Conservation of energy

Drop a stone and watch it fall. At the top it is perfectly still, yet it carries a hidden capacity to do work — potential energy, stored purely because of its height. Let it go, and that capacity does not vanish; it turns, moment by moment, into the energy of motion, kinetic energy, until the instant before it lands, when the stone is moving as fast as it ever will and has "spent" every joule it started with. Nothing was created and nothing was lost — energy only changed its form. This is the principle of conservation of energy, arguably the single most reliable bookkeeping rule in all of physics. This article verifies it for a falling body and a swinging pendulum, and shows why it applies just as surely to a battery lighting a torch bulb or food fuelling a sprinter's muscles.

The core insight

For an isolated system the total energy — kinetic plus potential plus every other form — never changes; it only converts between forms, so for a freely falling or swinging body $KE + PE$ stays constant throughout the motion.

The core idea

The principle, in one line. Energy can neither be created nor destroyed, only transformed from one form into another; the total energy of an isolated system stays fixed for all time. This is not a rule specific to mechanics — it governs chemical, electrical, thermal, nuclear, sound and light energy just as strictly.

Verifying it with a free fall. The cleanest classroom test uses a body falling under gravity alone, with air resistance ignored. At any instant during the fall, the sum of kinetic and potential energy — the total mechanical energy — stays the same:

E_{mech} = KE + PE = \frac{1}{2}mv^2 + mgh = \text{constant}

where: $m$ is the mass of the body (kg), $v$ is its speed at that instant (m/s), $g$ is the acceleration due to gravity ( $9.8\ \text{m/s}^2$ ), and $h$ is its height above the chosen reference level (m). The equation says that whatever $KE$ gains, $PE$ loses by exactly the same amount, so their total is frozen.

Three snapshots of a drop. Picture mass $m$ released from height $H$ , with the ground as the zero of $PE$ . At the top the body is at rest: $KE = 0$ , $PE = mgH$ , all energy potential. Partway down, at height $h$ , it has fallen $(H - h)$ and picked up speed: $PE = mgh$ and $KE = mg(H-h)$ — exactly the potential energy "lost" so far. Just before landing, $h = 0$ , so $PE = 0$ and $KE = mgH$ : every joule of potential energy has become kinetic. Add $KE + PE$ at any of the three snapshots and you get the same number, $mgH$ .

Where the energy actually goes in real life. Ideal free fall assumes nothing but gravity acts. Real falls, swings and slides always lose a little energy to friction or air resistance, which converts it into heat and sound. That energy is not destroyed — it is merely dissipated into forms no longer useful for the original motion — which is why a real pendulum swings a fraction lower on each pass and a dropped ball never bounces back to quite the height it fell from.

Figure

Bar chart of a falling ball showing kinetic and potential energy at the top middle and bottom of a twenty metre drop with their sum staying constant

Derivation ATotal mechanical energy of a freely falling body stays constant

The goal is to show, using nothing beyond the kinematics equation for constant acceleration and the definitions of $KE = \tfrac{1}{2}mv^2$ and $PE = mgh$ , that a body released from height $H$ has exactly the same value of $KE + PE$ at every point of its fall, so the "conservation" claim is not an assumption but a provable consequence of Newton's laws applied to gravity.

Step 1. Set up the drop. A body of mass $m$ is released from rest at height $H$ above the ground, which we take as the reference level for potential energy ( $PE = 0$ there). At the instant of release, $v = 0$ , so $KE = 0$ and $PE = mgH$ ; the total is $E_{mech} = mgH$ .

Step 2. Find the speed after falling a distance $(H-h)$ to reach height $h$ . Using the kinematics equation $v^2 = u^2 + 2as$ with initial speed $u = 0$ , acceleration $a = g$ and distance fallen $s = (H-h)$ , we get $v^2 = 2g(H-h)$ .

Step 3. Compute $KE$ at height $h$ using this speed. $KE = \frac{1}{2}mv^2 = \frac{1}{2}m \cdot 2g(H-h) = mg(H-h) = mgH - mgh$ .

