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A Dynamics Problem Regarding Friction.

  1. Apr 26, 2013 #1
    1. The problem statement, all variables and given/known data

    SSRnLDS.png

    That was the given information. Here are the questions they ask, I have solved a) and b), but I am completely stuck on c). I'll just put them here in case information from them are necessary.

    a) What is the coefficient of static friction between the 5.0 kg mass and the horizontal surface? (answer 0.33)

    b) What is the coefficient of static friction between the two masses?

    c) From the point when the top mass starts sliding how long will it take for the top mass to fall off the bottom mass? (0.97s)

    2. Relevant equations

    F=ma
    fric = μ|Fn|

    3. The attempt at a solution

    Block on Top (2.0kg)
    [tex]ƩFx=-fk[/tex]
    [tex]ƩFx=-(0.3)(2kg*9.81m/s^2)[/tex]
    [tex]ma=-(0.3)(2kg*9.81m/s^2)[/tex]
    [tex]2kg*a=-5.886N[/tex]
    [tex]a= \frac{-5.886N}{2kg}[/tex]
    [tex]a= -2.943m/s^2[/tex]

    then...
    [tex]Δd=viΔt+\frac{1}{2}aΔt^2[/tex]
    [tex]-1m=\frac{1}{2}2.943m/s^2*Δt^2[/tex]
    [tex]-1m=-1.4715m/s^2*Δt^2[/tex]
    [tex]-1m=-1.4715m/s^2*Δt^2[/tex]
    [tex]\frac{-2m}{-1.4715m/s^2}=Δt^2[/tex]
    [tex]Δt^2=0.68s^2[/tex]
    [tex]Δt=0.82s[/tex]^Which is wrong =(
     
    Last edited: Apr 26, 2013
  2. jcsd
  3. Apr 26, 2013 #2

    haruspex

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    You've found the 2kg's acceleration relative to the ground (except, not sure why you have it as negative; isn't positive to the right?). Is that the acceleration you need to answer the question?
     
  4. Apr 26, 2013 #3
    So does that mean I need to find the acceleration relative to the bottom block?
     
  5. Apr 27, 2013 #4

    haruspex

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    You want the time it takes the top block to move 1m, starting from 'rest'. 1m relative to what? "rest" relative to what?
     
  6. Apr 27, 2013 #5
    Oh... I see now... relative to the bottom block.
     
  7. Apr 28, 2013 #6

    haruspex

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    Quite so.
     
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