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

Given `y = x^(3)//6`
`(dy)/(dx) =(x^(2))/(2)`
The block is just placed on the surface at height where it does not slip The
`tan theta =(dy)/(dx) = (x^(2))/(2)`
At limiting equilibrium `mu = tan theta = 0.5`
or `0.5 = x^(2)//2` or `x^(2) =1` or `x = + - 1`
`y=(1)^(3)//6=(1//6)m`
.