Problem 97 - University Physics with Modern Physics

Question

It takes a force of 53 \(\mathrm{kN}\) on the lead car of a 16 -car passenger train with mass \(9.1 \times 10^{5} \mathrm{kg}\) to pull it at a constant 45 \(\mathrm{m} / \mathrm{s}\) \((101 \mathrm{mi} / \mathrm{h})\) on level tracks. (a) What power must the locomotive provide to the lead car? (b) How much more power to the lead car than calculated in part (a) would be needed to give the train an acceleration of 1.5 \(\mathrm{m} / \mathrm{s}^{2}\) , at the instant that the train has a speed of 45 \(\mathrm{m} / \mathrm{s}\) on level tracks? (c) How much more power to the lead car than that calculated in part (a) would be needed to move the train up a 1.5\(\%\) grade (slope angle \(\alpha=\arctan 0.015 )\) at a constant 45 \(\mathrm{m} / \mathrm{s} ?\)

Step-by-Step Solution

Answer

(a) 2385 kW; (b) Additional 61425 kW; (c) Additional 6008.85 kW.

Step 1: Determine Power for Constant Speed

To find the power emitted by the locomotive while pulling the train at a constant speed of 45 m/s, we use the formula for power: \( P = F \cdot v \), where \( F \) is the force applied and \( v \) is the velocity. Here, \( F = 53,000 \mathrm{N} \) and \( v = 45 \mathrm{m/s} \). Calculating gives:\[ P = 53,000 \times 45 = 2,385,000 \; \mathrm{W} \, (\mathrm{or} \; 2385 \, \mathrm{kW}) \].

Step 2: Calculate Additional Power for Acceleration

Next, we calculate the power required to accelerate the train. The additional force needed for acceleration is found using \( F = m \cdot a \), where \( m = 9.1 \times 10^5 \; \mathrm{kg} \) and \( a = 1.5 \; \mathrm{m/s^2} \).Find the force: \[ F_{\text{acc}} = 9.1 \times 10^5 \times 1.5 = 1.365 \times 10^6 \; \mathrm{N} \].The additional power is: \[ P_{\text{acc}} = F_{\text{acc}} \cdot v = 1.365 \times 10^6 \times 45 = 6.1425 \times 10^7 \; \mathrm{W} \, (\mathrm{or} \; 61425 \, \mathrm{kW}) \].

Step 3: Calculate Additional Power for Incline

Finally, calculate the power needed for maintaining 45 m/s speed up a 1.5% grade.The slope force is given by \(F_{\text{slope}} = m \cdot g \cdot \sin(\alpha)\), where \(g = 9.8 \; \mathrm{m/s^2}\).First, calculate \(\sin(\alpha)\):\(\alpha = \arctan(0.015)\), hence \(\sin(\alpha) \approx 0.015\) (since for small angles \(\sin(\alpha) \approx \tan(\alpha)\)).Calculate slope force:\[ F_{\text{slope}} = 9.1 \times 10^5 \times 9.8 \times 0.015 = 1.3353 \times 10^5 \; \mathrm{N} \].Additional power for slope:\[ P_{\text{slope}} = F_{\text{slope}} \cdot v = 1.3353 \times 10^5 \times 45 = 6.00885 \times 10^6 \; \mathrm{W} \, (\mathrm{or} \; 6008.85 \; \mathrm{kW}) \].

Key Concepts

Force and Motion

Force and motion are central concepts in physics, dealing with how objects move under various forces. Force is a vector quantity, meaning it has both magnitude and direction. Newton's laws of motion help us understand the relationship between an object's motion and the forces acting on it.
When a force acts on an object, the object will accelerate in the direction of the force, according to the equation:

• \( F = m imes a \)

This equation tells us that the force \( F \) is equal to the mass \( m \) of the object multiplied by its acceleration \( a \). For example, if we want to accelerate a train at a certain rate, we need to know both the train's mass and the additional force required to achieve that acceleration. Understanding these principles of force and motion is crucial for solving many physics problems.

Work-Energy Principle

The work-energy principle is a powerful concept in physics that relates the work done by forces on an object to its change in kinetic energy. Work is defined as the force applied to an object multiplied by the distance over which it is applied. It is expressed in the formula:

• \( W = F imes d \)

where \( W \) is the work done, \( F \) is the force applied, and \( d \) is the distance the object moves in the direction of the force.
When a force does work on an object, it can change the object's kinetic energy, given by \( KE = \frac{1}{2} m v^2 \), where \( KE \) is kinetic energy, \( m \) is mass, and \( v \) is velocity. This principle is used to calculate the power required to move a train under various conditions, such as maintaining constant speed or accelerating, as shown in the exercise.

Inclined Planes

Inclined planes are a classic example of how forces affect motion. An inclined plane is a flat surface tilted at an angle to the horizontal. When an object moves on an inclined plane, gravitational force causes the object to slide downward, while friction and other forces resist this motion.
To analyze motion on an incline, we often resolve forces into components parallel and perpendicular to the plane. The component of the gravitational force parallel to the incline causes the object to accelerate downward. This can be calculated using:

• \( F_{\text{slope}} = m imes g \times \sin(\alpha) \)

where \( \alpha \) is the angle of the incline. In this exercise, the power needed to move a train up a slope is determined by calculating the additional force required to overcome the incline.

Physics Problem Solving

Physics problem solving often involves identifying relevant principles and applying them to specific situations. Here are some steps that can help:

• **Identify Known Values:** Gather all the given information, such as mass, force, and speed. These values are crucial for calculations.


• **Use Appropriate Equations:** Choose the right physics equations that relate the known values to the unknowns, such as using \( P = F \cdot v \) for calculating power.


• **Break Down the Problem:** Divide the problem into smaller parts. For instance, calculate the constant speed power first, then additional power for acceleration, and lastly, for overcoming inclines.


• **Check Units and Signs:** Ensure all computations are consistent in terms of units and consider the direction of forces.

By following these strategies, one can approach complex physics problems systematically and find solutions effectively.

Newton's Second Law

Newton's Second Law is fundamental in understanding how objects move. It states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The law is summed up in the equation:

• \( F = m \times a \)

This means for a given force, a more massive object will accelerate less than a lighter one, and vice versa.
In the context of the exercise, we apply this law to find how much additional force is needed to accelerate the train. Once we know the force, it's possible to calculate how much extra power is needed to increase the train speed or overcome additional resistance like slopes. Newton's Second Law helps link the concepts of force, mass, and acceleration in a clear, mathematically precise way.