/*! 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

An inventive child wants to reach an apple in a tree without climbing the tree. Sitting in a chair connected to a rope that passes over a frictionless pulley (Fig. P4.81), the child pulls on the loose end of the rope with such a force that the spring scale reads \(250 \mathrm{~N}\). The child's true weight is \(320 \mathrm{~N}\), and the chair weighs \(160 \mathrm{~N}\). The child's feet are not touching the ground. (a) Show that the acceleration of the system is upward, and find its magnitude. (b) Find the force the child exerts on the chair.

The force exerted by the wind on the sails of a sailboat is \(390 \mathrm{~N}\) north. The water exerts a force of \(180 \mathrm{~N}\) east. If the boat (including its crew) has a mass of \(270 \mathrm{~kg}\), what are the magnitude and direction of its acceleration?

A \(1000-\mathrm{kg}\) car is pulling a \(300-\mathrm{kg}\) trailer. Together, the car and trailer have an acceleration of \(2.15 \mathrm{~m} / \mathrm{s}^{2}\) in the positive \(x\)-direction. Neglecting frictional forces on the trailer, determine (a) the net force on the car, (b) the net force on the trailer, (c) the magnitude and direction of the force exerted by the trailer on the car, and (d) the resultant force exerted by the car on the road.

(a) An elevator of mass \(m\) moving upward has two forces acting on it: the upward force of tension in the cable and the downward force due to gravity. When the elevator is accelerating upward, which is greater, \(T\) or \(w ?\) (b) When the elevator is moving at a constant velocity upward, which is greater, \(T\) or \(w\) ? (c) When the elevator is moving upward, but the acceleration is downward, which is greater, \(T\) or \(w ?\) (d) Let the elevator have a mass of \(1500 \mathrm{~kg}\) and an upward acceleration of \(2.5 \mathrm{~m} / \mathrm{s}^{2}\). Find \(T\). Is your answer consistent with the answer to part (a)? (e) The elevator of part (d) now moves with a constant upward velocity of \(10 \mathrm{~m} / \mathrm{s}\). Find T. Is your answer consistent with your answer to part (b)? (f) Having initially moved upward with a constant velocity, the elevator begins to accelerate downward at \(1.50 \mathrm{~m} / \mathrm{s}^{2}\). Find \(T\). Is your answer consistent with your answer to part (c)?

A 72-kg man stands on a spring scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of \(1.2 \mathrm{~m} / \mathrm{s}\) in \(0.80 \mathrm{~s}\). The elevator travels with this constant speed for \(5.0 \mathrm{~s}\), undergoes a uniform negative acceleration for \(1.5 \mathrm{~s}\), and then comes to rest. What does the spring scale register (a) before the elevator starts to move? (b) During the first \(0.80 \mathrm{~s}\) of the elevator's ascent? (c) While the elevator is traveling at constant speed? (d) During the elevator's negative acceleration?

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