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a chain is held on a frictionless table with(1/n)th of its length hanging over the edge.if the chain has a length L and mass M, how much work is required to slowly pull the hanging part back on the table?

a chain is held on a frictionless table with(1/n)th of its length hanging over the edge.if the chain has a length L and mass M, how much work is required to slowly pull the hanging part back on the table?

Grade:12

4 Answers

Saurabh Anand
38 Points
7 years ago

kinetic energy theorem i.e. change in kinetic energy= work done by gravity+work done by applied force

now since chain is slowly pulled up therefore change in kinetic energy will be zero...

and thus work done by applied force= -work done by gravity i.e. (MgL)/(2n2)

Russell khajuria
18 Points
7 years ago

since the chain is uniform the mass of chain hanging is M/nL. due to this mass a tension is developed against which work is to be done . tension T=Mg/nL. work done = T*n

vishal ojha
35 Points
7 years ago

kinetic energy theorem i.e. change in kinetic energy= work done by gravity+work done by applied force

now since chain is slowly pulled up therefore change in kinetic energy will be zero...

and thus work done by applied force= -work done by gravity i.e. (MgL)/(2n2)

 

SOURAV MISHRA
37 Points
7 years ago

since the chain is being pulled slowly that is with negligible velocity so its kinetic energy is also negligible compared to potential energy.

so by the conservation of energy we can say that the work done goes in increasing the potential energy of the hanging part.

to calculate this we need to perform integration.

the change in potential energy of the string in bringing up an element of length dx situated at a distance x below the table top is dmgx.

where dm = mass of the hanging part = (M/L)dx

so the work done to pull the hanging part to the table top is therefore the work done to pull up all such elements one by one slowly. that is the sum of the changes in the potential energies of all such elements.

= ∫0L/n(M/L)gxdx = MgL/2n2.

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