A block of mass 20 kg is acted upon by a force `F=30N` at an angle `53^@` with horizontal in downward direction as shown. The coefficient of friction between the block and the forizontal surface is `0.2`. The friction force acting on the block by the ground is `(g=10(m)/(s^2)`)

Options

A. `40.0 N`
B. `30.0 N`
C. `18.0 N`
D. `44.8 N`

Answer

C

Explanation

(Easy) Max. frictional force
`f_(max) = mu N`
`= mu (mg + F sin 53^(@))`
`= 0.2(20 xx 10 + 30 xx) = 44.8 N`
As applied horizontal force is `F cos 53^(@)`
`= 18 N lt f_(max)`, friction froce will also be 18 N.