Question

Question #b79c6

Answer 1

Let
`m->"mass of the block"`
`theta->"angle of inclination of the inclined plane "=29.7^@`
`g->"acceleration due to gravity"`
`mu_k->"coefficient of kinetic friction "`

Since in this case the block moves downward with constant speed the net force acting on it zero,
Here downward gravitational pull along the plane `color(blue)(mgsintheta)` acting on the block is just balanced by the upward frictional force `color(red)(mu_kmgcostheta)` along the plane

So `mgsintheta=mu_kmgcostheta`

`=>mu_k=tantheta=tan29.7^@=0.57`