Calculating Work: Electric Train's Ascent On An Incline

Hey guys! Let's dive into a physics problem that sounds a bit intimidating at first glance, but we'll break it down step by step to make it super clear. We're going to calculate the work done by the engines of an electric train as it climbs up an incline. This involves concepts like forces, acceleration, friction, and, of course, work. So, grab your favorite beverage, and let's get started! This problem is a fantastic example of how different forces interact to produce motion and how we can quantify the energy transfer involved. Understanding this will give you a solid grasp of mechanics. We'll go through the different forces at play, figure out the net force, and then use that information to determine the work done by the engines. It's like solving a puzzle, and trust me, it's pretty cool when everything clicks into place.

The Setup: Electric Train's Journey

Alright, here’s the scenario. We have an electric train with a mass of $1.2 imes 10^5 ext{ kg}$. This big boy is chugging along, going uphill on a slope. The slope has an angle of $\alpha = 10^ ext{o}$ relative to the horizontal. The train is moving upwards with an acceleration of $1.5 ext{ m/s}^2$. There’s also some friction to consider; the coefficient of friction is $0.05$. The train travels a distance of $100 ext{ m}$ along the incline. Our mission, should we choose to accept it, is to calculate the work done by the engines of the train over this distance. The setup is all about the real-world application of physics principles, and it's what makes these problems interesting. The key is to understand the forces at play and how they affect the train's motion. It is going to be a thrilling ride, trust me! Every component is interconnected and is going to make a difference in determining the overall work done. We are going to unravel this mystery, and the physics will all fall into place.

Let's go through the given parameters:

• Mass of the train, $m = 1.2 imes 10^5 ext{ kg}$
• Coefficient of friction, $\mu = 0.05$
• Distance along the incline, $s = 100 ext{ m}$
• Acceleration, $a = 1.5 ext{ m/s}^2$
• Angle of the incline, $\alpha = 10^ ext{o}$

Breaking Down the Forces at Play

Now, let's talk about the forces acting on the train. This is the critical part because forces determine the net force, and the net force, in turn, dictates the acceleration and, eventually, the work done. We have a few key players here: the force of gravity, the normal force, the force of friction, and the force exerted by the engine. Each force has a role, and they all work together (or against each other) to determine the train's movement. It's like a team, each member with their specific job! The force of gravity is always there, pulling the train downwards. The normal force is the support force from the incline, perpendicular to the surface. Friction opposes the motion, and the engine's force is what propels the train upwards.

1. Gravity: The force of gravity ($F_g$) acts downwards. We need to break it down into components parallel and perpendicular to the incline. The component parallel to the incline ($F_{g ext{,parallel}}$) is $mg imes ext{sin}(\alpha)$, and the component perpendicular to the incline ($F_{g ext{,perpendicular}}$) is $mg imes ext{cos}(\alpha)$, where $g$ is the acceleration due to gravity (approximately $9.8 ext{ m/s}^2$).
2. Normal Force: The normal force ($F_N$) is equal in magnitude and opposite in direction to the perpendicular component of gravity and also the component from friction. So, $F_N = mg imes ext{cos}(\alpha)$.
3. Friction: The force of friction ($F_f$) opposes the motion of the train and is calculated as $\mu imes F_N$. This force always works against the motion of the train.
4. Engine Force: The engine provides the force ($F_{engine}$) that propels the train upwards along the incline. This is what we are trying to find, we are calculating the work done by this force.

Calculating the Net Force

To figure out the work done, we need to find the engine's force. We can use Newton's second law of motion, which states that the net force ($F_{net}$) acting on an object is equal to its mass ($m$) times its acceleration ($a$): $F_{net} = ma$. The forces acting on the train along the incline are:

• The engine force ($F_{engine}$) acting upwards.
• The parallel component of gravity ($F_{g ext{,parallel}} = mg imes ext{sin}(\alpha)$) acting downwards.
• The force of friction ($F_f = \mu imes F_N = \mu imes mg imes ext{cos}(\alpha)$) acting downwards.

So, the net force can be written as: $F_{net} = F_{engine} - F_{g ext{,parallel}} - F_f$. Since $F_{net} = ma$, we can rearrange the equation to solve for $F_{engine}$: $F_{engine} = ma + mg imes ext{sin}(\alpha) + \mu imes mg imes ext{cos}(\alpha)$. Isn't it fascinating how a single equation can help us understand the interplay of multiple forces? Each term represents a different force acting on the train, and the equation puts them together to help us determine the engine force. Now, let's plug in the values and calculate the value of the engine force.

Plugging in the Numbers: Finding the Engine Force

Alright, time to get our hands dirty with some calculations. We have all the ingredients; let's bake a solution! We will insert the values we know into the equation for the engine force. Here we go! We will use $g = 9.8 ext{ m/s}^2$.

$F_{engine} = ma + mg imes ext{sin}(\alpha) + \mu imes mg imes ext{cos}(\alpha)$

$F_{engine} = (1.2 imes 10^5 ext{ kg}) imes (1.5 ext{ m/s}^2) + (1.2 imes 10^5 ext{ kg}) imes (9.8 ext{ m/s}^2) imes ext{sin}(10^ ext{o}) + 0.05 imes (1.2 imes 10^5 ext{ kg}) imes (9.8 ext{ m/s}^2) imes ext{cos}(10^ ext{o})$

First, calculating the gravitational components, you get the force from the engine.

Let's start with the first term: $(1.2 imes 10^5 ext{ kg}) imes (1.5 ext{ m/s}^2) = 1.8 imes 10^5 ext{ N}$.

Next, calculate the second term: $(1.2 imes 10^5 ext{ kg}) imes (9.8 ext{ m/s}^2) imes ext{sin}(10^ ext{o}) ext{ ≈ } 2.03 imes 10^5 ext{ N}$.

Then, calculate the third term: $0.05 imes (1.2 imes 10^5 ext{ kg}) imes (9.8 ext{ m/s}^2) imes ext{cos}(10^ ext{o}) ext{ ≈ } 0.58 imes 10^5 ext{ N}$.

Finally, add everything up:

$F_{engine} ext{ ≈ } 1.8 imes 10^5 ext{ N} + 2.03 imes 10^5 ext{ N} + 0.58 imes 10^5 ext{ N} ext{ ≈ } 4.41 imes 10^5 ext{ N}$.

So, the force provided by the engine is approximately $4.41 imes 10^5 ext{ N}$.

Calculating the Work Done

We're now at the final stretch! The work done ($W$) by the engine is given by the formula: $W = F_{engine} imes s$, where $s$ is the distance the train travels along the incline. This is a straightforward calculation, but it represents the culmination of all the force analysis we've done. It's how we quantify the energy transfer from the engine to the train, allowing it to move uphill. So, plug in the values of the force provided by the engine and the distance into the equation for work done to find the solution.

We have $F_{engine} ext{ ≈ } 4.41 imes 10^5 ext{ N}$ and $s = 100 ext{ m}$.

Therefore, $W = (4.41 imes 10^5 ext{ N}) imes (100 ext{ m}) = 4.41 imes 10^7 ext{ J}$.

The Grand Finale: Work Done by the Engine

And there you have it, guys! The work done by the engines of the electric train is approximately $4.41 imes 10^7 ext{ J}$. That's a significant amount of work, which is reasonable, considering the massive mass of the train, the incline, the acceleration, and the friction. I hope you enjoyed this adventure of calculating the work done. You can now apply this method to similar problems with different scenarios. Always remember to break down the forces, find the net force, and then use the work formula. Keep practicing, and these concepts will become second nature. And there you have it; you've successfully tackled a physics problem involving forces, work, and energy. Great job! Remember, it's all about understanding the forces and how they affect motion. You're doing great! Keep up the awesome work, and keep exploring the world of physics.