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Question

A uniform chain of length $$2$$ m and mass $$0.1$$ kg overchangs a smooth table with its two third part lying on the table. Find the kinetic energy of the chain as it completely slips-off the table.


Solution

$$W=\frac { mgl }{ 2n^{ 2 } } $$

Here $$\frac { 2 }{ 3 } th$$ part is lying on the table it means $$\frac { 1 }{ 3 } th$$is hanging

Remember n should always be greater than 1

Here n is 3

So $$W=\frac { mgl }{ 18 } $$

Now W=kinetic energy


$$W=\frac { mgl }{ 18 } =\frac { 0.1\times 9.81\times 2 }{ 18 } =0.10$$



Physics

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