Interconversion of energy forms

Rub your palms together and they warm up — kinetic energy has turned into heat. Strike a match and chemical energy becomes light and heat. Drop a stone and, as it falls, the height it loses is exactly matched by the speed it gains: potential energy is quietly turning into kinetic energy the whole way down. This constant switching of energy between forms, without any of it vanishing, is called the interconversion of energy, and it is the bridge between "forms of energy" and the law of conservation of energy in CBSE (NCERT) Class 9 Physics, Work and Energy. This article stays inside that syllabus scope: the mechanical exchange between kinetic energy ( $E_k = \tfrac12 mv^2$ ) and potential energy ( $E_p = mgh$ ) in falling objects, rising projectiles and swinging pendulums, plus a handful of everyday non-mechanical examples. We derive the falling-object speed formula from energy conservation, work two number examples fully, and flag the traps examiners love.

The core insight

Energy never disappears during a transformation, it only changes form, so in a falling, rising or swinging object (ignoring friction and air resistance) the sum $E_k + E_p$ stays fixed while $E_k$ and $E_p$ individually rise and fall.

The core idea

Energy does not stay locked in one form. It continually converts from one type to another as physical processes unfold, in every branch of physics — a battery turns chemical energy into electrical energy, a bulb turns electrical energy into light and heat, a microphone turns sound energy into electrical energy. This chapter's focus is the mechanical case: the exchange between kinetic energy and potential energy.

Free fall. As a raised object falls, its height $h$ shrinks, so $E_p$ shrinks, while its speed $v$ grows, so $E_k$ grows. Potential energy is converting into kinetic energy.
Rising projectile. As a thrown object climbs, its speed falls, so $E_k$ falls, while its height grows, so $E_p$ grows. Kinetic energy is converting into potential energy — the reverse process.
Swinging pendulum. The bob repeatedly trades $E_p$ (maximum at the two extreme, momentarily-still positions) for $E_k$ (maximum at the lowest point, where speed peaks), and back again, forever (in the ideal, frictionless case).

In every one of these cases, ignoring air resistance and friction, the total mechanical energy at any instant is

E_k + E_p = mgh_0 = \text{constant}. \qquad (1)

where: $E_k = \tfrac12 mv^2$ is kinetic energy in joules, $E_p = mgh$ is potential energy in joules, $m$ is mass in kg, $v$ is speed in m/s, $g = 9.8\ \text{m/s}^2$ , $h$ is height above a reference level in m, and $h_0$ is the starting height. The two terms constantly trade value between themselves, but their sum never changes — this constancy is exactly the idea that the law of conservation of energy (studied next) makes formal and general, for all forms of energy, not just mechanical ones.

Figure

A ball falling from a height with potential energy shrinking and kinetic energy growing so their sum stays constant

Derivation ASpeed of a freely falling object from energy conservation

The goal is to find how fast an object is moving after falling a height $h$ from rest, using only the constancy of $E_k + E_p$ established above and no separate motion formulas — this is the algebraic proof behind the number that appears in every falling-object worked example on this topic.

Step 1. Write the total mechanical energy at the start, where the object is at rest ( $v = 0$ ) at height $h$ above the ground: $E_{\text{total}} = E_k + E_p = 0 + mgh = mgh$ .

Step 2. Write the total mechanical energy just before it lands, where the height above the ground is zero: $E_{\text{total}} = E_k + E_p = \tfrac12 mv^2 + 0 = \tfrac12 mv^2$ .

Step 3. Because no friction or air resistance removes energy, the total at the start equals the total at the end: $mgh = \tfrac12 mv^2$ . The mass $m$ appears on both sides and cancels, which is why every object, light or heavy, falls at the same speed from the same height (ignoring air resistance).

Step 4. Solve for $v$ : multiply both sides by $2/m$ to get $v^2 = 2gh$ , then take the square root: $v = \sqrt{2gh}$ . This is the speed interconversion gives an object after it has fallen freely through height $h$ from rest. Check: units on the right are $\sqrt{(\text{m/s}^2)(\text{m})} = \sqrt{\text{m}^2/\text{s}^2} = \text{m/s}$ , matching a speed, and the mass cancelling out matches the everyday observation that a heavy stone and a light stone dropped together (in the absence of air resistance) hit the ground at the same time.

