In short

Potential energy (PE) is energy stored in a system because of its configuration. Gravitational PE near Earth's surface is U = mgh, measured from a chosen reference level. Elastic PE in a compressed or stretched spring is U = \tfrac{1}{2}kx^2. Both are derived from the work done against the relevant force. The zero of PE is a human choice — only changes in PE are physically meaningful. PE belongs to the system (body + field or body + spring), not to the single object alone.

A village in the Western Ghats has a dam high up in the hills. Behind that dam, water sits quietly — motionless, not doing anything visible. But open the sluice gates and that water rushes downhill, spins a turbine, and lights up an entire district. Where did all that energy come from? The water was not moving. It had no kinetic energy. Yet the energy was there, stored silently, waiting to be released.

That is potential energy — energy that a system possesses not because something is moving, but because of where things are relative to each other. Lift a bucket of water from a well and you do work against gravity. That work does not vanish — it gets stored in the Earth–bucket system as gravitational potential energy. Release the bucket and the stored energy converts back to kinetic energy as the bucket falls. Every joule you put in by lifting comes back out as motion.

The same idea applies to a spring. Compress the spring in a toy gun and it sits there, coiled and loaded. Pull the trigger, and the spring snaps back, launching a dart across the room. You did work compressing the spring; the spring stored that work as elastic potential energy; and when released, the stored energy became kinetic energy of the dart.

Potential energy is the universe's way of keeping accounts. Energy never disappears — it just changes address.

The key idea: PE is stored work

The work–energy theorem told you that the net work done on an object equals the change in its kinetic energy. But what happens when you lift a ball slowly upward at constant speed? You do positive work on the ball, gravity does negative work on the ball, and the net work is zero — so the kinetic energy does not change. Where did your work go?

It went into the gravitational field. The Earth–ball system now has more energy than it did before, stored in the arrangement of the ball being higher up. That stored energy is what you call potential energy.

Here is the crucial point: potential energy is defined through the work done against a force. You do not just declare that a ball "has" potential energy. You identify a force (gravity, a spring force), compute the work done against that force, and that work equals the gain in potential energy.

\Delta U = -W_{\text{by the force}}

Why the negative sign: when gravity does negative work on a ball you lift (the force is downward but the displacement is upward), the potential energy increases. The PE gained equals the work done against the force, which is -W_{\text{by the force}}.

This definition applies to any conservative force. For now, you will see it in action for two forces: gravity near Earth's surface, and the spring force.

Gravitational potential energy: deriving U = mgh

Take a ball of mass m near Earth's surface, where the gravitational acceleration is g = 9.8 m/s² downward. You lift the ball vertically from a height y_1 to a height y_2, where h = y_2 - y_1.

Lifting a ball from height y₁ to height y₂ against gravity A ball at height y₁ near the ground, with an arrow showing upward displacement h to a position y₂. Gravity acts downward as mg. The work done against gravity stores energy as gravitational PE. ground (reference level) y₁ y₂ h = y₂ − y₁ mg
A ball of mass $m$ is lifted from $y_1$ to $y_2$. Gravity ($mg$) pulls downward. The displacement is upward. The work done against gravity equals the gravitational PE gained.

Assumptions: The ball moves near Earth's surface, so g is constant. Air resistance is negligible. The ball moves slowly enough that changes in kinetic energy are negligible — all the work goes into PE.

Step 1. Write the gravitational force as a function of position. Take upward as positive.

F_g = -mg

Why negative: gravity pulls downward, and you chose upward as positive. The magnitude is mg, the direction is -\hat{y}, so the force is -mg.

Step 2. Compute the work done by gravity as the ball moves from y_1 to y_2.

W_{\text{gravity}} = \int_{y_1}^{y_2} F_g \, dy = \int_{y_1}^{y_2} (-mg) \, dy

Why an integral: the work done by a force over a displacement is \int \vec{F} \cdot d\vec{r}. Since the force is constant and the motion is along the y-axis, this becomes a simple integral of F_g with respect to y.

Step 3. Evaluate the integral. Since m and g are constants:

W_{\text{gravity}} = -mg \int_{y_1}^{y_2} dy = -mg(y_2 - y_1) = -mgh

Why: the integral of dy from y_1 to y_2 is just (y_2 - y_1) = h. The result is negative because gravity acts downward while the displacement is upward — gravity does negative work when you lift things.

Step 4. Apply the definition \Delta U = -W_{\text{by the force}}.

\Delta U = -W_{\text{gravity}} = -(-mgh) = mgh
\boxed{U = mgh}

Why: the negative of a negative gives a positive. Lifting the ball by height h increases the gravitational PE by mgh. This is the energy you stored in the Earth–ball system by doing work against gravity.

