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Conservation of Energy

Lesson ~10 min read 8 MCQs

In simple terms: In simple terms, conservation of energy means that in a closed system, total energy isn't created or destroyed, it just changes form—like converting the energy of height into the energy of speed.

Why this matters

Remember the feeling at the top of a roller coaster? That moment of silence right before the plunge, where you're holding your breath, high above the ground. Then, whoosh! You trade all that height for incredible speed as you rush to the bottom. You didn't get a push from a giant engine mid-drop; the ride just converted one thing (height) into another (speed).

This trade-off is the heart of one of the most powerful ideas in all of physics: the conservation of energy. It's like a fundamental rule of the universe's accounting system. In this lesson, we'll learn how to track this energy, predict speeds and heights, and see why choosing your "system" is the most important decision you'll make.

Concept overview

flowchart TD
    A[Start: Analyze a motion problem] --> B{Is work done by non-conservative forces like friction?};
    B -->|No| C{Is work done by external forces?};
    B -->|Yes| D[Use Work-Energy Theorem: <br> W_nc = ΔE_mech];
    C -->|No| E[Mechanical Energy is Conserved! <br> E_initial = E_final];
    C -->|Yes| F[Use Work-Energy Theorem: <br> W_ext = ΔE_mech];
    E --> G[K_i + U_gi + U_si = K_f + U_gf + U_sf];
A flowchart shows how to decide which energy principle to use. It starts by asking if non-conservative forces like friction are present. If yes, use the work-energy theorem. If no, ask if external forces do work. If no, use conservation of mechanical energy. If yes, use the work-energy theorem.

Core explanation

Before we can say energy is "conserved," we have to answer a critical question: conserved where? The first and most important step in any energy problem is to define your system. A system is just the collection of objects we've decided to pay attention to.

Choosing Your System Changes Everything

Let's think about that roller coaster cart.

Scenario 1: The System is "Just the Cart" If our system is only the cart, what's acting on it? The track pushes up (normal force), and the entire planet Earth pulls down on it (gravity). From the cart's perspective, gravity is an external force. As the cart falls, this external force does positive work, increasing the cart's kinetic energy. So, for a system of "just the cart," its energy is not constant; it changes because the environment (Earth) is doing work on it. A system with only one object can only have kinetic energy.

Scenario 2: The System is "The Cart + Earth" Now, let's be smarter. Let's define our system as the cart and the Earth together. In this larger system, the gravitational pull between the cart and Earth is an internal force. It's part of the system's interaction. This interaction stores energy, which we call gravitational potential energy (Ug).

Because the force is now internal, it can't change the total energy of the Cart-Earth system. It can only convert energy that's already there. This is the magic choice! When the only forces doing work are internal, conservative forces (like gravity), the total mechanical energy is conserved.

The Big Equation: E = K + U

Mechanical energy (E) is the total energy of motion and position. It's the sum of kinetic energy and potential energy.

E = K + U

  • Kinetic Energy (K): The energy of motion, calculated as K = ½mv².
  • Potential Energy (U): Stored energy. For now, we'll focus on gravitational potential energy, Ug = mgh.

The law of conservation of mechanical energy says that for an isolated system with no non-conservative forces (like friction), the initial total energy is equal to the final total energy.

E_initial = E_final K_initial + U_initial = K_final + U_final

This is your golden ticket for solving a huge number of physics problems without dealing with forces and acceleration directly.

Back to the Roller Coaster

Let's trace the energy on a frictionless ride, just like in the Shrutam visual. Our system is the Cart + Earth.

  • Point A (Top)
    The cart is at its maximum height (h_max) and is momentarily at rest (v=0).
    • K_A = ½m(0)² = 0
    • U_gA = mgh_max
    • Total Energy: E_A = 0 + mgh_max = mgh_max
  • Point B (Bottom)
    The cart is at its lowest point. Let's define this as our h=0 level. It's moving at its fastest speed (v_max).
    • K_B = ½mv_max²
    • U_gB = mg(0) = 0
    • Total Energy: E_B = ½mv_max² + 0 = ½mv_max²

Since energy is conserved, E_A = E_B. mgh_max = ½mv_max²

  • Point C (Mid-Hill): The cart is at some intermediate height (h_C) and has some speed (v_C).
    • K_C = ½mv_C²
    • U_gC = mgh_C
    • Total Energy: E_C = ½mv_C² + mgh_C

Again, energy is conserved, so E_A = E_C. mgh_max = ½mv_C² + mgh_C

The total energy E is the same at every single point on the ride. It's like you have a $100 budget. At the top, it's all in your bank account (potential). At the bottom, you've spent it all on something you're using (kinetic). In the middle, you have some cash in your wallet and some in the bank, but the total is still $100.

What Happens When Energy Isn't Conserved?

These are non-conservative forces. They take mechanical energy out of the system and turn it into thermal energy (heat). The track gets a little warmer, the air gets a little warmer. The total mechanical energy (K+U) decreases.

In this case, the change in mechanical energy is equal to the work done by these non-conservative forces.

W_nc = ΔE = E_final - E_initial

If an external force adds energy (like a booster on the track), W_ext would be positive and the system's total energy would increase. The key is that energy is never truly "lost"—it's just transferred out of our system or converted into a form we're not tracking, like heat. But for AP Physics 1, when a problem says "frictionless" or "negligible air resistance," that's your cue to use the simple conservation equation.

Worked examples

Example 1

Speed at the Bottom of the Hill

Problem: A 500 kg roller coaster cart starts from rest at the top of a frictionless hill that is 40 m high. How fast is it traveling when it reaches the bottom of the hill?

Solution Walkthrough:

  1. 1
    Define the System
    Our system is the cart + Earth. This allows us to use conservation of mechanical energy because gravity is an internal force.
  2. 2
    Identify Initial and Final States
    • Initial (i): At the top of the hill. h_i = 40 m, v_i = 0 m/s.
    • Final (f): At the bottom of the hill. We'll set our reference level here, so h_f = 0 m. We want to find v_f.
  3. 3
    Set up the Conservation Equation
    Since the track is frictionless, mechanical energy is conserved. E_i = E_f K_i + U_gi = K_f + U_gf
  4. 4
    Substitute the Formulas and Values
    ½mv_i² + mgh_i = ½mv_f² + mgh_f ½(500 kg)(0)² + (500 kg)(9.8 m/s²)(40 m) = ½(500 kg)v_f² + (500 kg)(9.8 m/s²)(0)
  5. 5
    Simplify and Solve
    0 + 196,000 J = ½(500 kg)v_f² + 0 196,000 J = (250 kg)v_f² v_f² = 196,000 / 250 = 784 v_f = √784 = 28 m/s

Example 2

The Skateboarder in the Half-Pipe

Problem: Priya, a 60 kg skateboarder, drops into a frictionless half-pipe from one edge, starting from rest. The edge is 3.0 m higher than the bottom of the pipe. At the very bottom of the pipe, she is moving at her maximum speed. How high up the opposite side will she go?

Solution Walkthrough:

  1. 1
    Define the System
    Priya + Earth.
  2. 2
    Identify States
    • Initial (i): At the starting edge. h_i = 3.0 m, v_i = 0 m/s.
    • Final (f): At the highest point on the opposite side. Her speed there will be momentarily zero, so v_f = 0 m/s. We need to find h_f.
  3. 3
    Set up the Conservation Equation
    No friction means we can use conservation. K_i + U_gi = K_f + U_gf
  4. 4
    Substitute and Simplify
    ½mv_i² + mgh_i = ½mv_f² + mgh_f 0 + mgh_i = 0 + mgh_f mgh_i = mgh_f h_i = h_f h_f = 3.0 m

Try it yourself

Problem 1: The Pendulum Swing

A heavy 2.0 kg bob is attached to a 1.5 m long string, making a pendulum. You pull it back so the string is horizontal and release it from rest. What is the speed of the bob as it passes through the lowest point of its swing?

Problem 2: Spring Launch

A 0.5 kg block is placed against a horizontal spring that has been compressed by 0.2 m. The spring constant is k = 200 N/m. After the block is released, it slides across a frictionless surface. What is the block's speed after it leaves the spring?