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Finding speed of block

  • #1

Homework Statement


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].

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.
 

Answers and Replies

  • #2
BruceW
Homework Helper
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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.
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?
 
  • #3
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2
Try writing
$$ a(x)=\frac{d^2 x}{dt^2} $$
and solve the differential equation for ##x(t)##... then final speed is
$$ \frac{dx}{dt}|_{x=L} $$
 
  • #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]
 
  • #5
BruceW
Homework Helper
3,611
119
yep. nice work!
 

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