Calculating Speed of a Block Pushed Across a Horizontal Surface

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A block of mass m is pushed across a surface with a decreasing coefficient of kinetic friction, transitioning from μ_0 to zero. The net force acting on the block is expressed as F_net = F_0 - mgμ_0 + (mgx/L). To find the block's speed at position L, a differential equation for acceleration a(x) can be established, leading to the calculation of kinetic energy changes. The final speed is derived using the work/kinetic energy method, resulting in the expression v = √(2F_0L/m - Lgμ_0). The solution effectively demonstrates the relationship between force, friction, and speed in this context.
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Homework Statement


A block of mass m is at rest at the origin at t=0. It is pushed with constant force F_0 from x=0 to x=L across a horizontal surface whose coefficient of kinetic friction is \mu_k=\mu_0(1-x/L). That is, the coefficient of friction decreases from \mu_0 at x=0 to zero at x=L.
Find an expression for the block's speed as it reaches position L.

The Attempt at a Solution


I ended up with F_{net}=F_0 - mg\mu_0 + \frac{mgx}{L}
I can use this to find a in terms of x, but I don't know what use that would be.
 
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crimpedupcan said:
I ended up with F_{net}=F_0 - mg\mu_0 + \frac{mgx}{L}
I can use this to find a in terms of x, 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?
 
Try writing
$$ a(x)=\frac{d^2 x}{dt^2} $$
and solve the differential equation for ##x(t)##... then final speed is
$$ \frac{dx}{dt}|_{x=L} $$
 
Okay, I've solved the problem using the work/kinetic energy method implied by BruceW. Thanks!
If you're curious I got v=\sqrt{\frac{2F_0L}{m}-Lg\mu_0}
 
yep. nice work!
 
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