Question

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

Answer 1

Answer: (A)

11.9 N

Answer 2

The frictional force acting on the block is
${{f}_{s}}={{\mu }_{s}}R$
But, $R=mg\cos {{30}^{o}}$
$\therefore$ ${{f}_{s}}=\mu mg\cos {{30}^{o}}$
$\Rightarrow$ ${{f}_{s}}=0.7\times 2\times 9.8\times 0.866$
$\Rightarrow$ ${{f}_{s}}=11.9\,N$