question_answer

A block of mass \[2\text{ }kg\]is lying on an inclined plane, inclined to the horizontal at\[{{30}^{o}}\]. If the coefficient of friction between the block and the plane is \[0.7,\] then magnitude of frictional force acting on the block will be

A)  \[11.9\text{ }N\]

B)  \[1.19N\]

C)  \[0.19\text{ }N\]

D)  \[11.0\text{ }N\]

Correct Answer: A

Solution :

The various forces acting on the block are as shown Frictional force \[=\mu R\]          ...(i) where \[\mu \] is coefficient of friction and R the normal reaction of the surface on the block.    Also,       \[R=mg\,\,\cos {{30}^{o}}\] ... (ii) From Eqs. (i) and (ii), we get \[F=\mu \,mg\,\cos \,{{30}^{o}}\] Given, \[\mu =0.7,\,\,m=2kg,\,g=9.8m/{{s}^{2}},\] \[\cos \,{{30}^{o}}=\frac{\sqrt{3}}{2}\] \[\therefore \] \[F=0.7\times 2\times 9.8\times \frac{\sqrt{3}}{2}=11.9N\]