question_answer

A block of mass mi rests on a horizontal table. A string tied to this block is passed over a friction less pulley fixed at one end of the table and another block of mass \[{{m}_{2}}\] is hung to the other end of the string, the acceleration a of the system will be :

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) \[{{m}_{1}}{{m}_{2}}g\]

Correct Answer: B

Solution :

The free body diagram of the system is shown. T is tension in the string and a is acceleration of the blocks. From Newtons second law \[{{m}_{2}}a={{m}_{2}}g-T\] ... (i) \[T={{m}_{1}}\,\,a\] ? (ii) Solving Eqs. (i) and (ii), we get \[a=\frac{{{m}_{2}}\,g}{{{m}_{1}}+{{m}_{2}}}\]