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

We consider the curve that makes the supporting surface. Its equation in Cartesian form is

$$y=\dfrac{x^{3}}{6}$$

Here the ground is the $$x$$-axis ($$y=0$$) and the point $$P(x,y)$$ on the curve is the position where the block is placed. At $$P$$ the surface is inclined at an angle $$\theta$$ with the horizontal. For a curve $$y=f(x)$$ the tangent makes an angle whose tangent is the derivative of the curve, that is

$$\tan\theta=\left|\dfrac{dy}{dx}\right|.$$

So we first differentiate $$y$$ with respect to $$x$$:

$$\dfrac{dy}{dx}=\dfrac{d}{dx}\!\left(\dfrac{x^{3}}{6}\right)=\dfrac{3x^{2}}{6}=\dfrac{x^{2}}{2}.$$

Hence the local slope and therefore the tangent of the angle of inclination are

$$\tan\theta=\dfrac{x^{2}}{2}.$$

Now the block will remain at rest provided that the component of its weight trying to pull it down the incline is not larger than the maximum static friction which resists that tendency. The usual condition is

$$\text{(down-slope component)}\;\le\;\text{(maximum static friction)}.$$

Writing the forces explicitly:

• The component of the weight along the incline is $$mg\sin\theta.$$

• The normal reaction from the surface is $$N=mg\cos\theta.$$

• The maximum possible static friction is $$f_{\max}=\mu N=\mu mg\cos\theta,$$ where $$\mu$$ is the coefficient of static friction.

The limiting (just-about-to-slip) case therefore satisfies

$$mg\sin\theta=\mu mg\cos\theta.$$

Dividing both sides by $$mg\cos\theta$$ we obtain

$$\tan\theta=\mu.$$

This is the standard result: the block is safe as long as $$\tan\theta\le\mu.$$ To find the highest position that is still safe we set the equality

$$\tan\theta=\mu.$$

Given $$\mu=0.5=\dfrac12,$$ and substituting $$\tan\theta=\dfrac{x^{2}}{2},$$ we write

$$\dfrac{x^{2}}{2}=\dfrac12.$$

Multiplying both sides by $$2$$:

$$x^{2}=1.$$

Taking the positive root because height increases with positive $$x$$:

$$x=1\;\text{m}.$$

Now we substitute this $$x$$ back into the equation of the curve to find the corresponding height $$y$$:

$$y=\dfrac{x^{3}}{6}=\dfrac{1^{3}}{6}=\dfrac16\;\text{m}.$$

This is the greatest vertical height above the ground at which the block can be placed without beginning to slide.

Hence, the correct answer is Option A.