In the adjacent figure, x-axis has been taken down the inclined plane and y-axis perpendicular to the inclined plane. Te coefficient friction varies with x as `mu = kx`, where `k = tan theta`. A block is released at O.

Frictional force acting on the block just before it comes to rest

Options

A. `mg sin theta`
B. `2mg sin theta`
C. `(mg sin theta)/(2)`
D. `2 mg cos theta`

Answer

B

Explanation

From figure
`mg sin theta - mu mg cos theta = ma`
`a = g (sin theta - kx cos theta)`
`= g (sin theta - sin theta . X)`
`:. a = g sin theta (1 - x)`

for `v_("max"), v (dv)/(dx) =g sin theta (1 - x)`......(i)
Speed will be maximum when acceleration = 0
`g sin theta (1 - x) = 0 rArr x = 1`
From equation (i)
`rArr int v xx dv = int g sin theta (1 - x) dx`
`rArr (v^(2))/(2) = g sin theta xx x - g sin theta xx (x^(2))/(2)`.....(ii)
`:. (v_("max")^(2))/(2) = (g sin)/(2) rArr v_("max") = sqrt(g sin theta)`
`F_(x) = kx xx mg cos theta = tan theta x xx mg sin theta = x xx mg sin theta`...(iii)
From equation (ii)
`(v^(2))/(2) = g sin theta xx x - g sin theta xx (x^(2))/(2)`
when block will be at rest then `v = 0`
`0 = g sin theta xx x - g sin theta xx (x^(2))/(2) rArr x = 2`
Putting value of x in equation (iii) we get
`F_(x) = 2 mg sin theta`