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Finding speed of block

  1. Mar 19, 2013 #1
    1. The problem statement, all variables and given/known data
    A block of mass [itex]m[/itex] is at rest at the origin at [itex]t=0[/itex]. It is pushed with constant force [itex]F_0[/itex] from [itex]x=0[/itex] to [itex]x=L[/itex] across a horizontal surface whose coefficient of kinetic friction is [itex]\mu_k=\mu_0(1-x/L)[/itex]. That is, the coefficient of friction decreases from [itex]\mu_0[/itex] at [itex]x=0[/itex] to zero at [itex]x=L[/itex].
    Find an expression for the block's speed as it reaches position [itex]L[/itex].

    3. The attempt at a solution
    I ended up with [itex]F_{net}=F_0 - mg\mu_0 + \frac{mgx}{L}[/itex]
    I can use this to find [itex]a[/itex] in terms of [itex]x[/itex], but I don't know what use that would be.
     
  2. jcsd
  3. Mar 19, 2013 #2

    BruceW

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    Homework Helper

    That is almost correct. The last term is not quite right. Try checking it, I think you just made a slight mistake. Also, about what to do next - you have an equation for the force on the block, so how could you find the change in kinetic energy of the block?
     
  4. Mar 19, 2013 #3
    Try writing
    $$ a(x)=\frac{d^2 x}{dt^2} $$
    and solve the differential equation for ##x(t)##... then final speed is
    $$ \frac{dx}{dt}|_{x=L} $$
     
  5. Mar 19, 2013 #4
    Okay, I've solved the problem using the work/kinetic energy method implied by BruceW. Thanks!
    If you're curious I got [itex]v=\sqrt{\frac{2F_0L}{m}-Lg\mu_0}[/itex]
     
  6. Mar 19, 2013 #5

    BruceW

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    yep. nice work!
     
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