As a 40 N block slides down a plane that is inclined at 21 degrees to the horizontal, its acceleration is 0.89 m/s^2, directed up the plane. What is the coefficient of kinetic friction between the block and the plane?

To determine the coefficient of kinetic friction (\( \mu_k \)) between the block and the inclined plane, we can use the following steps:

1. **Identify the forces acting on the block**:
   - Weight of the block (\( W = 40 \, \text{N} \))
   - Normal force (\( N \))
   - Frictional force (\( f_k \))
   - Component of weight parallel to the incline (\( W_{\parallel} \))
   - Component of weight perpendicular to the incline (\( W_{\perpendicular} \))

2. **Calculate the components of the weight**:
   - \( W_{\parallel} = W \sin(\theta) = 40 \sin(21^\circ) \)
   - \( W_{\perpendicular} = W \cos(\theta) = 40 \cos(21^\circ) \)

3. **Using Newton's second law**:
   The net force (\( F_{\text{net}} \)) acting on the block along the incline can be expressed as:
   \[
   F_{\text{net}} = W_{\parallel} - f_k
   \]
   where \( f_k = \mu_k N \).

   The normal force \( N \) is equal to the perpendicular component of the weight:
   \[
   N = W_{\perpendicular} = W \cos(21^\circ)
   \]

4. **Express the net force in terms of acceleration**:
   \[
   F_{\text{net}} = ma = m(0.89)
   \]

5. **Set up the equation**:
   Using the equations, we can write:
   \[
   W \sin(21^\circ) - \mu_k W \cos(21^\circ) = ma
   \]

6. **Plug in the values**:
   We know \( m = \frac{W}{g} = \frac{40 \, \text{N}}{9.81 \, \text{m/s}^2} \).

   Let's calculate each term step by step:

   - Calculate \( W_{\parallel} \):
     \[
     W_{\parallel} = 40 \sin(21^\circ) \approx 40 \times 0.3584 \approx 14.336 \, \text{N}
     \]

   - Calculate \( W_{\perpendicular} \):
     \[
     W_{\perpendicular} = 40 \cos(21^\circ) \approx 40 \times 0.9336 \approx 37.344 \, \text{N}
     \]

   - Substitute \( N \) and \( a \) into the equation:
     \[
     14.336 - \mu_k \times 37.344 = \frac{40}{9.81} \times 0.89
     \]

   - Calculate \( m \times a \):
     \[
     m \times a \approx \frac{40}{9.81} \times 0.89 \approx 3.608 \, \text{N}
     \]

7. **Now substitute and solve for \( \mu_k \)**:
   \[
   14.336 - \mu_k \times 37.344 = 3.608
   \]

   Rearranging gives:
   \[
   \mu_k \times 37.344 = 14.336 - 3.608
   \]
   \[
   \mu_k \times 37.344 = 10.728
   \]
   \[
   \mu_k = \frac{10.728}{37.344} \approx 0.287
   \]

Thus, the coefficient of kinetic friction (\( \mu_k \)) is approximately **0.287**.

If you need any further details or have questions, let me know! Here are some related questions:

1. How does the angle of inclination affect the frictional force?
2. What would happen to the coefficient of friction if the mass of the block increased?
3. Can you explain the difference between static and kinetic friction?
4. How would the problem change if the block was stationary?
5. What is the significance of the acceleration being directed up the plane? 

**Tip**: Always sketch a free-body diagram to visualize forces acting on an object.

Explore how to find the coefficient of kinetic friction for a 40 N block sliding down a 21-degree incline with an acceleration of 0.89 m/s² directed up the plane. This solution involves analyzing forces and applying Newton's second law, suitable for high school physics students.

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