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Work: Pulling a crate attached to a rope

  1. Oct 22, 2008 #1
    1. The problem statement, all variables and given/known data

    Given:
    A 306 kg crate hangs from the end of a 10.3 m long rope. You pull horizontally with a varying force to move it a distance d = 5.7 m to the right. The magnitude of the applied force, F, when the crate is at rest in its final position is 1992.4 N.

    Question:
    What is the work you do on the crate?

    Associated Diagram:
    [​IMG][​IMG]

    2. Relevant equations

    Work = Force x Displacement x Cos (angle between force and displacement)
    Pythagorean Theorem = a^2 + b^2 = c^2


    3. The attempt at a solution

    To attempt to solve this problem, I assumed 1992.4 N is the force I will be using. I then assumed the displacement would be a right-angled triangle with 5.7 m as the width and the difference between initial height (length of the rope 10.3 m) and final height (which I calculated using Pythagorean theorem to be 8.58m). This right-angled triangle’s hypotenuse, 5.954 m, would be used as my displacement. I then calculated the angle between displacement and force and found that to be 16.8 degrees. Since the motion was pulling and given the displacement, the work done will be a positive value, which I found to be 11356 J or 1.14E+04 J, but this is incorrect. I cannot seem to find a flaw in my reasoning, can you spot anything I’ve forgotten?

    I then tried a different approach, calculated the work done by gravitational potential energy to get the work done by the crate. I took this work and assumed it must be equal to the work done by the tension in the y-axis of the string. I used the angle I found with my length dimensions to calculate work done by tension in the x-axis of the string. I assumed it must be equal to the work done by the person. With this I got 3428.3 J, but it was incorrect.
     
    Last edited: Oct 22, 2008
  2. jcsd
  3. Oct 22, 2008 #2

    LowlyPion

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    Homework Helper

    That's good. You now can figure the difference in height.

    I would observe at this point though that all you need now to determine the Work is what the weight is of your object, because the change in Potential Energy will be your Work and that can be found neatly by m*g*h, where h is your change in height.
     
  4. Oct 22, 2008 #3

    Doc Al

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    Staff: Mentor

    The force varies as the crate moves, so you can't just assume this final value throughout.
     
  5. Oct 22, 2008 #4
    After I calculate the work done by the crate (change in gravitational energy)
    which is -5.16E+03 J
    What is the next step.
    I assume this is equal to my work done by tension in the string.
    I multiplied this value by the tan 33.6 to get work done in the x direction, and calculated
    3,428.3 J.
    I tried this value but it was incorrect. Is it supposed to be negative, or did I do something wrong?
     
  6. Oct 22, 2008 #5

    LowlyPion

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    There is no next step. Work done = Change in Potential Energy.

    Draw a force diagram. If you know the angle of the string and how much force it takes to hold it there ... then how much does the box weigh at that point? When you figure out its weight, then weight*change in height is work done.
     
  7. Oct 22, 2008 #6
    but, the " Work done = Change in Potential Energy "
    is the work done by the crate. We're looking for the work done by the person.
     
  8. Oct 22, 2008 #7

    LowlyPion

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    Is the crate at a higher potential energy state?

    If so how did it get there?

    I don't think the crate has done any work. Its increase in potential energy is from the work of the person.
     
  9. Oct 22, 2008 #8

    Doc Al

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    Staff: Mentor

    Since the crate rises, the change in PE should be positive.
    There is no next step. You're done.
    Since the tension in the string is always perpendicular to the motion of the crate, it does no work.
    As LowlyPion explained, the work done by the person will equal the change in PE.

    As an exercise, if you've done a little calculus, calculate the work done directly by figuring out the force at every point as the crate is pulled. Since the force is horizontal, you'll need to take the horizontal component of the displacement at each point. (Essentially, you'd be calculating W = ∫F*dx.) That's one way to verify your answer.
     
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