1. Limited time only! Sign up for a free 30min personal 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!

Pulleys and Conservation of Energy

  1. Feb 9, 2010 #1
    1. In the system shown in the figure, the pulleys on the left and right are fixed, but the pulley in the center can move left or right. The two hanging masses are identical, and the pulleys and ropes are all massless. Find the upward acceleration of the mass on the left, in terms of g.



    2. I have to solve this problem using only Conservation of Energy. The following equations are relevant: K = (1/2)mv^2 and U = mgy
    25umnaf.jpg



    3. I began by setting up a general Conservation of Energy equation.

    I'm calling the left mass m1 and the right mass m2
    m1 = m2 = m

    ΔU1 + ΔU2 + ΔK2 = ΔK1
    mgΔy1 + mgΔy2 + (1/2)mv2^2 = (1/2)m1^2

    (( v^2 = 2aΔy)) Substitute kinematic equation for velocity

    gΔy1 + gΔy2 + aΔy2 = aΔy1

    (( Δy1 = -2Δy2)) Substitute Δy1 in terms of Δy2

    -2gΔy2 + gΔy2 + aΔy2 = -2aΔy2
    -2g + g = -2a - a
    -g = -3a
    (1/3)g = a

    This however, doesn't seem to be correct. Any hints on where I may have gone wrong? Thanks!
     
  2. jcsd
  3. Feb 9, 2010 #2

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Looking at this quickly, a couple of points:
    (1) the acceleration of the masses, and hence their speeds, are not the same; one of them accelerates at twice the rate of the other.
    (2) Your conservation of energy equation is wrong. It's delta K_total plus delta U_total =0. I'm not sure why you have one of the delta K's on the other side of the equation.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook