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Step-by-Step Solution
Step 1: Understand the Problem
We have a surface described by the equation
$$y = \frac{x^3}{6},$$
and a block of mass $m$ is placed on this surface. The coefficient of friction between the block and the surface is
$$\mu = 0.5.$$
We want to find the maximum height $y$ above the ground such that the block just remains at rest without slipping.
Step 2: Condition for Limiting Equilibrium
When a block is on the verge of slipping on an inclined surface, we generally use the condition
$$\mu = \tan \theta,$$
where $\theta$ is the angle of inclination of the surface at the point where the block is placed. This condition tells us that friction is at its maximum possible value (limiting friction).
Step 3: Relate the Angle of Inclination to the Slope
The slope of the given curve
$$y = \frac{x^3}{6}$$
at any point $(x,\,y)$ is given by
$$\frac{dy}{dx} = \frac{x^2}{2}.$$
This slope is the tangent of the angle $\theta$ the surface makes with the horizontal at that point. Hence,
$$\tan \theta = \frac{dy}{dx} = \frac{x^2}{2}.$$
For the limiting equilibrium condition, we set
$$\mu = \tan \theta = \frac{x^2}{2}.$$
Step 4: Solve for x
Given
$$\mu = 0.5,$$
we have
$$0.5 = \frac{x^2}{2}.$$
Multiplying both sides by 2:
$$x^2 = 1 \quad \Longrightarrow \quad x = \pm 1.$$
Since we are interested in a real physical position above the ground, we take
$$x = 1 \quad \bigl(\text{the negative value symmetrically gives the same height}\bigr).$$
Step 5: Calculate the Corresponding Height y
For the value $x = 1$, substitute into the curve equation:
$$y = \frac{x^3}{6} = \frac{1^3}{6} = \frac{1}{6}.$$
Hence, the maximum height above the ground at which the block can be placed without slipping is
$$\frac{1}{6} \, \text{(in the same units as }x\text{ and }y\text{, here presumably meters}).$$
Final Answer
The correct answer is
$$\frac{1}{6}m.$$
