A uniform chain of mass m and length l overhangs a table. Find the work to be done by the person to put the hanging part back on the table

							

The Question says that initially the chain was hanging from the table to the ground which means that initially N(normal force)=mg .......... Now mass of length l of the chain is m.
so mass of 1 unit length of the chain will be m/l
so mass of elemental part is m/l * dx
no when we pull the chain back to the table it goes up by a distance 'x' onto the table,
length of chain on the table is also 'x'
so the normal force is m*g
because it is on the table the normal force on every point is the same i.e mg
dx part is pulled up at a particular rate
so N = m/l*g * dx
integrating from 0 to x we get
N = m/l*g * ∫dx
N = mgx / l
work done is N*g
Work is mg2x/ l
 
my apologies if the answer is incorrect but this is the procedure . If i have made some calculation error please excuse.
 
Best Of luck
Please approve !




						

							

mass of 1/3rd section of chain will be  M/3
now consider the center of mass of hanging part will be at center of hanging part...
workdone in pulling hanging part  above the table is equal to workdone in pulling center of mass of hanging part  to the table...
displacement of center of mass (h)= l/6
workdone = (dm)gh
              =Mgl/18

						

							
mgL/2 is work performed
						

							
Dear student,

A hanging flexible  chain suspended from its ends is subject to tension. The tension  depends on the horizontal and vertical distances between its ends and on  the length of the chain. Use the slider to move the point M to see how  this tension varies along the chain.
Note  that the tension is calculated for a linear weight of the chain of 1  N/m (the tension varies linearly with this linear weight). The color of  the chain gives the following qualitative information: green corresponds  to the minimum and magenta to the maximum of the tension.























































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