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

Question

shootingstar28 · Answer

l and m are length and mass of part of chain hanging down. M and L are the total length and mass of the chain. 
M = 4 kg 
L = 2 m 
l = 0.6 m 
m = l / L * 4 = (0.6 / 2) * 4 = 1.2 kg 
The tension in the chain hanging down = mg = 12 N (g=10 m/s^2) 
The Force required = Tension 
The Displacement = 0.6 m 
W= Force * Displacement = 12 * 0.6 = 7.2 J 
The answer would be 7.2 J.

irishboy123 · Answer

pulling it where?  question incomplete, i reckon?

and is there friction?

vijeya3 · Answer

@ShootingStar28  I don't understand this part-m = l / L * 4 = (0.6 / 2) * 4 = 1.2 kg

shootingstar28 · Answer

That is the chain.

vijeya3 · Answer

Yeah..okay I don't understand the formula that u have applied here...I am not familiar with such a formula @ShootingStar28