Calculating Energy and Work

1. Kinetic Energy: The Energy of Motion

Imagine you're standing on a sidewalk. A bicycle rolls past you slowly. No big deal. Now, imagine a massive truck roars past at 60 miles per hour. You feel the ground shake a little, you hear the noise—you can intuitively sense the immense energy it carries. That "energy of motion" is what we call kinetic energy.

The word "kinetic" comes from the Greek word kinetikos, meaning "moving." It's that simple. Anything that has mass and is moving has kinetic energy.

The two factors that determine an object's kinetic energy are its mass ($m$) and its speed ($v$). The mathematical relationship is:

Let's break that down:

• $m$ is the mass in kilograms (kg).
• $v$ is the speed in meters per second (m/s).
• $KE$ is the kinetic energy in Joules (J).

Notice the speed is squared ($v^2$). This is incredibly important. It means that if you double an object's speed, you don't just double its kinetic energy—you quadruple it ($2^2=4$)! This is why driving at high speeds is so dangerous; the energy involved in a potential crash increases exponentially, not linearly.

To see a clear explanation and some great examples of kinetic energy in action, let's watch a segment from this introductory video.

2. Potential Energy: Stored Energy

Now, let's talk about the other side of the coin: potential energy. This is energy that is stored in an object due to its position or its configuration. It's called "potential" because it has the potential to be converted into another form of energy, like kinetic energy.

Think about these examples:

• A rock perched at the top of a cliff. It's not moving, but it has stored energy because of its height. If it falls, that stored energy becomes energy of motion (kinetic energy).
• A stretched rubber band. It has energy stored in its stretched state. Let it go, and it snaps back, converting that potential energy into kinetic energy.
• A compressed spring in a toy gun. The spring's compression stores energy, which is released to launch the projectile.

In thermodynamics, the most common type we deal with is Gravitational Potential Energy, the energy stored due to an object's height in a gravitational field. The formula is:

Where:

• $m$ is the mass (kg).
• $g$ is the acceleration due to gravity (a constant, approximately $9.81{\text{m/s}}^2$ on Earth).
• $h$ is the height (m) relative to a chosen zero point.
• $PE$ is the potential energy in Joules (J).

An important point here is that height $h$ is relative. You have to define a "zero" level, or a datum. For example, is the height of a book measured from the floor or from the tabletop it's sitting on? The choice is yours, but you must be consistent throughout the problem. In most cases, it's the change in potential energy that we care about, not its absolute value.

Let's watch a video that builds this intuition for potential energy.

Quick Recap

So far, we have two types of mechanical energy:

• Kinetic Energy ($KE=\frac{1}{2}m{v}^2$): Energy of motion.
• Potential Energy ($PE=mgh$): Stored energy of position.

The sum of these two is called the Total Mechanical Energy: $E_{mech}=KE+PE$.

But how do we change an object's energy? How do we give a stationary object kinetic energy, or lift it to give it potential energy? The answer is: we have to do work.

3. Mechanical Work: Transferring Energy by Force

In everyday language, "work" means any kind of effort. In physics and engineering, the definition is much more precise: Work is the energy transferred to or from an object by applying a force along a displacement.

Let's break that down with the classic example: pushing a heavy box.

• You push on the box, but it's too heavy and doesn't move. You're exerting a force, you're sweating, but since there is no displacement, you have done zero work on the box.
• You push harder, and the box slides across the floor. Now you have a force and a displacement. You have done work on the box. You have transferred energy to it, which shows up as an increase in its kinetic energy (it started moving) and as heat generated by friction.

The formula for work done by a constant force is:

Where:

• $W$ is the work done (in Joules).
• $F$ is the magnitude of the constant force applied (in Newtons).
• $d$ is the magnitude of the displacement (in meters).
• $\theta$ (theta) is the angle between the direction of the force and the direction of the displacement.

The $\cos (\theta )$ part is key:

• If you push in the exact direction of motion, $\theta =0^{\circ }$, and $\cos (0^{\circ })=1$. The work is positive and maximal: $W=Fd$. You are adding energy to the object.
• If you carry a suitcase horizontally at a constant velocity, the force you apply is upward (to counteract gravity), but the displacement is horizontal. The angle is $\theta =90^{\circ }$, and $\cos (90^{\circ })=0$. You do zero work on the suitcase, even though you might get tired!
• The force of friction always opposes motion, so for friction, $\theta =180^{\circ }$, and $\cos (180^{\circ })=-1$. The work done by friction is negative: $W_{friction}=-f_{k}d$. Friction removes mechanical energy from a system (converting it to heat).

A Special Kind of Work: Boundary Work (PV Work)

In thermodynamics, we are often interested in the work done by a gas expanding or being compressed in a piston-cylinder device. This is fundamental to how engines and compressors work. This is called boundary work because it's done by the system on its boundary (the moving piston).

Let's think about it.

• Pressure ($P$) is Force per unit Area ($A$), so $F=P\cdot A$.
• When the gas expands, it pushes the piston a small distance $dx$. The tiny amount of work it does is $dW=F\cdot dx=(P\cdot A)\cdot dx$.
• But $A\cdot dx$ is just the small change in volume, $dV$.
• So, $dW=P\cdot dV$.

To find the total work for a process, we need to sum up all these tiny bits of work. Since you're comfortable with calculus, you'll recognize this sum as an integral:

This is one of the most important equations we'll encounter. It tells us that the work done by a gas is the area under the process curve on a Pressure-Volume (P-V) diagram.

• Expansion ($V_2>V_1$): The gas pushes the piston out. Work is done by the system. $W$ is positive.
• Compression ($V_2<V_1$): The surroundings push the piston in. Work is done on the system. $W$ is negative.

For a deep dive into the different types of work and the crucial connection between work and energy, let's go back to our main video resource.

Let's Practice: Problems and Questions

Time to solidify these concepts with some calculations. Engineering is an applied science, so let's apply this!

Worked Problems

Problem 1: Kinetic Energy of a Car

A car with a mass of 1200 kg is traveling at a speed of 90 km/h. What is its kinetic energy in Joules?

Solution:

1. Convert Units: Our formula needs speed in m/s. We'll use the factor-label method from last lesson.

   $$v=\left(\frac{90\text{ km}}{1\text{ h}}\right)\times \left(\frac{1000\text{ m}}{1\text{ km}}\right)\times \left(\frac{1\text{ h}}{3600\text{ s}}\right)=25\text{m/s}$$

2. Calculate Kinetic Energy: Now plug the values into the KE formula.

   $$KE=\frac{1}{2}m{v}^2=\frac{1}{2}\cdot (1200\text{ kg})\cdot (25\text{m/s}{)}^2$$ $$KE=(600\text{ kg})\cdot (625{\text{m}}^2/{\text{s}}^2)=375,000\text{ J}\text{or}375\text{ kJ}$$

   That's a lot of energy!

Problem 2: Potential Energy of a Diver

A diver with a mass of 60 kg stands on a 10-meter high diving platform. What is her potential energy relative to the surface of the water?

Solution:

1. Identify the Datum: We are asked for the PE relative to the water surface, so the water is our $h=0$ level.
2. Calculate Potential Energy: $$PE=mgh=(60\text{ kg})\cdot (9.81{\text{m/s}}^2)\cdot (10\text{ m})$$ $$PE=5886\text{ J}\text{or}5.89\text{ kJ}$$ This is the amount of energy that will be converted to kinetic energy as she falls.

Problem 3: Work Done by a Gas

A gas in a piston expands at a constant pressure of 200 kPa from an initial volume of 0.01 m³ to a final volume of 0.03 m³. How much boundary work did the gas do?

Solution:

1. Identify the Process: This is a constant pressure expansion.
2. Apply the Work Formula: Since pressure $P$ is constant, it can come out of the integral. $$W_{boundary}={\int }_{V_1}^{V_2}PdV=P{\int }_{V_1}^{V_2}dV=P(V_2-V_1)$$
3. Convert Units and Calculate: Pressure must be in Pascals (Pa) to get Joules. 1 kPa = 1000 Pa. $$P=200\text{ kPa}=200,000\text{ Pa}(\text{or }200,000{\text{N/m}}^2)$$ $$W_{boundary}=(200,000{\text{N/m}}^2)\cdot (0.03{\text{m}}^3-0.01{\text{m}}^3)$$ $$W_{boundary}=(200,000{\text{N/m}}^2)\cdot (0.02{\text{m}}^3)=4000\text{ N}\cdot \text{m}=4000\text{ J}\text{or}4\text{ kJ}$$ Since the result is positive, this confirms that work was done by the system.

Conceptual Questions

Lesson Summary

Fantastic job today! We've taken a huge step from abstract concepts to concrete calculations. These are the tools you'll use constantly as we move forward.

Here are the key takeaways:

• Kinetic Energy ($KE=\frac{1}{2}m{v}^2$) is the energy of an object in motion, and it's highly sensitive to speed.
• Gravitational Potential Energy ($PE=mgh$) is stored energy based on an object's height relative to a chosen zero point.
• Mechanical Work ($W=Fd\cos \theta$) is the transfer of energy via a force acting over a distance. It's the mechanism for changing an object's mechanical energy.
• Boundary Work ($W=\int PdV$) is the specific work done by an expanding or contracting gas, which is the heart of many thermodynamic devices.
• Energy Conservation: In a simple mechanical system without friction, energy is not lost but merely converted between kinetic and potential forms. Work done by non-conservative forces like friction removes mechanical energy from the system.

Preview of the Next Lesson

We have now officially completed the "Foundations" module. We can define systems, describe their state, understand their properties, and calculate the mechanical energy they possess and the mechanical work they can do.

What's missing? Think about heating a pot of water on the stove. Its temperature rises, but it's not moving faster ($KE=0$) and its height isn't changing ($PE=0$). Where did the energy from the stove go?

In our next lesson, we will enter Module 2 and answer that question by exploring the concepts of pressure, temperature, and the state of a gas. We'll look at pressure from a molecular level and define temperature in a new, more powerful way, which will lead us straight to the Zeroth Law of Thermodynamics. This will set the stage for understanding the hidden, internal energy of a system.