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Acceleration of a block on the floor

  1. Mar 3, 2008 #1
    [​IMG]1. The problem statement, all variables and given/known data
    The figure below shows an initially stationary block of mass m on a floor. A force of magnitude F=0.540mg is then applied at upward angle θ = 21°.



    (a) What is the magnitude of the acceleration of the block across the floor if (a) s = 0.600 and k = 0.500?


    (b) What is the magnitude of the acceleration of the block across the floor if s = 0.400 and k = 0.300?



    2. Relevant equations
    F=ma and F=s(N) or F=k(N)


    3. The attempt at a solution

    Well i drew a Free Body Diagram and found the forces in the x and y direction:

    x: F(s)-Fcos(21)=0

    y: mg-Fsin(21)-N=0

    Then i solved for F(s)=Fcos(21)=0.54mg(cos(21))=4.94m

    then i used F=s(N) to solve for N
    N=8.233m

    Then i set F=ma and F=s(N) equal and solved for a

    a=8.233(.500)=4.11

    I repeated the same steps for B...but the answers are not correct. Any help at all will be appreciated!
     
  2. jcsd
  3. Mar 3, 2008 #2

    Doc Al

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    Staff: Mentor

    This assumes equilibrium, that a = 0. Not good.

    Good. Use this to solve for N and thus compute the friction force.

    The first thing to figure out is: Does it move? Compare the applied horizontal force with the maximum static friction.

    If it accelerates, then use the kinetic friction to find the acceleration.
     
  4. Mar 3, 2008 #3

    Kurdt

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    Staff Emeritus
    Science Advisor
    Gold Member

    Why have you set up the initial equation to be zero? For it to accelerate there has to be some sort of force there.
     
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