Question

A man places a vertical uniform chain (of mass m and length l) on a table slowly. Initially, the lower end of the chain just touches the table. The man drops the chain when half of the chain is in a vertical position. Then the work done by the man in this process is
A) \[ - \dfrac{{mgl}}{2}\]
B) \[ - \dfrac{{mgl}}{4}\]
C) \[ - \dfrac{{3mgl}}{8}\]
D) \[ - \dfrac{{mgl}}{8}\]

Answer 1

Hint: Recall that Gravitational potential energy of the object is the energy possessed by it due to change in its position while it is placed in the gravitational field. Also, gravitational potential energy depends on the height, mass, acceleration, and distance of the object from the ground. That is if the height is doubled then gravitational potential energy is also doubled.

Complete step by step solution:
Step I:
Let the object be placed at any point in the plane. The potential energy at height ‘h’ is equal to the work done to lift the object to that height without any change in kinetic energy.
Step II:
Given m is the mass and l is the length
When the lower end touches the table, then at the point of the center of mass its length will be \[\dfrac{l}{2}\]
Initial Potential Energy \[{u_i} = mg\dfrac{l}{2}\]
Step III:
When half of the chain is in a vertical position, then final potential energy becomes
\[{u_f} = \dfrac{m}{2} \times g \times \dfrac{l}{4}\]
\[{u_f} = \dfrac{{mgl}}{8}\]
Step IV:
Net potential energy \[\Delta u = {u_f} - {u_i}\]
\[\Delta u = \dfrac{{3mgl}}{8}\]
Step V:
As per the work-energy theorem, work done by all the forces in a system is equal to change in kinetic energy. If the motion of an object is known and the values of one or more forces acting on it are not known then the work-energy theorem is used. Since the man is dropping the chain downwards, so change in kinetic energy will be zero.
Step VI:
Work done by man in dropping chain
\[{w_{man}} + {w_{gravity}} = \Delta K.E.\]
\[{w_{man}} + {w_{gravity}} = 0\]
\[{w_{man}} = - {w_{gravity}}\]
\[{w_{man}} = - \Delta v\]
\[{w_{man}} = - \dfrac{{3mgl}}{8}\]

$\therefore $ Work done by man is \[ - \dfrac{{3mgl}}{8}\]. Option (C) is the correct answer.

Note:
Potential energy is the energy stored in an object due to its position with respect to some other object in motion. This potential energy of the object depends only upon the path followed by that object. Gravitational potential energy is a conservative force, but it depends on the initial and final position. Always remember, for conservative forces, energy is independent of path. But for non-conservative forces, potential energy is dependent on the path.