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, \[R=mg\cos {{30}^{o}}\] \[\therefore \] \[{{f}_{s}}=\mu mg\cos {{30}^{o}}\] \[\Rightarrow \] \[{{f}_{s}}=0.7\times 2\times 9.8\times 0.866\] \[\Rightarrow \] \[{{f}_{s}}=11.9\,N\]