question_answer

The masses of the blocks A and B are 0.5 kg and 1 kg respectively. These are arranged as shown in the figure and are connected by a massless string. The coefficient of friction between all contact surfaces is \[0.4\]. The force, necessary to move the block B with constant velocity, will be \[(g=10\,m/{{s}^{2}})\]

A)  \[5\,N\]                        

B)  \[10\,N\]

C)  \[15\,N\]                      

D)  \[20\,N\]

Correct Answer: B

Solution :

 Here, \[{{m}_{A}}=0.5\,kg;\,\,\,\,\,\,{{m}_{B}}=1\,kg\] Force on block A \[T=\mu {{m}_{A}}g\]                                        ......(1) Force acting on block B \[F=T+\mu {{m}_{A}}g+\mu ({{m}_{A}}+{{m}_{B}})g\]         ??(2) From (1) & (2), \[F=\mu {{m}_{A}}g+\mu {{m}_{A}}g+\mu {{m}_{A}}g+\mu {{m}_{B}}g\] \[F=3\mu {{m}_{A}}g+\mu {{m}_{B}}g+=\mu g\,(3{{m}_{A}}+{{m}_{B}})\]