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 :

Taking \[{{m}_{A}}=0.5kg;\,\,{{m}_{B}}=1Kg\]

Force on block A                         ... (1)

Force acting on block B

\[F=T+\mu {{m}_{A}}g+\mu ({{m}_{A}}+{{m}_{B}})g\]                    ... (2)

From 1 & 2

\[F=\mu {{m}_{A}}g+\mu {{m}_{A}}g+\mu {{m}_{A}}g+\mu {{m}_{B}}g\]

\[F=3\mu {{m}_{A}}g+\mu {{m}_{B}}g=\mu g(3{{m}_{A}}+{{m}_{B}})\]

\[=0.4\times 10\times (3\times 0.5+1)=10N\]