Step 4. Add $PE = mgh$ at that same height. $E_{mech} = KE + PE = (mgH - mgh) + mgh = mgH$ . The $mgh$ terms cancel exactly, leaving $mgH$ — the same value found in Step 1, regardless of which height $h$ was chosen. Since $h$ was arbitrary, $KE + PE = mgH$ holds at every point of the fall, proving the total mechanical energy is constant.

Worked example

A ball of mass 2 kg is dropped from a height of 20 m above the ground (take $g = 9.8\ \text{m/s}^2$ , air resistance ignored). Find its kinetic energy and speed when it has fallen to a height of 5 m above the ground.

Step 1. Find the total mechanical energy at release. At the top, all energy is potential, so $E_{mech} = mgH = 2 \times 9.8 \times 20 = 392\ \text{J}$ . This value stays fixed for the whole fall.

Step 2. Find the potential energy at height 5 m. $PE_{5m} = mgh = 2 \times 9.8 \times 5 = 98\ \text{J}$ .

Step 3. Apply conservation of energy to get the kinetic energy at 5 m. $KE_{5m} = E_{mech} - PE_{5m} = 392 - 98 = 294\ \text{J}$ .

Step 4. Convert kinetic energy to speed. $294 = \frac{1}{2} \times 2 \times v^2 = v^2$ , so $v = \sqrt{294} \approx 17.1\ \text{m/s}$ .

At 5 m the ball has $KE = 294\ \text{J}$ and speed $v \approx 17.1\ \text{m/s}$ , while $PE = 98\ \text{J}$ — the two together still total 392 J.

Competitive stretch

A pendulum bob of mass 0.4 kg is pulled aside so it rises 0.2 m above its lowest point, then released. Its highest point gives $PE = mgh = 0.4 \times 9.8 \times 0.2 = 0.784\ \text{J}$ , with $KE = 0$ there. By conservation of energy, at the lowest point ( $h=0$ ) all of this becomes kinetic: $KE_{lowest} = 0.784\ \text{J}$ , so $\frac{1}{2}(0.4)v^2 = 0.784 \implies v^2 = 3.92 \implies v \approx 1.98\ \text{m/s}$ . Ideally the bob swings back up to the same 0.2 m on the other side, converting kinetic energy back into potential energy.

Common traps

Thinking KE and PE are each separately constant — it is their sum that is fixed; they continuously trade off against each other as height and speed change.
Forgetting real falls and swings lose a little energy to friction or air resistance — that energy is not destroyed, only converted to heat/sound, so measured KE is always slightly less than the ideal value.
Switching the reference level for height mid-calculation — fix one reference level (usually the ground or the lowest point) for the entire problem.
Confusing "conservation of energy" (true for all forms, always) with "conservation of mechanical energy" (true only when friction and air resistance are absent).

Ask the tutor: want to work through a bouncing ball that loses height each bounce, or build a KE/PE table for a swinging pendulum step by step?

If energy is always conserved, why does a swinging pendulum eventually stop?

Conservation of energy still holds — the pendulum's mechanical energy (KE + PE) is not what stays constant here, because air resistance and friction at the pivot are steadily converting some of that mechanical energy into heat and sound. The total energy across all forms (mechanical plus heat plus sound) is exactly conserved, but the useful mechanical energy keeps shrinking, so each swing reaches a slightly lower height until the pendulum comes to rest.

Why do we ignore air resistance in these problems if it is always present in real life?

Ignoring air resistance is a simplifying assumption that makes the KE plus PE equals constant relationship exactly true, which is what the syllabus wants you to verify and calculate with. In reality, air resistance removes a small amount of energy as heat, so a real falling ball's kinetic energy just before landing is always a little less than mgH. For short falls or swings in air, the effect is small enough that ideal conservation of energy gives a very good approximation.

Does conservation of energy apply outside mechanics, like to a torch battery?

Yes — the principle covers every form of energy, not just kinetic and potential. In a torch, chemical energy stored in the battery converts into electrical energy, which then converts into light and heat energy in the bulb. Add up the chemical energy used and the light plus heat energy produced, and the totals match exactly, just as KE plus PE matches mgH for a falling ball — only the forms of energy differ.

Conservation of energy

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads