1. Given curve and friction condition

The surface has vertical cross-section $y=\frac{x^3}{6}.$

The block will remain at rest if the component of gravity along the tangent does not exceed the maximum static friction.

For an incline making angle $\theta$ with the horizontal, equilibrium requires $mg\sin\theta \le \mu mg\cos\theta$ which gives $\tan\theta \le \mu.$

Here $\mu=0.5=\frac12$.

2. Find slope of the surface

For the curve, $\frac{dy}{dx}=\frac{x^2}{2}.$

But for a curve, the slope of the tangent is $\tan\theta=\frac{dy}{dx}.$

So the no-slipping condition becomes $\frac{x^2}{2} \le \frac12.$

Hence, $x^2 \le 1 \quad \Rightarrow \quad |x|\le 1.$

The maximum allowed height occurs at the largest allowed $x$ (taking positive side), i.e. $x=1.$

3. Compute the corresponding height

Substitute $x=1$ into $y=\frac{x^3}{6}:$

$y_{\max}=\frac{1^3}{6}=\frac16.$

So the maximum height above the ground is $\boxed{\frac16\text{ m}}.$

4. Check options

• A: $\frac16$ m ✅
• B: $\frac23$ m ❌
• C: $\frac13$ m ❌
• D: $\frac12$ m ❌

Therefore, the correct option is A.

5. Comparison with stored answer

Stored correct answer: A

My derived answer: A

They agree.