Question

A chain of mass m and length l rests on a rough table with part overhanging. The chain starts sliding down by itself if overhanging part is l/3. What will be the work performed by the friction force acting on the chain by the moment it slides completely off the table.

• A. $\frac { \mathrm { Mgl } } { 3 }$
• B.
• C. $\frac { 2 \mathrm { Mgl } } { 9 }$
• D.

Answer 1

Answer: (D)

Answer 2

Assume at any instant the length of the chain on the table is x then force friction = μN = $\frac { \mu \mathrm { M } } { 1 }$ xg

Work done against friction $\int _ { 0 } ^ { 2 ^ { 1 / 3 } } \mu \frac { M } { 1 }$ xg. dx

= $\left. \mu \frac { \mathrm { Mg } } { 1 } \frac { \mathrm { x } ^ { 2 } } { 2 } \right| _ { 0 } ^ { 2 ^ { 1 / 3 } } = \frac { 1 } { 2 } \frac { \mathrm { M } } { 1 } \frac { \mathrm { g } } { 2 } \left( \frac { 41 ^ { 2 } } { 9 } \right) = \frac { \mathrm { Mgl } } { 9 }$