A) 0.6
B) 0.5
C) 1.25
D) None of these
Correct Answer: A
Solution :
The free body diagram of the set up is shown: The forces acting on the block are: (i) weight of block (ii) reaction of surface (iii) frictional force The net force on the block parallel to the surface is \[(mg\,\sin {{30}^{o}}+{{f}_{k}})\] and perpendicular to surface \[R-mg\,\cos \,{{30}^{o}}\] From Newton's second law, we have \[mg\,\sin \,{{30}^{o}}+{{f}_{k}}=ma\] Since, there is no motion perpendicular to surface, so we, have \[R-mg\cos \,{{30}^{o}}=0\] \[\Rightarrow \] \[R=mg\cos \,{{30}^{o}}\] Frictional force \[{{f}_{k}}={{\mu }_{k}}R={{\mu }_{k}}mg\,\cos {{30}^{o}}\] \[\therefore \] \[mg\sin {{30}^{o}}+{{\mu }_{k}}\,mg\cos {{30}^{o}}=ma\] \[\Rightarrow \] \[a=g[sin{{30}^{o}}+\mu \cos {{30}^{o}}]=g\left( \frac{1}{2}+\frac{\mu \sqrt{3}}{2} \right)\]From equation of motion \[v=u+at\] Since, block comes to rest so \[v=0,u=5\text{ }m/s\]and a' = - a \[\therefore \] \[0=5-\frac{g}{2}(1+\mu \sqrt{3})\frac{1}{2}\] \[\Rightarrow \] \[\frac{g}{2}(1+\mu \sqrt{3})\frac{1}{2}=5\] Given, \[g=10\,\,m/{{s}^{2}}\] \[\therefore \] \[\frac{1+\mu \sqrt{3}}{2}=1\] \[\mu \sqrt{3}=1\] \[\Rightarrow \] \[\mu =\frac{1}{\sqrt{3}}\approx 0.6\]You need to login to perform this action.
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