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Chapter 6: Problem 99

Automotive Power II. (a) If 8.00 hp are required to drive a \(1800-\mathrm{kg}\) automobile at 60.0 \(\mathrm{km} / \mathrm{h}\) on a level road, what is the total retarding force due to friction, air resistance, and so on? (b) What power is necessary to drive the car at 60.0 \(\mathrm{km} / \mathrm{h}\) up a 10.0\(\%\) grade (a hill rising 10.0 \(\mathrm{m}\) vertically in 100.0 \(\mathrm{m}\) horizontally)? (c) What power is necessary to drive the car at 60.0 \(\mathrm{km} / \mathrm{h}\) down a 1.00\(\%\) grade? (d) Down what percent grade would the car coast at 60.0 \(\mathrm{km} / \mathrm{h} ?\)

Short Answer

Expert verified
(a) 358 N, (b) 47.4 hp, (c) 4.06 hp, (d) 2.04% grade.

Step by step solution

01

Calculate Total Retarding Force (Part a)

The power required to maintain a constant speed is used to overcome the retarding forces. Using the formula for power \( P = F \times v \), where \( F \) is the retarding force and \( v \) is the speed, we first convert the speed from km/h to m/s: \[ v = 60.0 \text{ km/h} \times \frac{1000 \text{ m/km}}{3600 \text{ s/h}} = 16.67 \text{ m/s} \]. Also, convert horsepower to watts: \[ 8.00 \text{ hp} = 8.00 \times 746 \text{ W} = 5968 \text{ W} \]. Thus, \[ F = \frac{P}{v} = \frac{5968 \text{ W}}{16.67 \text{ m/s}} \approx 358 \text{ N} \].
02

Calculate Power Needed Uphill (Part b)

When driving uphill, additional power is required to overcome the gravitational force component along the slope. The grade expressed in decimal form is 0.10, and the gravitational force is \( F_{g} = m \times g \times \sin(\theta) \approx m \times g \times 0.10 \). Therefore, the total force is \( F_{total} = F + F_{g} = 358 \text{ N} + (1800 \text{ kg} \times 9.8 \text{ m/s}^2 \times 0.10) = 358 \text{ N} + 1764 \text{ N} = 2122 \text{ N} \). The power required is \( P_{up} = F_{total} \times v = 2122 \text{ N} \times 16.67 \text{ m/s} \approx 35380 \text{ W} \approx 47.4 \text{ hp} \).
03

Calculate Power Needed Downhill (Part c)

For downhill motion, the gravitational component aids in the motion, reducing the needed power. For a 1% grade, \( F_{g} = m \times g \times 0.01 \). The total effective retarding force is \( F_{total} = F - F_{g} = 358 \text{ N} - 1800 \text{ kg} \times 9.8 \text{ m/s}^2 \times 0.01 = 358 \text{ N} - 176.4 \text{ N} = 181.6 \text{ N} \). The power required is \( P_{down} = F_{total} \times v = 181.6 \text{ N} \times 16.67 \text{ m/s} \approx 3026 \text{ W} \approx 4.06 \text{ hp} \).
04

Calculate Grade for Coasting (Part d)

To find the grade where no power is needed, balance the retarding force with the gravitational component: \( F_{g} = F \). Therefore, \( m \times g \times \sin(\theta) = 358 \text{ N} \), solving for \( \sin(\theta) \) gives \( \sin(\theta) = \frac{358 \text{ N}}{1800 \text{ kg} \times 9.8 \text{ m/s}^2} \approx 0.0204 \). The grade as a percentage is approximately 2.04%.

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Key Concepts

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

Retarding Forces
Retarding forces are the challenges a vehicle faces while moving, like friction and air resistance that attempt to slow it down. These forces have a significant impact on vehicle motion, especially on level roads. Retarding forces are calculated using the power equation:
  • The power required (\( P \)) to maintain a consistent speed combats these forces.
  • The formula used is \( P = F \times v \).
  • Here, \( F \) represents the total retarding force, and \( v \) is the velocity.
By using given values and transforming units where necessary, we can find the retarding force magnitude which affects how much energy is needed to keep the car moving at a steady pace.
Gravitational Force Components
When a vehicle is driving uphill or downhill, gravitational forces either aid or hinder its motion.
  • These forces are determined by the slope's angle. This angle is expressed as a percentage, known as the grade.
  • The component of gravitational force along the slope is calculated using \( F_{g} = m \times g \times \sin(\theta) \).
When moving uphill, this component adds to the retarding forces, requiring more power. Conversely, moving downhill, this force assists the vehicle, reducing the needed power.
Power Calculations
Power calculations for vehicles consider both maintaining speed and overcoming additional forces. Different scenarios require different calculations:
  • On a flat road, power combats the retarding forces using \( P = F \times v \).
  • Uphill travel needs more power due to gravitational force, calculated by summing retarding and gravitational forces.
  • Downhill travel calculates a reduced power requirement, as gravity aids motion.
Understanding these power requirements helps in designing efficient automotive systems that can tackle various terrains.
Vehicle Dynamics
Vehicle dynamics involve the interplay of forces and movement, explaining how a car moves under different conditions. Key dynamics to consider include:
  • Understanding the forces at play helps in calculating the necessary power to maintain or adjust speed.
  • Dynamic analysis allows engineers to optimize vehicle performance on different grades and road conditions.
  • Adjustments in design can be made to lower energy wastage due to unwanted forces like friction.
Grasping vehicle dynamics is crucial for manufacturing vehicles that deliver better fuel efficiency and performance.

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Most popular questions from this chapter

A car is stopped in a distance \(D\) by a constant friction force that is independent of the car's speed. What is the stopping distance (in terms of \(D\)) (a) if the car's initial speed is tripled, and (b) if the speed is the same as it originally was but the friction force is tripled? (Solve using the work-energy theorem.)

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of \(1.80 \times 10^{6} \mathrm{N}\) , one \(14^{\circ}\) west of north and the other \(14^{\circ}\) east of north, as they pull the tanker 0.75 \(\mathrm{km}\) toward the north. What is the total work they do on the supertanker?

A pump is required to lift 800 \(\mathrm{kg}\) of water (about 210 gallons) per minute from a well 14.0 \(\mathrm{m}\) deep and eject it with a speed of 18.0 \(\mathrm{m} / \mathrm{s}\) , (a) How much work is done per minute in lifting the water? (b) How much work is done in giving the water the kinetic energy it has when ejected? (c) What must be the power output of the pump?

Crash Barrier. A student proposes a design for an auto- mobile crash barrier in which a \(1700-\mathrm{kg}\) sport utility vehicle moving at 20.0 \(\mathrm{m} / \mathrm{s}\) crashes into a spring of negligible mass that slows it to a stop. So that the passengers are not injured, the acceleration of the vehicle as it slows can be no greater than 5.00\(g\) . (a) Find the required spring constant \(k\) , and find the distance the spring will compress in slowing the vehicle to a stop. In your calculation, disregard any deformation or crumpling of the vehicle and the friction between the vehicle and the ground. (b) What disadvantages are there to this design?

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.
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