A sled loaded with children starts from rest and slides down a snowy $25^\circ$ (with respect to the horizontal) incline traveling $85$ meters in $17$ seconds.  Ignore air resistance. What is the coefficient of kinetic friction between the sled and the slope? 

$ \textbf {(A) } 0.36 \qquad \textbf {(B) } 0.40 \qquad \textbf {(C) } 0.43 \qquad \textbf {(D) } 1.00 \qquad \textbf {(E) } 2.01 $
1. **Determine the acceleration of the sled:**
   The sled starts from rest and travels 85 meters in 17 seconds. We can use the kinematic equation:
   \[
   \Delta x = v_0 t + \frac{1}{2} a t^2
   \]
   Given that \( v_0 = 0 \), \( \Delta x = 85 \) meters, and \( t = 17 \) seconds, we have:
   \[
   85 = \frac{1}{2} a (17)^2
   \]
   Solving for \( a \):
   \[
   85 = \frac{1}{2} a \cdot 289 \implies 85 = 144.5a \implies a = \frac{85}{144.5} \approx 0.588 \, \text{m/s}^2
   \]

2. **Analyze the forces acting on the sled:**
   - The gravitational force \( F_g = mg \) acts downward.
   - The normal force \( F_n \) acts perpendicular to the incline.
   - The frictional force \( F_f \) acts opposite to the direction of motion along the incline.

3. **Resolve the gravitational force into components:**
   - Parallel to the incline: \( F_{g,\parallel} = mg \sin(25^\circ) \)
   - Perpendicular to the incline: \( F_{g,\perp} = mg \cos(25^\circ) \)

4. **Apply Newton's second law along the incline:**
   The net force along the incline is the difference between the gravitational component down the incline and the frictional force:
   \[
   \sum F = F_{g,\parallel} - F_f = ma
   \]
   Substituting the expressions for the forces:
   \[
   mg \sin(25^\circ) - F_f = ma
   \]
   Since \( F_f = \mu F_n \) and \( F_n = mg \cos(25^\circ) \), we have:
   \[
   mg \sin(25^\circ) - \mu mg \cos(25^\circ) = ma
   \]
   Dividing through by \( m \):
   \[
   g \sin(25^\circ) - \mu g \cos(25^\circ) = a
   \]
   Substituting \( g = 9.8 \, \text{m/s}^2 \) and \( a = 0.588 \, \text{m/s}^2 \):
   \[
   9.8 \sin(25^\circ) - \mu 9.8 \cos(25^\circ) = 0.588
   \]

5. **Solve for the coefficient of kinetic friction \( \mu \):**
   \[
   \mu = \frac{9.8 \sin(25^\circ) - 0.588}{9.8 \cos(25^\circ)}
   \]
   Using the values \( \sin(25^\circ) \approx 0.4226 \) and \( \cos(25^\circ) \approx 0.9063 \):
   \[
   \mu = \frac{9.8 \cdot 0.4226 - 0.588}{9.8 \cdot 0.9063} \approx \frac{4.14148 - 0.588}{8.88174} \approx \frac{3.55348}{8.88174} \approx 0.40
   \]

The final answer is \( \boxed{0.40} \).