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Incline problem

  1. Feb 4, 2007 #1
    1. The problem statement, all variables and given/known data

    Two objects are connected by a light string that passes over a frictionless pulley, as in Figure P5.26. The incline is frictionless, m1 = 2.00 kg, m2 = 6.00 kg, and = 50.0°.

    (?) Find the speed of each object 2.00 s after being released from rest.

    2. Relevant equations

    I already found the acceleration to be 3.18 m/s2 using the equation:
    a=(m2)(g)(sin50.0)-(m1)(g) / m1 + m2

    I already found the tension in the string to be 25.9 using the equation:
    T=(m1)(m2)(g)(sin50.0+1) / (m1+m2)

    3. The attempt at a solution

    I cannot figure out how to find the speed. I'm not even sure what equation to use for this type of problem. I was thinking S=mass/acceleration. But that equation doesn't take into account the friction.
  2. jcsd
  3. Feb 4, 2007 #2


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    "But that equation doesn't take into account the friction."

    It's a frictionless pulley and a frictionless incline, so you don't need to worry about this.

    Since the acceleration is constant, you can just use basic kinematics to get the speed. They've given you the time interval, and you know the masses were released from rest. Just find an equation that links those things.
  4. Feb 5, 2007 #3
    I'm so lost as to how to relate all those things. Do you know of an example...??
  5. Feb 5, 2007 #4


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    Don't think of it as a mass-pulley problem anymore. Just think of it as:

    A block initially at rest undergoes a constant acceleration of 3.18 m/s^2 for 2 seconds, what is the speed at this time?

    Think of your basic kinematic equations. What's the definition of acceleration for linear motion?
  6. Feb 5, 2007 #5
    So the equation looks like:
    V=Vo + at
    So, for this problem I know what the a is 3.18 m/s^2 . The t is 2 seconds and Vo is equal to zero. So substituting those factors in, V should equal 6.36 seconds correct?
  7. Feb 5, 2007 #6


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    If you mean 6.36 meters/second, then assuming you found "a" correctly (which I think you did), your speed is right.
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