A small chunk of ice falls from rest down a frictionless parabolic ice sheet shown in the figure. At the point labeled $\mathbf{A}$ in the diagram, the ice sheet becomes a steady, rough incline of angle $30^\circ$ with respect to the horizontal and friction coefficient $\mu_k$. This incline is of length $\frac{3}{2}h$ and ends at a cliff. The chunk of ice comes to a rest precisely at the end of the incline. What is the coefficient of friction $\mu_k$?

[asy]
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[/asy]

$ \textbf{(A)}\ 0.866\qquad\textbf{(B)}\ 0.770\qquad\textbf{(C)}\ 0.667\qquad\textbf{(D)}\ 0.385\qquad\textbf{(E)}\ 0.333 $
1. **Determine the initial potential energy:**
   The chunk of ice starts from rest at height \( h \). The potential energy at this height is given by:
   \[
   PE = mgh
   \]
   where \( m \) is the mass of the ice chunk, \( g \) is the acceleration due to gravity, and \( h \) is the height.

2. **Calculate the work done by friction:**
   As the ice chunk slides down the rough incline, friction does work against its motion. The work done by friction \( W_{\text{fr}} \) is given by:
   \[
   W_{\text{fr}} = F_{\text{fr}} \cdot d
   \]
   where \( F_{\text{fr}} \) is the frictional force and \( d \) is the distance traveled along the incline. The distance \( d \) is given as \( \frac{3}{2}h \).

3. **Determine the frictional force:**
   The frictional force \( F_{\text{fr}} \) is given by:
   \[
   F_{\text{fr}} = \mu_k F_N
   \]
   where \( \mu_k \) is the coefficient of kinetic friction and \( F_N \) is the normal force. For an incline at an angle \( \theta = 30^\circ \), the normal force \( F_N \) is:
   \[
   F_N = mg \cos \theta = mg \cos 30^\circ = mg \cdot \frac{\sqrt{3}}{2}
   \]

4. **Substitute the normal force into the frictional force equation:**
   \[
   F_{\text{fr}} = \mu_k \cdot mg \cdot \frac{\sqrt{3}}{2}
   \]

5. **Calculate the total work done by friction:**
   \[
   W_{\text{fr}} = F_{\text{fr}} \cdot d = \mu_k \cdot mg \cdot \frac{\sqrt{3}}{2} \cdot \frac{3}{2}h = \frac{3\sqrt{3}}{4} mgh \mu_k
   \]

6. **Set up the energy conservation equation:**
   The initial potential energy is converted into work done by friction:
   \[
   mgh - \frac{3\sqrt{3}}{4} mgh \mu_k = 0
   \]

7. **Solve for \( \mu_k \):**
   \[
   mgh = \frac{3\sqrt{3}}{4} mgh \mu_k
   \]
   \[
   1 = \frac{3\sqrt{3}}{4} \mu_k
   \]
   \[
   \mu_k = \frac{4}{3\sqrt{3}}
   \]
   Simplify the expression:
   \[
   \mu_k = \frac{4}{3\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{4\sqrt{3}}{9}
   \]
   \[
   \mu_k \approx 0.770
   \]

The final answer is \(\boxed{0.770}\)