Question

A chain is held on a frictionless table with one-third of its length hanging over the edge. if the chain has a length l and mass m, how much work is required to pull the hanging part back on the table?

Answer 1

Length of chain hanging from the table = L/4Mass of chain (only the hanging part)  = m/4Gravitation force on hanging part of the chain = mg/4Center of mass of the hanging part of the chain is at L/8 from end.( i.e, half of the hanging length)therefore, Work done = F.s                               = mg/4 * L/8 cos180(degrees)                               =-mgL/32 Hence, external work done required to pull the hanging length of the chain =1/32mgL

Answer 2

Answer:

Length of chain hanging from the table = L/4Mass of chain (only the hanging part)  = m/4Gravitation force on hanging part of the chain = mg/4Center of mass of the hanging part of the chain is at L/8 from end.( i.e, half of the hanging length)therefore, Work done = F.s                               = mg/4 * L/8 cos180(degrees)                               =-mgL/32 Hence, external work done required to pull the hanging length of the chain =1/32mgL

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