A chain of mass m and length l is placed on a table with one sixth of it hanging freely from the table edge. What is the amount of work done to pull the chain on the table?
Answers
First of all calculate linear mass density of chain that is equal to mass/length. next since 1/6 length is hanging the centre of mass will rise to a height of 1/12length on pulling the chain up.next work done =m*g*h=(1/6L*m/L)*g*1/12L=1/72 mgl.
The amount of work done to pull the chain on the table is mgl/72.
Given, the total mass of chain = m
Length of chain = l
So, mass per unit length is (m/l)
If we consider a small element dx at a distance 'x' from the table, then the small mass element dm is given as:
dm = (m/l) dx
We know, work done against gravity = mass of object×Acc. due to gravity×Distance traversed.
So, work done against gravity in lifting the small mass element dm through x distance is :
dW = dm×g×x
=(m/l) gx dx
Now, total work done is obtained by integrating dW from limits x = 0 to x = l/n
Here, n = 6 as one-sixth of the chain hangs from table.
So, W = integration of (m/l) gx dx from limits x = 0 to x = l/n
= (mg/l)[x²/2], with limits x = 0 to x = l/n [integration of xdx = x²/2]
= (mg/2l)(l²/n²)
Replacing n by 6, we get:
W = (mgl/72)
This is the total work done.