Q: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

(a) 1/6m

(b)2/3m

(c)1/3m

(d) 1/2m

Ans: (a)

Sol:$\large y = \frac{x^3}{6}$

$\large \frac{dy}{dx}=\frac{x^2}{2}$

$\large tan\theta = \frac{x^2}{2}$

$\large 0.5 = \frac{x^2}{2}$

x^{2} = 1

$x = \pm 1$

y = 1/6