The formula is beautifully simple: gravitational PE equals mass times gravitational acceleration times height above your chosen reference level. Double the height and you double the stored energy. Double the mass and you double it again. This is why a dam stores enormous energy — even at a modest height of 100 m, millions of tonnes of water represent billions of joules of stored gravitational PE.

The reference level: your choice, your accounting

Notice that the formula says U = mgh, where h is the height above a reference level. But which reference level? The floor of the room? The ground outside? Sea level? The centre of the Earth?

The answer: you choose. The reference level is a human decision, not a law of nature. You pick a convenient level and call it h = 0. All heights are measured from there.

This might feel unsettling — how can a physical quantity depend on a human choice? The key insight is that only changes in PE are physically meaningful. When a ball falls from height h_1 to h_2, the PE lost is:

\Delta U = mg(h_2 - h_1)

This difference does not depend on where you put the zero. If you shift every height up by 50 m, both h_1 and h_2 increase by 50, but their difference stays the same. The physics — how much kinetic energy the ball gains — depends only on \Delta U, never on U alone.

Same ball, two different reference levels — the PE change is identical Two side-by-side diagrams. Left: reference level at the ground, ball at h=3m has U=mgh. Right: reference level at the table (1m high), same ball has U=mg(2). In both cases, when the ball falls to the ground, the change in PE is the same: mg times 3. Reference: ground h = 0 table 3 m U = mg(3) Reference: table ground h = 0 2 m U = mg(2) −1 m
Left: with the ground as reference, the ball at 3 m has $U = 3mg$. Right: with the table as reference, the same ball has $U = 2mg$, and the ground is at $h = -1$ m. In both cases, if the ball falls to the ground, $\Delta U = -3mg$ — the same KE is gained. The reference level changes the numbers, not the physics.

The practical rule: choose the reference level that makes the problem easiest. For a ball dropped from a building, set h = 0 at the ground. For a roller coaster, set h = 0 at the lowest point of the track. For a pendulum, set h = 0 at the lowest point of the swing. The physics does not care — pick whatever makes the algebra cleanest.

Elastic potential energy: deriving U = \tfrac{1}{2}kx^2

A spring exerts a restoring force proportional to its displacement from the natural (unstretched) length. This is Hooke's law:

F_{\text{spring}} = -kx

where k is the spring constant (in N/m) and x is the displacement from the natural length. The negative sign means the force always opposes the displacement — stretch the spring to the right and it pulls you back to the left.

You compress a spring by a distance x from its natural length. How much energy is stored?

Spring compressed by distance x from natural length A horizontal spring attached to a wall on the left. Top: spring at natural length, end at position 0. Bottom: spring compressed by distance x, with a block pushing from the right. The spring force kx points to the right (opposing compression). Natural length x = 0 Compressed by x m x kx
Top: the spring at its natural length, end at $x = 0$. Bottom: compressed by distance $x$. The restoring force $kx$ points to the right — the spring pushes back against the compression. The block stores elastic PE.

Step 1. Compute the work done by the spring as the block is pushed from x = 0 (natural length) to x = x (compressed position).

W_{\text{spring}} = \int_0^x F_{\text{spring}} \, dx' = \int_0^x (-kx') \, dx'

Why: the spring force is -kx' at each position x'. As you compress the spring from 0 to x, the force grows stronger — this is why the integral is essential. Unlike gravity, the spring force is not constant.

Step 2. Evaluate the integral.

W_{\text{spring}} = -k \int_0^x x' \, dx' = -k \left[\frac{x'^2}{2}\right]_0^x = -k \cdot \frac{x^2}{2} = -\frac{1}{2}kx^2

Why: the integral of x' from 0 to x is x^2/2. The result is negative because the spring force opposes the motion during compression — the spring does negative work as you compress it.

Step 3. Apply \Delta U = -W_{\text{by the force}}.

\Delta U = -W_{\text{spring}} = -\left(-\frac{1}{2}kx^2\right) = \frac{1}{2}kx^2
\boxed{U = \frac{1}{2}kx^2}

Why: the elastic PE stored equals the work you did against the spring force. Since the spring force grows linearly with displacement, the stored energy grows as the square of the displacement. Compress twice as far and you store four times the energy.

Notice that x^2 is always positive regardless of whether you stretch or compress the spring. A spring compressed by 3 cm stores the same energy as a spring stretched by 3 cm. The natural length (x = 0) is the automatic reference point for elastic PE — you do not need to choose one.

Why the ½ matters: area under the force-displacement graph

The factor of \frac{1}{2} has a clear geometric meaning. Plot the spring force magnitude (kx) against displacement (x). The graph is a straight line through the origin with slope k. The work done is the area under this line from 0 to x, which is a triangle with base x and height kx:

\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times x \times kx = \frac{1}{2}kx^2
Force vs displacement for a spring with $k = 100$ N/m. The shaded area under the line — a triangle — equals the elastic PE stored: $\frac{1}{2} \times 0.05 \times 5 = 0.125$ J. The $\frac{1}{2}$ in the formula is literally the $\frac{1}{2}$ of a triangle's area.

This is why elastic PE has the \frac{1}{2} but gravitational PE (near the surface) does not. Gravity is constant, so the force-displacement graph is a horizontal line and the area is a rectangle (mgh, no half). The spring force grows linearly, so the area is a triangle (\frac{1}{2}kx^2, with the half).

Drag the compression in the figure below to feel the quadratic growth:

Interactive: elastic PE grows as the square of compression A parabolic curve showing PE = ½kx² for k = 800 N/m. Drag the red point to change compression and watch the PE grow quadratically. compression x (m) elastic PE (J) 0 1 2 3 4 0 0.025 0.05 0.075 0.10 U = ½kx² drag the red point along the axis
Drag the red point to change the compression of a spring ($k = 800$ N/m). Notice the parabolic shape: doubling the compression from 2.5 cm to 5 cm quadruples the PE from 0.25 J to 1 J. The energy grows as $x^2$, not $x$.

PE as stored energy — convertible to KE

The deepest reason potential energy matters is that it can be converted to kinetic energy and back again, with no loss (in the absence of friction). This is the principle behind every roller coaster, every pendulum, every ball thrown upward and caught again.

When you lift a ball to height h and release it:

The total mechanical energy E = \text{KE} + \text{PE} stays constant throughout. This is the conservation of mechanical energy, and it only works because gravity is a conservative force — the work it does depends only on the starting and ending positions, not on the path taken.

Drag the height in the figure below to watch a 0.5 kg ball's energy split at any position (g = 10 m/s², released from 2 m):

Interactive: PE and KE trade off as height changes Two crossing lines show PE = mgh rising and KE falling as the ball's height increases. Drag to explore the energy split at any height. Total energy stays at 10 J. height h (m) energy (J) 0 2 4 6 8 10 0 0.5 1.0 1.5 2.0 total = 10 J PE = mgh KE drag the red point along the axis
Drag the red point to set the ball's height. At $h = 2$ m (the top), all 10 J is PE. At $h = 0$ (the ground), all 10 J is KE. The dashed line confirms: the total never changes — energy only changes form, not amount.

The same logic applies to a spring. Compress a spring and release a block:

This interconversion is what makes PE useful. It is not just a bookkeeping trick — it is the mechanism by which energy is stored and released in every physical system, from ISRO rockets climbing out of Earth's gravity to the compressed spring in a retractable pen.

PE belongs to the system, not the single object

A subtle but important point: when you say "a ball at height h has potential energy mgh," you are being slightly imprecise. The potential energy does not belong to the ball alone. It belongs to the Earth–ball system.

Why? Because potential energy arises from the interaction between two things — the ball and Earth's gravitational field. If there were no Earth (no gravitational field), lifting the ball would not store any energy. The energy is in the arrangement of the ball relative to the Earth, not in the ball itself.

The same is true for a spring. The elastic PE \frac{1}{2}kx^2 belongs to the block–spring system. Remove the spring and the block has no elastic PE. The energy is in the interaction — the compressed spring pushing against the block — not in either object alone.

In everyday language, you will hear "the ball has potential energy" and that is fine as shorthand. But when the physics requires precision — when you draw energy diagrams or apply conservation of energy — remember that PE is a system property. This becomes especially important in energy diagrams and equilibrium, where you plot PE as a function of position for the entire system.

Worked examples

Example 1: Ball lifted and released

A rubber ball of mass 500 g is lifted 2 m above the floor and released from rest. Taking the floor as the reference level (h = 0) and g = 10 m/s², find: (a) the PE at the top, (b) the KE at the bottom (just before hitting the floor), and (c) the speed at the bottom.

Energy bar chart showing PE converting to KE as a ball falls Three positions of a ball: top (h=2m), middle (h=1m), and bottom (h=0). At each position, bar charts show PE and KE. At the top, PE=10J and KE=0. At the middle, PE=5J and KE=5J. At the bottom, PE=0 and KE=10J. Total energy bar stays constant at 10J. floor (h = 0) h = 2 m h = 1 m h = 0 PE 10 J KE=0 PE 5 J KE 5 J PE=0 KE 10 J Top Midway Bottom
Energy bar chart at three positions. At the top: all 10 J is PE (red bar). At the midpoint: PE and KE split equally (5 J each). At the bottom: all 10 J has converted to KE (dark bar). The total energy (red + dark) is 10 J at every position.

