1. Equation of the curved surface

The vertical cross-section of the surface is $y=\frac{x^2}{4}$ where $y$ is vertical and $x$ is horizontal.

We need the maximum height at which a block placed at rest will not slide downward.

2. Condition for equilibrium on a rough incline

At any point on the curve, the block experiences:

• weight $mg$ vertically downward,
• normal reaction $N$ perpendicular to the surface,
• static friction $f$ along the tangent.

For the block to remain at rest, $mg\sin\theta \le \mu N$ where $\theta$ is the angle the tangent makes with the horizontal.

Since $N=mg\cos\theta,$ the condition becomes $mg\sin\theta \le \mu mg\cos\theta$ $\tan\theta \le \mu$

So the limiting condition for the highest possible point is $\tan\theta = \mu = 0.5$

3. Slope of the curve

For the curve $y=\frac{x^2}{4},$ its slope is $\frac{dy}{dx}=\frac{x}{2}$

But slope of tangent is also $\tan\theta = \frac{dy}{dx}$

Hence at limiting equilibrium, $\frac{x}{2}=0.5$ $x=1$

4. Find the corresponding height

Substitute $x=1$ into the curve equation: $y=\frac{1^2}{4}=\frac{1}{4}$

Thus the maximum vertical height is $y=0.25\ \text{m}$

Converting to cm: $0.25\times 100=25\ \text{cm}$

5. Final answer

The maximum height at which the block will not slip is $\boxed{25\ \text{cm}}$