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

Explanation

To solve the problem, we need to determine the maximum height at which a block can be placed on a surface described by the equation \( y = \frac{x^3}{6} \) without slipping, given that the coefficient of friction is \( \mu = 0.5 \).

### Step-by-Step Solution:

1. **Understand the Surface Equation**:
The surface is defined by the equation:
\[ y = \frac{x^3}{6} \]
This represents a curve in the xy-plane.

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

3. **Relate the Slope to the Coefficient of Friction**:
The maximum angle of inclination before slipping occurs can be determined using the coefficient of friction. The condition for slipping is given by:
\[ \tan(\theta) = \mu \]
where \( \theta \) is the angle of the slope. For small angles, we can approximate \( \tan(\theta) \) by the slope:
\[ \frac{dy}{dx} = \mu \]
Substituting the values we have:
\[ \frac{x^2}{2} = 0.5 \]

4. **Solve for \( x \)**:
Rearranging the equation gives:
\[ x^2 = 0.5 \times 2 = 1 \]
Taking the square root:
\[ x = \pm 1 \]
We will consider \( x = 1 \) since we are looking for the maximum height.

5. **Calculate the Maximum Height \( y \)**:
Now, substitute \( x = 1 \) back into the equation for \( y \):
\[ y = \frac{(1)^3}{6} = \frac{1}{6} \]

6. **Final Result**:
The maximum height above the ground at which the block can be placed without slipping is:
\[ \boxed{\frac{1}{6}} \]