A block of mass 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:

Options

A. (a)`(1)/(3) m`
B. (b)`(1)/(2) m`
C. (c)`(1)/(6) m`
D. (d)`(2)/(3) m`

Explanation

To solve the problem step by step, we will analyze the situation of a block placed on a surface defined by the equation \( y = \frac{x^3}{6} \) and determine the maximum height at which it can be placed without slipping, given that the coefficient of friction is 0.5.

### Step 1: Understand the Geometry of the Problem
The surface is described by the equation \( y = \frac{x^3}{6} \). We can visualize this as a curve in the xy-plane. The block will be placed on this curve, and we need to find the maximum height (y-coordinate) at which the block can rest without slipping.

**Hint:** Draw the curve \( y = \frac{x^3}{6} \) to visualize how the block will rest on it.

### Step 2: Calculate the Slope of the Surface
To find the slope of the surface at any point, we need to differentiate the equation with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}\left(\frac{x^3}{6}\right) = \frac{3x^2}{6} = \frac{x^2}{2} \]

**Hint:** Remember that the slope of the surface at any point is given by the derivative of the height function.

### Step 3: Relate the Slope to the Angle of Inclination
The slope of the surface can be expressed as:
\[ \tan(\theta) = \frac{dy}{dx} = \frac{x^2}{2} \]
where \( \theta \) is the angle of inclination of the surface.

**Hint:** The angle of inclination is crucial for understanding the forces acting on the block.

### Step 4: Apply the Condition for Limiting Equilibrium
At the point of slipping, the maximum static friction force equals the component of gravitational force acting down the slope. The condition for limiting equilibrium is:
\[ \mu_s = \tan(\theta) \]
where \( \mu_s \) is the coefficient of static friction. Given \( \mu_s = 0.5 \), we have:
\[ 0.5 = \frac{x^2}{2} \]

**Hint:** Set the coefficient of friction equal to the tangent of the angle to find the critical point.

### Step 5: Solve for \( x \)
From the equation \( 0.5 = \frac{x^2}{2} \), we can solve for \( x \):
\[ x^2 = 1 \implies x = \pm 1 \]

**Hint:** Remember that \( x \) can be positive or negative, but we are interested in the height, which will be the same for both.

### Step 6: Calculate the Corresponding Height \( y \)
Now, substitute \( x = 1 \) back into the original equation to find the height \( y \):
\[ y = \frac{x^3}{6} = \frac{1^3}{6} = \frac{1}{6} \]

**Hint:** The height is derived from the original equation of the surface.

### Final Answer
The maximum height above the ground at which the block can be placed without slipping is:
\[ \boxed{\frac{1}{6} \text{ meters}} \]