Physics, asked by varsiniduraisamy, 7 months ago

Hint : W=MGL/2n^2 Plz solve no nonsense answers

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Answered by Atαrαh
3

A uniform change held on a frictionless table with half of its length hanging over the edge

We know that ,

work done by agent is a sum of of potential and kinetic energy

 \boxed{W ag = \Delta  KE + \Delta  U }

As the initial and final value of KE is zero

\boxed {W ag =  \Delta U }

Initial potential energy ,

 \implies{U \: initial =  - mgh}

( - ve sign is denotes that the hanging part of the chain is below the reference line )

here ,

  • mass of the hanging portion = M / 2

  • length of the hanging portion = L / 2

( COM = L / 4 )

\implies{U \: initial =  -  \frac{m g L}{2 \times 4} }

\implies{U \: initial =  -  \frac{m g L}{8} }

Final potential energy,

\implies{U \: final=  0}

( as the chain is kept on the table )

Now substituting the above values in the equation we get

\implies{W ag =   U final - U \: initial}

\implies{W ag =   0- (-  \frac{m g L}{8})}

\boxed{W ag =     \frac{m g L}{8}}

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