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Automotive power. 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 \(60.0 \mathrm{~km} / \mathrm{h}(37.7 \mathrm{mi} / \mathrm{h})\) 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 kilowatts and in horsepower.

Step by step solution

01

Calculate Resisting Force

The power, \( P \), used by the truck is related to the resisting force, \( F \), and velocity, \( v \), by the formula \( P = F \cdot v \). Using the given values: \[ P = F \cdot v, \]\[ 28.0 \text{ kW} = F \times \left( \frac{60.0 \text{ km/h}}{3.6 \text{ m/s}} \right). \]\[ 28,000 = F \times 16.67, \] solving for \( F \) gives us \( F = \frac{28,000}{16.67} \approx 1680 \text{ N}. \) Thus, the resisting force is 1680 N.

02

Calculate Rolling Friction Force

According to the problem, 65% of the resisting force is due to rolling friction. Calculate the rolling friction force as follows: \[ F_{\text{rolling}} = 0.65 \times 1680 \approx 1092 \text{ N}. \]

03

Calculate Air Resistance Force

The air resistance force constitutes the remaining 35% of the total resisting force. Thus, \[ F_{\text{air}} = 0.35 \times 1680 \approx 588 \text{ N}. \]

04

Relationship between Air Resistance and Speed

The force of air resistance is proportional to the square of the speed \( v \). Thus, \[ F'_{\text{air}} = F_{\text{air}} \left( \frac{v'}{v} \right)^2, \] where \( v' \) is the new speed and \( v \) is the original speed.

05

Power Calculation at 30 km/h

At 30 km/h, velocity \( v' \) is \( \frac{30}{3.6} = 8.33 \) m/s. The new air resistance is \[ F'_{\text{air}} = 588 \left( \frac{8.33}{16.67} \right)^2 \approx 147 \text{ N}. \] The total resisting force at this speed is \( 1092 + 147 = 1239 \text{ N}. \) Power required is \[ P' = 1239 \cdot 8.33 \approx 10309 \text{ W} = 10.3 \text{ kW} \approx 13.8 \text{ hp}. \]

06

Power Calculation at 120 km/h

At 120 km/h, velocity \( v'' \) is \( \frac{120}{3.6} = 33.33 \) m/s. The new air resistance is \[ F''_{\text{air}} = 588 \left( \frac{33.33}{16.67} \right)^2 \approx 2352 \text{ N}. \] The total resisting force is \( 1092 + 2352 = 3444 \text{ N}. \) Power required is \[ P'' = 3444 \cdot 33.33 \approx 114765 \text{ W} = 114.8 \text{ kW} \approx 153.9 \text{ hp}. \]

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Key Concepts

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

Resisting Force

The resisting force is like an invisible hand that tries to slow down your truck. When a vehicle moves at a constant speed, the engine must work against this resisting force to keep going. This force is a combination of different factors such as friction, drag, and other obstacles to movement. For a truck traveling at a steady speed, this resistance can be calculated using the power and velocity:

• The formula we use to find the resisting force is: \( P = F \cdot v \)
• Where \( P \) is power, \( F \) is the resisting force, and \( v \) is velocity.

In our example, the calculation shows that the resisting force is 1680 N.

Rolling Friction

Rolling friction is one of the components that make up the resisting force. Picture the wheels of a truck rolling over the road. As they turn, there is a small force acting against them called rolling friction. This force makes it a bit harder for the truck to move forward.

• In our problem, 65% of the total resisting force is attributed to rolling friction.
• To find this force, multiply the total resisting force by 0.65: \( F_\text{rolling} = 0.65 \times 1680 \approx 1092 \text{ N} \).

Understanding rolling friction helps us realize how important tire quality and road conditions are for vehicle efficiency.

Air Resistance

Air resistance is another significant part of the resisting force faced by a vehicle, also known as drag. Think of it as the invisible push of air you feel when sticking your hand out of a moving car window. Unlike rolling friction, air resistance depends on the speed of the vehicle.

• 35% of the resisting force in our scenario is due to air resistance.
• To find the air resistance force: \( F_\text{air} = 0.35 \times 1680 \approx 588 \text{ N} \).
• It's important to note that air resistance increases with speed and is proportional to the square of the speed.

Proportional Relationships

Proportional relationships are crucial in understanding how variables interact in physics. In our scenario, air resistance has a square proportional relationship with velocity, meaning that if you double the speed, the air resistance increases fourfold.

• We use this relationship to adjust calculations based on changing speeds.
• The formula is: \( F'_\text{air} = F_\text{air} \left( \frac{v'}{v} \right)^2 \).
• This helps us determine what portion of the resisting force comes from air resistance at different speeds.

Grasping these relationships enables engineers to predict how changes in speed affect resisting forces, which is vital for designing efficient vehicles.

Power Calculation

Calculating power lets us know how much effort an engine needs to exert to overcome resisting forces and maintain a specific speed. Power is a measure of work done over time, often described in kilowatts (kW) or horsepower (hp).

• In our example, the engine produces power to match the resisting force at any given speed.
• At 30 km/h, the power required is \( 10.3 \text{ kW} \approx 13.8 \text{ hp} \).
• At 120 km/h, the power needed is \( 114.8 \text{ kW} \approx 153.9 \text{ hp} \).

Knowing how to calculate power accurately is essential for designing engines that are both powerful and efficient for various driving conditions.