question_answer

A block A of mass \[{{m}_{1}}\] rests on a horizontal table. A light string connected to it passes over a frictionless pulley at the edge of table and from its other end another block B of mass \[{{m}_{2}}\] is suspended. The coefficient of kinetic friction between the block and the table is\[{{\mu }_{k}}\]. When the block A is sliding on the table, the tension in the string is                                                                                                   

A)  \[\frac{({{m}_{2}}+{{\mu }_{k}}{{m}_{1}})g}{({{m}_{1}}+{{m}_{2}})}\]

B)  \[\frac{({{m}_{2}}+{{\mu }_{k}}{{m}_{1}})g}{({{m}_{1}}+{{m}_{2}})}\]

C)         \[\frac{{{m}_{1}},{{m}_{2}}(1+{{\mu }_{k}})g}{({{m}_{1}}+{{m}_{2}})}\]

D)         \[\frac{{{m}_{1}}{{m}_{2}}(1-{{\mu }_{k}})g}{({{m}_{1}}+{{m}_{2}})}\]

Correct Answer: C

Solution :

FED of block A, \[T={{m}_{1}}a={{f}_{k}}\,\]        ?(i)                        FBD of block B                             \[{{m}_{2}}g-T={{m}_{2}}a\]                       ?(ii) Adding Eqs. (i) and (ii), we get      \[{{m}_{2}}g\,-{{m}_{1}}a\,={{m}_{2}}a\,+{{f}_{k}}\] Þ \[{{m}_{2}}g\,-{{m}_{1}}a\,={{m}_{2}}a\,+{{\mu }_{k}}\,{{m}_{1}}g\] \[\Rightarrow \,\,\,\,\,a=\frac{\left( {{m}_{2}}-{{\mu }_{k}}{{m}_{1}} \right)g}{{{m}_{1}}+{{m}_{2}}}\] From Eq. (ii), \[T={{m}_{2}}(g-a)\] \[={{m}_{2}}\,\left[ 1-\frac{({{m}_{2}}-{{\mu }_{k}}{{m}_{1}})}{{{m}_{1}}+{{m}_{2}}} \right]g\] \[T=\frac{{{m}_{1}}{{m}_{2}}\,(1+{{\mu }_{k}})}{{{m}_{1}}+{{m}_{2}}}g\]