Conservation of Energy: How to Solve Energy Problems

Pick a system, then set initial energy equal to final energy plus losses: Ki + Ui = Kf + Uf + energy lost to friction. Kinetic energy is (1/2)mv^2, gravitational potential energy change is mgh, and spring energy is (1/2)k times stretch squared. Friction is energy leaving the system as heat.

AP Physics: Unit 3 (topics 3.3 Potential Energy, 3.4 Conservation of Energy). Covers Topics 3.3 and 3.4 of AP Physics 1 Unit 3 (Work, Energy, and Power), weighted 18 to 23 percent of the multiple-choice section. AP Physics C: Mechanics runs the same energy accounting with calculus in its own Unit 3, weighted 15 to 25 percent.

The energy accounting method

To solve a conservation of energy problem, treat energy like money in a budget: whatever the system starts with must equal whatever it ends with plus whatever left along the way.

K_i + U_i = K_f + U_f + \Delta E_{thermal}

Work through every problem with the same five moves:

Pick the system. For most problems that means the object plus the Earth, and the spring if there is one.
Choose two snapshots: an initial moment and a final moment, usually the place where you know the most and the place the question asks about.
Set a zero height for gravitational potential energy. The lowest point in the problem is usually the cleanest choice.
Write down every energy present in each snapshot. Anything at rest has zero kinetic energy; anything at zero height has zero gravitational potential energy.
Set initial equal to final plus losses, then solve for the unknown.

If nothing rubs, skids, or collides, the loss term is zero and mechanical energy is conserved.

The three energies you track: K, Ug, and Us

AP Physics 1 energy problems only ever involve three forms.

Energy	Formula	Nonzero when
Kinetic	$K = \frac{1}{2}mv^2$	the object is moving
Gravitational potential	$\Delta U_g = mg\Delta y$	the height changes
Spring (elastic) potential	$U_s = \frac{1}{2}k(\Delta x)^2$	a spring is stretched or compressed

To find kinetic energy, square the speed first, then multiply by half the mass. A 2.0 kg cart at 3.0 m/s carries $K = \frac{1}{2}(2.0)(3.0)^2 = 9.0$ J. The kinetic energy calculator checks that arithmetic in both directions, including solving for speed when you know the energy.

The gravitational potential energy formula on the AP equation sheet is written as a change, $\Delta U_g = mg\Delta y$ , because only height differences matter. Use $g = 9.8\ \text{m/s}^2$ in every calculation. For springs, $\Delta x$ is measured from the natural, unstretched length, not from wherever the spring happens to start.

Pick the system and the zero height

The system choice decides whether gravity shows up as potential energy or as work. Include the Earth in your system and gravity becomes an internal interaction, so you track $U_g$ . Leave the Earth out and gravity is an external force doing work on the object, which is the work-energy theorem picture. Both give the same answer; for problems with height changes and springs, the accounting version is usually faster to set up.

Zero height is also your call, and no choice is wrong. Pick the lowest point the object reaches so every $U_g$ value stays positive. For a pendulum that means the bottom of the swing; for a ramp, the floor.

Two forces never move energy in these problems: the normal force on a sliding object and the tension in a pendulum string. Each stays perpendicular to the motion, so each does zero work. That is why a pendulum drop is a pure trade of $U_g$ for $K$ even though the string pulls on the bob the whole time.

Friction is energy leaving the system

Friction does not destroy energy. It converts mechanical energy into thermal energy, and that amount leaves your K and U accounting. The loss equals the friction force times the distance the object slides:

\Delta E_{thermal} = F_f d

Here $d$ is the path length, not the displacement. A block dragged 3.0 m out and 3.0 m back ends where it started but pays for 6.0 m of sliding. On a slope, the friction force follows from $|F_f| \leq |\mu F_N|$ with the normal force taken from a free-body diagram; inclined plane problems shows why $F_N = mg\cos\theta$ on a ramp rather than $mg$ .

Read the problem statement for the signal words. Rough surface, the block skids, or a given number of joules converted to thermal energy all mean you need the loss term. Frictionless, smooth, and light string are permission to set $\Delta E_{thermal} = 0$ and move on. The thermal energy that friction produces is itself the domain of thermodynamics, where this same conservation principle becomes the first law for gases; see thermodynamics and PV diagrams.

Energy method or kinematics: how to choose

Reach for energy when the question asks for a speed at a position, a height reached, or a spring compression, and especially when the path is curved or time is never mentioned. Energy accounting ignores the path entirely: a bead sliding down a frictionless wire of any shape reaches the bottom at the same speed as one dropped straight down from the same height.

Reach for the kinematic equations when the question involves time, and for Newton's second law when it asks for a force or an acceleration. Plenty of AP problems chain the methods: energy to get the launch speed at the bottom of a ramp, then kinematics for the projectile flight that follows.

Oscillating systems are the energy method on repeat. A mass on a spring trades $K$ and $U_s$ back and forth twice per cycle, and simple harmonic motion covers how that trade sets the maximum speed (reached at equilibrium) for the amplitude you chose.

Where this sits on the AP exam

Conservation of energy is Topics 3.3 (Potential Energy) and 3.4 (Conservation of Energy) in Unit 3: Work, Energy, and Power, which carries 18 to 23 percent of the multiple-choice section, tied with Unit 2 for the heaviest weight in AP Physics 1. The exam runs 3 hours: 42 multiple-choice questions in 85 minutes, then 4 free-response questions in 95 minutes, with a calculator allowed on both sections.

