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Let me solve this problem step-by-step:

### Given Data:
- Mass of the airplane, \( m = 89,320 \, \text{kg} \)
- Initial velocity, \( v_i = 83.5 \, \text{m/s} \)
- Final velocity, \( v_f = 0 \, \text{m/s} \) (since the plane stops)
- Stopping distance, \( d = 415 \, \text{m} \)
- Reverse thrust force, \( F_{\text{reverse}} = 313 \, \text{kN} = 313,000 \, \text{N} \)

We need to calculate:
1. The deceleration (\( a \)).
2. The additional force of friction required to stop the plane (\( F_{\text{friction}} \)).
3. The minimum coefficient of friction (\( \mu_{\text{min}} \)).

---

### 1. **Finding Deceleration (\( a \)):**
We can use the kinematic equation:
\[
v_f^2 = v_i^2 + 2ad
\]
Rearranging for \( a \):
\[
a = \frac{v_f^2 - v_i^2}{2d}
\]
Substituting the values:
\[
a = \frac{0^2 - (83.5)^2}{2 \times 415}
\]
\[
a = \frac{-6972.25}{830} \approx -8.4 \, \text{m/s}^2
\]
Thus, the deceleration is \( a = -8.4 \, \text{m/s}^2 \).

---

### 2. **Finding Additional Force of Friction (\( F_{\text{friction}} \)):**
Using Newton's Second Law, the total force required to stop the plane is:
\[
F_{\text{total}} = ma
\]
Substituting the values:
\[
F_{\text{total}} = 89,320 \times (-8.4) \approx -750,288 \, \text{N}
\]

The reverse thrust provides \( F_{\text{reverse}} = -313,000 \, \text{N} \). The additional force of friction required is:
\[
F_{\text{friction}} = F_{\text{total}} - F_{\text{reverse}}
\]
\[
F_{\text{friction}} = -750,288 - (-313,000) \approx -437,288 \, \text{N}
\]
Thus, \( F_{\text{friction}} = 437,288 \, \text{N} \) (acting opposite to the motion).

---

### 3. **Finding Minimum Coefficient of Friction (\( \mu_{\text{min}} \)):**
The force of friction is related to the normal force (\( F_N \)) and the coefficient of friction:
\[
F_{\text{friction}} = \mu F_N
\]
Here, \( F_N = mg \), where \( g = 9.8 \, \text{m/s}^2 \):
\[
F_N = 89,320 \times 9.8 \approx 875,336 \, \text{N}
\]

The minimum coefficient of friction is:
\[
\mu_{\text{min}} = \frac{F_{\text{friction}}}{F_N}
\]
Substituting the values:
\[
\mu_{\text{min}} = \frac{437,288}{875,336} \approx 0.5
\]

---

### Final Results:
1. Deceleration of the airplane: \( 8.4 \, \text{m/s}^2 \)
2. Additional force of friction required: \( 437,288 \, \text{N} \)
3. Minimum coefficient of friction: \( 0.5 \)

---

Would you like further details on any step? Here are some follow-up questions for further exploration:

1. How would the stopping distance change if the reverse thrust were doubled?
2. What if the coefficient of friction was only \( 0.3 \)—would the plane still stop in time?
3. How would the problem change if the mass of the airplane were increased?
4. Can you derive the deceleration formula from basic principles of kinematics?
5. How does air resistance influence stopping forces in real-world scenarios?

**Tip:** Always check your units to ensure consistency throughout calculations.

An 89,320 kg airplane landing at 83.5 m/s must stop in a distance of 415 m. The engines can generate 313 kN of reverse thrust. What must be the deceleration of the airplane? What additional force of friction from the tires does the plane need to stop? What is the minimum coefficient of friction that must exist between the tires of the plane and the runway?

Learn how to calculate the deceleration, additional friction force, and minimum coefficient of friction for an airplane landing at 83.5 m/s and stopping within 415 meters. This problem involves kinematics, forces, and friction, suitable for Grades 11-12 physics students.

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