question_answer

A block of mass 10 kg is moving up an inclined plane of inclination \[30{}^\circ \] with an , initial speed of 5 m/s. It stops after 0.5 s, what is the value of coefficient of kinetic friction?                            

A)  0.5                                        

B)  0.6           

C)  \[\sqrt{3}\]                       

D)         \[\frac{1}{\sqrt{3}}\]       

Correct Answer: D

Solution :

For equilibrium, normal to plane \[N=mg\,\cos \,\]                                                            ? (1) Net force along the plane downward \[F=mg\,\sin \,\,\text{+}{{f}_{k}}\]                                          ...(2) where\[{{f}_{k}}\]is kinetic friction            but\[{{f}_{k}}\]\[\mu N=\mu \,mg\,\cos \,\,\]                                   ...(3) from eq. (1), (2), and (3) we get \[\therefore \]\[F=mg\sin \text{ + }\mu \,\text{me cos }\] According to Newtons \[\text{I}{{\text{I}}^{\text{nd}}}\] law F = ma \[\therefore \]         \[ma=mg\,\sin \text{+}\mu \,mg\,\cos \,\] \[\therefore \]Retardation, \[a=g\,\sin \theta +\mu \,g\,\cos \theta \] From equation \[v=u+at,\] we have \[\,\text{=}\,\text{u-(g}\,\text{sin}\,\text{+}\mu \,\text{g}\,\text{cos}\,\text{)t}\] \[\Rightarrow \]\[g\,\sin \,\,\text{+}\mu \,\text{g}\,\text{cos}\,\text{=}\frac{u}{t}\] \[\Rightarrow \]\[10\times \sin \,30{}^\circ +\mu \times 10\,\cos \,30{}^\circ =\frac{5}{0.5}\] \[\Rightarrow \]\[10\times \frac{1}{2}+10\mu \times \frac{\sqrt{3}}{2}=10\] \[\Rightarrow \]               \[5\sqrt{3}\mu =5\] or            \[\mu =\frac{1}{\sqrt{3}}\]