A block of mass m 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:

- 1/6 m
- 1/3 m
- 1/2 m
- 2/3 m

**Solution**

mg sin θ = μ mg cosθ

tan θ = μ

dy/dx = tan θ = μ = 1/2

y = x^{3}/6

dy/dx = x^{2}/2 = 1/2

x = 1

y = 1/6 m

**The correct option is A.**