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!

Solving for different masses in pulley system with KE

  1. Feb 1, 2015 #1
    1. The problem statement, all variables and given/known data
    So I am stuck on this homework problem. I understand the general direction I have to take, but my algebra and physics aren't good. Here's the problem:
    A simple Atwood's machine uses two masses, m1 and m2. Starting from rest, the speed of the two masses is 10.0 m/s at the end of 6.0 s. At that instant, the kinetic energy of the system is 90 J and each mass has moved a distance of 30.0 m. Determine the values of m1 and m2.


    2. Relevant equations
    Wnet = change in KE
    total KE of the system = 0.5(m1)v^2 + 0.5(m2)v^2
    W = force * distance

    3. The attempt at a solution
    My work so far has not really gotten me anywhere, any tips would be very helpful.

    First, I saw that the KE of the system would be the KE equation but with m 1 and m 2 , like this: KE = 0.5(m1+m2 )v^2, or 90=0.5(m1+m2 )v^2, or 90=0.5(m1+m2 )100

    So then then m1+m2 would be equal to 9/5.

    I'm not sure about this next part I've done: The net work is equal to the change in KE. Since the system starts from rest, the initial KE is 0, and the final KE is 90.

    So I did: W=0.5(m1+m2 )100. Then using the work equation: FΔX=0.5(m1+m2)100, and since ΔX is equal to 30m: 30F=0.5(m1+m2)100

    I tried substituting in the acceleration in order to get the force of the masses, 10/6=5/3m/s/s, so weight 1 = m1*5/3, and weight 2 = m2*5/3.

    With these, I know there is supposed to be 2 equations for 2 unknowns, but I can't seem to figure them out. I'm kind of lost at his point.

    I've uploaded a picture of the problem. I believe I'm headed in the right direction. Any tips?
     

    Attached Files:

    • q11.jpg
      q11.jpg
      File size:
      28.8 KB
      Views:
      125
  2. jcsd
  3. Feb 1, 2015 #2

    Nathanael

    User Avatar
    Homework Helper

    And what is F?
    edit: You actually don't need to worry about F; what is the kinetic energy?

    This is not right. The weight 1 = m1g and the weight 2 = m2g
     
  4. Feb 1, 2015 #3
    The total kinetic energy is 90J. So would it be 90=0.5(m1+m2)v^2, which solving for m1+m2 would get me 9/5kg, like I had before. I'm failing to see the connection between this and the force.
    I think the net force on m1 should be T-m1g, where T is the tension. And the net force on m2 should be T-m2g? And would T be equal to the sum of the weights of the blocks?
     
  5. Feb 1, 2015 #4

    Nathanael

    User Avatar
    Homework Helper

    Good. Do you know a relationship between the net force and acceleration?
     
  6. Feb 1, 2015 #5
    Ok, so then by F=ma, the net force for object 1 would be T-m1g=m1a, and the net force for object 2 would be T-m2g = m2a.
    Then since a is the same for each object, (T-m1g)/m1 = (T-m2g)/m2?
     
  7. Feb 1, 2015 #6

    Nathanael

    User Avatar
    Homework Helper

    Right, but there's no need to eliminate the acceleration a
    Use the two equations to eliminate the tension T
    Then you will you have your second relationship between m1 and m2
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Solving for different masses in pulley system with KE
  1. Two mass pulley system (Replies: 5)

Loading...