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To solve this problem, we need to apply concepts of work and energy, specifically the work-energy theorem, and analyze the forces acting on the object as it slides on the inclined plane and then on the horizontal plane.

### Problem Breakdown:
1. **Inclined Plane**: The object slides down an incline of \(30^\circ\) and reaches the horizontal plane with velocity \(v\).
2. **Horizontal Plane**: The object continues to move on a horizontal plane where it eventually comes to a stop due to friction.
3. **Objective**: Find the coefficient of kinetic friction (\(\mu_k\)) on the horizontal plane.
4. **Given Condition**: The object travels the same distance on the inclined plane and the horizontal plane.

### Solution Approach:
We will solve this problem using the concepts of energy and work done by friction.

1. **Energy Analysis on the Inclined Plane**:
   - The object starts from rest, so the initial kinetic energy \(KE_i = 0\).
   - The potential energy lost by the object as it descends is transformed into kinetic energy at the bottom of the incline.

   Let:
   - \(d\) be the distance traveled along the incline.
   - \(m\) be the mass of the object.
   - \(g\) be the acceleration due to gravity.

   **Potential Energy Lost**:
   \[
   \Delta PE = m \cdot g \cdot d \cdot \sin(30^\circ)
   \]

   Since \(\sin(30^\circ) = 0.5\), this becomes:
   \[
   \Delta PE = 0.5 \cdot m \cdot g \cdot d
   \]

   **Kinetic Energy at the Bottom**:
   By conservation of energy:
   \[
   KE = \Delta PE
   \]
   \[
   \frac{1}{2} m v^2 = 0.5 \cdot m \cdot g \cdot d
   \]

   Simplifying, we find:
   \[
   v^2 = g \cdot d
   \]

2. **Work Done by Friction on the Horizontal Plane**:
   - On the horizontal plane, the object decelerates due to kinetic friction until it stops.
   - The work done by friction is equal to the initial kinetic energy at the start of the horizontal plane.

   Let \( \mu_k \) be the coefficient of kinetic friction, and \(d\) the distance over which friction acts (equal to the distance traveled on the incline).

   **Friction Force**:
   \[
   F_f = \mu_k \cdot m \cdot g
   \]

   **Work Done by Friction**:
   \[
   W_f = F_f \cdot d = \mu_k \cdot m \cdot g \cdot d
   \]

   Since the object comes to rest, this work is equal to the kinetic energy at the bottom of the incline:
   \[
   \mu_k \cdot m \cdot g \cdot d = \frac{1}{2} m v^2
   \]

   Using the expression \(v^2 = g \cdot d\):
   \[
   \mu_k \cdot m \cdot g \cdot d = \frac{1}{2} m \cdot g \cdot d
   \]

   Canceling \(m\), \(g\), and \(d\) (since \(d \neq 0\)):
   \[
   \mu_k = \frac{1}{2}
   \]

### Final Answer:
The coefficient of kinetic friction \(\mu_k\) on the horizontal plane is:
\[
\mu_k = 0.5
\]

Would you like further details on any step of this solution?

### Related Questions:
1. How would the solution change if the angle of the incline were different?
2. What if the friction coefficient were different on the inclined and horizontal planes?
3. How would you approach this problem if the object had an initial velocity?
4. What role does the object's mass play in determining the coefficient of friction?
5. How would you solve this problem using forces instead of energy?

### Tip:
Always check if you can simplify the problem by canceling common variables across equations—this often simplifies the math significantly, as seen here where mass canceled out in the final equation.

Un cuerpo desliza a lo largo de un plano inclinado con un ángulo de 30° y luego continúa moviéndose sobre el plano horizontal. Determinar el coeficiente de rozamiento si se sabe que el cuerpo recorre en el plano inclinado la misma distancia que en el horizontal. Resolver por trabajo y energía.

Learn how to calculate the coefficient of kinetic friction for a body sliding down an inclined plane with a 30° angle and moving along a horizontal plane. This solution involves using energy conservation, work-energy theorem, and kinetic friction concepts. Suitable for high school students.

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