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Work-Energy Principle: Pulley System

  1. Jul 22, 2014 #1
    1. The problem statement, all variables and given/known data
    (see attatchment)

    We know masses M and m, and I am trying to describe the downward speed of m in terms of the hight from the ground. There is no friction

    2. Relevant equations

    [itex] \sum \text{F} = ma [/itex]
    [itex] W_{\text{total}} = \Delta K = \int \text{F} \cdot d\text{s} [/itex]

    3. The attempt at a solution

    I first need to note that the distance moved by [itex] M [/itex] is half of that moved by [itex] m[/itex] and thus the acceleration of [itex] M [/itex] is half that of [itex] m [/itex], or [itex] a_M = a_m/2 [/itex].

    I first solved the problem using only newtons laws, so from the FBD I solved for the acceleration of [itex] m [/itex]:

    $$a_m = v\frac{dv}{dh} = \frac{mg}{M/4 + m} \implies v = \sqrt{\frac{2mgh}{M/4 + m}}$$

    Now if I use the Work-Energy principle, I get that [itex] W_{\text{total}} = W_{\text{tension}_x} + W_{\text{tension}_y} + W_g [/itex].

    This is where I think I am not understanding something:

    Work done by tension on [itex]M[/itex] is equal to the work done by tension on [itex] m [/itex] even though force of tension on [itex] M [/itex] is twice that of the force of tension on [itex] m [/itex] because it travels half the distance, and since tension acting on [itex] m [/itex] is acting opposite the direction of motion, it is negative, so [itex] W_{\text{tension}_x} = -W_{\text{tension}_y} [/itex], so

    $$W_{\text{total}} = W_g = mgh = \frac{1}{2}(m + M)v^2 = \Delta K $$

    [itex] v = \sqrt{\frac{2mgh}{m+M}}[/itex]

    I am not sure what I am missing that gives me [itex]m+M[/itex] instead of [itex] m + M/4 [/itex] in the denominator. I would appreciate some help!


    [EDIT] The answer in the back of the book is the same as the answer I got using Newtons laws

    Attached Files:

    Last edited: Jul 22, 2014
  2. jcsd
  3. Jul 22, 2014 #2
    The masses are moving with different speeds .
    Last edited: Jul 22, 2014
  4. Jul 22, 2014 #3
    Hah, wow. Total brain fart there. Thank you very much!
  5. Jul 22, 2014 #4
    Even the answer you are getting by force approach is incorrect .
  6. Jul 22, 2014 #5
    How is that? It's the same as in the book as well
  7. Jul 22, 2014 #6
    Sorry ... its correct .
  8. Jul 22, 2014 #7
    Thanks for your help.
  9. Jul 22, 2014 #8
    You are welcome :)
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