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 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:
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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.
$W = -\frac{1}{2} \mu MgL f^2$.
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.
Updated On: Jul 22, 2026
- $-\frac{2}{9} \mu MgL$
- $-\frac{6}{9} \mu MgL$
- $-\frac{1}{9} \mu MgL$
- $-\frac{4}{9} \mu MgL$
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
A uniform chain has a fraction of its length on a rough table.
As it slips off completely, the normal force on the table decreases, changing the friction force.
We need to calculate the work done by this variable friction force.
Step 2: Key Formula and Approach:
Let $x$ be the length of the chain remaining on the table at any instant.
Mass per unit length is $\lambda = \frac{M}{L}$.
The mass on the table is $m(x) = \lambda x$.
The variable friction force is:
\[ f = \mu N = \mu m(x) g = \mu \frac{M}{L} x g \] The incremental work done by friction as the chain slides by an amount $dy$ (where $x$ decreases by $dy$, so $dx = -dy$) is $dW = -f dy$.
Step 3: Detailed Explanation:
• Formulate the integral:
Let $y$ be the distance the chain has slipped.
The length of the chain on the table is:
\[ x = \frac{2}{3}L - y \] The friction force at this position is:
\[ f(y) = \mu \frac{Mg}{L} \left(\frac{2}{3}L - y\right) \]
• Integrate to find total work:
The limits of integration for $y$ are from $0$ to $\frac{2}{3}L$:
\[ W = -\int_{0}^{\frac{2}{3}L} f(y) dy = -\mu \frac{Mg}{L} \int_{0}^{\frac{2}{3}L} \left(\frac{2}{3}L - y\right) dy \] Evaluate the integral:
\[ \int_{0}^{\frac{2}{3}L} \left(\frac{2}{3}L - y\right) dy = \left[ \frac{2}{3}L y - \frac{y^2}{2} \right]_{0}^{\frac{2}{3}L} \] \[ = \frac{2}{3}L \left(\frac{2}{3}L\right) - \frac{1}{2}\left(\frac{4}{9}L^2\right) = \frac{4}{9}L^2 - \frac{2}{9}L^2 = \frac{2}{9}L^2 \]
• Multiply by constants:
\[ W = -\mu \frac{Mg}{L} \left(\frac{2}{9}L^2\right) = -\frac{2}{9} \mu MgL \]
Step 4: Final Answer:
The work done by friction is $-\frac{2}{9} \mu MgL$, which corresponds to Option (A).
A uniform chain has a fraction of its length on a rough table.
As it slips off completely, the normal force on the table decreases, changing the friction force.
We need to calculate the work done by this variable friction force.
Step 2: Key Formula and Approach:
Let $x$ be the length of the chain remaining on the table at any instant.
Mass per unit length is $\lambda = \frac{M}{L}$.
The mass on the table is $m(x) = \lambda x$.
The variable friction force is:
\[ f = \mu N = \mu m(x) g = \mu \frac{M}{L} x g \] The incremental work done by friction as the chain slides by an amount $dy$ (where $x$ decreases by $dy$, so $dx = -dy$) is $dW = -f dy$.
Step 3: Detailed Explanation:
• Formulate the integral:
Let $y$ be the distance the chain has slipped.
The length of the chain on the table is:
\[ x = \frac{2}{3}L - y \] The friction force at this position is:
\[ f(y) = \mu \frac{Mg}{L} \left(\frac{2}{3}L - y\right) \]
• Integrate to find total work:
The limits of integration for $y$ are from $0$ to $\frac{2}{3}L$:
\[ W = -\int_{0}^{\frac{2}{3}L} f(y) dy = -\mu \frac{Mg}{L} \int_{0}^{\frac{2}{3}L} \left(\frac{2}{3}L - y\right) dy \] Evaluate the integral:
\[ \int_{0}^{\frac{2}{3}L} \left(\frac{2}{3}L - y\right) dy = \left[ \frac{2}{3}L y - \frac{y^2}{2} \right]_{0}^{\frac{2}{3}L} \] \[ = \frac{2}{3}L \left(\frac{2}{3}L\right) - \frac{1}{2}\left(\frac{4}{9}L^2\right) = \frac{4}{9}L^2 - \frac{2}{9}L^2 = \frac{2}{9}L^2 \]
• Multiply by constants:
\[ W = -\mu \frac{Mg}{L} \left(\frac{2}{9}L^2\right) = -\frac{2}{9} \mu MgL \]
Step 4: Final Answer:
The work done by friction is $-\frac{2}{9} \mu MgL$, which corresponds to Option (A).
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