question_answer

A block of mass \[2\,\,kg\] rests on a plane inclined at an angle of \[{{30}^{o}}\] with the horizontal. The coefficient of friction between the block and surface is \[0.7\]. The frictional force acting on the block is:

A) \[11.9\,\,N\]                     

B) \[25\,\,N\]

C) \[50\,\,N\]                         

D)  \[22.9\,\,N\]

Correct Answer: A

Solution :

The frictional force acting on the block is                 \[{{f}_{s}}={{\mu }_{s}}R\] But,\[R=mg\cos {{30}^{o}}\] \[\therefore \]  \[{{f}_{s}}=\mu mg\cos {{30}^{o}}\] \[\Rightarrow \]               \[{{f}_{s}}=0.7\times 9.8\times 0.866\] \[\Rightarrow \]               \[{{f}_{s}}=11.9\,\,N\]