A block of mass m slides down the plane inclined at an angle `60^(@)` with an acceleration `(g)/(2)` . Find the coefficient of kinetic friction?

Explanation

Kinetic friction comes to play as the block is moving on the surface.
The forces acting on the mass are the normal force perpendicular to surface, downward gravitational force and kinetic friction `f_(k)` along the surface.

`mg sin theta- f_(k)=ma`
But `a=g//2`
`mg sin 60^(@)-f_(k)=mg//2`
`(sqrt(3))/(2) mg -f_(k)= mg//2`
`f_(k)=mg (( sqrt(3))/(2)-(1)/(2))`
`f_(k)=((sqrt(3)-1)/(2))mg`
There is no motion along the y-direction as normal force is exactly balanced by the `mg cos theta`.
`mg cos theta=Nmg//2`
`f_(k)=mu_(k)N=mu_(k)=mg//s`
`mu_(k)=(((sqrt(3)-1)/(2))mg)/((mg)/(2))`
`mu_(k)= sqrt(3)-1`