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What is the acceleration of the block?

  1. Nov 16, 2016 #1
    1. The problem statement, all variables and given/known data
    A 50.0-kg block is being pulled up a 16.0° slope by a force of 250 N that is parallel to the slope. The coefficient of kinetic friction between the block and the slope is 0.200. What is the acceleration of the block?

    2. Relevant equations
    FkkN
    F=MA

    3. The attempt at a solution
    answer is .412 m/s2. I know the answer I just don't know the steps to solve this equation.
    It is saying it is being pulled up, does that mean it has tension instead of normal force? it is not sitting on anything so it can not have a normal force right?
     
  2. jcsd
  3. Nov 16, 2016 #2

    Simon Bridge

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    You don't even have an equation to solve, and how do you know that is the right answer?
    Surely the block is sittig on a slope? A tension is a force. Is the block being pulled by a rope or has someone just grabbed it and pulled it? Does it matter to the question?

    Your first step is to sketch a free body diagram ... you will have course notes about this: so how many forces are on the block, and what are they?
     
  4. Nov 16, 2016 #3
    In that case, there is a normal force. So the forces on the block are the Normal force, MG and 250cos16, 250 sin16
     
  5. Nov 16, 2016 #4

    Simon Bridge

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    Identify the forces themselves - you have a normal force to the slope, gravity, friction, and the pulling force.
    GIve them names - so you have ##F_g=mg## pointing vertically down, ##N## points upwards at 90deg to the slope, ##f=\mu N## points down the slope and ##F=250##N points up the slope. F is only one force, not two like you wrote. Don't do the cosine and sine stuff untilyou decide how to orient your axes - hint: point x axis up the slope.
     
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