A uniform chain of length l is placed on the table in such a manner that its l' part is hanging
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unit length of rope is M/L
given L/3 part of rope is hanging
we have to solve using integration methods, lets say 'x' part of rope is consided with an element 'dx' which is pulled up with limits from 0 to L/3
Potential energy = mgh
here dE= integral of ( Mx/L*g*dx)
E = integral of ( M/L*g*x.dx) limits from 0 to L/3
P.E = MgL/18
@Physics GuruDev
given L/3 part of rope is hanging
we have to solve using integration methods, lets say 'x' part of rope is consided with an element 'dx' which is pulled up with limits from 0 to L/3
Potential energy = mgh
here dE= integral of ( Mx/L*g*dx)
E = integral of ( M/L*g*x.dx) limits from 0 to L/3
P.E = MgL/18
@Physics GuruDev
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