A) 0.15 mg
B) 0.05 mg
C) 0.1 mg
D) 0.45 mg
Correct Answer: D
Solution :
[d] : Here, \[{{m}_{A}}=\frac{m}{2},{{m}_{B}}=m\]\[{{\mu }_{A}}=0.2,{{\mu }_{B}}=0.1\] Let both the blocks are moving with common acceleration a. Then, \[a=\frac{{{\mu }_{A}}{{m}_{A}}g}{{{m}_{A}}}={{\mu }_{A}}g=0.2g\] and\[F-{{\mu }_{B}}({{m}_{B}}+{{m}_{A}})g=({{m}_{B}}+{{m}_{A}})a\] \[F=({{m}_{B}}+{{m}_{A}})g+{{\mu }_{B}}({{m}_{B}}+{{m}_{A}})g\] \[=\left( m+\frac{m}{2} \right)(0.2g)+(0.1)\left( m+\frac{m}{2} \right)g\] \[=\left( \frac{3}{2}m \right)(0.2g)+\left( \frac{3}{2}m \right)(0.1g)=\frac{0.9}{2}mg=0.45mg\]You need to login to perform this action.
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