question_answer

 A block of mass m1 rests on a horizontal table. A string tied to this block is passed over a frictional pulley fixed at one end of the table and another block of mass m1 is hung to the other end of the string. The acceleration (a) of the system is :

A) \[\frac{{{m}_{1}}{{m}_{2}}g}{{{m}_{1}}-{{m}_{2}}}\]                       

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

C) \[\frac{{{m}_{1}}g}{{{m}_{1}}+{{m}_{2}}}\]                         

D) \[g\]

Correct Answer: B

Solution :

Key Idea: Force acting on block which causes acceleration is F = ma. The free body diagram depicting the situation is shown. Resultant force acting on hanging block is  \[({{m}_{2}}g-T).\] Force acting on block \[{{m}_{2}}\]which causes acceleration is \[{{m}_{2}}a\] Hence,    \[{{m}_{2}}g-T={{m}_{2}}a\]     ?(i) Also,        \[T={{m}_{1}}a\]            ?(ii) From Eqs. (i) and (ii), we get \[a=\frac{{{m}_{2}}g}{{{m}_{1}}+{{m}_{2}}}\]