Problem 101 - University Physics with Modern Physics

Question

Automotive Power II. (a) If 8.00 hp are required to drive a \(1800-\) -kg automobile at 60.0 \(\mathrm{km} / \mathrm{h}\) on a level road, what is the total retarding force due to friction, air resistance, and so on? (b) What power is necessary to drive the car at 60.0 \(\mathrm{km} / \mathrm{h}\) up a 10.0\(\%\) grade (a hill rising 10.0 \(\mathrm{m}\) vertically in 100.0 \(\mathrm{m}\) horizon- tally)? (c) What power is necessary to drive the car at 60.0 \(\mathrm{km} / \mathrm{h}\) down a 1.00\(\%\) grade? (d) Down what percent grade would the car coast at 60.0 \(\mathrm{km} / \mathrm{h}\) ?

Step-by-Step Solution

Answer

(a) 357.8 N (b) 47.4 hp (c) 4.06 hp (d) 2.03% grade.

Step 1: Understanding the Power Conversion

1 horsepower (hp) is equivalent to 746 Watts (W). Thus, 8 hp can be converted to Watts as follows: \[8 \text{ hp} = 8 \times 746 \text{ W} = 5968 \text{ W}\] This is the required power to keep the car moving at 60.0 km/h on a level road.

Step 2: Calculating Retarding Force

The power formula is given by \( P = F \times v \), where \( F \) is the force and \( v \) is the velocity in m/s. First, convert 60.0 km/h to m/s: \[v = \frac{60.0 \text{ km/h} \times 1000 \text{ m/km}}{3600 \text{ s/h}} = 16.67 \text{ m/s} \] Solve for \( F \): \[ F = \frac{P}{v} = \frac{5968 \text{ W}}{16.67 \text{ m/s}} \approx 357.8 \text{ N} \] This is the total retarding force due to friction and air resistance.

Step 3: Power for Uphill Driving

A 10.0% grade means the hill rises 10 m for every 100 m. The additional force required to overcome gravity is \( F_{g} = m \cdot g \cdot \sin(\theta) \). With \( \sin(\theta) \approx 0.1 \) for small angles: \[ F_{g} = 1800 \text{ kg} \times 9.8 \text{ m/s}^2 \times 0.1 = 1764 \text{ N}\] Total force \(F_{\text{total}} = 357.8 \text{ N} + 1764 \text{ N} = 2121.8 \text{ N} \). Required power: \[ P = F_{\text{total}} \times v = 2121.8 \text{ N} \times 16.67 \text{ m/s} = 35374 \text{ W} \] Convert to hp: \[ \frac{35374 \text{ W}}{746 \text{ W/hp}} \approx 47.4 \text{ hp}\]

Step 4: Power for Downhill Driving

For a 1.00% grade, the additional force due to gravity is \( F_{g} = 1800 \text{ kg} \times 9.8 \text{ m/s}^2 \times 0.01 \approx 176.4 \text{ N} \). Net retarding force \( F_{\text{net}} = 357.8 \text{ N} - 176.4 \text{ N} = 181.4 \text{ N} \). Required power: \[ P = 181.4 \text{ N} \times 16.67 \text{ m/s} = 3024 \text{ W}\] Convert to hp: \[ \frac{3024 \text{ W}}{746 \text{ W/hp}} \approx 4.06 \text{ hp} \]

Step 5: Calculating Grade for Coasting

When coasting, the power is zero, meaning the gravitational force equals retarding force. Using \( F_{g} = 357.8 \text{ N} \), solve for \( \sin(\theta) \): \[ \sin(\theta) = \frac{F_{g}}{m \cdot g} = \frac{357.8 \text{ N}}{1800 \text{ kg} \cdot 9.8 \text{ m/s}^2} \approx 0.0203 \] Converting to percent: \( 0.0203 \times 100 \approx 2.03 \text{%} \).

Key Concepts

Power Conversion

Power conversion is the fundamental step in understanding how much work a machine, such as a car, can do based on its power source. It involves converting units of horsepower (hp) into watts (W), which are more commonly used in scientific calculations. Knowing that 1 horsepower is equivalent to 746 watts is essential. For example, if a car engine is using 8 hp, this translates to \(8 \times 746\) W, or 5968 W. This conversion is crucial because it provides us with a common unit that can then be used in further calculations such as force and energy needs, especially when performing tasks like going uphill or fighting drag and friction on a flat surface.

Retarding Forces

Retarding forces are the forces that work against the motion of the car, slowing it down. These include friction from the road and resistance from the air. To calculate the total retarding force when a car is cruising at a constant speed, we can use the formula \( P = F \times v \). Here, \( P \) is the power in watts, \( F \) is the force in newtons, and \( v \) is the speed in meters per second (converted from kilometers per hour). For our exercise, with a known power of 5968 W and a speed of 16.67 m/s, the retarding force comes out to about 357.8 N. This force needs to be constantly overcome by the engine to maintain a steady speed on a level road.

Uphill and Downhill Driving

Driving uphill and downhill introduces additional forces into the power calculation. When driving uphill at a 10% grade, the car has to work against gravity. Calculating this involves recognizing that the hill’s rise adds an additional force component \( F_{g} = m \cdot g \cdot \sin(\theta) \). For a 10% grade, \( \sin(\theta) \approx 0.1 \), leading to an uphill gravitational force of 1764 N. Combined with the retarding force, the total force becomes 2121.8 N. The power needed is then calculated as \( P = F_{\text{total}} \times v \), which results in a significant increase in power requirements to about 47.4 hp.
On the other hand, when driving downhill, gravity aids in propulsion. This reduces the net retarding force, as calculated for a 1% grade, resulting in lower necessary power, around 4.06 hp.

Coasting Calculations

Coasting occurs when the car moves at a steady speed without additional power from the engine. This happens when the gravitational force matches the retarding forces perfectly, requiring no extra energy from the engine. To find the downhill slope where coasting occurs, set the retarding force equal to the gravitational force component. The equation \( F_{g} = 357.8 \) N is used to solve for \( \sin(\theta) \), leading to approximately 0.0203. Converting this to a percentage gives a downhill grade of about 2.03%. This grade indicates a balance between gravity and the retarding forces, allowing the car to maintain speed without engine power.