A block of mass `'m'` is placed on a rough surface with a vertical cross section of `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 .

Explanation

`Tan theta=(dy)/(dx)=(d)/(dx)((x^(3))/(6)) rArr Tan theta=(x^(2))/(2)`
At limiting equilibrium, we get `mu=Tan theta rArr 0.5=(x^(2))/(2)`
`x^(2)=1 rArr x=pm1`
Now putting the values of .x. in `y=(x^(3))/(6)`, we get
When `x=1 rArr y=(1)/(6), x=-1 rArr y=-(1)/(6)`
So the maximum height above the ground at which the block can be placed without slipping is `y=(1)/(6)m`