A uniform chain of mass m and length l initially 1/nth part is hanging from the edge of table . If 1/mth (m<n) part is pulled in to the table then what is the work done?
Question
A uniform chain of mass m and length l initially 1/nth part is hanging from the edge of table . If 1/mth (m<n) part is pulled in to the table then what is the work done?
Solution
The work done in pulling a chain onto a table is given by the formula:
W = mgh
where:
- m is the mass of the chain,
- g is the acceleration due to gravity, and
- h is the height the chain is lifted.
In this case, the chain is initially 1/nth on the table, so the height the chain is lifted is (1 - 1/n) * l, where l is the length of the chain.
The mass of the chain that is lifted is (1 - 1/m) * m, where m is the total mass of the chain.
Substituting these values into the formula gives:
W = [(1 - 1/m) * m] * g * [(1 - 1/n) * l]
Simplifying this expression gives:
W = (m - 1) * g * (l - l/n)
This is the work done in pulling the chain onto the table.
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