1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

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

    [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!

    Thanks

    [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 :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted