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Automotive Power I. A truck engine transmits 28.0 \(\mathrm{kW}\) \((37.5 \mathrm{hp})\) to the driving wheels when the truck is traveling at a constant velocity of magnitude 60.0 \(\mathrm{km} / \mathrm{h}\) (37.3 \(\mathrm{mi} / \mathrm{h} )\) on a level road. (a) What is the resisting force acting on the truck? (b) Assume that 65\(\%\) of the resisting force is due to rolling friction and the remainder is due to air resistance. If the force of rolling friction is independent of speed, and the force of air resistance is proportional to the square of the speed, what power will drive the truck at 30.0 \(\mathrm{km} / \mathrm{h}\) ? At 120.0 \(\mathrm{km} / \mathrm{h}\) ? Give your answers in kilo-watts and in horsepower.

Step by step solution

01

Convert Velocity to Meters per Second

To solve the problem, it is helpful to convert the truck's speed from kilometers per hour to meters per second for easier calculations using SI units.To convert from km/h to m/s, use the conversion factor: \[ \text{speed (m/s)} = \text{speed (km/h)} \times \frac{1000}{3600} \]For 60 km/h:\[ v = 60 \times \frac{1000}{3600} = 16.67 \text{ m/s} \]Repeat the conversion for 30 km/h and 120 km/h to get 8.33 m/s and 33.33 m/s, respectively.

02

Use Power Formula to Find Resisting Force

Power (P) is related to force (F) and velocity (v) by the formula: \[ P = F \times v \]Given that the power transmitted is 28 kW, or 28,000 watts, and the velocity is 16.67 m/s, solving for force gives:\[ F = \frac{P}{v} = \frac{28,000}{16.67} = 1680 \text{ N} \]Thus, the resisting force acting on the truck is 1680 N.

03

Determine Components of Resisting Force

65% of the resisting force is due to rolling friction \((F_r)\), and 35% is due to air resistance \((F_a)\).Calculate the rolling friction:\[ F_r = 0.65 \times 1680 = 1092 \text{ N} \]And calculate the air resistance:\[ F_a = 0.35 \times 1680 = 588 \text{ N} \]

04

Calculate Power at 30 km/h

At 30 km/h (8.33 m/s), rolling friction remains constant, and air resistance is proportional to \(v^2\):First, find the new air resistance:\[ F_a' = 588 \times \left(\frac{8.33}{16.67}\right)^2 = 147 \text{ N} \]The new total force \(F'\):\[ F' = F_r + F_a' = 1092 + 147 = 1239 \text{ N} \]The power required:\[ P' = F' \times v = 1239 \times 8.33 = 10315 \text{ W} \approx 10.3 \text{ kW} \]Convert to horsepower:\[ \text{HP} = 10.3 \times 0.134 = 13.8 \text{ hp} \]

05

Calculate Power at 120 km/h

At 120 km/h (33.33 m/s), repeat a similar calculation process:New air resistance:\[ F_a'' = 588 \times \left(\frac{33.33}{16.67}\right)^2 = 2352 \text{ N} \]The new total force \(F''\) is:\[ F'' = F_r + F_a'' = 1092 + 2352 = 3444 \text{ N} \]The power required is:\[ P'' = F'' \times v = 3444 \times 33.33 = 114736 \text{ W} \approx 114.7 \text{ kW} \]Convert to horsepower:\[ \text{HP} = 114.7 \times 0.134 = 153.7 \text{ hp} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Resisting Force Analysis

When a vehicle travels at a constant speed on a level road, it faces a resisting force. This force must be overcome by the engine to maintain a steady velocity. Resisting forces are typically composed of rolling friction and air resistance.

• **Rolling Friction:** This occurs due to the deformation of the wheels and the road surface where they contact, consuming energy. It's generally proportional to the load of the vehicle.
• **Air Resistance:** As a vehicle moves through the air, it has to push air molecules out of its way, causing drag. This resistance grows exponentially as speed increases.

Understanding these components allows engineers to optimize vehicle designs for better fuel efficiency and performance by minimizing these resistive forces.

Rolling Friction and Air Resistance

Two crucial sources of resistance impede a vehicle's motion: rolling friction and air resistance. These forces are important in calculating the total resistive force a truck experiences.

• **Rolling Friction:** It remains relatively stable and independent of the truck's speed. Even when the truck slows down or speeds up, this force is mostly constant, being primarily a function of weight and road conditions.
• **Air Resistance:** Unlike rolling friction, air resistance increases with the velocity. In fact, it increases with the square of the velocity. This means it's four times higher when a vehicle doubles its speed.

This knowledge helps in calculating how different speeds impact the required power. For a truck, if 65% of the resisting force is rolling friction, the remaining 35% is subject to change with speed due to air resistance, as demonstrated in the power calculations for various velocities.

Velocity Conversion in Physics

In physics, converting velocities into a consistent unit system simplifies calculations. In most cases, using meters per second (m/s) is preferable in physics for clarity and consistency with other SI units.

• **Conversion Factor:** The standard conversion factor from kilometers per hour (km/h) to meters per second (m/s) is \left( \frac{1000}{3600} \right). This factor accounts for the number of meters in a kilometer and the number of seconds in an hour.
• **Example Calculations:** For instance, a speed of 60 km/h would be converted as follows: \[ v = 60 \times \frac{1000}{3600} = 16.67 \text{ m/s} \]
• **Practical Tech Use:** This conversion is crucial for using equations involving power, force, and efficiency. It is especially useful in automotive problems where power requirements are calculated based on velocity, such as those involving different resistance forces.

This approach ensures calculations are manageable and integrates seamlessly into physics equations and real-world applications.