A block sliding down on an inclined plane forms an angle `theta` with the horizontal. The friction coefficient depends on the distance `x` covered as `mu = kx` , where `k` is a constant. Find the distance covered by the block till it stops and its maximum velocity over this distance. Also , sketch a graph between square of velocity and distance covered `x`.

Explanation

`mu = kx`
`N = mg cos theta , f = mu N = mu mg cos theta`
Along the plane : `mg sin theta - mu mg cos theta = ma`
`a = g sin theta - mu g cos theta`
`mu = kx`
`a = g sin theta - kgx cos theta`
`v(dv)/(dx) = g sin theta - kgx cos theta`
`int_(0)^(v) v dv = int_(0)^(x) ( g sin theta - kgx cos theta) dx`
`v^(2) = 2 g sin theta x - kgx^(2) cos theta`
When the block stops , `v = 0 `
`0 = 2f sin theta x - kgx^(2) cos theta`
When the block stops , `v = 0`
`0 = 2g sin theta x - kgx^(2) cos theta`
`x = x_(0) = ( 2 tan theta)/(k)`
Velocity will be maximum when `a = 0`
`a = g sin theta - kgx cos theta`
`0 = g sin theta - kgx cos theta rArr x = ( tan theta)/(k) = (x_(0))/(2)`
`v_(max)^(2) = 2 g sin theta (tan theta)/(k) - kg cos theta (tan^(2) theta)/(k^(2))`
`v_(max) = sqrt((g)/(k) tan theta sin theta)`
`v^(2) v//s x`
`v^(2) = 2g sin theta x - kg cos x^(2)`
Graph : `y = 2g sin theta x - kg cos theta x^(2)` ( parabola , coefficient of `x^(2) is -ve` , parabola open downward)
Note : (1). From `x = 0 to x_(0)//2 , mg sin theta gt f` , acceleration its increasing.
2. At `x = x_(0)//2 , mg sin theta = f` , net force = 0 , the block is in equilibrium , velocity is maximum.
3. `x = x_(0)//2 to x_(0) , mg sin theta lt f` , retardation , velocity is decreasing and ultimately teh blocks stops.