In short

Mechanical energy is the sum of kinetic energy (\frac{1}{2}mv^2) and potential energy (mgh or \frac{1}{2}kx^2). When only conservative forces (gravity, springs) do work — no friction, no air resistance — the total mechanical energy stays constant: KE_i + PE_i = KE_f + PE_f. This lets you connect speed and height directly, solving many problems in a single line that would otherwise need multiple kinematics equations.

A child sits in a roller coaster car at the top of the first hill — 30 metres above the ground. The chain releases. From this point on, no engine pushes the car. Yet it plunges into the valley at terrifying speed, climbs the next hill, slows, crests it, and dives again. The entire ride — every burst of speed, every slowdown — happens without a single joule of new energy entering the system.

At the hilltop, the car is high but barely moving. At the valley floor, the car is low but racing. The car does not gain energy on the way down — it converts one form into another. Height becomes speed. Speed becomes height. Back and forth, hill after valley, the total never changes.

This is conservation of mechanical energy. It is one of the most powerful shortcuts in all of physics — connecting speed and height directly, without tracking forces, accelerations, or time. If you know where something starts and where it ends up, energy conservation tells you how fast it is going.

Mechanical energy: two ledgers, one balance

You have met both pieces already. Kinetic energy is the energy of motion:

KE = \tfrac{1}{2}mv^2

Potential energy is energy stored by configuration — height in a gravitational field, or compression in a spring:

U_{\text{grav}} = mgh \qquad U_{\text{spring}} = \tfrac{1}{2}kx^2

Mechanical energy is the sum:

E_{\text{mech}} = KE + PE = \tfrac{1}{2}mv^2 + U

Think of it as a bank account with two ledgers. One ledger records "motion energy" (KE). The other records "position energy" (PE). When the ball rises, kinetic energy debits and potential energy credits — by exactly the same amount. When the ball falls, the transfer reverses. The total balance never changes, as long as no energy leaks out of the system through friction or air resistance.

Here is a cricket ball (mass 0.16 kg) thrown straight up at 20 m/s. Track its energy at several heights:

Height (m) Speed (m/s) KE (J) PE (J) Total (J)
0 (launch) 20.0 32.0 0 32.0
5.0 17.4 24.2 7.8 32.0
10.0 14.3 16.3 15.7 32.0
15.0 10.3 8.5 23.5 32.0
20.4 (peak) 0 0 32.0 32.0

The right-hand column is the same number every single row. That is conservation of mechanical energy — kinetic energy flows into potential energy and back, but the total is locked at 32.0 J.

Energy bar chart at five heights for a ball thrown upward Five stacked bar charts showing KE and PE at heights 0, 5, 10, 15, and 20.4 metres. At ground level, all energy is kinetic (red bar). At increasing heights, the red bar shrinks and the dark bar grows. At the peak, all energy is potential. The combined height stays the same. height above ground total E 32 J 0 m 5 m 10 m 15 m 32 J 20.4 m KE PE
Energy bar chart for a cricket ball thrown straight up at 20 m/s. Red bars are kinetic energy; dark bars are potential energy. The dashed line marks the total mechanical energy — it stays at 32 J at every height. As the ball rises, KE converts to PE; on the way back down, the process reverses.

The bar chart tells the whole story at a glance: at the bottom, all energy is kinetic. At the peak, all energy is potential. In between, the split shifts smoothly — but the total never budges.

Deriving conservation from the work-energy theorem

The pattern in the table is not a coincidence. It is a mathematical consequence of the forces involved.

Start from the work-energy theorem:

W_{\text{net}} = KE_f - KE_i \tag{1}

Why: the net work done on an object equals the change in its kinetic energy. This was derived from Newton's second law in the previous chapter.

Step 1. Split the net work into two groups: work done by conservative forces (gravity, springs — forces that have an associated potential energy) and work done by non-conservative forces (friction, air resistance, applied pushes).

W_{\text{cons}} + W_{\text{nc}} = KE_f - KE_i \tag{2}

Why: the net work is the sum of work by all forces. Separating them lets you use the definition of potential energy for the conservative group.

Step 2. For any conservative force, the work done equals the negative of the change in potential energy. This is the definition of potential energy:

W_{\text{cons}} = -\Delta PE = -(PE_f - PE_i) = PE_i - PE_f \tag{3}

Why: when gravity does positive work (the object falls, gaining speed), the potential energy decreases by the same amount. The minus sign connects the two: work done by gravity = -\Delta U_{\text{grav}}.

