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

Explanation

`N =mg cos 60^(@) =(6)(10)((1)/(2))=30`newton
`mu _(s)N =18`newton
`mu _(k)N =12`newton
Driving force `F_(0)=mg sin 60^(@)=(6) (10)(sqrt(3)/(2))=52N`
(a) Force needed to keep the block stationary is
`F_(1) =F_(0)-mu _(s)N` (upwards)
`=52-18`
`=34N` (upwards)

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

(c) In this case, Kinetic friction will act in downward direction
`F_(3)-F_(0)-mu _(k)N =ma`
or `F_(3)-52-12=ma =(6)(4)`
`F_(3)=88N` (upwards )