• # question_answer 33) 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)  $20\,N$

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), $F=\mu {{m}_{A}}g+\mu {{m}_{A}}g+\mu {{m}_{A}}g+\mu {{m}_{B}}g$ $F=3\mu {{m}_{A}}g+\mu {{m}_{B}}g+=\mu g\,(3{{m}_{A}}+{{m}_{B}})$