Law of conservation of energy

Drop a ball and watch it fall. At the top it has height but no speed; just before it lands it has speed but no height. Somewhere in between, both height and speed exist together — and if you do the sums, something remarkable happens: kinetic energy plus potential energy always adds up to the same total, at every single point of the fall. That is not a coincidence of falling balls. It is the law of conservation of energy, one of the deepest statements in all of physics: energy is never created and never destroyed, only converted between forms. This article states the law precisely, proves it algebraically for a falling body, illustrates it with a swinging pendulum, and works through the numbers so you can apply it confidently in exam problems.

The core insight

For an isolated system with no friction or air resistance, energy only changes form — never total amount — so kinetic energy plus potential energy stays fixed at every instant: $E_k + E_p = \text{constant}$ .

The core idea

The law of conservation of energy is stated as: energy can neither be created nor destroyed; it can only be transformed from one form to another, and the total energy of an isolated system before and after any change remains the same. This is a general law covering all forms of energy — chemical, electrical, thermal, nuclear, mechanical — not just the mechanical energy studied in this chapter. For mechanical energy specifically, with no friction or air resistance to drain energy away as heat or sound, the law takes a compact, testable form: the sum of kinetic energy $E_k$ and potential energy $E_p$ is the same at every point of the motion.

E_k + E_p = \text{constant}

where: $E_k = \tfrac12 mv^2$ is kinetic energy (mass $m$ , speed $v$ ) and $E_p = mgh$ is gravitational potential energy (height $h$ above a reference level, $g$ the acceleration due to gravity). The constant's actual value depends on the situation — it is fixed by whatever mass, height and speed the problem gives you, and must be recalculated for each new problem. Two everyday systems make the law concrete. A freely falling body trades potential energy for kinetic energy as it drops: it starts with maximum $E_p$ and zero $E_k$ , and ends (just before impact) with maximum $E_k$ and zero $E_p$ , the sum staying fixed throughout. A simple pendulum does the same trade sideways: at each extreme of its swing it is momentarily at rest ( $E_k = 0$ , $E_p$ maximum), and at the lowest point it moves fastest ( $E_k$ maximum, $E_p = 0$ ). With no friction, a pendulum would swing forever to the exact same height on both sides. In reality, air resistance and friction at the support slowly convert mechanical energy into heat and sound, so a real pendulum loses height over many swings — the total energy is still conserved, but usable mechanical energy is not, because some of it has leaked into forms we no longer track as $E_k$ or $E_p$ .

Figure

A falling ball at three heights showing potential energy converting to kinetic energy while the sum stays the same

Derivation

Derivation AEk + Ep stays constant for a freely falling body

The goal is to show algebraically, using only $E_k = \tfrac12 mv^2$ , $E_p = mgh$ and the kinematic relation $v^2 = 2g(h-x)$ for a body dropped from rest, that the sum of kinetic and potential energy is the same fixed value $mgh$ at every height during the fall — proving the law of conservation of energy for this simple, friction-free case rather than merely asserting it.

Step 1. Set up the drop. A body of mass $m$ is released from rest ( $u = 0$ ) at height $h$ above the ground. At the instant of release, all its energy is potential: $E_k = 0$ and $E_p = mgh$ , so the total is $E_k + E_p = mgh$ .

Step 2. Find its speed after falling to a lower height $x$ (so it has fallen a distance $h - x$ ). Using $v^2 = u^2 + 2a s$ with $u = 0$ , $a = g$ , $s = h - x$ , gives $v^2 = 2g(h-x)$ .

Step 3. Compute both energies at height $x$ . Kinetic energy is $E_k = \tfrac12 mv^2 = \tfrac12 m \times 2g(h-x) = mg(h-x)$ . Potential energy at that height is $E_p = mgx$ .

