A `6 kg` block is kept on an inclined rough surface as shown in figure.Find the force `F` requiredto
(a) keep the block stationary.
(b) move the block downwards with contact velocity and
( c) move the block upwards with an acceleration `4 m//s^(2) (Take g = 10 m//s^(2))`

Explanation

`N = mg cos 60^(@)`
`= (6) (10) ((1)/(2)) = 30 N`
`mu_(s)N = 18 N`
`mu_(k)N = 12 N`
During force`F = mg sin 60^(@)`
`=` force responsible to move the body downwards
`= (6) (10) (sqrt((3)/(2)))`
`= 52 N`
(a) Force needed to keep the block stationary is
`F_(1) = F - mu_(s)N` (upwards)
`= 52 - 18`
`= 34 N` (upwards)

(b) If the move block downwards with constant velocity `(a = 0,F_("net") = 0)`, then kinetic friction will act in upward direction.
:. Force needed `F_(2) = F -mu_(k)N`
`= 52 - 12`
`= 40 N` (upwards)

(c) Kinetic friction will act in downward direction.
`F_(3) - 52 - 12 = ma = (6) (4)`
`F_(3) = 88 N` (upwards)