Question

A rope of mass m uniformly distributed along it's length l can slide without friction on a horizontal table. At instant t=0 the hanging part of the rope has a length b, and the rope is held at rest. When it is left, the rope begins to slide. At instant t the part of the rope that hangs has a length x and a velocity v. Take the horizontal table as a reference.
Calculate the mechanical energy of the system at t=0 in terms of m,g,l,b

Answer 1

I guess potential energy is zero when entire rope lies on table :)
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Since no Joules are lost in this problem, the total mechanical energy remains ZERO
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That is the end but if you really want to make a problem:
is x not b ?

rho = mass/unit length = m/L

force pulling = rho x g = m g x/L
so
a = d^2x/dt^2 = g x/L
but for energy
change in potential =- mg(x/2)(x/L)
ke = (11/2) m v^2
find v by integrating a from x = 0 to x = b