/*! 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 55 An older-model car accelerates f... [FREE SOLUTION] | 91Ó°ÊÓ

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An older-model car accelerates from 0 to speed \(v\) in \(10 \mathrm{~s}\). A newer, more powerful sports car of the same mass accelerates from 0 to \(2 v\) in the same time period. Assuming the energy coming from the engine appears only as kinetic encrgy of the cars, compare the power of the two cars.

Short Answer

Expert verified
The sports car's engine produces four times the power of the older model's engine.

Step by step solution

01

Calculation of final kinetic energies

Start off by calculating each car’s kinetic energy at the end of the 10-second interval. The kinetic energy \(K\) is given by the formula: \(K = \frac{1}{2} m v^2\), where \(m\) is the car's mass and \(v\) is its speed. Here, the older car reaches speed \(v\), and the sports car reaches speed \(2v\). Hence, the kinetic energy of the older car is \(K1 = \frac{1}{2} m v^2\), and the kinetic energy of the sports car is \(K2 = \frac{1}{2} m (2v)^2 = 2 m v^2\).
02

Calculation of power

The power \(P\) produced by the car's engine equals to the work done (which is the change in kinetic energy) divided by the time \(t\). For each car, this will be the kinetic energy computed in step 1 divided by the time of 10 seconds. Thus, power of the older car is \(P1 = \frac{K1}{10} = \frac{\frac{1}{2} m v^2}{10}\) and the power of the sports car is \(P2 = \frac{K2}{10} = \frac{2 m v^2}{10}\).
03

Comparison of the two powers

To compare the two power outputs, form the ratio \(P2 / P1 = \frac{2 m v^2 / 10}{\frac{1}{2} m v^2 / 10} = 4\). This shows that the sports car's engine provides four times the power of the older model's engine.

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Key Concepts

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

Kinetic Energy Formula
Kinetic energy, a form of energy associated with the motion of an object, plays a pivotal role in physics. It quantifies the energy that an object possesses due to its movement. The kinetic energy formula is a key equation that enables us to calculate this energy. It is expressed as \( K = \frac{1}{2} mv^2 \) where \( K \) represents the kinetic energy, \( m \) is the mass of the object, and \( v \) is its velocity. This equation implies that kinetic energy is directly proportional to the mass of the object and the square of its velocity, meaning that even a slight increase in speed results in a considerable increase in kinetic energy.

Let's apply this to the given exercise. Two cars with the same mass, one older and one a sports car, accelerate to different speeds, \( v \) and \(2v\) respectively. Using the kinetic energy formula, we observe that for the sports car, which reaches double the velocity of the older car, the kinetic energy is quadrupled, as velocity is squared in the formula. This is a central concept that informs us about how different velocities translate into different kinetic energies, despite the mass remaining constant.
Power Calculation
Power in physics is a measure of how quickly work is done or energy is transferred. The calculation of power is fundamental in understanding how efficient machines or engines are. Power is defined as the work done over time, with the unit of measurement being the watt (W), where 1 watt equals 1 joule of work done per second. The power formula is given by \( P = \frac{W}{t} \), where \( W \) is work done and \( t \) is the time taken to do the work.

In our exercise, power calculation is used to compare the engines of two cars. We determine the work done by the engines by observing the change in kinetic energy, then divide this by the time interval of 10 seconds to find the power output for both cars. This gives us a tangible comparison of the engines' performances, as we are focusing on the rate at which each one is doing work, which is essentially what power reflects.
Work-Energy Principle
The work-energy principle is a core concept in physics that connects work (the effort exerted on an object to cause displacement) and energy (the capacity to do work). In its essence, the principle states that the work done on an object results in a change in the object's kinetic energy. Mathematically, it is expressed as \( W = \Delta K \), where \( W \) is the work done and \( \Delta K \) represents the change in kinetic energy.

In the context of our example with the two cars, the work done by the engines of both vehicles over 10 seconds is used to accelerate them from 0 to their respective speeds. This work translates into kinetic energy, thereby exemplifying the work-energy principle. By finding the change in kinetic energy (from zero at rest to \( K1 \) or \( K2 \) while moving), we effectively determine the work done by each car's engine. The sports car does more work in the same period – it generates more kinetic energy, supporting the claim that its engine is capable of delivering higher power, as concluded from the exercise's steps.

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Most popular questions from this chapter

While running, a person dissipates about \(0.60 \mathrm{~J}\) of mechanical energy per step per kilogram of body mass. If a \(60-\mathrm{kg}\) person develops a power of \(70 \mathrm{~W}\) during a race, how fast is the person running? (Assume a running step is \(1.5 \mathrm{~m}\) long.)

A loaded ore car has a mass of \(950 \mathrm{~kg}\) and rolls on rails with negligible friction. It starts from restand is pulled up a mine shaft by a cable connected to a winch. The shaft is inclined at \(30.0^{\circ}\) above the horizontal. The car accelerates uniformly to a speed of \(2.20 \mathrm{~m} / \mathrm{s}\) in \(12.0 \mathrm{~s}\) and then continues at constant speed. (a) What power must the winch motor provide when the car is moving at constant speed? (b) What maximum power must the motor provide? (c) What total energy transfers out of the motor by work by the time the car moves off the end of the track, which is of length \(1250 \mathrm{~m} ?\)

When a \(2.50-\mathrm{kg}\) object is hung vertically on a certain light spring described by Hooke's law, the spring stretches \(2.76 \mathrm{em}\). (a) What is the force constant of the spring? (b) If the \(2.50-\mathrm{kg}\) object is removed, how far will the spring stretch if a \(1.25-\mathrm{kg}\) block is hung on it? (c) How much work must an external agent do to stretch the same spring \(8.00 \mathrm{~cm}\) from its unstretched position?

A sledge loaded with bricks has a total mass of \(18.0 \mathrm{~kg}\) and is pulled at constant speed by a rope inclined at \(20.0^{\circ}\) above the horizontal. The sledge moves a distance of \(20.0 \mathrm{~m}\) on a horizontal surface. The coefficient of kinetic friction between the sledge and surface is \(0.500 .\) (a) What is the tension in the rope? (b) How much work is done by the rope on the sledge? (c) What is the mechanical energy Iost due to friction?

A \(60.0\) -kg athlete leaps straight up into the air from a trampoline with an initial speed of \(9.0 \mathrm{~m} / \mathrm{s}\). The goal of this problem is to find the maximum height she attains and her speed at half maximum height. (a) What are the interacting objects and how do they interact? (b) Select the height at which the athlete's speed is \(9.0 \mathrm{~m} / \mathrm{s}\) as \(y=0 .\) What is her kinetic energy at this point? What is the gravitational potential energy associated with the athlete? (c) What is her kinetic energy at maximum height? What is the gravitational potential energy associated with the athlete? (d) Write a general equation for energy conservation in this case and solve for the maximum height. Substitute and obtain a numerical answer. (c) Write the general equation for energy conservation and solve for the velocity at half the maximum height. Substitute and obtain a numerical answer.

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