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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#