A block of mass is placed on a surface with a vertical cross section given by `y=x^3/6`. If the coefficient of friction is 0.5, the maximum height above the ground at which the block can be placed without slipping is:

Options

A. `(1)/(3) m`
B. `(1)/(2) m`
C. `(1)/(6) m`
D. `(2)/(3) m`

Answer

C

Explanation

Block is under limiting friction, so
`mg sin theta = mu m g cos theta implies mu = tan theta`
Equation of the surface, `y = (x^(2))/(6)`
Slope of the curved surface
`tan theta = (dy)/(dx) = (x^(2))/(2)`
From equations (i) and (ii) we get
`0.5 = (x^(2))/(2) implies 0.5 = (x^(2))/(2) implies x^(2 = 1` or x = 1 so, `y = (x^(3))/(6) = ((1)^(2))/(6) = (1)/(6)`