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\(\bullet\) \(\bullet\) \(\bullet\) 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 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 horse- power.

Step by step solution

01

Understand the Problem

We need to find the resisting force acting on the truck when traveling at 60 km/h, and then calculate the power required to drive the truck at 30 km/h and 120 km/h. We know that the engine transmits 28 kW of power at 60 km/h.

02

Calculate Resisting Force

Using the power formula, power \( P = F \times v \), where \( F \) is the force in newtons and \( v \) is the speed in meters per second. Convert 60 km/h to m/s: \( 60 \; \text{km/h} = \frac{60 \times 1000}{3600} = 16.67 \; \text{m/s} \). Solve for \( F \): \( 28,000 = F \times 16.67 \). Thus, \( F = \frac{28,000}{16.67} \approx 1680 \; \text{N} \).

03

Determine Components of Resisting Force

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

04

Force Equations for Different Speeds

Rolling friction \( F_r \) remains the same (1092 N) regardless of speed. The force of air resistance \( F_a \) is proportional to velocity squared, \( F_a = 588 \times \left(\frac{v}{60}\right)^2 \).

05

Calculate Power at 30 km/h

Convert 30 km/h to m/s: \( 30 \; \text{km/h} = 8.33 \; \text{m/s} \). Calculate total resisting force: \( F_{total} = F_r + F_a = 1092 + 588 \times \left(\frac{8.33}{16.67}\right)^2 \approx 1180.5 \; \text{N} \). Power required: \( P = F_{total} \times v = 1180.5 \times 8.33 \approx 9835 \; \text{W} \).

06

Calculate Power at 120 km/h

Convert 120 km/h to m/s: \( 120 \; \text{km/h} = 33.33 \; \text{m/s} \). Calculate total resisting force: \( F_{total} = F_r + F_a = 1092 + 588 \times \left(\frac{33.33}{16.67}\right)^2 \approx 2805 \; \text{N} \). Power required: \( P = F_{total} \times v = 2805 \times 33.33 \approx 93510 \; \text{W} \).

07

Convert Power to Horsepower

Convert power from watts to kilowatts and to horsepower. At 30 km/h, \( 9835 \; \text{W} = 9.835 \; \text{kW} \approx 13.19 \; \text{hp} \). At 120 km/h, \( 93510 \; \text{W} = 93.51 \; \text{kW} \approx 125.44 \; \text{hp} \).

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Key Concepts

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

Resisting Force

In the context of automotive physics, resisting force plays a crucial role in understanding how vehicles move. Resisting force refers to all forces acting against a vehicle's motion. When a truck moves at a constant speed, like the given 60 km/h, the resisting force is equal and opposite to the engine's thrust.
This balance of forces allows the truck to maintain its speed. In our problem, the engine's power output is 28 kW, which we use alongside speed to calculate the total resisting force using the formula:
\( F = \frac{P}{v} \)where \( P \) is power in watts (28,000 W) and \( v \) is velocity in meters per second (16.67 m/s). The resulting resisting force comes out to be approximately 1680 N.

Rolling Friction

Rolling friction is a part of the overall resisting force acting against the vehicle's motion. It arises from the deformation of the wheels and the contact surface. Unlike other forms of friction, rolling friction is relatively small.
In our truck scenario, 65% of the total resisting force is attributed to rolling friction. This is calculated as follows:

• Total resisting force \( = 1680 \) N
• Rolling friction force \( F_r = 0.65 \times 1680 \) N \( = 1092 \; \text{N} \)

The force due to rolling friction remains constant regardless of the vehicle's speed, simplifying calculations for changing speeds.

Air Resistance

Air resistance, also known as drag, includes effects of air pushing against the vehicle as it moves. Unlike rolling friction, air resistance increases with speed—specifically, it is proportional to the square of velocity.
In our exercise, air resistance accounts for 35% of the total resisting force at 60 km/h, calculated by:

• Total resisting force \( = 1680 \) N
• Air resistance force \( F_a = 0.35 \times 1680 \) N \( = 588 \; \text{N} \)

For different speeds, the force due to air resistance is calculated using:
\( F_a = 588 \times \left(\frac{v}{60}\right)^2 \)This relationship shows how air resistance rapidly increases with higher speeds.

Power Calculation

Calculating power gives insights into the amount of work done by the vehicle's engine to overcome resisting forces. Power in this context can be solved via the formula:
\( P = F_{total} \times v \)Our challenge was to determine the power needed at two different speeds, considering the shift in air resistance and constant rolling friction:

• At 30 km/h: Convert to 8.33 m/s. Calculate resisting force from friction and air resistance: \( F_{total} \approx 1180.5 \; \text{N} \). Compute power: \( P \approx 9835 \; \text{W} \) or 9.835 kW.
• At 120 km/h: Convert to 33.33 m/s. Determine resisting force: \( F_{total} \approx 2805 \; \text{N} \). Power becomes \( P \approx 93510 \; \text{W} \) or 93.51 kW.

Power calculation helps in understanding the energy requirements for varying driving conditions.

Velocity

Velocity is the speed and direction of an object's motion. In physics, understanding how velocity affects forces like air resistance and power is vital. Here, speed is the main component of velocity, given in kilometers per hour which is often converted to meters per second for calculations.
In our problem, we had different velocities:

• 60 km/h converted to 16.67 m/s
• 30 km/h converted to 8.33 m/s
• 120 km/h converted to 33.33 m/s

Each change in velocity affects how air resistance is computed, showing its quadratic relationship with velocity. This directly impacts the power needed, demonstrating the importance of velocity in automotive physics.