Conservation of energy

The principle, restated precisely. Energy can neither be created nor destroyed; it can only be transformed from one form to another, and the total amount of energy in an isolated system stays constant. When you see potential energy "disappearing," look for where it reappears — usually as kinetic energy, sometimes as heat.

Two forms that matter here. At this level, the interconversion you study is between gravitational potential energy (energy due to height) and kinetic energy (energy due to motion). Their formulas are:

where: $PE$ is potential energy in joules (J), $m$ is mass in kilograms (kg), $g$ is the acceleration due to gravity (taken as $10\ \text{m/s}^2$ unless stated otherwise), and $h$ is height above a chosen reference level, in metres (m).

where: $KE$ is kinetic energy in joules (J), $m$ is mass in kilograms, and $v$ is speed in metres per second (m/s).

The conservation statement for a falling or swinging body. If friction and air resistance are negligible, the total mechanical energy — potential plus kinetic — stays the same at every point of the motion:

where: this holds specifically when the object starts at rest at the top (so it has zero kinetic energy there) and all its potential energy has converted to kinetic energy by the time it reaches the reference height at the bottom.

Pendulum and falling body, in formula terms. At the two extreme ends of a pendulum's swing, $v = 0$ so $KE = 0$ and $PE$ is maximum; at the lowest point, $h = 0$ (by choice of reference) so $PE = 0$ and $KE$ is maximum. A falling body follows the same trade continuously: as $h$ decreases, $PE = mgh$ decreases, and as $v$ increases, $KE = \tfrac{1}{2}mv^2$ increases, always by a matching amount.

There is no formal proof required at this level — the reasoning is physical, not algebraic. Think of potential energy as energy "banked" by lifting a mass against gravity: someone (or something) had to do work to raise it, and that work is stored, ready to be released. When the object is allowed to fall or swing down, gravity does that stored work back on the object, speeding it up. Since work done equals energy transferred, exactly as much kinetic energy appears as potential energy disappears — nothing is left over, and nothing extra is created. This is why the process is reversible in an ideal (frictionless) world: a ball rolling down a smooth ramp and back up a symmetrical ramp on the other side would, in principle, rise to exactly its starting height, because every joule of $KE$ at the bottom converts back to $PE$ on the way up. In the real world, some energy always leaks away as heat through friction and air resistance, so the ball never quite reaches its original height — but that "lost" energy has not vanished; it has simply spread into the surroundings as heat, which is why conservation of energy still holds for the universe as a whole even when useful mechanical energy is not conserved.

Question: A ball of mass $2\ \text{kg}$ is released from rest at the top of a smooth ramp of height $5\ \text{m}$. Taking $g = 10\ \text{m/s}^2$ and ignoring friction, find (a) the ball's potential energy at the top, and (b) its speed at the bottom of the ramp.

Step 1. Find the potential energy at the top, where the ball is at rest. Using $PE = mgh$, with $m = 2\ \text{kg}$, $g = 10\ \text{m/s}^2$ and $h = 5\ \text{m}$: $PE = 2 \times 10 \times 5 = 100\ \text{J}$.

Step 2. Apply conservation of energy. At the top, the ball is at rest, so its kinetic energy is zero and all its energy is potential: total mechanical energy $= 100\ \text{J}$. Since the ramp is smooth (no friction), this total stays the same all the way down. At the bottom, taking the reference height as zero, $PE = 0$, so all $100\ \text{J}$ has converted into kinetic energy: $KE_{\text{bottom}} = 100\ \text{J}$.

Step 3. Find the speed from the kinetic energy. Using $KE = \tfrac{1}{2}mv^2$: $100 = \tfrac{1}{2} \times 2 \times v^2$, so $v^2 = 100$, giving $v = 10\ \text{m/s}$.

The ball has $100\ \text{J}$ of potential energy at the top, and this converts entirely into kinetic energy by the bottom, giving it a speed of $10\ \text{m/s}$.

Competitive stretch

If the same ramp were not perfectly smooth and the ball reached the bottom at only $8\ \text{m/s}$ instead of $10\ \text{m/s}$, the kinetic energy actually gained would be $\tfrac{1}{2} \times 2 \times 8^2 = 64\ \text{J}$, so the "missing" $100 - 64 = 36\ \text{J}$ was converted to heat by friction — the total energy is still conserved, just not all of it stayed mechanical.

• Thinking conservation of energy means the object keeps the same form of energy throughout — it is the total that stays constant, while the form (potential, kinetic, heat) keeps changing.
• Forgetting that when a ball is released "from rest," its kinetic energy at the very start is exactly zero, not some small leftover value.
• Believing that friction destroys energy — it only converts mechanical energy into heat, which spreads into the surroundings; the overall total energy is unchanged.
• Mixing up $PE = mgh$ and $KE = \tfrac{1}{2}mv^2$ — remember $PE$ depends on height, $KE$ depends on speed squared, and both give an answer in joules.

Ask the tutor: Want to try conservation of energy on a roller coaster, a bouncing ball, or a problem with an incline of a different height — ask the AI tutor to work through it with you.