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F=MA Exam 2010 Question 10

  1. Jan 17, 2013 #1
    1. The problem statement, all variables and given/known data
    A block of mass m1 is on top of a block of mass m2. The lower block is on a horizontal surface, and a rope can pull horizontally on the lower block. The coefficient of kinetic friction for all surfaces is μ. What is the resulting acceleration of the lower block if a force F is applied to the rope? Assume that F is sufficiently large so that the top block slips on the lower block.


    2. Relevant equations

    F = ma
    Frictional Force = Normal Force * μ

    3. The attempt at a solution

    Normal Force (for lower block) = (m1 + m2) g
    Frictional Force (for lower block) = μ * (m1 + m2) g
    F - (μ * (m1 + m2) g) = m2 * a
    a = ((F - (μ * (m1 + m2) g)) / m2

    Correct Answer should be: a2 = (F - μg(2m1 + m2)) / m2
     
  2. jcsd
  3. Jan 17, 2013 #2

    tiny-tim

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    welcome to pf!

    hi vjnmath! welcome to pf! :smile:
    … sooo, you'll need to include the friction force from the top block :wink:
     
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