Step 3. Substitute equation (3) into equation (2):

(PE_i - PE_f) + W_{\text{nc}} = KE_f - KE_i

Rearrange by collecting all initial energies on the left and final energies on the right:

KE_i + PE_i + W_{\text{nc}} = KE_f + PE_f \tag{4}

Why: this is the general energy equation. It works whether friction is present or not. The non-conservative work W_{\text{nc}} accounts for any energy entering or leaving the mechanical system.

Step 4. Now impose the key condition: if the only forces doing work are conservative — no friction, no air resistance, no external push — then W_{\text{nc}} = 0:

\boxed{KE_i + PE_i = KE_f + PE_f} \tag{5}
\boxed{E_{\text{mech, initial}} = E_{\text{mech, final}}}

Why: with no non-conservative work, the left and right sides of equation (4) are equal. The total mechanical energy at the start equals the total at the end. Energy is neither created nor destroyed — only shuffled between kinetic and potential.

Go back and check the table: at h = 0, PE = 0 and KE = 32 J. At h = 20.4, PE = 32 J and KE = 0. At every point in between, KE + PE = 32 J. The theorem explains why the total was constant — the only force doing work is gravity, which is conservative.

Conservation of Mechanical Energy

When only conservative forces do work on a system, the total mechanical energy remains constant:

\tfrac{1}{2}mv_i^2 + U_i = \tfrac{1}{2}mv_f^2 + U_f

The potential energy U can include gravitational PE (mgh), elastic PE (\frac{1}{2}kx^2), or both.

Why energy conservation is so powerful

You might wonder: what does this give you that kinematics does not?

Consider this problem: a ball is released from rest at the top of a smooth, curved ramp of height 8 m. What is its speed at the bottom?

Using Newton's laws: you need the angle of the ramp at every point (which changes if the ramp is curved), the normal force (which varies with curvature), the component of gravity along the ramp, and then an integral of the acceleration along the entire path. For a curved ramp, this is a calculus problem.

Using energy conservation: the ball starts from rest at height h. At the bottom, height is zero.

mgh = \tfrac{1}{2}mv^2

The mass cancels:

v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 8} = \sqrt{156.8} = 12.5 \text{ m/s}

One line. No angles, no forces, no path shape, no calculus. A straight slide, a spiral chute, a parabolic curve — the answer is the same. Energy conservation cares only about where you start and where you end up, not about how you get there.

This is the deep reason energy methods are powerful: the conservation equation is scalar and connects initial and final states directly. Newton's second law is a vector equation that forces you to track every point along the path. Energy conservation lets you skip the path entirely.

Explore: height determines speed

The relationship v = \sqrt{2gh} means the speed at the bottom depends only on the release height. Drag the red point in the figure below and watch the speed respond.

Interactive: release height determines speed at the bottom of a frictionless ramp A curve showing speed at the bottom of a ramp as a function of release height. The curve is v equals the square root of 2gh. Drag the red point along the height axis to explore different release heights and see the resulting speed. release height h (m) speed at bottom v (m/s) 0 5 10 15 20 5 10 15 20 drag the red dot along the x-axis
Drag the red point to change the release height $h$. The curve shows $v = \sqrt{2gh}$ — the speed of an object released from rest and falling through height $h$ on a frictionless surface. Notice the curve flattens as $h$ increases: doubling the height does not double the speed. To double the speed, you need four times the height.

Worked examples

Example 1: The roller coaster valley

A roller coaster car (mass 500 kg) starts from rest at the top of a 25 m hill. The track is frictionless. Find the car's speed at the bottom of the valley (ground level) and at the top of the next hill, which is 15 m high.

Roller coaster track profile with two hills A roller coaster track starting at 25 m height, descending to ground level, then climbing to 15 m. Car positions marked at each location with velocity labels. 25 m ground 15 m v = 0 v = ? v = ?
The car starts from rest at 25 m. What is its speed at the valley floor and at the 15 m hilltop?

Step 1. Set the reference level at the valley floor (h = 0). The car starts from rest, so KE_i = 0.

Initial state: v_i = 0, h_i = 25 m.

