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Homework Help: Work done using a pulley system

  1. Oct 20, 2008 #1
    1. The problem statement, all variables and given/known data

    There is a pulley system, identical to system (ii) in the following image:

    A 1.24 kg canister hangs from one pulley, you are lifting with constant speed, what is the work you do while lifting the canister 1.60 cm and a force 6.08 N?

    2. Relevant equations

    W = F x D (displacement) x Cos theta (where theta is the angle between work and displacement

    3. The attempt at a solution

    I originally just plugged in my values into the formula and yielded -0.09728 J. When I got it wrong I did a little looking around and tried both multiplying and dividing by 2. Here are all my attempts: 0.0486 J, -0.0486 J, -0.09728 J, 0 J, 0.09728 J, 379.75 J, 0.218 J, -0.218 J
  2. jcsd
  3. Oct 20, 2008 #2

    Doc Al

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    Staff: Mentor

    Show what you did. What force and displacement did you use? Note that the applied force is over a different displacement than the load.

    You can also just figure the change in gravitational PE.
  4. Oct 20, 2008 #3
    Here was my attempt:

    Work done by the person = 0.016m x 6.08 N x cos(180) = -0.09728 J.
    Unfortunately, that's incorrect.
    So I figured that it was because the crate was being raised 0.016m (1.6cm) and therefore the amount of displacement in his string must be different. I first assumed it was half the amount, and tried -0.0486 J, when that was incorrect I tried double the amount -0.218 J, but that was also incorrect.
    Now I'm stuck
  5. Oct 20, 2008 #4
    Oh my, I didn't multiply by two correctly, problem solved, but thanks for your help it was as follows:

    Work done by the person = 0.016m x 6.08 N x cos(0) x 2 = 0.19456 J
  6. Oct 20, 2008 #5

    Doc Al

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    Staff: Mentor

    Right. That's incorrect because (1) you used the wrong displacement, and (2) you have the wrong sign. (The applied force acts down, but the displacement of that force is also down--so the angle is wrong.)
    Well, it's either half or twice, I'll tell you that much. See if you can figure it out by staring at the diagram. For example, when the load moves up by 1 m, how much extra rope passes over the left pulley? (Use a piece of string to figure it out.)

    As a check, figure out by how much the PE of the load must change.

    Edit: Oops... Looks like I was too slow... or you were too fast! :approve:
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