a uniform chain of mass m and length l is lying on the table such that its one-fifth part is hanging from the edge of the table. what maximum work is done in lifting up the chain?
Answers
Answered by
3
mass = m
length = l
mass per unit length, M = (m/l)
One fifth is hanging from the edge.
So length of hanging part = l/5
Centre of mass of the hanging part will be at mid point of hanging part
So CM will be at = (l/5) / 2 = l/10.
mass of hanging part = M×(l/5) = (m/l) × (l/5) = m/5
So the problem is equivalent to lifting an object of mass m/5 to a height of l/10.
Work done = mgh = (m/5)×g×(l/10) = mgl/50
length = l
mass per unit length, M = (m/l)
One fifth is hanging from the edge.
So length of hanging part = l/5
Centre of mass of the hanging part will be at mid point of hanging part
So CM will be at = (l/5) / 2 = l/10.
mass of hanging part = M×(l/5) = (m/l) × (l/5) = m/5
So the problem is equivalent to lifting an object of mass m/5 to a height of l/10.
Work done = mgh = (m/5)×g×(l/10) = mgl/50
Similar questions