Problem 96 - University Physics with Modern Physics

Question

Six diesel units in series can provide 13.4 \(\mathrm{MW}\) of power to the lead car of a freight train. The diesel units have total mass \(1.10 \times 10^{6} \mathrm{kg}\) . The average car in the train has mass \(8.2 \times 10^{4} \mathrm{kg}\) and requires a horizontal pull of 2.8 \(\mathrm{kN}\) to move at a constant 27 \(\mathrm{m} / \mathrm{s}\) on level tracks. (a) How many cars can be in the train under these conditions? (b) This would leave no power for accelerating or climbing hills. Show that the extra force needed to accelerate the train is about the same for a \(0.10-\mathrm{m} / \mathrm{s}^{2}\) acceleration or a 1.0\(\%\) slope (slope angle \(\alpha=\arctan 0.010 )\) . (c) With the 1.0\(\%\) slope, show that an extra 2.9 \(\mathrm{MW}\) of power is needed to maintain the \(27-\mathrm{m} / \mathrm{s}\) speed of the diesel units. (d) With 2.9 \(\mathrm{MW}\) less power available, how many cars can the six diesel units pull up a 1.0\(\%\) slope at a constant 27 \(\mathrm{m} / \mathrm{s} ?\)

Step-by-Step Solution

Answer

(a) 177 cars, (b) Extra force is about the same, (c) 2.9 MW needed, (d) 138 cars.

Step 1: Calculate power needed for one car

The question states that each car requires a horizontal pull of 2.8 kN to move at a constant speed of 27 m/s. To find the power needed for one car, use the equation for power: \( P = F \times v \).Substitute the given values: \( P = 2800 \times 27 \, \text{W} = 75600 \, \text{W or} \ 75.6 \, \text{kW} \). Thus, each car requires 75.6 kW of power to maintain the speed of 27 m/s.

Step 2: Determine number of cars the diesel units can pull

The total power available is 13.4 MW, which is equivalent to 13400 kW. Each car requires 75.6 kW to maintain the speed.Find the number of cars by dividing the total power by the power required per car: \( \text{Number of cars} = \frac{13400}{75.6} \approx 177.65 \).Since the number of cars must be a whole number, the train can pull 177 cars.

Step 3: Calculate extra force for acceleration

To find the force required for acceleration, use Newton's second law: \( F = ma \),where \( a = 0.10 \, \text{m/s}^2 \) is the acceleration.The total mass of the train with \( n \) cars is \( 1.10 \times 10^6 + 8.2 \times 10^4 n \), and the force required is \( F = (1.10 \times 10^6 + 8.2 \times 10^4 n) \times 0.10 \, \text{N} \). With \( n = 177 \), the force is approximately 1.897 MN. Compare this to the force needed to overcome a 1.0% slope.

Step 4: Calculate force needed for 1.0% slope

The force required to move all cars on a slope involves gravitational force: \( F = mg \sin(\alpha) \).With \( \alpha = \arctan(0.010) \approx 0.01 \) and \( g = 9.8 \, \text{m/s}^2 \), the force due to gravity is: \( F = (1.10 \times 10^6 + 8.2 \times 10^4 \times 177) \times 9.8 \times 0.01 \approx 1.897 \, \text{MN} \).This result matches the force needed for acceleration, showing they are about the same.

Step 5: Calculate extra power for 1.0% slope

The power required to maintain speed with an extra slope force is calculated using \( P = F \times v \):\( P = 1.897 \times 10^6 \times 27 \, \text{W} \approx 2.9 \, \text{MW} \).This is consistent with the given additional power needed.

Step 6: Determine cars with reduced power

With 2.9 MW less power (13.4 MW - 2.9 MW = 10.5 MW available), determine how many cars the train can pull:Using the same method, \( \text{Number of cars} = \frac{10500}{75.6} \approx 138.89 \).Thus, the train can pull 138 cars up a 1.0% slope at a constant speed of 27 m/s.

Key Concepts

Diesel Power Units

Diesel power units are key components when it comes to powering large trains. These powerful engines are capable of producing significant amounts of power, which in our scenario is 13.4 megawatts (MW). Diesel units operate by converting the chemical energy stored within diesel fuel into mechanical energy, which drives the train. The amount of power these engines produce is crucial for meeting the demands of moving heavy freight, including accelerating, maintaining speed, and overcoming resistance forces like friction and gravity. When assessing the capabilities of diesel power units, it's essential to consider both their maximum power output and efficiency under different conditions.

Freight Train Dynamics

Understanding the dynamics of a freight train involves examining the forces at play as it moves along the tracks. In this context, the term "dynamics" refers to the relations between motion and the forces affecting it. Freight trains are typically composed of multiple cars, each having a mass that influences its motion. The total mass, including the diesel units and each freight car, impacts the amount of force required to move the train. Dynamics consider factors like resistance to motion due to friction and inertia, which is the tendency of an object to resist changes in its state of motion, and is related directly to the mass of the train. This is especially pertinent when discussing how many cars a train can pull at a given speed or when analyzing the effects of inclines or accelerations.

Newton's Second Law

Newton's second law forms a fundamental basis for analyzing train motion and predicting how the train will respond to various forces. The law is expressed as \( F = ma \), where \( F \) is the net force applied to an object, \( m \) is the object's mass, and \( a \) is the acceleration produced. For the freight train, every car added increases the total mass, which in turn requires more force to achieve the same acceleration. Similarly, this law helps calculate the extra force needed for the train to climb a slope or change speed. The train’s force requirements due to changes in velocity and elevation can be directly assessed using Newton’s second law, underscoring its importance in determining operational capabilities.

Power Calculations

Power calculations provide insight into the capacity of the train to maintain its speed under different conditions. Power, denoted in watts, is the rate at which work is done or energy is transformed. For the train, the critical equation is \( P = F \times v \), where \( P \) is power, \( F \) is force, and \( v \) is velocity. In practice, these calculations help determine how much power is required for the train to maintain its constant speed of 27 m/s. This includes overcoming horizontal pull and potential additional forces due to acceleration or climbing slopes. In our scenario, assessing the power needs accurately ensures that the train can operate effectively without exceeding the available power of the diesel units, which is particularly important when working on slopes or facing increased inertia.

Force and Acceleration

Force and acceleration are closely linked concepts, significant in analyzing how a train moves and responds to changes in speed or track incline. Force is a vector quantity that causes an object to change its motion, and it is directly related to acceleration through Newton's second law. When a force is applied to the train, the acceleration is determined by the equation \( a = \frac{F}{m} \), where \( m \) is the mass. Acceleration can also refer to changes in velocity, whether speeding up, slowing down, or changing direction. In the case of our freight train, understanding the force required for a particular acceleration helps in designing solutions to ensure the train's capability to handle varying track conditions, increasing or decreasing speed efficiently while maintaining safety and fuel efficiency.