A block of mass 2 kg rests on a plane inclined at an angle of `30^(@)` with the horizontal the coefficient of friction between the block and the surface is 0.7 what will be the frictional force acting on the block

Explanation

Here f =`mu R = mu m g cos theta`
`=0.7 xx 2 xx 9.8 cos 30^(@)`
`=0.7 xx 9.8 xx 0.866 = 11.9N`