Initially, a flexible chain of length l and mass m is slowly pulled at constant speed up over the edge of a table by a force parallel to the surface of the table. Assuming that there is no friction between the table and chain, calculate the work done by force until the chain reaches to the horizontal surface of the table.

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Solution

Initial Potential energy, $U_{1} = - m g \frac{l}{2}$

And final potential energy $U_{2} = 0$

Since, chain in pulled slowly, there is no change in kinetic energy.

Hence, Work done by force, $W = Δ U = U_{2} - U_{1}$

$⟹ W = m g \frac{l}{2}$

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Initial Potential energy, U1=−mgl2And final potential energy U2=0Since, chain in pulled slowly, there is no change in kinetic energy.Hence, Work done by force, W=ΔU=U2−U1⟹W=mgl2

Initial Potential energy, $U_{1} = - m g \frac{l}{2}$

And final potential energy $U_{2} = 0$

Since, chain in pulled slowly, there is no change in kinetic energy.

Hence, Work done by force, $W = Δ U = U_{2} - U_{1}$

$⟹ W = m g \frac{l}{2}$