Question

A uniform chain of length 2 m is kept on a table such that a length of 80 cm hangs freely from the edge of the table. the total mass of the chain is 4 kg. then the work done in pulling the entire chain on the table is

Answer 1

M and L are total length and mass of the chain.
M = 4 kg
L = 2 m

And let length and mass of part of chain hanging down are l and m respectively

l = 0.6 m

considering mass per unit length =>

M/L = m/l

m = l / L * 4 = (0.6 / 2) * 4 = 1.2 kg

Tension in chain hanging down = mg = 11.76 N (g=9.8 m/s^2)

Force required = Tension
Displacement = 0.6 m (part of chain hanging comes up)
W= Force * Displacement = 11.76 * 0.6 = 7.056 J