E_i = KE_i + PE_i = 0 + mgh_i = 500 \times 9.8 \times 25 = 122{,}500 \text{ J}

Why: the car is at rest at the top, so all 122,500 J of mechanical energy is gravitational PE.

Step 2. At the valley floor (h_f = 0), all PE has converted to KE.

\tfrac{1}{2}mv_f^2 + 0 = 122{,}500
v_f^2 = \frac{2 \times 122{,}500}{500} = 490 \qquad v_f = \sqrt{490} = 22.1 \text{ m/s}

Why: at ground level the potential energy is zero (by our reference choice), so all mechanical energy is now kinetic. You can also write v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 25} directly — the mass cancels.

Step 3. At the top of the second hill (h = 15 m):

\tfrac{1}{2}mv^2 + mgh = 122{,}500
\tfrac{1}{2}(500)v^2 + 500 \times 9.8 \times 15 = 122{,}500
250\,v^2 + 73{,}500 = 122{,}500
v^2 = \frac{49{,}000}{250} = 196 \qquad v = 14.0 \text{ m/s}

Why: at 15 m, the car has used 73,500 J of its kinetic energy to "buy" potential energy. The remaining 49,000 J stays as KE. Notice this is equivalent to the car falling through a height of 25 - 15 = 10 m: v = \sqrt{2g \times 10} = \sqrt{196} = 14 m/s.

Result: Speed at the valley floor: 22.1 m/s (about 80 km/h). Speed at the 15 m hilltop: 14.0 m/s (about 50 km/h).

What this shows: The car's speed at any point depends only on the height difference from the start, not on the shape of the track. A spiral, a straight slide, a corkscrew loop — the answer is the same. This is why energy conservation is so much faster than Newton's laws for curved-path problems.

Example 2: Spring launcher

A toy dart gun has a spring with spring constant k = 200 N/m compressed by x = 0.1 m. The dart has mass 20 g. The gun fires the dart vertically upward. How high does the dart go?

Spring launcher: compressed spring fires a dart upward Left panel shows a compressed spring with a dart on top, labelled with elastic PE of 1 J. Right panel shows the dart at maximum height with the spring relaxed, labelled with gravitational PE. Before 20 g x = 0.1 m ½kx² = 1 J → After v = 0 h = ? mgh
The compressed spring stores 1 J of elastic PE. When fired, that PE converts to KE (as the dart leaves) and then to gravitational PE (as the dart rises). At maximum height, all energy is gravitational PE.

Step 1. Identify the initial and final states.

Initial: spring compressed by x = 0.1 m, dart at rest. The energy is entirely elastic PE.

E_i = \tfrac{1}{2}kx^2 = \tfrac{1}{2}(200)(0.1)^2 = 1.0 \text{ J}

Final: dart at maximum height h, momentarily at rest (v = 0), spring relaxed (x = 0). The energy is entirely gravitational PE.

Why: at maximum height, KE = 0 (the dart has stopped) and elastic PE = 0 (the spring is at its natural length). All 1 J has been converted to gravitational PE.

Step 2. Apply conservation of mechanical energy. Elastic PE at start = gravitational PE at top.

\tfrac{1}{2}kx^2 = mgh
h = \frac{kx^2}{2mg} = \frac{200 \times (0.1)^2}{2 \times 0.020 \times 9.8} = \frac{2.0}{0.392}

Why: set elastic PE equal to gravitational PE and solve for h. You never need to compute the dart's launch speed — energy conservation connects initial spring compression directly to final height.

Step 3. Compute.

h = 5.1 \text{ m}

Why: a spring storing just 1 joule of energy can launch a 20 g dart over five metres high. The small mass is what makes this possible — the same 1 J would lift a 1 kg object only 0.10 m.

Result: The dart rises to 5.1 m above the launch point.

What this shows: Energy conservation handles transitions between two different forms of PE (elastic → gravitational) just as smoothly as KE ↔ PE transitions. You can skip the intermediate kinetic energy entirely — though if you wanted the dart's launch speed, it would be v = \sqrt{2E/m} = \sqrt{2 \times 1.0/0.020} = 10 m/s.

When mechanical energy is NOT conserved

The derivation made one critical assumption: W_{\text{nc}} = 0. When friction or air resistance acts, non-conservative work is negative (these forces always oppose motion), and mechanical energy decreases.

