Question

A chain of mass m and length L is over hanging from the edge of a smooth horizontal table such that 3/4th of its length is lying on the table. The work done in pulling the chain completely on to the table is:
(A) $\dfrac{{mgl}}{{16}}$
(B) $\dfrac{{mgl}}{{32}}$
(C) $\dfrac{{3mgl}}{{32}}$
(D) $\dfrac{{mgl}}{8}$

Answer 1

Hint: We are looking for the work done in this problem. This could be calculated with the basic formula given by ${\text{dW = f}}{\text{.ds}}$ and integrating it up to the limit as force is not constant throughout.
Formula used: To calculate the amount of force, we use the equation:
${\text{dW = f}}{\text{.ds}}$
Here, $dw$ is the small amount of work being done,
${\text{f}}$ is the force applied.
${\text{ds}}$ is the small displacement,

Complete step by step answer:
It is already known that the weight of the hanging part $\left( {\dfrac{L}{4}} \right)$ of chain is $\dfrac{{Mg}}{4}$ .
The weight acts at the centre of gravity of the hanging part which is at a distance of $\left( {\dfrac{{3L}}{4}} \right)$ from the table.
${\text{dW = f}}{\text{.ds = }}\Delta {\text{P}}{\text{.E}}$
On solving further,
$W = \dfrac{{Mg}}{4} \times \dfrac{L}{8} = \dfrac{{MgL}}{{32}}$
So, we need to see from the above options, and select the correct value.

Thus, the correct answer is option B.

Additional Information: A force is said to do work when it acts on a body so that there is a displacement of the point of application in the direction of the force. Thus, a force does work when it results in movement. The same integration approach can be also applied to the work done by a constant force. This suggests that integrating the product of force and distance is the general way of determining the work done by a force on a moving body.

Note: The examples of variable force can be seen everywhere as all forces in the world are not constant. Constant force is just an ideal situation used in solving questions. Work may still be calculated as the area under the force- displacement curve.