A) \[\frac{({{m}_{2}}+{{\mu }_{k}}{{m}_{1}})g}{({{m}_{1}}+{{m}_{2}})}\]
B) \[\frac{({{m}_{2}}+{{\mu }_{k}}{{m}_{1}})g}{({{m}_{1}}+{{m}_{2}})}\]
C) \[\frac{{{m}_{1}},{{m}_{2}}(1+{{\mu }_{k}})g}{({{m}_{1}}+{{m}_{2}})}\]
D) \[\frac{{{m}_{1}}{{m}_{2}}(1-{{\mu }_{k}})g}{({{m}_{1}}+{{m}_{2}})}\]
Correct Answer: C
Solution :
FED of block A, \[T={{m}_{1}}a={{f}_{k}}\,\] ?(i) FBD of block B \[{{m}_{2}}g-T={{m}_{2}}a\] ?(ii) Adding Eqs. (i) and (ii), we get \[{{m}_{2}}g\,-{{m}_{1}}a\,={{m}_{2}}a\,+{{f}_{k}}\] Þ \[{{m}_{2}}g\,-{{m}_{1}}a\,={{m}_{2}}a\,+{{\mu }_{k}}\,{{m}_{1}}g\] \[\Rightarrow \,\,\,\,\,a=\frac{\left( {{m}_{2}}-{{\mu }_{k}}{{m}_{1}} \right)g}{{{m}_{1}}+{{m}_{2}}}\] From Eq. (ii), \[T={{m}_{2}}(g-a)\] \[={{m}_{2}}\,\left[ 1-\frac{({{m}_{2}}-{{\mu }_{k}}{{m}_{1}})}{{{m}_{1}}+{{m}_{2}}} \right]g\] \[T=\frac{{{m}_{1}}{{m}_{2}}\,(1+{{\mu }_{k}})}{{{m}_{1}}+{{m}_{2}}}g\]You need to login to perform this action.
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