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Answer 1 · 2016-11-13T20:10:06

$W_x = mu(3x+2h)g$

First, some asumptions.

i) The chain falls on the floor, resting at the floor level.
ii) The rope remaining above the table, keeps straigth.

For an excursion $x$ the weigth supported on the floor is given by

$W_x=m_x g + I$

where $m_x = mu x$ is the mass landed and $I$ is the impulse exerted by the $dm$ arriving at that instant.

$I = ((dm)/(dt)) v = (mu v dt)/(dt)v = mu v^2$

but we know that for any $dm$ after the excursion $x$

$1/2v^2dm=dm (x+h)g$ or

$v^2=2(x+h)g$ Now, substituting into $I$ we have

$I=2mu(x+h)g$ and finally

$W_x =mu x g + 2mu(x+h)g = mu(3x+2h)g$