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Question: A chain of mass ‘M’ and length ‘L’ is put on a rough horizontal surface and is pulled by constant ho...

A chain of mass ‘M’ and length ‘L’ is put on a rough horizontal surface and is pulled by constant horizontal force ‘F’ as shown in figure. Velocity of chain as it turns completely: (Coefficient of friction = m)

A

{2(FM−μg)L}12\left\{ 2 \left( \frac { F } { M } - \mu g \right) L \right\} ^ { \frac { 1 } { 2 } }

B

{(2FM−μg)L2}12\left\{ \left( \frac { 2 F } { M } - \mu g \right) \frac { L } { 2 } \right\} ^ { \frac { 1 } { 2 } }

C

{2(2FM−μg)L}12\left\{ 2 \left( \frac { 2 F } { M } - \mu \mathrm { g } \right) \mathrm { L } \right\} ^ { \frac { 1 } { 2 } }

D

{(4 FM−μg)L2}12\left\{ \left( \frac { 4 \mathrm {~F} } { \mathrm { M } } - \mu \mathrm { g } \right) \frac { \mathrm { L } } { 2 } \right\} ^ { \frac { 1 } { 2 } }

Answer

{2(2FM−μg)L}12\left\{ 2 \left( \frac { 2 F } { M } - \mu \mathrm { g } \right) \mathrm { L } \right\} ^ { \frac { 1 } { 2 } }

Explanation

Solution

Let point of application of force have moved by distance 'dx'.

\ Work done by friction

Wf = ∫02L−μ(ML⋅x)gdx\int _ { 0 } ^ { 2 L } - \mu \left( \frac { M } { L } \cdot x \right) g d x

̃ Wf = – Mgl

Wf = 2Fl

\Wnet = 2FL – mMgL

̃ 12\frac { 1 } { 2 } Mv2 = 2FL – mMgL

̃ v = {2(2FM−μg)L}12\left\{ 2 \left( \frac { 2 F } { M } - \mu \mathrm { g } \right) \mathrm { L } \right\} ^ { \frac { 1 } { 2 } }