/*! 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 6 A freight train has a mass of \(... [FREE SOLUTION] | 91Ó°ÊÓ

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A freight train has a mass of \(1.5 \times 10^{7} \mathrm{~kg}\). If the locomotive can exert a constant pull of \(7.5 \times 10^{5} \mathrm{~N}\), how long does it take to increase the speed of the train from rest to \(80 \mathrm{~km} / \mathrm{h}\) ?

Short Answer

Expert verified
It takes approximately 444.4 seconds for the train to increase its speed from rest to 80 km/hr.

Step by step solution

01

Conversion of Speed to SI Units

Firstly, convert the speed from kilometers per hour to meters per second by multiplying by \(\frac{5}{18}\). So, \(80 \mathrm{km/hr} = 80 \times \(\frac{5}{18}\) = 22.22 \mathrm{m/s}\)
02

Calculation of Acceleration

According to Newton's second law, force equals mass times acceleration. Rearranging for acceleration:\( a = \frac{F}{m}\) where \(F = 7.5 \times 10^{5} N\) and \(m = 1.5 \times 10^{7} kg\). Substituting these values, we get \( a = \frac{7.5 \times 10^{5}}{1.5 \times 10^{7}} = 0.05 \mathrm{m/s^2}\)
03

Calculation of Time

We know that acceleration equals the change in velocity divided by change in time. Rearranging for time, \(t = \frac{Vf - Vi}{a}\) where \(Vf = 22.22 m/s\) is final velocity, \(Vi = 0 m/s\) is initial velocity and \(a = 0.05 m/s^2\) is acceleration. Substituting these values, we get \(t = \frac{22.22 - 0}{0.05} = 444.4 \mathrm{s}\).

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Key Concepts

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

Understanding Newton's Second Law
Newton's second law of motion is fundamental when solving problems involving forces and acceleration. It states that the force (\textbf{F}) applied to an object is equal to the mass (\textbf{m}) of the object multiplied by its acceleration (\textbf{a}). In formula terms, it's expressed as: \[ F = m \times a \]
In our problem with the freight train, the second law allows us to connect the pull from the locomotive, which is the force exerted, with the resulting acceleration of the train. Since we have both the force exerted and the mass of the train, we can rearrange the formula to solve for acceleration.

It’s also essential for students to understand how Newton’s second law relates to real-world scenarios, such as the increase in speed of vehicles or the effect of forces on motion. This law is not just a mathematical formula; it describes how objects in our universe interact with each other.
Conversion of Units Simplified
Conversion of units is a critical skill in physics, as working with consistent units is key to accurate calculations. In physics problems, we typically use the International System of Units (SI). One common task is converting speed from kilometers per hour (\textbf{km/h}) to meters per second (\textbf{m/s}), as seen in the freight train problem.
The conversion rate between \textbf{km/h} and \textbf{m/s} is based on the number of meters in a kilometer (1000) and the number of seconds in an hour (3600). The simplified conversion factor is \( \frac{5}{18} \).

To perform the conversion, you multiply the speed in \textbf{km/h} by \( \frac{5}{18} \), resulting in the equivalent speed in \textbf{m/s}. This step is crucial; otherwise, you may end up with an incorrect time calculation since acceleration's standard unit is \textbf{m/s}^2.
Calculating Acceleration with Ease
Calculating acceleration involves understanding the object's change in velocity over a certain time period. In our case, we first needed to find the train's acceleration using the provided force and mass. Following Newton's second law, we find acceleration by dividing the force by the mass: \[ a = \frac{F}{m} \]
The result tells us how quickly the train's speed will increase per second. When you grasp this concept, you will appreciate how different masses and forces affect how fast an object can speed up or slow down. This understanding is not only vital in solving textbook problems but also in appreciating everyday phenomena, such as what happens when you step on the gas pedal in a car.
Determining Time from Acceleration
Once you have calculated acceleration, determining the time it takes for an object to reach a certain speed is straightforward. You use the formula for acceleration, which relates change in velocity to the time taken: \[ t = \frac{Vf - Vi}{a} \]
Here, \( Vf \) is the final velocity, \( Vi \) is the initial velocity, and \( a \) is the acceleration. Arranging the formula to solve for time is a skill that will serve you well not just in physics, but also in understanding everyday motion, such as timing a runner's sprint or calculating the duration of a car's acceleration phase. By practicing this calculation, students develop their problem-solving skills and become more familiar with the kinematic equations that describe motion.

In regards to the freight train problem, after converting the train’s final velocity to \textbf{m/s}, and using the calculated acceleration, we found the time needed to reach the train’s final speed.

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Most popular questions from this chapter

A \(15.0\)-lb block rests on a horizontal floor. (a) What force does the floor exert on the block? (b) A rope is tied to the block and is run vertically over a pulley. The other end is attached to a free-hanging \(10.0\)-lb object. What now is the force exerted by the floor on the 15.0-lb block? (c) If the 10.0-lb object in part (b) is replaced with a \(20.0\)-lb object, what is the force exerted by the floor on the \(15.0\)-lb block?

A van accelerates down a hill (Fig. P4.73), going from rest to \(30.0 \mathrm{~m} / \mathrm{s}\) in \(6.00 \mathrm{~s}\). During the acceleration, atoy \((m=0.100 \mathrm{~kg})\) hangs by a string from the van's ceiling. The acceleration is such that the string remains perpendicular to the ceiling. Determine (a) the angle \(\theta\) and (b) the tension in the string.

An inquisitive physics student, wishing to combine pleasure with scientific inquiry, rides on a roller coaster sitting on a bathroom scale. (Do not try this yourself on a roller coaster that forbids loose, heavy packages.) The bottom of the seat in the roller-coaster car is in a plane parallel to the track. The seat has a perpendicular back and a seat belt that fits around the student's chest in a plane parallel to the bottom of the seat. The student lifts his feet from the floor so that the scale reads his weight, \(200 \mathrm{lb}\), when the car is horizontal. At one point during the ride, the car zooms with negligible friction down a straight slope inclined at \(30.0^{\circ}\) below the horizontal. What does the scale read at that point?

The parachute on a race car of weight \(8820 \mathrm{~N}\) opens at the end of a quarter-mile run when the car is traveling at \(35 \mathrm{~m} / \mathrm{s}\). What total retarding force must be supplied by the parachute to stop the car in a distance of \(1000 \mathrm{~m}\) ?

After falling from rest from a height of \(30 \mathrm{~m}\), a \(0.50-\mathrm{kg}\) ball rebounds upward, reaching a height of \(20 \mathrm{~m}\). If the contact between ball and ground lasted \(2.0 \mathrm{~ms}\), what average force was exerted on the ball?

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