question_answer

A block moving up at \[\theta \] = 30° with a velocity 5 m/s, stops after 0.5 s then what is value of friction (\[\mu \]) ?

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,      \[g=10\,\,m/{{s}^{2}}\] \[\therefore \]           \[\frac{1+\mu \sqrt{3}}{2}=1\]            \[\mu \sqrt{3}=1\] \[\Rightarrow \]    \[\mu =\frac{1}{\sqrt{3}}\approx 0.6\]