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Homework Help: Two mass pulley system

  1. Jul 18, 2015 #1
    1. The problem statement, all variables and given/known data

    https://int.erlace.com/uploads/teacher/1533/images/3modifiedatwood(1).jpg [Broken]
    Ignore friction and the the rope and the pulley and the rope have negligible mass.
    1. When the hanging block gif.latex?M_2.gif has descended a distance gif.latex?%5CDelta%20h.gif , how much work has the gravitational force by the earth done on the system?
    2. That work went into the kinetic energy of the masses. What fraction of that kinetic energy went into the motion of the hanging block?
    3. Suppose now there's a coefficient of kinetic friction gif.latex?%5Cmu_k.gif between gif.latex?M_1.gif and the table surface. Find the speed of the masses when gif.latex?M_2.gif has descended that distance gif.latex?%5CDelta%20h.gif .
    2. Relevant equations


    3. The attempt at a solution

    1. I figured the work done was equal to change in potential energy, so

    gif.latex?W%3Dmg%5CDelta%20h.gif gif.latex?%3D%28m_%7B2%7D+m_%7B1%7D%29gh.gif

    2. I'm not sure how to go about this one. I tried to take what I found for the work done and relate it the kinetic energy of the masses, so something like 7B1%7D%29gh%20%3D%20%5Cfrac%7B1%7D%7B2%7Dm_%7B2%7Dv%5E2+%5Cfrac%7B1%7D%7B2%7Dm_%7B1%7Dv%5E2.gif , but I don't know if this would help, or if it is right at all.

    3. Not sure. I derived an expression for acceleration of the system, and again I don't know if it would help.


    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Jul 18, 2015 #2
    Looks like two pictures didn't post

    For part 1, (m1 + m2)gh

    For part 3, a = (m2g - μm1g)/(m1 + m2)
  4. Jul 18, 2015 #3
    Your solution :

    (1.) is wrong .
    Does gravity do work on both blocks ?

    *Remember - Work done is ∫F.dx

    (2.) - What is acceleration of each block ? So what is velocity after moving a distance Δh ?

    (3.) - Repeat as previous part .

    I hope this helps .
  5. Jul 18, 2015 #4


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    You got that part right.
  6. Jul 18, 2015 #5
    Yes, this helps. Gravity would only do work on m2 so the work done would be equal to m2gh. The acceleration of the blocks are the same, so each would have an an acceleration a = (m2g)/(m1+m2 ), without friction. For the m2, the velocity would be (m2v2)/2 = m2gh, so v = √(2gh). Does this tell me anything in terms of how much kinetic energy went into moving m2, would it be all of it?
  7. Jul 18, 2015 #6
    I think my expression for velocity was wrong, since I didn't account for m1, it would be v=√2h(m2g)/(m1+m2 )
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