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Chapter 8: Problem 118

Resistance to the motion of an automobile consists of road friction, which is almost independent of speed, and air drag, which is proportional to speed- squared. For a certain car with a weight of \(12000 \mathrm{~N},\) the total resistant force \(F\) is given by \(F=300+1.8 v^{2}\) with \(F\) in newtons and \(v\) in meters per second. Calculate the power (in horsepower) required to accelerate the car at \(0.92 \mathrm{~m} / \mathrm{s}^{2}\) when the speed is \(80 \mathrm{~km} / \mathrm{h}\)

Short Answer

Expert verified
Approximately 65.97 horsepower is required.

Step by step solution

01

Convert Speed to Meters per Second

The given speed is 80 km/h. Convert this to meters per second (m/s) using the conversion factor: \[ 1 \text{ km/h} = \frac{1}{3.6} \text{ m/s} \]Therefore, \[ v = 80 \times \frac{1}{3.6} \approx 22.22 \text{ m/s} \]
02

Calculate the Total Resistance Force

Use the formula for total resistance: \[ F = 300 + 1.8v^2 \]Substituting the converted speed:\[ F = 300 + 1.8 \times (22.22)^2 \approx 1088.89 \text{ N} \]
03

Calculate the Net Force Required for Acceleration

To find the force for acceleration, use the formula: \[ F_{\text{net}} = ma \]where mass \( m \) is weight divided by gravity \( g = 9.8 \text{ m/s}^2 \):\[ m = \frac{12000}{9.8} \approx 1224.49 \text{ kg} \]\[ F_{\text{net}} = 1224.49 \times 0.92 \approx 1127.73 \text{ N} \]
04

Calculate the Total Force Required

The total force to overcome both resistance and accelerate is:\[ F_{\text{total}} = F + F_{\text{net}} = 1088.89 + 1127.73 \approx 2216.62 \text{ N} \]
05

Calculate Power in Watts

Power is calculated using the formula:\[ P = F_{\text{total}} \times v \]Substituting the values:\[ P = 2216.62 \times 22.22 \approx 49212.43 \text{ W} \]
06

Convert Power to Horsepower

Use the conversion factor:\[ 1 \text{ horsepower} = 746 \text{ W} \]Therefore, \[ \text{Power in horsepower} = \frac{49212.43}{746} \approx 65.97 \text{ horsepower} \]

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Key Concepts

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

Understanding Road Friction
Road friction is one of the main forces working against the motion of an automobile. It primarily arises due to the contact between the car's tires and the road surface. Unlike other types of resistance, road friction tends to remain constant across different speeds. This is because it largely depends on the quality and type of the road surface and the tires themselves. In our example, road friction is a part of the equation that remains unaffected by changes in velocity, contributing a constant component to the total resistant force.
Exploring Air Drag
Air drag, often referred to as aerodynamic drag, is another force that resists the forward movement of vehicles. Unlike road friction, air drag increases with speed, more specifically, with the square of the speed. This means that as the car moves faster, the air resistance it encounters grows significantly.
The formula used for air drag in our problem, \( F = 1.8v^2 \), reflects its dependence on speed squared. As cars travel at higher velocities, the impact of air drag becomes more notable, affecting fuel efficiency and overall power requirements.
Steps in Power Calculation
Calculating the power required to maintain or change a car's speed involves understanding both the forces at play and the speed at which the car is traveling. Power is the rate at which work is done, and it鈥檚 calculated by multiplying the total force exerted on the car by its velocity.
  • Identify the total forces acting on the vehicle. This includes both road friction and air drag, as well as any additional forces needed for acceleration.
  • Combine these forces to find the total force. This total will determine how much power is needed from the engine.
  • Multiply by the speed. Since power is force times velocity, we use the calculated total force with the car鈥檚 speed to find the work rate, measured in watts.
The Concept of Acceleration
Acceleration refers to the rate at which the velocity of an automobile changes over time. It鈥檚 a key factor in determining the additional force needed to increase the speed of the vehicle.
Using the formula \( F_{\text{net}} = ma \), where \( m \) is the mass of the car and \( a \) is the given acceleration, we can compute the net force required. Understanding acceleration helps in designing how much extra power is needed, beyond overcoming resistances, to achieve the desired speed change.
Importance of Unit Conversion
Unit conversion is often necessary in solving physics problems because different components of a problem may be expressed in different units. In our example, converting speed from kilometers per hour to meters per second allows for consistent use of units in equations involving force and power, which typically employ metric units.
The conversion given, \( v = 80 \times \frac{1}{3.6} \approx 22.22 \text{ m/s} \), ensures that all units harmonize, enabling accurate calculations. Such clarity prevents errors that could arise from unit mismatches and simplifies the mathematical process, making the solution more reliable and easier to understand.

