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Chapter 6: Problem 59

A 1000 -kg auto travels up a \(3.0\) percent grade at \(20 \mathrm{~m} / \mathrm{s}\). Find the cruising power required, neglecting friction.

Short Answer

Expert verified
The required power is 5886 watts.

Step by step solution

01

Understand the Problem

We need to find the power required to maintain the speed of the car as it climbs a 3.0% grade hill at a speed of 20 m/s. We neglect all forms of friction and only consider the gravitational force component acting against the car.
02

Calculate the Grade's Angle

A 3.0% grade means that for every 100 units of horizontal distance, the elevation changes by 3 units. Therefore, the sine of the angle (\(\theta)\) of the incline can be approximated as 0.03. We'll use this to calculate the force acting against the car due to gravity.
03

Determine the Gravitational Force Component

The force due to gravity acting down the slope is given by \(F_{gravity} = m \cdot g \cdot \sin(\theta)\), where \(m\) is the mass of the car, \(g\) is the acceleration due to gravity (9.81 m/s虏), and \(\sin(\theta) = 0.03\).Substitute the known values:\[ F_{gravity} = 1000 \times 9.81 \times 0.03 \]
04

Calculate the Gravitational Force

Plug in the values and calculate:\[ F_{gravity} = 1000 \cdot 9.81 \cdot 0.03 = 294.3 \text{ N} \]
05

Calculate the Power Required

The power required (\(P\)) to move the car up the slope at constant speed is given by the product of the opposing gravitational force and the speed:\[ P = F_{gravity} \cdot v \]Substitute \( F_{gravity} = 294.3 \text{ N} \) and \( v = 20 \text{ m/s} \):\[ P = 294.3 \cdot 20 \]
06

Evaluate the Power

Calculate the power:\[ P = 294.3 \times 20 = 5886 \text{ W} \]
07

Conclusion

The cruising power required to maintain a speed of 20 m/s up a 3.0% incline is 5886 watts (or approximately 5.89 kilowatts).

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Key Concepts

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

Mechanics
Mechanics is a substantial branch of physics dedicated to studying motion and the forces that produce it. It's crucial for understanding how objects move and interact with their environment. In our scenario, mechanics help explain how a car can travel up an inclined plane. When considering this motion, it's essential to focus on a few key principles.
  • **Forces**: The car is subjected to several forces, though in this example, we only consider the force due to gravity due to the problem's specifications.

  • **Mass**: The mass of the car influences how much force is needed to move it. Here, the car's given mass is 1000 kg.

  • **Gravity**: This is a constant force pulling objects toward the earth, and has a standard value of 9.81 m/s虏.

These parts of mechanics come together when calculating how much power the car requires to overcome the gravitational force while moving up the slope. Ignoring friction makes it simpler to see how the force transferred through the wheels is nearly solely focused on counteracting gravity.
Power Calculation
Power is the rate at which work is done or energy is transferred over time. When dealing with power calculations, especially in mechanics, it helps determine how efficiently a machine, like a car in this example, can perform work over time. Calculating power involves a few steps:
  • **Determine the Force**: Use the formula for the gravitational force component parallel to the incline, which is determined by the car's weight and the incline's angle.

  • **Velocity**: In this problem, the car's velocity is a given constant, 20 m/s.

  • **Power Formula**: Power can be calculated with the formula \[ P = F \times v \]where F is force and v is velocity.

  • **Calculating Power**: By multiplying the gravitational force (294.3 N) by the car's velocity (20 m/s), you find the power needed: \[ P = 294.3 \times 20 = 5886 \, \text{W} \]

This process allows for the calculation of the power required, showing how much energy per second is necessary to maintain a specific speed in light of the slope.
Inclined Plane
The inclined plane is an essential part of mechanics, often used to simplify the understanding of forces. When an object moves on an inclined plane, like a car on a hill, its weight components affect how much work is needed to move it.
  • **Inclined Plane Description**: A slope or hill forms an inclined plane. In our case, it has a 3% grade, meaning a slope of 3 units up for every 100 units across.

  • **Sine of the Angle**: This small angle can be approximated to better understand the forces involved. The sine of this angle is about 0.03, helping to calculate the opposition force from gravity smoothly.

  • **Work and Inclines**: The angle affects how much of gravity's force acts parallel to the incline. Understanding this is crucial in calculating power and force, as seen in our exercise.

Inclined planes like this provide a practical model of how to dissect forces in real-world scenarios, balancing gravity's pulling effect with the horizontal movement needed to overcome it.

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Most popular questions from this chapter

A \(0.25\) -hp motor is used to lift a load at the rate of \(5.0 \mathrm{~cm} / \mathrm{s}\). How great a load can it raise at this constant speed? Assume the power output of the motor to be \(0.25 \mathrm{hp}=186.5 \mathrm{~W}\). In \(1.0 \mathrm{~s}\), the load \(m g\) is lifted a distance of \(0.050 \mathrm{~m}\). Therefore, Work done in \(1.0 \mathrm{~s}=(\) weight \()(\) height change in \(1.0 \mathrm{~s})=(\mathrm{mg})(0.050 \mathrm{~m})\) By definition, Power = Work / Time, and so $$ 186.5 \mathrm{~W}=\frac{(m g)(0.050 \mathrm{~m})}{1.0 \mathrm{~s}} $$ Using \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\), we find that \(m=381 \mathrm{~kg}\). The motor can lift a load of about \(0.38 \times 10^{3} \mathrm{~kg}\) at this speed.

A force of \(1.50 \mathrm{~N}\) acts on a \(0.20\) -kg cart so as to uniformly accelerate it along a straight air track. The track and force are horizontal and in line. How fast is the cart going after acceleration from rest through \(30 \mathrm{~cm}\), if friction is negligible? The work done by the force causes, and is equal to, the increase in KE of the cart. Therefore, Substituting gives $$ (1.50 \mathrm{~N})(0.30 \mathrm{~m})=\frac{1}{2}(0.20 \mathrm{~kg}) v_{F}^{2} $$ from which \(v_{f}=2.1 \mathrm{~m} / \mathrm{s}\).

A \(1200-\mathrm{kg}\) car coasts from rest down a driveway that is inclined \(20^{\circ}\) to the horizontal and is \(15 \mathrm{~m}\) long. How fast is the car going at the end of the driveway if \((a)\) friction is negligible and \((b)\) a friction force of \(3000 \mathrm{~N}\) opposes the motion?

A 4.0-kg object is lifted \(1.5 \mathrm{~m}\). (a) How much work is done against the Earth's gravity? (b) Repeat if the object is lowered instead of lifted.

A 900 -kg car whose motor delivers a maximum power of \(40.0\) hp to its wheels can maintain a steady speed of \(130 \mathrm{~km} / \mathrm{h}\) on a horizontal roadway. How large is the friction force that impedes its motion at this speed?
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