Physics Chain Problem 
A chain of length d lies on a table, $d/4$ of it hangs loose on the side of it. Friction coefficient is 0.2. It is released and begins to slide. What is the final velocity with which it falls of the table?


My attempt:
$Wf =   \bigtriangleup Ug = -      \int_0^{0.75d} \!   \mu_k  g \frac{m}{d}(\frac{d}{4}+x) \, \mathrm{d}x = -0.4687\mu_k  mgd$
Which is the change and loss in energy during the slide. My plan was to equate the initial gravitational potential energy to the final kinetic energy plus the energy lost from friction (calculated above), however there is no way to define the friction for the rope since it has two parts, with apparently distinct potential energies. Is there another way to do it? 
 A: The conceptual problem seems to be

however there is no way to define the friction for the rope since it has two parts, with apparently distinct potential energies.

Don't worry about the "distinct potential energies". You can just compute the gain in potential energy for each part separately. For the bit already hanging down, as the rope slides by a certain distance, the center of mass moves by that same distance. For the bit of rope that starts off horizontal, when the last part goes over the edge its center of mass has moved half as far.
That leaves the calculation of the work done by friction. If you consider your chain like a train with lots of cars, then for each car in the train you can consider the work done due to friction. In other words - for an infinitesimal element $d\ell$ that starts out at a distance $x$ from the edge, the force of friction is $\mu \frac{d\ell}{L} m g$ and the work done is $F \cdot x$.
Summing the work for all these elements will give you the total work done. That is a simple integral.
In the spirit of "homework like questions", I will leave you with these hints.
A: Yes, the work-energy method can be used.  The particular difficulties you noticed can be dealt with by considering the chain in 2 parts : that part hanging down (which loses PE but experiences no friction) and that part remaining on the table (which does not lose PE but experinces friction).
Suppose the mass of the chain is $M$.  In the initial position, the CM of the overhaning section is $\frac{d}{8}$ below the table. The section remaining on the table has $0$ PE.  So the initial PE of the chain relative to the table is $-\frac14Mg(\frac{d}{8}) = -\frac{1}{32} Mgd$.  When the last link of the chain slips off the table then the CM of the whole chain is $\frac{d}{2}$ below the table so the PE is $-\frac12 Mgd$. The decrease in PE is
$(\frac12-\frac{1}{32})Mgd = \frac{15}{32}Mgd.$  
When length y of the chain remains on the table then the friction force on it is $\frac15Mg\frac{y}{d}$ where $\frac15$ is the coefficient of friction.  The work done against friction is
$\int \frac15 Mg\frac{y}{d}dy = \frac15 Mg[\frac{1}{2d}y^2] = \frac{9}{160}Mgd$
where the limits of integration are from $y=\frac34d$ to $0$.  
The KE of the chain when the last link slips off the table is therefore
$\frac12MV^2 = (\frac{15}{32}-\frac{9}{160})Mgd$
$V^2=\frac{33}{40}gd$
where V is the velocity at this point.  
Alternatively you can apply Newton's 2nd Law :
$F = M\frac{dv}{dt} = M\frac{dv}{dy}\frac{dy}{dt} = -Mv\frac{dv}{dy}$.
When length y remains on the table the force accelerating the chain is $Mg\frac{d-y}{d}$, while the friction force retarding motion is $\frac15Mg\frac{y}{d}$.  
The equation of motion is
$-Mv\frac{dv}{dy} = Mg(1-\frac{y}{d}) - \frac15Mg\frac{y}{d} = (1-\frac65\frac{y}{d})Mg$
$2v\frac{dv}{dy} =  (2\frac65\frac{y}{d}-2)g$
$v^2 =  (\frac65\frac{y^2}{d}-2y)g$.  
The limits of integration are $v=0$ to $V$ and $y=\frac34d$ to $0$.  So
$V^2 = (\frac65(0-\frac{9}{16}d)-2(0-\frac34d))g = \frac{33}{40}gd$.  
