A block of mass 2 kg rests on a rough inclined plane making an angle of 30 degree with the horizontal. The coefficient of static friction between he block and the plane is 0.7. The frictional force on the block is :

0.7 × 9.8 N

The force applied on the body that is on the inclined plane is given as,

F = mgsinθ

F = 2 × 9.8 × sin30^∘

 = 9.8N

The limiting friction force between the block and the inclined plane is given as,

f = μmgcosθ

f = 0.7 × 2 × 9.8 cos30^∘

  =11.88N

Since the limiting friction force is greater than the force that tends to slide the body.

Thus, the body will be at rest and the force of friction on the block is 9.8N.