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What is the speed of block one when block 2 has fallen h?

  1. Oct 19, 2016 #1
    1. The problem statement, all variables and given/known data
    A block with a mass m1 is held in place on a table and is attached to a block with mass m2 by a massless inextensible string that passes over a massless pulley with frictionless bearings. The coefficient of static friction between block 1 and the table is μs and kinetic is μk. When block 1 is released, what must be true for block 2 to fall? If that requirement is met, what is the speed of block 1 when block 2 has fallen a distance h?

    2. Relevant equations
    Conservation of energy
    Ko +Uo -W = Kf + Uf


    3. The attempt at a solution
    In order for block 2 to fall its force due to tension and gravity has to be greater than the difference between the force of block one due to tension and its force of static friction. This is what I think. Thoughts? Also, how would I put this in terms of conservation of energy?

    For the speed I used (1/2)(m1 +m2)V2km1gh = (1/2)(m1+m2)gh and solved for V. I used h for the part of work because the same distance block 2 falls is the distance block 1 moves.
     
  2. jcsd
  3. Oct 19, 2016 #2
    Here is the set up. Because I defined the potential energy being 0 as the initial height of mass 2 I think there should be a potential on the left side of the equation, but then it would be negative.
     

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  4. Oct 19, 2016 #3

    haruspex

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    That is not right. You might have confused yourself in the process of writing it out in words. Write it as an inequality in algebra.
    No. According to that, the greater the friction, the greater the velocity.
     
  5. Oct 19, 2016 #4
    Okay so if I changed the signs in the equation is it then valid? Or am I missing terms still?
    And the total energy of block 2 has to be greater than the addition of kinetic energy and work rather than the difference between its force and friction. How is that?
     
  6. Oct 19, 2016 #5

    haruspex

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    No. Explain the potential energy term.
    For part 1? No. Take what you wrote initially and turn it into an algebraic inequality.
     
  7. Oct 19, 2016 #6
    Wow holy cow I'm so sorry for that... I've got the right equation - thank you again for your help. I've written the inequality and saw what was not right.
     
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