A body of mass 2 kg rests on a rough inclined plane making an angle of `30^(@)` with the horizontal. The coefficient of static friction between the block and the plane is 0.7. The frictional force on the block is :

Options

A. 9.8 N
B. `9.8 sqrt3N`
C. `0.7 xx 9.8 N`
D. `0.7 xx 9.8 xx 3N`

Answer

c

Explanation

Here force of limiting friction`F_(f)=muR=mu mg cos 30^(@)`
`=0.7xx2xx9.8xx(sqrt3)/(2)\=11.88N`

Downward force responsible for motion
`=mg sin 30^(@) = 2 xx 9.8 xx(1)/(2)=9,8N`
Since mg sin is less than `F_(f)` the block does not move. Thus force of friction = force down the inclined plane
`:. F_(f)=  9.8 N`
Hence correct choice is (a).