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!

Two masses, frictionless pulley

  1. Oct 8, 2011 #1
    1. The problem statement, all variables and given/known data
    Two masses are connected by a light string that goes over a light, frictionless pulley, as shown in the figure. The 8.7-kg mass m1 is released and falls through a vertical distance of h = 1.09 m before hitting the ground. Use conservation of mechanical energy to determine:
    *picture attached*

    a) how fast the 3.9-kg mass m2 is moving just before the 8.7-kg mass hits the ground; and
    b) the maximum height attained by the 3.9-kg mass.



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 8, 2011 #2
    So what exactly are you having trouble with? If you haven't tried to solve the problem or you don't tell us where you're stuck, you won't get much help.

    What are the relevant equations for conservation of energy problems?
     
  4. Oct 8, 2011 #3
    i'm stuck. period.
    i have NO idea where to even start.
     
  5. Oct 9, 2011 #4
    Okay so:
    MEi = MEf
    m1=8.7 hi=1.09 hf = 0
    m2 = 3.9 hi = 0 hf = ?
    KEi + PEi = KEf + PEf
    1/2mv^2 + mgh = 1/2mv^2 + mgh
    (1/2)(m??1 or 2??) V^2 + m?g0 = 1/2 m? 0 + mgh
     
  6. Oct 10, 2011 #5
    Yeah you're on the right path. Just remember that you have two masses, so you need KE and PE for both masses, before and after (8 terms). If you want to write out the complete equation, it's like this:

    (KE of m1 before) + (KE of m2 before) + (PE of m1 before) + (PE of m2 before) = (KE of m1 after) + (KE of m2 after) + (PE of m1 after) + (PE of m2 after)

    (1/2)m1*v1i^2 + (1/2)m2*v2i^2 + m1*g*h1i + m2*g*h2i = (1/2)m1*v1f^2 + (1/2)m2*v2f^2 + m1*g*h1f + m2*g*h2f

    And like you said, h2i = 0, h1f = 0, and you also have v1i = 0 and v2i = 0 (they both start from rest). Moreover, you know that both velocities are the same since they're connected by a string, so v1f = v2f = V (we'll call both final velocites V). Now your equation reduces to:

    m1*g*h1i = (1/2)m1*V^2 + (1/2)m2*V^2 + m2*g*h2f

    Now that's not so bad. The rest is algebra, solving for V.

    Edit: That does it for part (a). For part (b), simply use the V you got from part (a) as an initial upward velocity against gravity (ignore the other mass and string and everything else).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook