Problem 44
Question
Locomotive The union Pacific's Big Boy locomotive could pull 6000 -ton trains with a tractive effort (pull) of \(602,148 \mathrm{N}(135,375\) lb). At this level of effort, about how much work did \(\operatorname{Big}\) Boy do on the (approximately straight) \(605-\mathrm{km}\) journey from San Francisco to Los Angeles?
Step-by-Step Solution
Verified Answer
The Big Boy locomotive did approximately \( 3.640,958 \times 10^{11} \text{ Joules} \) of work.
1Step 1: Understand the formula for work
Work is calculated using the formula: \ \[ \text{Work} = \text{Force} \times \text{Distance} \times \cos(\theta) \] where \( \theta \) is the angle between the force and the direction of motion. For a straight line journey with force parallel to motion, \( \theta = 0 \), so \( \cos(\theta) = 1 \). This simplifies the formula to: \ \[ \text{Work} = \text{Force} \times \text{Distance} \]
2Step 2: Identify the given values
From the problem statement, the tractive effort (force) \( F \) is given as \( 602,148 \) N. The distance \( d \) of the journey is \( 605 \) km. Before calculating work, make sure to convert the distance into meters: \ \[ 605 \text{ km} = 605,000 \text{ meters} \]
3Step 3: Calculate the work done using the formula
Substitute the given values into the simplified work formula: \ \[ \text{Work} = 602,148 \text{ N} \times 605,000 \text{ m} \] Calculating this gives: \ \[ \text{Work} = 3.640,958 \times 10^{11} \text{ Joules} \]
4Step 4: Verify the units and result
Ensure the units are consistent: Force in Newtons and Distance in meters means the result will be in Joules (since work is generally measured in Joules). The calculated work is \( 3.640,958 \times 10^{11} \text{ J} \). This checks out with our understanding of physics principles.
Key Concepts
Physics calculationsTractive effortUnit conversionCalculating work in physics
Physics calculations
In physics, calculations often involve fundamental concepts like force, distance, and work. These are crucial when solving problems related to motion and energy. One of the basic principles is understanding how force interacts with motion. Force, measured in Newtons (N), is the push or pull exerted by an object, influencing its movement.
Calculating work involves using the work formula, \[ \text{Work} = \text{Force} \times \text{Distance} \times \cos(\theta) \]where \(\theta\) is the angle between the direction of the force and the path of motion. For straight-line movements where the force is parallel to the motion, \(\theta\) is 0, simplifying \(\cos(\theta)\) to 1. Thus, the formula becomes simply: \[ \text{Work} = \text{Force} \times \text{Distance} \]
Grasping this formula is fundamental to many physics problems, where you need to quantify how much work a force does over a specific distance.
Calculating work involves using the work formula, \[ \text{Work} = \text{Force} \times \text{Distance} \times \cos(\theta) \]where \(\theta\) is the angle between the direction of the force and the path of motion. For straight-line movements where the force is parallel to the motion, \(\theta\) is 0, simplifying \(\cos(\theta)\) to 1. Thus, the formula becomes simply: \[ \text{Work} = \text{Force} \times \text{Distance} \]
Grasping this formula is fundamental to many physics problems, where you need to quantify how much work a force does over a specific distance.
Tractive effort
Tractive effort refers to the pulling power of a locomotive or vehicle. It's the force transmitted through the wheels to move the train forward on the tracks. As given, the Union Pacific's Big Boy locomotive has a tractive effort of 602,148 N, a substantial force required to move a train weighing 6000 tons.
This concept is pivotal in understanding how locomotives function to overcome resistive forces such as friction and gravity. Knowing the tractive effort helps in assessing how much work the locomotive can perform over a certain distance. Tractive effort enables engineers to calculate the efficiency and capacity of trains and other similar vehicles to transport heavy loads over long distances.
Each locomotive's tractive effort is affected by factors such as wheel adhesion, the weight of the train, and track conditions, which engineers consider to optimize performance.
This concept is pivotal in understanding how locomotives function to overcome resistive forces such as friction and gravity. Knowing the tractive effort helps in assessing how much work the locomotive can perform over a certain distance. Tractive effort enables engineers to calculate the efficiency and capacity of trains and other similar vehicles to transport heavy loads over long distances.
Each locomotive's tractive effort is affected by factors such as wheel adhesion, the weight of the train, and track conditions, which engineers consider to optimize performance.
Unit conversion
Unit conversion plays a vital role in solving physics problems accurately. In the context of the given exercise, the task involves converting kilometers into meters, as work is calculated using force in Newtons and distance in meters.
To convert distance into meters, remember the conversion factor:- 1 kilometer equals 1000 meters.
Hence, converting 605 km is done by multiplying:\[ 605 \text{ km} = 605 \times 1000 = 605,000 \text{ m} \]
It’s essential to perform such conversions to ensure consistency in units across all elements of your calculations, thereby avoiding errors that arise from mixing units.
Precision in unit conversions prevents miscalculations and ensures results are correctly expressed in standard scientific units like Joules for energy.
To convert distance into meters, remember the conversion factor:- 1 kilometer equals 1000 meters.
Hence, converting 605 km is done by multiplying:\[ 605 \text{ km} = 605 \times 1000 = 605,000 \text{ m} \]
It’s essential to perform such conversions to ensure consistency in units across all elements of your calculations, thereby avoiding errors that arise from mixing units.
Precision in unit conversions prevents miscalculations and ensures results are correctly expressed in standard scientific units like Joules for energy.
Calculating work in physics
Calculating work in physics gives us insight into how energy is transferred when a force moves an object. In this problem, we use the formula \[ \text{Work} = \text{Force} \times \text{Distance} \]to find out how much work the Big Boy locomotive performs.
Given:- Force \( F = 602,148 \text{ N} \)- Distance \( d = 605,000 \text{ m} \)
Substituting these into the formula, the calculation becomes:\[ \text{Work} = 602,148 \text{ N} \times 605,000 \text{ m} \ = 3.640,958 \times 10^{11} \text{ J} \]
This result, expressed in Joules, indicates the vast amount of energy required to pull such heavy loads over long distances. Understanding the calculation helps in appreciating the enormous power locomotives need to accomplish their tasks. This verification of units and calculations are crucial in ensuring the accuracy of physics solutions.
Given:- Force \( F = 602,148 \text{ N} \)- Distance \( d = 605,000 \text{ m} \)
Substituting these into the formula, the calculation becomes:\[ \text{Work} = 602,148 \text{ N} \times 605,000 \text{ m} \ = 3.640,958 \times 10^{11} \text{ J} \]
This result, expressed in Joules, indicates the vast amount of energy required to pull such heavy loads over long distances. Understanding the calculation helps in appreciating the enormous power locomotives need to accomplish their tasks. This verification of units and calculations are crucial in ensuring the accuracy of physics solutions.
Other exercises in this chapter
Problem 43
Find the centers and radii of the spheres in Exercises \(41-44\) $$ (x-\sqrt{2})^{2}+(y-\sqrt{2})^{2}+(z+\sqrt{2})^{2}=2 $$
View solution Problem 44
Triangle area Find a concise formula for the area of a triangle with vertices \(\left(a_{1}, a_{2}\right),\left(b_{1}, b_{2}\right),\) and \(\left(c_{1}, c_{2}\
View solution Problem 44
In Exercises 39–44, find the distance from the point to the plane. $$ (1,0,-1), \quad-4 x+y+z=4 $$
View solution Problem 44
Force vector A kite string exerts a 12 -lb pull \((|\mathbf{F}|=12)\) on a kite and makes a \(45^{\circ}\) angle with the horizontal. Find the horizontal and ve
View solution