A 1900-kg car experiences a combined force of air resistance and friction that has the same magnitude whether the car goes up or down a hill at 27 m/s. Going up a hill, the car's engine needs to produce 47 hp more power to sustain the constant velocity than it does going down the same hill. At what angle is the hill inclined above the horizontal ?

Options

A. `1.4^(@)`
B. `2.5^(@)`
C. `1.1^(@)`
D. `2.0^(@)`

Answer

To solve the problem, we need to determine the angle of inclination of the hill (θ) based on the power difference required when the car is moving uphill versus downhill. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Forces Acting on the Car
When the car is moving uphill, the forces acting on it include:
- The gravitational force component acting down the slope: \( mg \sin \theta \)
- The friction and air resistance force \( F_R \) acting opposite to the direction of motion.

When the car is moving downhill, the forces acting on it include:
- The gravitational force component acting down the slope: \( mg \sin \theta \)
- The friction and air resistance force \( F_R \) acting up the slope.

### Step 2: Write the Equations for Power
The power required to move the car uphill (P1) and downhill (P2) can be expressed as:
- \( P_1 = (F_R + mg \sin \theta) \cdot V \)
- \( P_2 = (mg \sin \theta - F_R) \cdot V \)

Where:
- \( V = 27 \, \text{m/s} \) (the constant velocity of the car)

### Step 3: Set Up the Power Difference Equation
According to the problem, the power difference when going uphill and downhill is given as:
\[ P_1 - P_2 = 47 \, \text{hp} \]
Converting horsepower to watts:
\[ 47 \, \text{hp} = 47 \times 746 \, \text{W} = 35022 \, \text{W} \]

Substituting the expressions for \( P_1 \) and \( P_2 \):
\[ (F_R + mg \sin \theta) \cdot V - (mg \sin \theta - F_R) \cdot V = 35022 \]

### Step 4: Simplify the Equation
Expanding and simplifying the equation:
\[ F_R V + mg \sin \theta V - mg \sin \theta V + F_R V = 35022 \]
\[ 2 F_R V = 35022 \]
Thus, we can express \( F_R \):
\[ F_R = \frac{35022}{2V} \]

### Step 5: Substitute Values
Substituting \( V = 27 \, \text{m/s} \):
\[ F_R = \frac{35022}{2 \times 27} = \frac{35022}{54} \approx 648.5 \, \text{N} \]

### Step 6: Relate Forces to Find θ
Now, we can relate \( F_R \) to the angle of inclination:
Using the equation for the forces when the car is going uphill:
\[ F_R + mg \sin \theta = 0 \]
Thus:
\[ mg \sin \theta = F_R \]
Substituting \( F_R \):
\[ mg \sin \theta = 648.5 \]

### Step 7: Solve for sin θ
Substituting \( m = 1900 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \):
\[ 1900 \cdot 9.8 \cdot \sin \theta = 648.5 \]
\[ \sin \theta = \frac{648.5}{1900 \cdot 9.8} \]
Calculating:
\[ \sin \theta = \frac{648.5}{18620} \approx 0.0348 \]

### Step 8: Find θ
Taking the inverse sine:
\[ \theta = \sin^{-1}(0.0348) \approx 2.0^\circ \]

### Final Answer
The angle of inclination of the hill above the horizontal is approximately **2 degrees**.

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