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The mass of a car is 1500 kg. The shape of the body is such that its aerodynamic drag coefficient is \(D=0.330\) and the frontal area is \(2.50 \mathrm{m}^{2} .\) Assuming that the drag force is proportional to \(v^{2}\) and neglecting other sources of friction, calculate the power required to maintain a speed of \(100 \mathrm{km} / \mathrm{h}\) as the car climbs a long hill sloping at \(3.20^{\circ}\).

Short Answer

Expert verified
The car requires a power of 30.167 kW to maintain its speed against the aerodynamic drag and gravitational force as it climbs up the hill.

Step by step solution

01

Convert Velocity

Convert the car's velocity to m/s from km/h using the conversion factor 1 km/h = 0.2778 m/s. Thus, \( v = 100 \times 0.2778 = 27.78 \, m/s \).
02

Calculate the Drag Force

Use the formula for the drag force \( F_d = 0.5 \times D \times A \times \rho \times v^2 \). Here \( D = 0.330 \), \( A = 2.50 \, m^2 \), \( \rho \) (air density) is approximately \( 1.2 \, kg/m^3 \) under normal conditions at sea level, and \( v = 27.78 \, m/s \). So, \( F_d = 0.5 \times 0.330 \times 2.50 \times 1.2 \times (27.78)^2 = 267.8 \, N \).
03

Calculate the Gravitational Force

Use the formula for the gravitational force \( F_g = m \times g \times \sin(ο) \) with \( m = 1500 \, kg \), \( g = 9.81 \, m/s^2 \), and \( ο = 3.20^{\circ} \). Convert \( ο \) to radian before using it in calculations using the relation \( 1 \, rad = 57.3 \, deg \). Hence, \( F_g = 1500 \times 9.81 \times \sin(3.20/57.3) = 819.0 \, N \).
04

Calculate the Total Force

The total force on the car is the sum of the drag force and the gravitational force. Hence, \( F_t = F_d + F_g = 267.8 + 819.0 = 1086.8 \, N \).
05

Calculate the Power Required

The power required can be found by multiplying the total force with the velocity. Hence, \( P = F_t \times v = 1086.8 \times 27.78 = 30167 \, W \) or \( 30.167 \, kW \).

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Key Concepts

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

Aerodynamic Drag Force
When objects move through the air, they experience a resistive force known as the aerodynamic drag force. This force is crucial to understand as it greatly affects the performance and fuel efficiency of vehicles such as cars and airplanes.

The drag force, often symbolized as \( F_d \), opposes the direction of an object's motion through a fluid, which in most cases is air. It's the result of collisions between the molecules in the air and the object's surface. The drag force can be calculated using the equation: \[ F_d = 0.5 \times C_d \times A \times \rho \times v^2 \] where \( C_d \) is the drag coefficient that represents an object’s aerodynamic shape, \( A \) is the cross-sectional area facing the flow, \( \rho \) is the air density, and \( v \) is the velocity of the object relative to the fluid.

In our car example, the drag coefficient is given, and so are the frontal area and the velocity. These parameters influence \( F_d \) greatly; a high drag coefficient or a large frontal area increases the aerodynamic drag and thus, the power requirement to maintain speed. The equation shows that the drag force is proportional to the square of the velocity, meaning that even a modest increase in speed can result in a significant rise in drag. This quadratic relationship plays a critical role in designs aiming to improve vehicles' aerodynamic efficiency.
Power Calculation
Power in physics is the rate at which work is done or energy is transferred. It's an important concept in understanding how systems like engines and motors operate. Power calculation is especially pertinent when determining the capability of a vehicle to maintain certain speeds against resistive forces.

Mathematically, power \( P \) is expressed as: \[ P = F \times v \] where \( F \) is the force applied to the object and \( v \) is its velocity. This formula implies that the power required to maintain a specific speed is directly proportional to both the force opposing the motion and the velocity itself.

For example, driving a car up a hill at a constant speed requires more power than driving on a flat surface due to the additional gravitational force. The same applies if the car faces a headwind, increasing the aerodynamic drag. Therefore, understanding power requirements allows engineers to design engines and other components that can adequately supply the necessary force at the desired speed to overcome such challenges.
Gravitational Force
Gravitational force is a fundamental concept in physics, acting as the attractive force between two bodies due to their masses. On Earth, this force is what gives us weight and keeps us grounded. In practical applications like driving a car up a hill, gravitational force must be considered because it will affect the total force the car's engine needs to overcome.

The force of gravity along a slope can be calculated using the component of gravitational acceleration parallel to the slope. This is found using the formula: \[ F_g = m \times g \times \text{sin}(\theta) \] where \( m \) is the mass of the body, \( g \) is the acceleration due to gravity (approximately \( 9.81 \text{m/s}^2 \) on Earth's surface), and \( \theta \) is the slope angle in radians.

It is essential to convert the angle to radians when performing the calculation, as trigonometric functions in most calculators and mathematical equations require angle measures in radians, not degrees. The gravitational force increases as the slope becomes steeper, which, combined with aerodynamic drag, can significantly increase the power requirement for a vehicle—illustrating the substantial role that gravitational force plays in scenarios such as driving uphill.

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