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Most popular questions from this chapter

A \(5.0 \mathrm{~g}\) marble is fired vertically upward using a spring gun. The spring must be compressed \(8.0 \mathrm{~cm}\) if the marble is to just reach a target \(20 \mathrm{~m}\) above the marble's position on the compressed spring. (a) What is the change \(\Delta U_{g}\) in the gravitational potential energy of the marble-Earth system during the \(20 \mathrm{~m}\) ascent? (b) What is the change \(\Delta U_{s}\) in the elastic potential energy of the spring during its launch of the marble? (c) What is the spring constant of the spring?

A pendulum consists of a \(2.0 \mathrm{~kg}\) stone swinging on a \(4.0 \mathrm{~m}\) string of negligible mass. The stone has a speed of \(8.0 \mathrm{~m} / \mathrm{s}\) when it passes its lowest point. (a) What is the speed when the string is at \(60^{\circ}\) to the vertical? (b) What is the greatest angle with the vertical that the string will reach during the stone's motion? (c) If the potential energy of the pendulum-Earth system is taken to be zero at the stone's lowest point, what is the total mechanical energy of the system?

Figure 8 - 34 shows a thin rod, of length \(L=2.00 \mathrm{~m}\) and negligible mass, that can pivot about one end to rotate in a vertical circle. A ball of mass \(m=5.00 \mathrm{~kg}\) is attached to the other end. The rod is pulled aside to angle \(\theta_{0}=30.0^{\circ}\) and released with initial velocity \(\vec{v}_{0}=0 .\) As the ball descends to its lowest point, (a) how much work does the gravitational force do on it and (b) what is the change in the gravitational potential energy of the ball-Earth system? (c) If the gravitational potential energy is taken to be zero at the lowest point, what is its value just as the ball is released? (d) Do the magnitudes of the answers to (a) through (c) increase, decrease, or remain the same if angle \(\theta_{0}\) is increased?

We move a particle along an \(x\) axis, first outward from \(x=1.0 \mathrm{~m}\) to \(x=4.0 \mathrm{~m}\) and then back to \(x=1.0 \mathrm{~m},\) while an external force acts on it. That force is directed along the \(x\) axis, and its \(x\) component can have different values for the outward trip and for the return trip. Here are the values (in newtons) for four situations, where \(x\) is in meters: $$ \begin{array}{ll} \hline \text { Outward } & \text { Inward } \\ \hline \text { (a) }+3.0 & -3.0 \\ \text { (b) }+5.0 & +5.0 \\ \text { (c) }+2.0 x & -2.0 x \\ \text { (d) }+3.0 x^{2} & +3.0 x^{2} \\ \hline \end{array} $$ Find the net work done on the particle by the external force for the round trip for each of the four situations. (e) For which, if any, is the external force conservative?

A \(20 \mathrm{~kg}\) object is acted on by a conservative force given by \(F=-3.0 x-5.0 x^{2},\) with \(F\) in newtons and \(x\) in meters. Take the potential energy associated with the force to be zero when the object is at \(x=0 .\) (a) What is the potential energy of the system associated with the force when the object is at \(x=2.0 \mathrm{~m} ?\) (b) If the object has a velocity of \(4.0 \mathrm{~m} / \mathrm{s}\) in the negative direction of the \(x\) axis when it is at \(x=5.0 \mathrm{~m},\) what is its speed when it passes through the origin? (c) What are the answers to (a) and (b) if the potential energy of the system is taken to be \(-8.0 \mathrm{~J}\) when the object is at \(x=0 ?\)
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