A small bar starts sliding down an inclined plane forming an angle `alpha` with the horizontal. The friction coefficient depends on the distance x covered as `k=ax`, where a is a constant. Find the distance covered by the bar till it stops, and its maximum velocity over this distance.

Explanation

From Newton's second law for the bar in projection from, `F_x=mw_x` along x direction we get
`mg sin alpha-kmg cos alpha=mw`
or, `v(dv)/(dx)=g sin alpha-ax g cos alpha`, (as `k=ax`),
or, `vdv=(g sin alpha-axg cos alpha)dx`
or, `underset(0)overset(v)intvdv=gunderset(0)overset(x)int(sin alpha-x cos alpha)dx`
So, `v^2/2=g(sin alphax-x^2/2a cos alpha)` (1)
From (1) `v=0` at either
`x=0`, or `x=2/atanalpha`
As the motion of the bar is unidirectional it stops after going through a distance of `2/atan alpha`.
From (1), for `v_(max)`,
`(d)/(dx)(sin alphax-x^2/2a cos alpha)=0`, which yields `x=1/atanalpha`
Hence, the maximum velocity will be at the distance, `x=tanalpha//a`
Putting this value of x in (1) the maximum velocity,
`v_(max)=sqrt((gsin alphatan alpha)/(a))`