The general equation (4) still works:

KE_i + PE_i + W_{\text{nc}} = KE_f + PE_f

For kinetic friction with force f_k acting over a distance d along the path:

W_{\text{friction}} = -f_k d

So the energy equation becomes:

KE_i + PE_i - f_k d = KE_f + PE_f

The "missing" mechanical energy — the difference between E_i and E_f — is the energy that friction has converted to thermal energy. It heats up the surfaces, the air, the surroundings. The energy has not vanished. It has left the mechanical ledger and entered the thermal ledger.

Here is a concrete example. A 25 kg child slides down a 5 m high playground slide. On a frictionless slide, energy conservation predicts v = \sqrt{2 \times 9.8 \times 5} = 9.9 m/s at the bottom. But the child actually arrives at 7 m/s. Where did the missing energy go?

E_i = mgh = 25 \times 9.8 \times 5 = 1225 J

E_f = \tfrac{1}{2}mv^2 = \tfrac{1}{2}(25)(7)^2 = 612.5 J

Energy lost to friction: 1225 - 612.5 = 612.5 J. Half the initial energy went into heating the slide surface and the child's clothes. That is a lot of friction — and it is exactly why metal slides at a park get warm on a busy day.

The rule of thumb: if a problem says "smooth" or "frictionless," use conservation of mechanical energy directly. If it mentions "rough" or gives a friction coefficient, use the generalised equation with W_{\text{nc}} = -f_k d.

Common confusions

If you came here to use KE_i + PE_i = KE_f + PE_f and solve problems, you have everything you need. What follows is for readers who want the deeper connections — why energy conservation works at a fundamental level, and what it has to do with the structure of the universe.

Proving conservation from Newton's second law

Conservation of mechanical energy is not an independent axiom — it is a theorem that follows from Newton's laws. Here is the proof in one dimension.

A particle of mass m moves under a conservative force F(x) with potential energy U(x) such that F = -dU/dx.

Start from Newton's second law:

F = ma = m\frac{dv}{dt}

Use the chain rule to rewrite dv/dt:

m\frac{dv}{dt} = m\frac{dv}{dx}\cdot\frac{dx}{dt} = mv\frac{dv}{dx}

Why: dx/dt = v, so the chain rule converts acceleration with respect to time into a product involving velocity and acceleration with respect to position. This is a standard trick in mechanics.

Now substitute F = -dU/dx:

-\frac{dU}{dx} = mv\frac{dv}{dx}

Recognise that mv\frac{dv}{dx} = \frac{d}{dx}\left(\frac{1}{2}mv^2\right):

-\frac{dU}{dx} = \frac{d}{dx}\left(\tfrac{1}{2}mv^2\right)
\frac{d}{dx}\left(\tfrac{1}{2}mv^2 + U\right) = 0

Why: if the derivative of a quantity with respect to position is zero everywhere, the quantity is constant. The quantity \frac{1}{2}mv^2 + U does not change as the particle moves — that is conservation of mechanical energy, derived purely from Newton's second law and the definition of potential energy.

Noether's theorem: the symmetry behind conservation

There is a reason even deeper than Newton's laws: the laws of physics do not change with time. An experiment done today gives the same result as the same experiment done tomorrow. This time-translation symmetry is what guarantees energy conservation.

The connection was proved by Emmy Noether in 1918. Her theorem states: every continuous symmetry of the laws of physics corresponds to a conserved quantity.

Symmetry Conserved quantity
Time translation (laws don't change over time) Energy
Space translation (laws are the same everywhere) Linear momentum
Rotational symmetry (laws are the same in every direction) Angular momentum

Noether's theorem is one of the deepest results in theoretical physics. It tells you why conservation laws exist — not as separate facts to memorise, but as consequences of the symmetries of nature. If the universe were to suddenly change its gravitational constant tomorrow, energy conservation would break. The constancy of the laws is what makes the conservation work.

Beyond mechanical energy: the first law of thermodynamics

When friction converts mechanical energy to heat, the mechanical energy is not conserved — but the total energy (mechanical + thermal + all other forms) is. This is the first law of thermodynamics: energy can change form but can never be created or destroyed.

Conservation of mechanical energy is a special case of this broader law — the special case where the only energy conversions are between kinetic and potential. The general law encompasses chemical energy, nuclear energy, electromagnetic radiation, and every other form. It is one of the most thoroughly tested principles in all of science, and no violation has ever been observed.

Where this leads next