity was acting for on the 6kg mass. I get it now, really silly mistake.

    So, Work(net)=-Work(friction)+Work(gravity)
    <=> -5.67=-u(8)(-9.81)(2)+(6)(-9.81)(2)

    and solve from there?

    Also, could you give me another example of an internal force, not in the system perhaps, but in another general situation?
     
  11. Oct 23, 2009 #10

    Doc Al

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    Much better... almost correct. You are still making sign errors:
    (1) Your net work is negative, since the blocks lose KE. You have that correct.
    (2) Since friction opposes the motion of the 8 kg block, the work it does is negative: -μmgd.
    (3) Since gravity acts in the same direction as the 6 kg block moves, the work it does is positive: +mgd.

    Whenever a system is composed of multiple parts (such as A and B), the force they exert on each other is internal to the system. Of course, how you define your system depends on what you are trying to find. Example: Say you have two bricks, A and B, sitting on top of each other. The normal force between them--the forces that A and B exert on each other--are internal forces to the system composed of both bricks. Depending on the problem, sometimes you care about that force, sometimes you don't.

    If the concept confuses you, forget about it for now--you don't need it. In this problem, your system is the two blocks, so just consider the work done by all forces. In addition to the forces you already considered, you'll have the tension in the connecting rope doing work. As explained before, the work done on one mass by the rope tension exactly cancels the work done on the other mass. (Td - Td = 0) You don't even need to figure out what that tension is.
     
    Last edited: Oct 23, 2009
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