On a free response, make the accounting explicit. Name your system, state your zero height, and write the energy equation in symbols before substituting numbers; graders award reasoning, not just the final value. Practice translating between representations too: the same 2.0 J can appear as a bar chart, a number in a table, or a term in an equation, and questions that ask you to explain without calculating usually reduce to a sentence like the drop height is the same, so the final kinetic energy is the same.

Ramp with friction: speed at the bottom

A 2.0 kg box is released from rest at the top of a ramp, 1.5 m above the floor. As it slides down, friction converts 8.0 J of mechanical energy to thermal energy. How fast is the box moving at the bottom?

Define the setup. System: box plus Earth. Initial snapshot: at rest at the top, so $K_i = 0$ . Final snapshot: moving at the bottom. Zero height: the floor. The accounting equation is $mgh = \frac{1}{2}mv^2 + \Delta E_{thermal}$ .
Compute the starting energy: $U_g = mgh = (2.0\ \text{kg})(9.8\ \text{m/s}^2)(1.5\ \text{m}) = 29.4\ \text{J}$ .
Subtract the loss to get the final kinetic energy: $K_f = 29.4\ \text{J} - 8.0\ \text{J} = 21.4\ \text{J}$ .
Solve for speed: $v = \sqrt{2K_f/m} = \sqrt{2(21.4\ \text{J})/(2.0\ \text{kg})} = \sqrt{21.4} = 4.63\ \text{m/s}$ .

The box reaches the bottom at about $4.6\ \text{m/s}$ . Without friction it would arrive at $v = \sqrt{2gh} = \sqrt{2(9.8)(1.5)} = 5.4\ \text{m/s}$ , so the 8.0 J loss costs nearly 0.8 m/s. Notice the ramp angle never entered the calculation: energy accounting does not care about the path, only the height drop and the sliding loss.

Pendulum drop: speed at the lowest point

A 0.50 kg pendulum bob hangs from a light string. You pull it to the side until it sits 0.40 m above its lowest point, then release it from rest. How fast is the bob moving at the bottom of the swing?

Define the system: bob plus Earth. The string tension stays perpendicular to the bob's motion, so it does zero work, and $\Delta E_{thermal} = 0$ . Zero height: the lowest point of the swing.
Write the accounting: $mgh = \frac{1}{2}mv^2$ . The mass appears on both sides and cancels, leaving $v = \sqrt{2gh}$ .
Substitute: $v = \sqrt{2(9.8\ \text{m/s}^2)(0.40\ \text{m})} = \sqrt{7.84} = 2.8\ \text{m/s}$ .
Check the books balance: $U_g = (0.50)(9.8)(0.40) = 1.96\ \text{J}$ at the start, and $K = \frac{1}{2}(0.50)(2.8)^2 = 1.96\ \text{J}$ at the bottom.

The bob passes the lowest point at $2.8\ \text{m/s}$ . Neither the mass nor the string length mattered, only the 0.40 m drop height. A pendulum drop is a ramp problem where the friction is guaranteed to be zero.

Spring launcher: speed and maximum height

A spring with $k = 400\ \text{N/m}$ is compressed 0.10 m against a 0.25 kg cart on a frictionless track. The track runs flat, then curves uphill. Find the cart's speed just after it leaves the spring and the maximum height it reaches.

Energy stored in the spring: $U_s = \frac{1}{2}k(\Delta x)^2 = \frac{1}{2}(400\ \text{N/m})(0.10\ \text{m})^2 = 2.0\ \text{J}$ .
Spring to flat track: all 2.0 J becomes kinetic energy. $v = \sqrt{2K/m} = \sqrt{2(2.0\ \text{J})/(0.25\ \text{kg})} = \sqrt{16.0} = 4.0\ \text{m/s}$ .
Flat track to highest point: all 2.0 J becomes gravitational potential energy. $mgh = 2.0\ \text{J}$ , so $h = 2.0/[(0.25)(9.8)] = 2.0/2.45 = 0.82\ \text{m}$ .
Notice one number, 2.0 J, flowed through three forms: spring, kinetic, gravitational. You could skip the middle snapshot and set $U_s = mgh$ directly.

The cart leaves the spring at $4.0\ \text{m/s}$ and climbs to a maximum height of $0.82\ \text{m}$ above its starting level.

Frequently asked questions

What is the conservation of energy equation?

Initial kinetic plus initial potential equals final kinetic plus final potential plus any energy converted to thermal: Ki + Ui = Kf + Uf + E_thermal. When there is no friction or other loss, the last term is zero and mechanical energy is conserved.

How do you find kinetic energy?

Use K = (1/2)mv^2 with mass in kilograms and speed in meters per second; the result comes out in joules. Square the speed before multiplying, since doubling the speed quadruples the kinetic energy. A 2.0 kg cart moving at 3.0 m/s has K = 0.5 times 2.0 times 9.0 = 9.0 J. The kinetic energy calculator solves for any of the three variables.

What is the gravitational potential energy formula?

The AP Physics 1 equation sheet writes it as a change: delta Ug = mg delta y, meaning mass times 9.8 m/s^2 times the change in height. Only height differences matter, so you can place the zero level anywhere; the lowest point in the problem is usually the most convenient choice.

Does friction break conservation of energy?

No. Friction converts mechanical energy into thermal energy, so total energy is still conserved; it is mechanical energy (K plus U) that decreases. In a problem, account for it by adding the loss, friction force times sliding distance, to the final side of the energy equation.

Why does mass cancel in ramp and pendulum problems?

When gravity is the only force moving energy, the initial potential energy mgh and the final kinetic energy (1/2)mv^2 both contain the mass, so it divides out and v = sqrt(2gh). A heavy bob and a light bob released from the same height reach the same speed. The cancellation fails when the problem gives a fixed energy loss in joules, because that loss does not scale with mass.