/*! 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 92 The engine of a car with mass \(... [FREE SOLUTION] | 91Ó°ÊÓ

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The engine of a car with mass \(m\) supplies a constant power \(P\) to the wheels to accelerate the car. You can ignore rolling friction and air resistance. The car is initially at rest. (a) Show that the speed of the car is given as a function of time by \(v=(2 P t / m)^{1 / 2}\) (b) Show that the acceleration of the car is not constant but is given as a function of time by \(a=(P / 2 m t)^{1 / 2} \cdot(\text { c) Show that the }\) displacement as a function of time is given by \(x-x_{0}=\) \((8 P / 9 m)^{1 / 2} t^{3 / 2}\).

Short Answer

Expert verified
(a) \(v=(2Pt/m)^{1/2}\); (b) \(a=(P/2mt)^{1/2}\); (c) \(x-x_0=(8P/9m)^{1/2} t^{3/2}\).

Step by step solution

01

Define Power and Work

Power is the rate at which work is done. Therefore, for a constant power \(P\), the work done in time \(t\) is given by \(W = Pt\). This work done contributes to the kinetic energy of the car.
02

Relate Work to Kinetic Energy

The work done on the car is equal to the change in its kinetic energy. Initially, the car is at rest, so its initial kinetic energy is zero. Thus, \(W = \frac{1}{2}mv^2 = Pt\).
03

Solve for Velocity

From the expression \(\frac{1}{2}mv^2 = Pt\), solve for \(v\) to get the speed as a function of time: \[v = \left( \frac{2Pt}{m} \right)^{1/2}\]
04

Differentiate Velocity to Find Acceleration

Differentiate the velocity function with respect to time to find the acceleration: \[v = \left( \frac{2Pt}{m} \right)^{1/2}\]Taking the derivative, we find:\[\frac{dv}{dt} = \frac{1}{2}\left( \frac{2P}{m}\right)^{1/2} t^{-1/2}\]Thus, \[a = \left( \frac{P}{2mt} \right)^{1/2}\]
05

Integrate Velocity to Find Displacement

Integrate the velocity function to find displacement:\[v = \left( \frac{2Pt}{m} \right)^{1/2}\]Integrate this with respect to time:\[x = \int \left( \frac{2Pt}{m} \right)^{1/2} dt\]Set \( u = t^{1/2} \) so that \( du = \frac{1}{2}t^{-1/2}dt \) leading to:\[x - x_0 = \frac{2}{3}\left( \frac{2P}{m} \right)^{1/2} t^{3/2}\]Rewriting, we get:\[(x - x_0) = \left( \frac{8P}{9m} \right)^{1/2} t^{3/2}\]

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Key Concepts

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

Power and Energy Relations
Power is essentially the rate of doing work. In physics, this definition is expressed as the amount of work done in a unit of time. When a car engine provides a constant power, it means it’s supplying a steady amount of energy every second. This power is used to increase the speed of the car by converting energy into motion, or kinetic energy.
As the formula \[ W = P imes t \]shows, the work done by the engine is directly proportional to the time that elapses and the power it outputs.
The energy supplied (which is measured as work done) builds up in the car as kinetic energy, which is the energy possessed due to motion. Initially, when the car is at rest, this kinetic energy is zero. But as power continues to be applied, the kinetic energy increases, leading to a change in the car's velocity.
Work-Energy Principle
The work-energy principle is a fundamental concept in mechanics that illustrates how work done on a body results in a change in its kinetic energy. In the case of our car, the work done by the engine translates directly to kinetic energy. This is mathematically expressed as:\[ W = \Delta KE = \frac{1}{2}mv^2 \]where \( \Delta KE \) represents the change in kinetic energy and \( v \) is the car's velocity.
At the start, with the car at rest, the initial kinetic energy is zero. So, when the engine outputs power over time, it increases the car's kinetic energy. This increase manifests as an increase in speed, as calculated in the solution.
By equating work done to kinetic energy change, we derive expressions showing how speed changes over time, illustrating the energy-to-motion transformation enacted by the engine.
Variable Acceleration
Acceleration typically implies a change in velocity. When acceleration isn't constant, it means this rate of change varies over time, often requiring more complex calculations. In this exercise, the acceleration due to the engine power is not constant; it changes with time as shown:\[ a = \left( \frac{P}{2mt} \right)^{1/2} \]This equation results from differentiating velocity with respect to time. Since the power is constant but the velocity follows a square root dependence on time, the acceleration becomes time-dependent, decreasing as time progresses.
This notion of variable acceleration means the car gets less additional speed per unit of time as time goes on, a realistic outcome for real-world scenarios where, over time, it becomes increasingly difficult to maintain high acceleration.
Ultimately, understanding this concept aids in comprehending how power affects movement and speed dynamically, especially in vehicles and machinery relying on engines with sustained power inputs.

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Most popular questions from this chapter

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?

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A \(2.50-\mathrm{kg}\) textbook is forced against a horizontal spring of negligible mass and force constant \(250 \mathrm{N} / \mathrm{m},\) compressing the spring a distance of 0.250 \(\mathrm{m}\) . When released, the textbook slides on a horizontal tabletop with coefficient of kinetic friction \(\mu_{k}=\) \(0.30 .\) Use the work-energy theorem to find how far the textbook moves from its initial position before coming to rest.

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