Worked example

A ball of mass 2 kg is dropped from a height of 5 m above the ground. Using $g = 9.8\ \text{m/s}^2$ and ignoring air resistance, find (a) its potential energy at the top, (b) its kinetic energy just before it hits the ground, and (c) its speed at that instant.

Step 1. Potential energy at the top (ball at rest, $v = 0$ , so all its mechanical energy is potential): $E_p = mgh = 2 \times 9.8 \times 5 = 98\ \text{J}$ .

Step 2. Just before impact, $h = 0$ , so $E_p = 0$ . With no friction or air resistance to remove energy, all 98 J has converted into kinetic energy: $E_k = 98\ \text{J}$ .

Step 3. Speed from the derivation above, $v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 5} = \sqrt{98} \approx 9.90\ \text{m/s}$ . Checking against Step 2: $E_k = \tfrac12 mv^2 = \tfrac12 \times 2 \times 9.90^2 \approx 98\ \text{J}$ , which matches.

The ball's potential energy of 98 J at the top converts entirely into 98 J of kinetic energy just before landing, giving an impact speed of about 9.90 m/s.

Competitive stretch

A pendulum bob of mass 0.5 kg is released from a point 0.2 m higher than its lowest point. Step 1. At the top the bob is at rest, so $E_p = mgh = 0.5 \times 9.8 \times 0.2 = 0.98\ \text{J}$ and $E_k = 0$ . Step 2. At the lowest point $h = 0$ , so by interconversion all $0.98\ \text{J}$ has become kinetic: $E_k = 0.98\ \text{J}$ . Step 3. Speed there: $v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 0.2} = \sqrt{3.92} \approx 1.98\ \text{m/s}$ . The bob's maximum speed, about 1.98 m/s, occurs exactly at the lowest point of the swing.

Common traps

Thinking energy "disappears" during a transformation — it only changes form; the total mechanical energy stays constant in an idealised, friction-free situation.
Assuming kinetic energy is maximum at the highest point of a pendulum's swing or a thrown ball's path — it is actually minimum, often zero, at the highest point, and maximum at the lowest point or the ground.
Forgetting real situations have friction or air resistance, which converts part of the mechanical energy into heat and sound — this is why a real pendulum eventually stops swinging.
Confusing interconversion (energy changing form) with the law of conservation of energy, which formally states that the total energy of an isolated system never changes, across all forms, not just mechanical ones.

Ask the tutor: Working out a falling-object or pendulum problem and want to check how much potential energy has turned into kinetic energy at some point? Share your numbers with the AI tutor and ask it to trace the interconversion.

Why does the mass cancel out when finding the speed of a falling object?

Because both the potential energy lost, $mgh$ , and the kinetic energy gained, $\tfrac12 mv^2$ , are directly proportional to mass $m$ . Setting them equal, $mgh = \tfrac12 mv^2$ , lets you divide both sides by $m$ , leaving $v = \sqrt{2gh}$ with no mass in it at all. That is the algebraic reason a heavy stone and a light stone, dropped from the same height with no air resistance, land at exactly the same speed.

Is the total energy really constant, or does the pendulum lose height over time?

In the idealised case taught here, with no friction at the support and no air resistance, the total mechanical energy $E_k + E_p$ stays exactly constant, and the pendulum would swing forever to the same height on each side. A real pendulum does lose a little height on each swing, because friction and air resistance convert a small part of its mechanical energy into heat and sound — that lost mechanical energy is not destroyed, it has simply interconverted into a form (heat) that no longer helps the bob rise.

How is interconversion of energy different from conservation of energy?

Interconversion just describes the everyday observation that energy keeps changing form — kinetic to potential, chemical to electrical, and so on. Conservation of energy is the stronger, formal law built from that observation: it says that for an isolated system, no matter how many forms the energy passes through, the total amount never increases or decreases. Interconversion is the process you watch happening; conservation is the accounting rule that process always obeys.

Interconversion of energy forms

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads