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Atwoods Machine - Mass Height Question

  1. Sep 18, 2003 #1
    The problem reads as follows:

    Two masses are each initially 1.80 m above the ground, and the massless frictionless pulley is 4.8m above the ground. What maximum height does the lighter object reach after the system is released?

    It gave me a hint to find the acceleration and velocity at the moment the heavier one hit the ground.

    To find 'a' I used ((M1-M2)/(M1+M2))*9.8m/s^2 and got the answer 1.815m/s^2.

    For velocity, I used the equation V=Vo+at, where time was determined by the distance to the ground, 1.8m divided by acceleration 1.815m/s^2. t=1.8/1.815m/s^2 and I got t=0.99s. Question here, since the a is squared, I would have to square my t, right?

    Now my real question is what does it want for maximum height? The distance to the pulley is 4.8m...and the heavier object is only going to travel 1.8m downward until it reaches the ground. So thinking like a lamen, maximum height upward should only be 3m....but it's more complicated than that, isn't it? Why and where do I need to use my determined acceleration and velocity?
  2. jcsd
  3. Sep 18, 2003 #2


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    Once you have found the acceleration of the more massive object toward the ground, you can find the time it takes to hit the ground and, of course, it's speed at that instant.

    That speed is the same as the upward speed of the lighter object (at that time). The point is that it doesn't just stop instantaneously! (The heavier object does because it hit the ground!)
    Treat it as an "object thrown upward problem". You know it's height (it will have moved upward from it's initial position exactly as much as the heavier object moved downward: 1.8 meters), it's speed is the speed you calculated for the heavier object as it hit the ground, AND ITS ACCELERATION IS -9.8 m/s^2 now. Use that to find an expression for the height of the object and use THAT to determine the highest point.
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