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The problem describes a block moving along a frictionless curved path, then entering a rough inclined plane with a kinetic friction coefficient \( \mu_k = 0.2 \), and eventually stopping at a height \( y_{\text{maks}} \). We are tasked with finding the equation for the maximum height \( y_{\text{maks}} \), where the block stops.

### Analysis:

1. **Energy Conservation (Curved Path)**: 
   - Initially, the block starts from height \( h \), so its potential energy is \( U_{\text{initial}} = mgh \) (where \( m \) is the mass of the block and \( g \) is the gravitational acceleration). Since the path is frictionless, no energy is lost in this phase.

2. **Energy Loss Due to Friction (Inclined Plane)**: 
   - On the rough inclined plane with friction coefficient \( \mu_k \), the work done by friction reduces the mechanical energy. The work done by friction as the block moves on the inclined plane is \( W_{\text{friction}} = \mu_k N d \), where \( N = mg \cos \theta \) is the normal force and \( d \) is the distance traveled along the incline.
   - The energy lost due to friction can be written as \( E_{\text{lost}} = \mu_k mg \cos \theta d \).

3. **Energy Balance**:
   - The block loses energy due to friction and eventually comes to rest. At height \( y_{\text{maks}} \), the remaining energy is \( mgy_{\text{maks}} \).
   - Therefore, applying the work-energy principle:
   \[
   mgh = mgy_{\text{maks}} + \mu_k mg \cos \theta \cdot d
   \]
   Canceling the mass \( m \) and simplifying gives:
   \[
   gh = gy_{\text{maks}} + \mu_k g \cos \theta \cdot d
   \]
   Simplifying further:
   \[
   h = y_{\text{maks}} + \mu_k \cos \theta \cdot d
   \]

4. **Relating Distance \( d \) to Height \( y_{\text{maks}} \)**:
   - Using geometry, the height \( y_{\text{maks}} \) on the inclined plane is related to the distance \( d \) as \( y_{\text{maks}} = d \sin \theta \). Solving for \( d \):
   \[
   d = \frac{y_{\text{maks}}}{\sin \theta}
   \]
   Substituting this into the energy equation gives:
   \[
   h = y_{\text{maks}} + \mu_k \cos \theta \cdot \frac{y_{\text{maks}}}{\sin \theta}
   \]
   Simplifying:
   \[
   h = y_{\text{maks}} \left( 1 + \mu_k \cot \theta \right)
   \]
   Solving for \( y_{\text{maks}} \):
   \[
   y_{\text{maks}} = \frac{h}{1 + \mu_k \cot \theta}
   \]

### Conclusion:
From the options given, the correct equation is:

- **Option D**: \( y_{\text{maks}} = \frac{h}{1 + \mu_k \cot \theta} \)

Would you like further clarification or details? 

Here are some related questions for further practice:
1. What happens to \( y_{\text{maks}} \) if the friction coefficient \( \mu_k \) is increased?
2. How does the angle \( \theta \) affect the maximum height \( y_{\text{maks}} \)?
3. What is the physical meaning of the term \( \mu_k \cot \theta \) in the equation for \( y_{\text{maks}} \)?
4. How would the problem change if the initial path was not frictionless?
5. How does the presence of kinetic friction affect the total mechanical energy of the system?

**Tip:** Always pay attention to how friction affects energy conservation principles!

A block moves along a smooth curved path and then enters a rough inclined plane with a kinetic friction coefficient of μ_k = 0.2. The block stops at a maximum height y_max as shown. Which equation correctly represents the final position of the block at y_max?

This physics problem involves energy conservation, friction, and trigonometry to calculate the maximum height y_max where a block stops after moving on a rough inclined plane. The kinetic friction coefficient μ_k = 0.2 is given, and the solution uses the work-energy principle.

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