A body is moving down a long inclined plane in inclination `45^(@)` with horizontal. The coefficient of friction between the body and the plane varies as `mu=x//2`, where `x` is the distance moved down the plane. Initially
`x=0 & v=0`.

Options

A. When `x=2` the velocity of the body is `sqrt(gsqrt(2))m//s`
B. The velocity of the body increases all the time
C. At an instant when `v!=0` the instantaneous acceleration of the body down the plane is
`(g(2-x))/(2sqrt(2))`
D. The body first accelerates and then decelerates

Answer

A, C, D

Explanation

When `v cancel=0` the aceeleration is
`vec(a)_("inst") = (1)/(m)(mg sin theta-mu mg cos theta)`
`= g(sin theta-(x)/(2) cos theta)-(g(2-x))/(2sqrt(2))`
`rArr v(dv)/(dx) = (g)/(2sqrt(2))(2-x)`
`rArr underset(0)overset(v)int v dv = (g)/(2sqrt(2))[2x-(x^(2))/(2)]_(0)^(2) rArr v = sqrt(gsqrt(2))m//s`
Also it is clear that for `x lt 2` the body accelerates, at `x = 2` the acceleration is zero and for `x gt 2` the body retards till it comes to rest.