A uniform chain of length 2 m is kept on a 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 on the table? please explain each and every step thank u

Dear aditya,

Given

M and L are total lenght 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

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this q can be easily solved by considering motion of center of mass of hanging part of chain....

center of mass of hanging part lies in the middle ...

distance of center of mass from table is 30cm or 0.3m.....

workdone in pulling hanging part of chain  is equivalent to workdone in pulling center of mass of the hanging part to the table...

mass of comlete chain =4kg

 mass per unit length =4/2=2kg/m

 mass of 60 cm of chain=0.6*2=1.2kg

now workdone=mgH(com)

                    =1.2 * 10 *0.3

                    =3.6J

Dear Student,
Please find below the solution to your problem.

Given
M and L are total lenght 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

Thanks and Regards