Question

A uniform chain of length 'L' and mass 'M' overhangs a horizontal table with its two-third part on the table. The coefficient of friction between the table and the chain is $\mu$. The work done by friction during the period the chain slips off the table is:

• A. $-\frac{2}{9} \mu MgL$
• B. $-\frac{6}{9} \mu MgL$
• C. $-\frac{1}{9} \mu MgL$
• D. $-\frac{4}{9} \mu MgL$

Hint:

For a chain of fraction $f$ on a table (here $f = 2/3$), the work done by friction as it slips off completely is always:
$W = -\frac{1}{2} \mu MgL f^2$.
Here, $f = \frac{2}{3}$, so $W = -\frac{1}{2} \mu MgL \left(\frac{4}{9}\right) = -\frac{2}{9} \mu MgL$.
This general formula is very useful for competitive exams.

Answer 1

The Correct Option is A

Step 1: Picture the friction force graph.
As the chain slides off, the length remaining on the table drops steadily from $\frac{2}{3}L$ to zero, and the friction force is proportional to that length, so a plot of friction force against the slipped distance $y$ is just a straight line from $f_0 = \mu\frac{2}{3}Mg$ down to zero.
Step 2: Use the area under that graph.
The work done against friction is $-1$ times the area under this force versus distance line, and since the line is straight, that area is a triangle.
Step 3: Calculate the triangle's area. \[ \text{Area} = \frac{1}{2}\times f_0\times\frac{2}{3}L = \frac{1}{2}\times\mu\frac{2}{3}Mg\times\frac{2}{3}L = \frac{2}{9}\mu MgL \]
Step 4: Attach the sign.
Friction opposes the sliding, so the work it does is negative.
\[ \boxed{W_f = -\frac{2}{9}\mu MgL} \]