/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 118 Resistance to the motion of an a... [FREE SOLUTION] | 91Ó°ÊÓ

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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
The power required is approximately 68.93 horsepower.

Step by step solution

01

Convert Speed to Meters per Second

First, convert the speed from kilometers per hour to meters per second.The speed is given as \(80\, \text{km/h}\). To convert to \( \text{m/s} \):\[80\, \text{km/h} = 80 \times \frac{1000 \, \text{m}}{3600 \, \text{s}} = \frac{80000}{3600} \, \text{m/s} \approx 22.22 \, \text{m/s}\]
02

Calculate Total Resistance Force

Use the given formula to calculate the total resistant force \(F\) when the speed \(v = 22.22 \, \text{m/s}\).\[F = 300 + 1.8 v^2 = 300 + 1.8 (22.22)^2\]\[F = 300 + 1.8 \times 493.94 = 300 + 889.092 = 1189.092 \, \text{N}\]
03

Calculate Acceleration Force

The acceleration force \(F_a\) needed to accelerate the car can be calculated using Newton's second law \(F = ma\).Given: - Weight of the car \(W = 12000 \, \text{N}\).- Gravitational acceleration \(g = 9.81 \, \text{m/s}^2\).First, find the mass \(m\):\[m = \frac{W}{g} = \frac{12000}{9.81} \, \text{kg} \approx 1223.75 \, \text{kg}\]Then calculate \(F_a\):\[F_a = ma = 1223.75 \cdot 0.92 = 1125.05 \, \text{N}\]
04

Calculate Total Force Required

The total force \(F_\text{total}\) required is the sum of the resistant force and the acceleration force.\[F_\text{total} = F + F_a = 1189.092 + 1125.05 = 2314.142 \, \text{N}\]
05

Calculate Power in Watts

Power \(P\) can be calculated using the formula \(P = F \times v\).\[P = 2314.142 \, \text{N} \times 22.22 \, \text{m/s} \approx 51415.19 \, \text{W}\]
06

Convert Power to Horsepower

Convert the power from watts to horsepower. 1 horsepower (hp) is equivalent to approximately 746 watts.\[\text{Power in hp} = \frac{51415.19}{746} \approx 68.93 \, \text{hp}\]

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Key Concepts

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

Newton's Second Law
Newton's Second Law of Motion is a cornerstone of classical mechanics and plays a crucial role in understanding how forces affect motion. This law states that the force acting on an object is equal to the mass of the object multiplied by its acceleration, written as \( F = ma \). It tells us that if you apply a force to an object, it will accelerate in the direction of the force.
If you think about an automobile, this law helps us determine how much force is needed to achieve a desired acceleration.
In the exercise, we calculated the force needed to accelerate the car using this principle. First, we determined the car's mass by using its weight and the acceleration due to gravity: \( m = \frac{W}{g} \). Then, we used this mass to find the force required for a specific acceleration: \( F_a = ma \). This allows us to understand how heavy and fast cars need specific force quantities to speed up.
Power Calculation
Power in physics is the rate at which work is done or energy is transferred. In the case of cars, power gives us an idea of how effectively an automobile's engine can move it against forces like air drag and road friction. Power is crucial for describing vehicle performance since it's essentially how quickly a car can apply force to keep or increase its speed.
To calculate power, we use the formula \( P = F \cdot v \), where \( P \) is power, \( F \) is force, and \( v \) is velocity. For this scenario, we combined the total resistant force with the acceleration force needed, and multiplied by the car's speed. The power was then converted from watts to horsepower because it's a more commonly used unit in the automotive industry.
Air Drag
Air drag, or aerodynamic drag, is a type of frictional force that a vehicle experiences as it moves through the air. It increases with speed and is dependent on the shape and size of the vehicle. Mathematically, air drag is often modeled as being proportional to the square of the velocity (\( v^2 \)).
In the problem, air drag is addressed in the formula \( F = 300 + 1.8v^2 \). The term \( 1.8v^2 \) represents the air drag component, showing how much resistance increases as the car speeds up. Understanding air drag is key to improving vehicle efficiency, as reducing air drag leads to less force needed to maintain higher speeds, thus saving fuel and improving performance.
Frictional Forces
When a vehicle moves, several forces act against its motion. Frictional forces are one of these and can be split into different components. Road friction is one kind of friction experienced by vehicles and usually remains nearly constant regardless of speed.
In the previous calculations, frictional forces were captured by the constant term in the resistance equation, which was 300 N. While air drag scales with speed, this road friction component does not.
  • Friction contributes to the overall resistance a vehicle has to overcome.
  • Reducing unnecessary friction can lead to better fuel efficiency.
  • Understanding and managing friction is essential for vehicle maintenance and design.
This comprehension helps engineers design better-performing and more fuel-efficient vehicles by minimizing the unwanted resistance, all thanks to understanding the role of friction and its constancy.

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Most popular questions from this chapter

A single conservative force \(F(x)\) acts on a \(1.0 \mathrm{~kg}\) particle that moves along an \(x\) axis. The potential energy \(U(x)\) associated with \(F(x)\) is given by $$U(x)=-4 x e^{-x / 4} \mathrm{~J}$$ where \(x\) is in meters. At \(x=5.0 \mathrm{~m}\) the particle has a kinetic energy of \(2.0 \mathrm{~J} .\) (a) What is the mechanical energy of the system? (b) Make a plot of \(U(x)\) as a function of \(x\) for \(0 \leq x \leq 10 \mathrm{~m}\), and on the same graph draw the line that represents the mechanical energy of the system. Use part (b) to determine (c) the least value of \(x\) the particle can reach and (d) the greatest value of \(x\) the particle can reach. Use part (b) to determine (e) the maximum kinetic energy of the particle and (f) the value of \(x\) at which it occurs. (g) Determine an expression in newtons and meters for \(F(x)\) as a function of \(x .(\mathrm{h})\) For what (finite) value of \(x\) does \(F(x)=0\) ?

What is the spring constant of a spring that stores \(25 \mathrm{~J}\) of elastic potential energy when compressed by \(7.5 \mathrm{~cm} ?\)

During a rockslide, a \(520 \mathrm{~kg}\) rock slides from rest down a hillside that is \(500 \mathrm{~m}\) long and \(300 \mathrm{~m}\) high. The coefficient of kinetic friction between the rock and the hill surface is \(0.25 .(\mathrm{a})\) If the gravitational potential energy \(U\) of the rock-Earth system is zero at the bottom of the hill, what is the value of \(U\) just before the slide? (b) How much energy is transferred to thermal energy during the slide? (c) What is the kinetic energy of the rock as it reaches the bottom of the hill? (d) What is its speed then?

The summit of Mount Everest is \(8850 \mathrm{~m}\) above sea level. (a) How much energy would a \(90 \mathrm{~kg}\) climber expend against the gravitational force on him in climbing to the summit from sea level? (b) How many candy bars, at \(1.25 \mathrm{MJ}\) per bar, would supply an energy equivalent to this? Your answer should suggest that work done against the gravitational force is a very small part of the energy expended in climbing a mountain.

The string is \(L=120 \mathrm{~cm}\) long, has a ball attached to one end, and is fixed at its other end. A fixed peg is at point \(P\). Released from rest, the ball swings down until the string catches on the peg; then the ball swings up, around the peg. If the ball is to swing completely around the peg, what value must distance \(d \mathrm{ex}\) ceed? (Hint: The ball must still be moving at the top of its swing. Do you see why?)

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