Question

A uniform chain of length $L$ and mass $M$ overhangs a horizontal table with its two-third part on the table. The friction coefficient between the table and the chain is $\mu $. The work done by the friction during the period, the chain slips off the table is:
a) $\dfrac{2}{9}\mu MgL$
b) $\dfrac{6}{7}\mu MgL$
c) $\dfrac{1}{4}\mu MgL$
d) $\dfrac{4}{9}\mu MgL$

Answer 1

Hint: Recall what is meant by force of friction and coefficient of friction. To calculate the answer to this question, we first get an equation for a small amount of work done and then integrate it within the definite limits of the length to find the total work done.

Complete answer:
The force required to keep the body moving or initiate the moving of the body by overcoming the friction, is known as the force of friction.
The force of friction against the slipping of the chain off the table
Consider $d{F_f}$ is the force on the small element $dx$ of mass $dm$ due to friction is $\mu \times dm \times g$
$d{F_f} = \mu \times dm \times g$
We can replace the mass element $dm$ with mass density as $dm = (\dfrac{m}{L})dx$
$ \Rightarrow d{F_f} = \mu g(\dfrac{m}{L})dx$
Element $dx$ is displaced by $x$ when the chain slides
Therefore, work done by friction on $dx$ is $W = d{F_f} \times x$
$ \Rightarrow dW = \mu g(\dfrac{m}{L})dx \times x$
Integrating this between $0$ and $\dfrac{{2L}}{3}$ to find the total work done, we get
$ \Rightarrow W = \int\limits_0^{\dfrac{{2L}}{3}} {\mu g(\dfrac{m}{L})dx \times x} $
$ \Rightarrow W = \int\limits_0^{\dfrac{{2L}}{3}} {\mu g(\dfrac{m}{L})dx \times x} $
Removing all the constants out of the integration,
$ \Rightarrow W = \mu g(\dfrac{m}{L})\int\limits_0^{\frac{{2L}}{3}} {xdx} $
$ \Rightarrow W = \mu g(\dfrac{m}{L})[\dfrac{{{x^2}}}{2}]_0^{\dfrac{{2L}}{3}}$
$ \Rightarrow W = \mu g(\dfrac{m}{L})[\dfrac{{{x^2}}}{2}]_0^{\dfrac{{2L}}{3}}$
$ \Rightarrow W = \mu g(\dfrac{m}{L})\dfrac{{{{(\dfrac{{2L}}{3} - 0)}^2}}}{2}$
$ \Rightarrow W = \mu g(\dfrac{m}{L})\dfrac{{4{L^2}}}{{18}}$
Simplifying and cancelling out all the common terms, we get,
$ \Rightarrow W = \dfrac{2}{9}\mu gML$
Thus, work done by the friction during the period when the chain slips off is $W = \dfrac{2}{9}\mu gML$

Therefore, option (a) is the correct option.

Note: The force that resists the motion of the object is known as friction. The four types of friction are: static, sliding, rolling and fluid friction. Static friction keeps the object at rest and is the strongest friction followed by the sliding friction which keeps the object moving or sliding and is moderate while rolling friction is the weakest and keeps the object rolling. Friction plays a very important role in our lives from making it possible to walk on ground to burning of asteroids in the atmosphere before reaching earth.