Question

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

Answer 1

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

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

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

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