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KE problem help

  1. Dec 12, 2004 #1
    Im having trouble with this problem. I know that KE=1/2mv^2 and the Wrok engergy theorm W=KFf-KEo but im having trouble relating the figure to KE.

    Two boxes are connected to eachother as show. The system is released from rest and the 1.00 kg box falls 1.00 m. The surface of the table is frictionless. What is the KE of the box before it reaches the floor?

    Diagram of Problem

    I was looking itover and I think I kind of know what to do...

    I need to find KEf.


    and to find W, W=(F cos 180) s

    is that on the right track?
    Last edited: Dec 12, 2004
  2. jcsd
  3. Dec 12, 2004 #2
    yuuup...sounds right!
  4. Dec 12, 2004 #3
    I got the first part down but now im having trouble with the basic stuff. Im failing the class right now b/c I know the concepts and the forumulas I just dont know how to work them in problems :frown:

    With the W=(F cos 180) 1m How do I find the F? I know F=ma but I dont have the a.

    For the Work Engery Theorem, where do I find the vo for 1/2mvo^2
  5. Dec 12, 2004 #4


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    Homework Helper

    Edit: Forgot about the mass of the other block :smile:

    I got this with conservation of mechanical energy

    [tex] m_{b}gh = \frac{1}{2}(m_{a} + m_{b})v^2 [/tex]

    If i use Work-Kinetic Energy Principle:

    [tex] W_{gravity} = \Delta K [/tex]

    Conservative forces work is equal to the change in potential energy caused by them in the system

    [tex] W_{gravity} = -\Delta \Omega [/tex]

    [tex] -\Delta \Omega = \Delta K [/tex]

    [tex] 0 = \Delta K + \Delta \Omega [/tex]

    which gives:

    [tex] 0 = \frac{1}{2}m_{a}v^2 + \frac{1}{2}m_{b}v^2 - 0 + 0 - m_{b}gh [/tex]

    [tex] m_{b}gh = \frac{1}{2}(m_{a} + m_{b})v^2 [/tex]

    [tex] \sqrt{2 \frac{m_{b}gh}{m_{a} + m_{b}}} = v [/tex]

    Alternatively, you could apply Newton's 2nd Law

    [tex] \sum_{i=1}^{n} \vec{F}_{i} = m \vec{a} [/tex]

    Analysing our B box it gives

    [tex] T - m_{b}g = -m_{b}a [/tex]

    [tex] m_{b}g - T = m_{b}a [/tex]

    Analysing our A box it gives

    [tex] T = m_{a}a [/tex]

    Adding both equations

    [tex] m_{b}g = a(m_{b} + m_{a}) [/tex]

    [tex] \frac{m_{b}g}{m_{b} + m_{a}} = a [/tex]

    Now using kinematics:

    [tex] v^2 = v_{o}^2 + 2a \Delta x [/tex]

    [tex] v^2 = 0 + 2 \frac{m_{b}g}{m_{b} + m_{a}} \Delta x [/tex]

    [tex] \Delta x = h [/tex]

    [tex] v^2 = 2 \frac{m_{b}gh}{m_{b} + m_{a}} [/tex]

    [tex] v = \sqrt{2 \frac{m_{b}gh}{m_{b} + m_{a}}} [/tex]
    Last edited: Dec 12, 2004
  6. Dec 12, 2004 #5
    I pluged everything in and found the KEf=-0.01
    I think its supposed to be 0, I just rounded wrong. But that can't be right, it has to have some KE before it hits the ground.
    and thanks for the help everyone :smile:
  7. Dec 12, 2004 #6


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    Homework Helper

    Why do you say kinetic energy is supposed to be 0??, if the final kinetic energy is 0 then the objet didn't move at all, because it's initial kinetic energy was 0. The gravitational potential energy turned into kinetic energy making the object accelerate and cover a displacement that has a magnitude of 1 meter.

    If you want to find the kinetic energy of the B box, just find the speed (magnitude of velocity) in any of the above equations, and then just use

    [tex] K_{b} = \frac{1}{2}m_{b}v^2 [/tex]
    Last edited: Dec 12, 2004
  8. Dec 12, 2004 #7
    I didnt mean for the KE=0 as in the correct answer, just that with my calculations it came out to -.01 b/c I rounded wrong.

    Thanks for the help.
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