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Homework Help: Two Boxes, a pulley, and KE

  1. Dec 1, 2006 #1
    1. The problem statement, all variables and given/known data

    Two boxes are set up in a system as shown in the attached picture.

    Box 1 - mass = 5 kg
    Box 2 - mass = 2 kg

    h = 5m

    The system is released, and box 2 hits the ground. With a frictionless system, find the total KE before box 2 hits the ground.

    2. Relevant equations

    I am fairly sure I have to use the conservation of energy, which in this case would be:

    (1/2)*m*vf^2=m*g*h

    ((1/2)*m*vf^2)-(m*g*h)=KE


    3. The attempt at a solution

    I dont quite understand how to go about this problem with the two box system. I can figure out the boxes force and acceleration, but not what I need.

    Thanks,
     

    Attached Files:

  2. jcsd
  3. Dec 1, 2006 #2

    radou

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    Homework Helper

    Unfortunately, the attachment doesn't seem to be approved yet. What exactly does 'h' represent?
     
  4. Dec 1, 2006 #3
    Oh, ok.

    Block 1 is sitting on a (frictionless) table attached to block 2 which is hanging off of the table. Both blocks are attached by a rope and a (frictionless) pully.

    Before the system is released, block 2 is 5m from the ground (h = 5m)

    Thanks for the obervation about the pic...
     
  5. Dec 1, 2006 #4

    radou

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    You were right when you said you have to use energy conservation. Since the block are connected with a rope, and since there is no friction, in what relation are the velocities of the blocks?
     
  6. Dec 1, 2006 #5
    The velocities should be the same. However, Ive calculated the acceleration and the Tension of the blocks in motion. The acceleration and T of block 2 can be illustrated as T-w2=m2(-a)
    And Block 1 can be illustrated as 0+T=m1*a

    The tension turnes out to be 14 N and the acceleration is 2.8 m/s^2

    Im not sure how to use the data...
     
    Last edited: Dec 1, 2006
  7. Dec 1, 2006 #6

    radou

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    Do you need the acceleration and the tension? Think about how the kinetic energy of the system is defined.
     
  8. Dec 1, 2006 #7
    So, KE = mgh + w
    KE = 2*-9.8*5+14
    KE = -84J? How can I reach a negative KE?
     
  9. Dec 1, 2006 #8

    radou

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    What exactly is w?

    Do you agree if we say that the total kinetic energy of the system equals 1/2 M v^2 + 1/2 m v^2 ?
     
  10. Dec 1, 2006 #9
    of course.... I was thinking w was the work done by the tension of the system.
     
  11. Dec 1, 2006 #10

    Doc Al

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    Staff: Mentor

    The tension is an internal force and does no net work on the system as a whole. If you were to analyze one mass by itself, then you would need to include the work done by the tension force.

    Note also: When calculating changes in gravitational PE using [itex]mg\Delta h[/itex], g is just a positive number (g = 9.8 m/s^2, not -9.8 m/s^2). g is just the magnitude of the acceleration due to gravity.
     
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