/*! 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 36 The electric motor of a model tr... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 36

The electric motor of a model train accelerates the train from rest to 0.620 \(\mathrm{m} / \mathrm{s}\) in 21.0 \(\mathrm{ms}\) . The total mass of the train is 875 \(\mathrm{g}\) . Find the average power delivered to the train during the acceleration.

Short Answer

Expert verified
The average power delivered to the train during the acceleration is approximately 0.1611 Watts.

Step by step solution

01

Convert Units

First, ensure all units are consistent. Convert the mass of the train into kilograms by dividing the given mass in grams by 1000. Convert the acceleration time into seconds by dividing the given time in milliseconds by 1000.
02

Calculate Acceleration

Use the formula for acceleration, which is final velocity minus initial velocity over time, to find the acceleration of the train.
03

Calculate Force

Apply Newton's second law, where force equals mass times acceleration, to calculate the force exerted on the train.
04

Calculate Work Done

Work done on the train is calculated by multiplying the force by the displacement, which is found using the formula for the average velocity times the time interval.
05

Calculate Average Power

Power is defined as work done over time. Use the work calculated in the previous step and divide it by the time interval to find the average power.

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 91影视!

Key Concepts

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

Convert Units
Converting units is an essential first step in many physics problems to ensure that all measurements are in consistent units. For example, when dealing with mass, it's common to convert grams to kilograms since the standard unit of mass in physics is the kilogram (kg). To convert grams to kilograms, you divide the number of grams by 1000, as there are 1000 grams in a kilogram. Similarly, time is often measured in seconds within the International System of Units (SI), so milliseconds should be converted to seconds by dividing by 1000, because there are 1000 milliseconds in a second. Ensuring units are consistent avoids confusion and errors in calculations.
Acceleration Formula
The acceleration formula is fundamental when you need to determine how quickly an object is speeding up or slowing down. It is defined as the change in velocity divided by the time over which that change occurs. Mathematically, it's expressed as \(a = \frac{\Delta v}{\Delta t}\), where \(a\) is acceleration, \(\Delta v\) is the change in velocity, and \(\Delta t\) is the change in time. In the case of our model train, the initial velocity is zero (since it starts from rest), and the final velocity is given, allowing for a straightforward application of this formula to find the train's acceleration.
Newton's Second Law
Newton's Second Law is a cornerstone of classical mechanics. This law relates the net force acting on an object to its mass and the acceleration it experiences through the formula \(F = ma\), with \(F\) representing force, \(m\) standing for mass, and \(a\) for acceleration. In practical terms, this tells us that the greater the mass of an object or the more acceleration you want to achieve, the more force you'll need to apply. For the model train in our exercise, we can use this law to determine the force required to accelerate the train by plugging in the mass and the previously calculated acceleration.
Work Done Calculation
The work done on an object is a measure of the energy transferred as the result of applying a force over a distance. To calculate the work done, the formula \(W = Fd\) is used, where \(W\) is work, \(F\) is the force applied, and \(d\) is the displacement of the object. Calculating the displacement of an accelerating object like our model train can be done by determining its average velocity over the time interval and then using that velocity to find the distance traveled during the acceleration. This distance is then multiplied by the calculated force to find the work done on the train.
Average Power Definition
Power in physics is defined as the rate at which work is done or energy is transferred. When we talk about average power, we're considering the total work done over a certain time period. The formula for average power is \(P = \frac{W}{t}\), where \(P\) is power, \(W\) is work done, and \(t\) is the time over which the work was done. For our exercise, by dividing the work done to accelerate our model train by the time it took to accelerate, we obtain the average power delivered by the electric motor to the train.

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

A baseball outfielder throws a \(0.150-\mathrm{kg}\) baseball at a speed of 40.0 \(\mathrm{m} / \mathrm{s}\) and an initial angle of \(80.0^{\circ} .\) What is the kinetic energy of the baseball at the highest point of its trajectory?

A sled of mass m is given a kick on a frozen pond. The kick imparts to it an initial speed of 2.00 m/s. The coefficient of kinetic friction between sled and ice is 0.100. Use energy considerations to find the distance the sled moves before it stops.

A 15.0 -kg block is dragged over a rough, horizontal sur- face by a 70.0 . N force acting at \(20.0^{\circ}\) above the horizontal. The block is displaced \(5.00 \mathrm{m},\) and the coefficient of kinetic friction is 0.300 . Find the work done on the block by (a) the \(70-\mathrm{N}\) force, (b) the normal force, and (c) the gravitational force. (d) What is the increase in internal energy of the block-surface system due to friction? (e) Find the total change in the block's kinetic energy.

1 force \(\mathbf{F}=(6 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}) \mathrm{N}\) acts on a particle that under- goes a displacement \(\Delta \mathbf{r}=(3 \mathbf{i}+\mathbf{j}) \mathrm{m} .\) Find \((\mathrm{a})\) the work done by the force on the particle and \((\mathrm{b})\) the angle between \(\mathbf{F}\) and \(\Delta \mathbf{r} .\)

The force acting on a particle is \(F_{x}=(8 x-16) \mathrm{N},\) where \(x\) is in meters. (a) Make a plot of this force versus \(x\) from \(x=0\) to \(x=3.00 \mathrm{m}\) (b) From your graph, find the net work done by this force on the particle as it moves from \(x=0\) to \(x=3.00 \mathrm{m} .\)
See all solutions

Recommended explanations on Physics Textbooks

Fields in Physics

Read Explanation

Medical Physics

Read Explanation

Torque and Rotational Motion

Read Explanation

Wave Optics

Read Explanation

Geometrical and Physical Optics

Read Explanation

Electricity and Magnetism

Read Explanation
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.