A small block starts sliding down an inclined plane subtending an angle `theta` with the horizontal . The coefficient of friction between the block and plane depends on the distance covered from rest along the plane as `mu = mu_(0)x` where `mu_(0)` is a constant . Find (a) the distance covered by the block down the plane till it stops sliding and (b) its maximum velocity during this journey .

Explanation

If a be the acceleration of the block sliding down the plane when it is at a distance x from the starting point , the equation fo motion will be

`ma = "mg sin" theta - mu R`
But R = mg cos `theta` and `mu = mu_(0)x`
So , a = g sin `theta - mu_(0)xg "cos" theta " " .........(i)`
Now as `a = (dv)/(dt) = (dv)/(dx) * (dx)/(dt) = v(dv)/(dx)`
So , Eqn . (i) becomes
`v(dv)/(dx) = g"sin" theta - mu_(0) xg "cos" theta`
or ` " " vdv = ( "g sin" theta - mu_(0) xg "cos" theta )dx`
The above equation on intergration yields
`(v^(2))/(2) = 2"gx sin" theta - mu_(0) x^(2) . "g cos" theta` ...... (ii)
From Eqn. (ii) it is clear that :
(a) v will be zero when
2gx sin `theta = mu_(0)x^(2) "g cos" theta = 0`
i.e., gx ( 2 sin `theta - mu_(0)x "cos" theta ) = 0`
i.e., x = 0 or x = 2 tan `theta//mu_(0) " " ......... (iii) `
(b) v will be maximum when
(dv/dx) = 0
But from Eqn. (ii) , i.e.,
`v^(2) = 2gx "sin" theta - mu_(0) x^(2)g "cos" theta`
i.e., `" " 2v ((dv)/(dx)) = 2 g "sin" theta - 2 mu_(0) x g "cos" theta `
For (dv/dx) = 0 , it yields x = tan `theta // mu_(0) " " ..... (iv)`
Substituting this value of x in Eqn . (ii) , we get
`v_("max")^(2) = 2 g "sin" theta (1)/(mu_(0)) "tan" theta - mu_(0) "g cos" theta (1)/(mu_(0)^(2)) tan^(2) theta`
which on simplification gives
`v_("max") = "sin" theta sqrt((g)/(mu_(0) "cos" theta)) " " ........ (v)`