Step 4. Add them: $E_k + E_p = mg(h-x) + mgx = mgh - mgx + mgx = mgh$ . The $mgx$ terms cancel exactly, leaving $mgh$ — the same total found at release in Step 1, for any height $x$ between $h$ and $0$ . Checking the ground ( $x = 0$ ): $E_k = mgh$ , $E_p = 0$ , sum $= mgh$ again. The sum $E_k + E_p = mgh$ holds throughout the fall, confirming the law of conservation of energy for this case.

Worked example

A ball of mass 0.4 kg is dropped from a height of 20 m. Taking $g = 9.8\ \text{m/s}^2$ and ignoring air resistance, find its kinetic energy and speed at a height of 8 m, and verify that $E_k + E_p$ matches the value at the start.

Step 1. Total energy at the start (at rest, height 20 m, so $E_k = 0$ ): $E_{total} = E_p = mgh = 0.4 \times 9.8 \times 20 = 78.4\ \text{J}$ .

Step 2. Potential energy at 8 m: $E_p = mgx = 0.4 \times 9.8 \times 8 = 31.36\ \text{J}$ .

Step 3. By conservation of energy, kinetic energy at 8 m is $E_k = E_{total} - E_p = 78.4 - 31.36 = 47.04\ \text{J}$ .

Step 4. Speed from $E_k = \tfrac12 mv^2$ : $v^2 = \dfrac{2E_k}{m} = \dfrac{2 \times 47.04}{0.4} = 235.2\ \text{m}^2/\text{s}^2$ , so $v \approx 15.3\ \text{m/s}$ .

Step 5. Verify: $E_k + E_p = 47.04 + 31.36 = 78.4\ \text{J}$ , exactly matching the starting total of 78.4 J. At 8 m the ball has kinetic energy 47.04 J and speed about 15.3 m/s, confirming energy conservation.

Common traps

Thinking "conserved" means "nothing changes" — the forms of energy change constantly (potential to kinetic and back); only the total amount stays fixed.
Applying $E_k + E_p = \text{constant}$ to real situations with noticeable friction or air resistance without accounting for energy lost as heat and sound — total energy is still conserved overall, but the mechanical sum $E_k + E_p$ will appear to shrink.
Mixing up conservation of energy with conservation of momentum — these are two separate laws from different topics; do not swap their formulas.
Assuming the constant value of $E_k + E_p$ is a fixed number that applies to every problem — it equals $mgh$ (or whatever the given data fix it to be) and must be worked out fresh for each situation.

Ask the tutor: Given a falling-body or pendulum problem with the energy known at one point, ask the AI tutor to help you set up $E_k + E_p = \text{constant}$ and solve for the missing speed, height, or energy at another point.

Why does a real pendulum eventually stop swinging if energy is always conserved

Energy is never destroyed, but it can move into forms you are not tracking. As a real pendulum swings, friction at its support and air resistance continuously convert a small amount of its mechanical energy ( $E_k + E_p$ ) into heat and sound, which spread into the surroundings. The total energy of everything involved — pendulum plus air plus support — is still exactly conserved. What decreases is only the mechanical energy of the pendulum itself, which is why it swings to a slightly lower height each time and eventually stops.

If Ek + Ep is always constant, why do I need to calculate it separately for every problem

Because the constant is not a universal number — it depends on the mass, the starting height or speed, and the value of g given in that particular problem. For a body dropped from rest at height h, the constant works out to mgh, but a different starting height or an initial push (nonzero starting speed) would give a different total. Always compute the total energy from the situation described at one known point, usually the start, and then use that same value at every other point in the same problem.

Does the law of conservation of energy apply outside falling objects and pendulums

Yes — the general law (energy can only be transformed, never created or destroyed) applies to every physical process: a battery converting chemical energy to electrical energy, a bulb converting electrical energy to light and heat, food converting chemical energy into the kinetic energy of your muscles. The simple form $E_k + E_p = \text{constant}$ used in this chapter is a special case that works only for mechanical motion without friction or air resistance. In Class 9 you apply it mainly to falling bodies and pendulums, but the underlying law is far more general.

Law of conservation of energy

The core idea

Derivation

Worked example

Common traps

Before this

Where this leads