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Homework Help: A problem on work and energy

  1. Nov 6, 2009 #1
    1. The problem statement, all variables and given/known data
    A light inelastic string passes round two small smooth pulleys A and B in the same horizontal line at a distance 2a apart, and carries masses m1 at each end and a mass m2(<2m1) at its middle point. The system is released from rest with m2 at the middle point of AB

    By applying the principle of conservation of mechanical energy, show that the system comes to instantaneous at rest when m2 has fallen a distance, d = (4am1m2)/(4m1^2-m2^2)
    There is an original version of the question attached, check it if you cant understand what i am tring to express

    2. Relevant equations
    by consevation of energy, initial potential energy=final potential energy

    3. The attempt at a solution
    what i am thinking is that the distance fallen, ie. PE loss of m2 is gain and of PE of 2m1
    and since both at the initial and final postion , the system is at rest so the
    initial KE = Final KE = 0
    But i just cant come up with the expression given by the question

    Attached Files:

  2. jcsd
  3. Nov 6, 2009 #2


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    I don't think you can solve this problem just using conservation of energy. At the resting position the string tensions need to balance the forces of gravity. You need to use that.
  4. Nov 7, 2009 #3
    thx for replying,

    i am very close to the answer, but i find that there is something wrong, i dont know if it is because of my calculation or my concept

    let the distance fallen of m2 be d
    and the distance raised of m1 be s

    by conservation of energy

    m2gd = 2m1gs---------1

    by newton's second law

    2m1sinA = m2----------2

    d/(a+s) = sinA---------3

    after i solve these three question,

    i found that d = 2am1m2/(4m1^2-m2^2) (this ans times 2 = the answer required)

    is it something wrong with my equations or my calculation?
  5. Nov 7, 2009 #4


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    Oh, I see what the question is. m2 will fall and potential energy will be converted to kinetic energy, it will stop falling when the potential energy reaches the same value it started with (hence no more kinetic energy). But at this point it's not in equilibrium so it will rebound, but you don't care about that, you just want to find out where it stops. So i) the forces aren't balanced there, so your newton's law relation is incorrect ii) PE isn't conserved, it's constantly changing. Your 'conservation of PE' relation is also incorrect, the relation between a and s is fixed by geometry. Write the total PE as 2*m1*g*s-m2*g*d. When d=0 the PE=0. Now what you want to do is find another value of d where PE=0. That will be the stopping point. So you will need to solve for s in terms of d.
  6. Nov 7, 2009 #5
    i am quite confuse here,

    1.Isnt that your "conservation of PE" equation is exactly like mine?

    m2gd = 2m1gs 2*m1*g*s-m2*g*d=0

    2.how can i know that it will rebound, and what is providing for the upward force?
    Last edited: Nov 7, 2009
  7. Nov 7, 2009 #6


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    Yes, it is. And that is the equation to solve. Sorry to confuse you. I only meant to point out that 'conservation' of PE is the wrong concept. It's not conserved. The system will oscillate around the point where the forces balance between the initial point where PE is equal to 0 and another point where PE equals 0. In between, PE is not equal to 0. I think you are trying to solve the correct equation. Can you show how you got the wrong answer?
  8. Nov 7, 2009 #7



    sinA= m2/(2*m1)----------2

    d/(a+s) = sinA------------3

    d=a*sinA + s*sinA--------4

    sub 1 and 2 into 4


    d= (m2a/2m1)+m2^2*d/4m1^2


    d= 2am1m2/(4m1^2-m2^2)

    sry the symbol and numbers may be confusing
  9. Nov 7, 2009 #8


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    I think your force balance equation involving sin(A) is wrong. Because, like I said, when the weights become stationary, the forces aren't balanced. Just solve m2*d=2*m1*s. I may have sent you down this track by misunderstanding the problem initially. Can you find an equation for s in terms of a and d?
    Last edited: Nov 7, 2009
  10. Nov 8, 2009 #9
    i cant find any equation relating s , a , d except the geometry one
    Last edited: Nov 8, 2009
  11. Nov 8, 2009 #10


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    Use the geometry one. (s+a)^2=a^2+d^2. Solve for s.
  12. Nov 8, 2009 #11
    i am trying , but one question

    if the "newtons law" is wrong, then i can never solve sinA

    that means i hv 2 equations only, equation 1 and (s+a)^2=a^2+d^2

    but i have 3 unknowns s, a, d is it possible to solve them?
  13. Nov 8, 2009 #12
    oh, sry my mistake "a" is given
  14. Nov 8, 2009 #13
    problem solved, thx a lot
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