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 |
\[F=\mu {{m}_{A}}g+\mu {{m}_{A}}g+\mu {{m}_{A}}g+\mu {{m}_{B}}g\] |
\[F=3\mu {{m}_{A}}g+\mu {{m}_{B}}g=\mu g(3{{m}_{A}}+{{m}_{B}})\] |
\[=0.4\times 10\times (3\times 0.5+1)=10N\] |
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