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Finding tension and the time taken

  1. Jun 26, 2009 #1
    1. The problem statement, all variables and given/known data

    A mass of 15 kg is supported at 2 meters from the ground by a light rope which passes over a light, smooth pulley and it carries at its over end a mass of 6kg which is held to the floor. Find the tension in the rope when the restraint is removed and the time taken for the 15kg mass to reach the floor.


    2. Relevant equations

    F net = T-m1g
    F net = T+M2g



    3. The attempt at a solution

    F net = T-m1g
    F net = T+m2g

    -T+m2g=m2a

    T=m1g+m1a

    -(m1g+m1a)+m2g=M2a

    a = (m2 - m1) ÷ (m1 + m2) × 9.8m/s^2

    If m2 = 15 kg , m1= 6kg, g = 9.8m/s^2

    a = (15kg - 6kg) ÷ (6kg + 15kg) × 9.8m/s^2

    The acceleration of the both blocks is 4.2m/s^2

    I would like to know how to work out the time taken for the 15kg block to hit the floor.

    I would like to know if the following equation would work for the tension question.
    Tension = m * g * sin + m * g
     
  2. jcsd
  3. Jun 26, 2009 #2

    tiny-tim

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    Science Advisor
    Homework Helper

    Hi speedy46! :smile:
    Yes, that's fine. :smile:
    uhh? :confused: you have the acceleration … so just use one of the standard constant acceleration equations. :wink:
    sin? sin of what?

    what are you copying from? :rolleyes:

    just use the equations you wrote above …

    -T+m2g=m2a

    T=m1g+m1a :wink:
     
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