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Let's solve the problem step-by-step.

### **Known Data:**
- **Mass of top box (m₁)** = 32.0 kg  
- **Mass of bottom box (m₂)** = 48.0 kg  
- **Coefficient of kinetic friction between bottom box and ramp (μₖ)** = 0.444  
- **Coefficient of static friction between the two boxes (μₛ)** = 0.800 (assumed sufficient to prevent slipping).  
- **Speed** = 15.0 cm/s = 0.150 m/s (but since it’s constant, acceleration = 0)  
- **Incline height (h)** = 2.50 m  
- **Incline length (L)** = 4.75 m  

We need to **find the force (F)** the man must apply to pull both boxes up the ramp at constant speed.

---

### **Step 1: Calculate the angle of the ramp (θ).**

Using trigonometry:
\[
\sin \theta = \frac{h}{L} = \frac{2.50}{4.75}
\]
\[
\theta = \sin^{-1}(0.5263) \approx 31.8^\circ
\]

---

### **Step 2: Find the forces acting on the system.**

Since the boxes move at **constant speed**, the net force along the ramp is **zero**. That means the force applied by the man (F) exactly balances the total friction and the components of weight along the ramp.

#### Forces to account for:
1. **Weight components along the incline** (parallel to the ramp):  
   \[ (m₁ + m₂) g \sin \theta \]
2. **Kinetic friction between the lower box and the ramp**:  
   \[ f_k = \mu_k N \]  
   where \( N = (m₁ + m₂) g \cos \theta \) is the normal force.

---

### **Step 3: Calculate the individual components.**

- **Total mass of both boxes**:  
  \[ m₁ + m₂ = 32.0 + 48.0 = 80.0 \, \text{kg} \]

- **Weight component along the incline**:  
  \[ (m₁ + m₂) g \sin \theta = 80.0 \times 9.81 \times \sin(31.8^\circ) \]  
  \[ = 80.0 \times 9.81 \times 0.5263 \approx 412.9 \, \text{N} \]

- **Normal force (N)**:  
  \[ N = (m₁ + m₂) g \cos \theta = 80.0 \times 9.81 \times \cos(31.8^\circ) \]  
  \[ = 80.0 \times 9.81 \times 0.8509 \approx 666.4 \, \text{N} \]

- **Kinetic friction force (fₖ)**:  
  \[ f_k = \mu_k N = 0.444 \times 666.4 \approx 295.8 \, \text{N} \]

---

### **Step 4: Calculate the total force required (F).**

Since the boxes move at a **constant speed**, the force the man applies must balance the sum of the weight component along the incline and the kinetic friction force:

\[
F = (m₁ + m₂) g \sin \theta + f_k
\]
\[
F = 412.9 + 295.8 = 708.7 \, \text{N}
\]

---

### **Final Answer:**

The force the man must apply to move the boxes up the ramp at a constant speed of 15.0 cm/s is:

\[
\boxed{708.7 \, \text{N}}
\]

---

Would you like further details on any part of this solution or help with related concepts?

---

### **5 Related Questions:**
1. How would the required force change if the coefficient of kinetic friction was reduced?
2. What would happen if the incline angle was increased?
3. How can we calculate the maximum force before the top box slips off the bottom one?
4. What is the minimum force needed if there was no friction at all?
5. How would acceleration affect the force calculation if the boxes were speeding up?

---

### **Tip:**  
When solving incline problems, always break forces into components parallel and perpendicular to the ramp to simplify the calculations.

A man is pulling two boxes, one on top of the other, up a ramp by pulling on a rope parallel to the surface of the ramp. The coefficient of kinetic friction between the ramp and the lower box is 0.444, and the coefficient of static friction between the two boxes is 0.800. With what force does the man have to pull in order to move the boxes up the ramp at a constant speed of 15.0 cm/s?

Learn how to calculate the force required to pull two stacked boxes up an inclined ramp at a constant speed of 15.0 cm/s. This problem involves trigonometry, frictional forces, and Newton's Second Law. Suitable for students in Grades 11-12.

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