A uniform chain of length `2m` is kept on a table such that a length of `60 cm` hangas freely from the adge of the table . The table . The total mass of the chain ia `4 kg` What is the work done in pulling the entire the chain the on the table ?

Options

A. `7.2 J`
B. `3.6J`
C. `120 J`
D. `1200 J`

Answer

B

Explanation

A chain of length `L` and mass `M` is held on a frictionless table with `y` of its length hanging over the edge.

Let `m = (M)/(L) = mass` per unit length of the chain and `y` is the length of the chain hanging over the edge. So the mass of the chain of length `y` will be `ym` and the froce acting on it due to gravity will be `mgy`. The work done in pulling the `dy` length of the chain on the table.
`dW = F(-dy)` [As `y` is decreasing]
i.e., `dW = mgy (-dy)`
So the work done in pulling the hanging portion on the table.
`W = - underset(y) overset(0) int mgydy = - mg [(y^(2))/(2)]_(y)^(0) = (Mgy^(2))/(2L)`
`[As m = M//L]`
`:. W = ( 4xx 10 xx (0.6)^(2))/(2 xx 2) = 3.6 J`.