question_answer

A block is moving up an inclined plane of inclination \[{{60}^{o}}\] with velocity of 20 m/s and stops after 2s. If \[g=10\,m/{{s}^{2}}\], then the approximate value of coefficient of friction is

A)  3             

B)  3.3

C)  0.27          

D)  0.33

Correct Answer: C

Solution :

 Key Idea: Frictional force acts in a direction opposite to direction of motion. The various forces acting on the block are as shown. Let a be the acceleration with which the block is moving upwards. Then from Newtons law, we have \[mg\sin {{60}^{o}}+f=ma\] ... (i) \[R-mg\cos {{60}^{o}}=0\] ... (ii) \[{{f}_{k}}=\mu \,R\] ... (iii) From Eqs. (i), (ii) and (iii), we get  \[ma=g\,(\sin {{60}^{o}}+\mu \cos {{60}^{o}})\,m\] \[ma=10\left( \frac{\sqrt{3}}{2}+\frac{\mu }{2} \right)\,m\] \[ma=5(\mu +\sqrt{3})m\] \[\Rightarrow \] \[a=5\,(\mu +\sqrt{3})\] Since, block stops, hence v = 0. From equation of motion \[v=u-at\] \[\therefore \] \[0=u-at\] \[\Rightarrow \] \[20=5(\sqrt{3}+\mu )\times 2\] \[\Rightarrow \] \[\mu =2-\sqrt{3}\] \[\Rightarrow \] \[\mu =2-1.732\] \[\mu =0.27\]