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Given this physics problem, let's go through the calculations step by step to determine the coefficient of kinetic friction between the wooden crate and the inclined plane.

### Problem Analysis
We have:
- Mass of the crate, \( m = 60 \, \text{kg} \)
- Applied force, \( F = 800 \, \text{N} \)
- Incline height, \( h = 2 \, \text{m} \)
- Length of the incline, \( L = 2.5 \, \text{m} \)
- Acceleration of the crate, \( a = 2 \, \text{m/s}^2 \)
- Gravitational acceleration, \( g = 10 \, \text{m/s}^2 \)

We are tasked with finding the coefficient of kinetic friction \( \mu_k \).

### Step 1: Calculate the Inclination Angle
The incline angle \( \theta \) can be determined from the height and the length of the incline using trigonometry:
\[
\sin \theta = \frac{h}{L} = \frac{2}{2.5} = 0.8
\]
\[
\theta = \arcsin(0.8)
\]

### Step 2: Resolve Forces along the Incline
The forces acting on the crate along the inclined plane are:
1. The gravitational component along the incline: \( F_{\text{gravity}} = mg \sin \theta \)
2. The applied force \( F = 800 \, \text{N} \)
3. The frictional force \( F_{\text{friction}} = \mu_k N \), where \( N \) is the normal force.

### Step 3: Calculate the Normal Force
The normal force \( N \) is given by:
\[
N = mg \cos \theta
\]

### Step 4: Apply Newton's Second Law along the Incline
Using Newton's second law, we get:
\[
F - F_{\text{gravity}} - F_{\text{friction}} = ma
\]
Substituting in the expressions:
\[
800 - (mg \sin \theta) - (\mu_k \cdot mg \cos \theta) = ma
\]

### Step 5: Solve for \( \mu_k \)
Isolating \( \mu_k \), we have:
\[
\mu_k = \frac{F - mg \sin \theta - ma}{mg \cos \theta}
\]

Let’s substitute the known values and calculate \( \mu_k \):

1. \( mg = 60 \times 10 = 600 \, \text{N} \)
2. Using \( \sin \theta = 0.8 \) and \( \cos \theta = \sqrt{1 - 0.8^2} = 0.6 \)
3. Plugging these values into the formula for \( \mu_k \):

\[
\mu_k = \frac{800 - 600 \times 0.8 - 60 \times 2}{600 \times 0.6}
\]

Calculating each term individually:
- \( 600 \times 0.8 = 480 \)
- \( 60 \times 2 = 120 \)
- \( 600 \times 0.6 = 360 \)

So,
\[
\mu_k = \frac{800 - 480 - 120}{360} = \frac{200}{360} = \frac{5}{9} \approx 0.56
\]

### Final Answer
The coefficient of kinetic friction \( \mu_k \approx 0.56 \).

Would you like further details on any of these steps or have additional questions?

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Here are 5 related questions for practice:

1. How would the coefficient of kinetic friction change if the incline length increased?
2. What would be the effect of a larger applied force on the acceleration?
3. How would you calculate the normal force if the incline angle was unknown?
4. What is the role of the frictional force in this problem?
5. How would the solution change if the crate moved down the incline instead?

**Tip:** When solving physics problems, always break down forces along and perpendicular to the incline to simplify calculations.

Sebuah peti kayu bermassa 60 kg didorong oleh seseorang dengan gaya 800 N ke atas sebuah truk menggunakan papan yang disandarkan membentuk bidang miring. Ketinggian bak truk tempat papan bersandar adalah 2 m dan panjang papan yang digunakan adalah 2,5 m. Jika peti bergerak ke atas dengan percepatan 2 m/s^2 dan g=10 m/s^2 maka tentukan koefisien gesek kinetis antara peti kayu dengan papan.

Learn how to calculate the coefficient of kinetic friction for a crate on an inclined plane. Given a 60 kg crate pushed with 800 N, inclined at an angle with height 2 m and length 2.5 m, and an acceleration of 2 m/s^2, this problem involves trigonometry and Newton's second law. Suitable for students in Grades 11-12.

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