/*! 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 43 To dramatize the loss of energy ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

To dramatize the loss of energy in an automobile, consider a car having a weight of \(25000 \mathrm{~N}\) that is traveling at \(56 \mathrm{~km} / \mathrm{h}\). If the car is brought to a stop, determine how long a \(100-\mathrm{W}\) light bulb must burn to expend the same amount of energy.

Short Answer

Expert verified
The equivalent energy spent by the car to stop from its initial speed could light a \(100-\mathrm{W}\) bulb for approximately 3140.86 seconds.

Step by step solution

01

Conversion of Speed from km/h to m/s

The car's speed needs to be converted from km/h to m/s using the conversion factor \(1 \mathrm{~km/h} = 0.27778 \mathrm{~m/s}\). So, \(56 \mathrm{~km/h}\) becomes \(56 \times 0.27778 = 15.56 \mathrm{~m/s}\).
02

Calculation of the Car's Kinetic Energy

The automotive kinetic energy (\(E_{kin}\)) can be determined by the formula \(E_{kin} = \frac{1}{2} m v^2\), where \(m\) is the mass and \(v\) is the velocity. Mass can be calculated from the given weight by using the formula \(m = \frac{W}{g}\), where \(g\) is the acceleration due to gravity (\(9.81 \mathrm{~m/s^2}\)). So, the mass of the car is \(\frac{25000}{9.81} = 2547.41 \mathrm{~kg}\). Substituting \(m\) and \(v\) into the kinetic energy equation gives \(E_{kin} = \frac{1}{2} \times 2547.41 \times (15.56)^2 = 314086.31 \mathrm{~J}\) (Joules).
03

Compute Energy Consumption per Second for the Light Bulb

The energy consumed by a \(100-\mathrm{W}\) light bulb in one second is simply \(100\mathrm{~J}\), because 1 Watt is equivalent to 1 Joule per second.
04

Calculate the required Time for the Lightbulb

We can now find the time it would take for the light bulb to consume the same amount of energy as was lost by the car when it stopped. The time (\(t\)) can be found by dividing the total energy to be consumed (\(E_{kin}\)) by the energy consumed per second. So, \(t = \frac{E_{kin}}{100} = \frac{314086.31}{100} = 3140.86 \mathrm{~seconds}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Kinetic Energy Calculation
Kinetic energy represents the energy that an object possesses due to its motion. It's a key concept in dynamics, especially when analyzing moving vehicles like cars, planes, or even simple objects like a rolling ball. To calculate the kinetic energy (\(E_{kin}\)) of an object, you use the formula \[E_{kin} = \frac{1}{2} m v^2\] where \(m\) is the mass of the object and \(v\) is its velocity.

For an object moving with a certain speed, its kinetic energy increases with the square of the speed. This means that if a car's speed doubles, its kinetic energy will increase by a factor of four. To relate this concept to everyday experiences, consider a moving car: as it speeds up, the energy required to bring it to a stop also increases, demonstrating the practical importance of the kinetic energy concept in designing safety features like brakes and crumple zones. When solving problems, it's essential to ensure that units are consistent—I'll touch more on this in the 'Unit Conversion' section.

Understanding kinetic energy calculation is not only crucial for solving physics problems but also plays a significant role in designing efficient energy systems and can help us develop safer vehicles.
Unit Conversion
Unit conversion is a fundamental skill you need to master in engineering and physics because different units of measurement can lead to confusion if not properly handled. In the exercise, we convert the car's speed from kilometers per hour (km/h) to meters per second (m/s), a more standard unit in physics. This conversion is done using the factor \(1 \text{km/h} = 0.27778 \text{m/s}\), ensuring that the velocity is compatible with the units used to express acceleration due to gravity (\(g\text{), typically m/s}^2\)).

Keep in mind that when performing kinetic energy calculations, it's important that all quantities are expressed in the base units of mass (kg), distance (m), and time (s), according to the International System of Units (SI). Failing to convert units properly can lead to incorrect answers and cause significant errors in engineering calculations. Accurate conversions are also necessary in various scenarios like calculating fuel economy in vehicles, adjusting recipes in culinary arts, or administering medication in healthcare.
Energy Consumption
Energy consumption refers to the amount of energy utilized by devices, systems, or processes. In our exercise, we're looking at a light bulb, measured in terms of power (\(P\text{, in Watts (W), which is defined as energy per unit time. Since the energy unit Joule (J) is equal to one Watt-second (W·s), a light bulb with a power rating of }\)100 W consumes 100 Joules per second. This measurement helps us understand the efficiency and energy use of electrical devices.

By calculating how long a \(100-W\) light bulb would need to be on to consume the same amount of energy as the car's kinetic energy, we can make comparisons that connect theoretical physics with practical energy utilization. This also has broader implications in terms of energy-saving strategies, where understanding energy consumption can lead to more efficient use of resources, reduced operating costs, and a lower carbon footprint for both individuals and businesses.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The conveyor belt delivers each \(12-\mathrm{kg}\) crate to the ramp at \(A\) such that the crate's velocity is \(v_{A}=2.5 \mathrm{~m} / \mathrm{s}\) directed down along the ramp. If the coefficient of kinetic friction between each crate and the ramp is \(\mu_{k}=0.3,\) determine the speed at which each crate slides off the ramp at \(B\). Assume that no tipping occurs.

The "flying car" is a ride at an amusement park which consists of a car having wheels that roll along a track mounted inside a rotating drum. By design the car cannot fall off the track, however motion of the car is developed by applying the car's brake, thereby gripping the car to the track and allowing it to move with a constant speed of the track, \(v_{t}=3 \mathrm{~m} / \mathrm{s}\). If the rider applies the brake when going from \(B\) to \(A\) and then releases it at the top of the drum, \(A,\) so that the car coasts freely down along the track to \(B(\theta=\pi\) rad), determine the speed of the car at \(B\) and the normal reaction which the drum exerts on the car at \(B\). Neglect friction during the motion from \(A\) to \(B\). The rider and car have a total mass of \(250 \mathrm{~kg}\) and the center of mass of the car and rider moves along a circular path having a radius of \(8 \mathrm{~m}\).

A rocket having a total mass of \(8 \mathrm{Mg}\) is fired vertically from rest. If the engines provide a constant thrust of \(T=300 \mathrm{kN},\) determine the power output of the engines as a function of time. Neglect the effect of drag resistance and the loss of fuel mass and weight.

If the jet on the dragster supplies a constant thrust of \(T=20 \mathrm{kN},\) determine the power generated by the jet as a function of time. Neglect drag and rolling resistance, and the loss of fuel. The dragster has a mass of \(1 \mathrm{Mg}\) and starts from rest.

The 2 -Mg car increases its speed uniformly from rest to \(25 \mathrm{~m} / \mathrm{s}\) in \(30 \mathrm{~s}\) up the inclined road. Determine the maximum power that must be supplied by the engine, which operates with an efficiency of \(\varepsilon=0.8 .\) Also, find the average power supplied by the engine.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.