Step 1. Convert mass to SI units.

m = 500 \text{ g} = 0.5 \text{ kg}

Step 2. Compute PE at the top.

U_{\text{top}} = mgh = 0.5 \times 10 \times 2 = 10 \text{ J}

Why: the ball is 2 m above the reference level (the floor), so h = 2 m. This is straightforward: mass times g times height.

Step 3. Apply conservation of energy to find KE at the bottom.

At the top, the ball is at rest, so \text{KE}_{\text{top}} = 0. At the bottom, h = 0, so \text{PE}_{\text{bottom}} = 0.

\text{KE}_{\text{top}} + \text{PE}_{\text{top}} = \text{KE}_{\text{bottom}} + \text{PE}_{\text{bottom}}
0 + 10 = \text{KE}_{\text{bottom}} + 0
\text{KE}_{\text{bottom}} = 10 \text{ J}

Why: total mechanical energy is conserved (no friction, no air resistance). Every joule of PE lost becomes a joule of KE gained.

Step 4. Find the speed at the bottom using \text{KE} = \frac{1}{2}mv^2.

10 = \frac{1}{2} \times 0.5 \times v^2
v^2 = \frac{10 \times 2}{0.5} = 40
v = \sqrt{40} \approx 6.32 \text{ m/s}

Why: rearrange the KE formula to isolate v. The ball arrives at the floor at about 6.3 m/s — roughly 23 km/h.

Result: PE at the top = 10 J. KE at the bottom = 10 J. Speed at the bottom \approx 6.32 m/s.

What this shows: Every joule of gravitational PE at the top becomes a joule of KE at the bottom. The energy bar chart makes this visible — the total height of the bars is the same at every position; only the split between PE and KE changes.

Example 2: Spring-loaded toy gun

A spring with spring constant k = 800 N/m is compressed by x = 5 cm. A plastic dart of mass 200 g is placed against the spring and released. Find: (a) the elastic PE stored in the compressed spring, and (b) the speed of the dart as it leaves the spring (when the spring returns to its natural length).

Before and after diagram for a spring launching a dart Top: spring compressed by 5 cm with a 200 g block against it, storing PE = 1 J. Bottom: spring at natural length, block moving to the right with KE = 1 J and velocity v. Before (compressed) 200 g x = 0 5 cm PE = ½kx² = 1 J KE = 0 After (released) 200 g v = ? PE = 0 KE = 1 J
Top: the spring is compressed by 5 cm, storing 1 J of elastic PE. The dart is stationary. Bottom: the spring has returned to its natural length, all PE has converted to KE, and the dart moves to the right at speed $v$.

Step 1. Convert to SI units.

k = 800 N/m, x = 5 cm = 0.05 m, m = 200 g = 0.2 kg.

Step 2. Compute the elastic PE stored.

U = \frac{1}{2}kx^2 = \frac{1}{2} \times 800 \times (0.05)^2 = \frac{1}{2} \times 800 \times 0.0025 = 1 \text{ J}

Why: the spring constant is large (800 N/m, a fairly stiff spring), but the compression is small (5 cm). The x^2 factor means that even a small compression stores nontrivial energy if k is large enough.

Step 3. Apply conservation of energy.

When the spring returns to its natural length, all elastic PE has been converted to KE of the dart (neglecting friction and the mass of the spring):

\frac{1}{2}kx^2 = \frac{1}{2}mv^2
1 = \frac{1}{2} \times 0.2 \times v^2
v^2 = \frac{1 \times 2}{0.2} = 10
v = \sqrt{10} \approx 3.16 \text{ m/s}

Why: set the initial elastic PE equal to the final KE. The spring transfers all its stored energy to the dart. Notice that the \frac{1}{2} appears on both sides — it cancels conceptually, but you need to keep it in the algebra to get the right numbers.

Result: Elastic PE stored = 1 J. Speed of the dart = \sqrt{10} \approx 3.16 m/s (about 11.4 km/h).

What this shows: A stiff spring (k = 800 N/m) compressed by just 5 cm stores enough energy to launch a 200 g dart at over 3 m/s. This is why spring-powered toy guns work — the spring stores energy during compression and releases it all at once. Double the compression to 10 cm and the PE quadruples to 4 J (because x^2), doubling the dart's speed.

Common confusions

If you came here to understand what potential energy is, derive the formulas, and solve problems, you have everything you need. What follows is for readers preparing for JEE Advanced who want to see the general framework and the connection between PE and conservative forces.

The general definition: PE from any conservative force

The gravitational and elastic PE formulas are specific instances of a general definition. For any conservative force \vec{F}, the potential energy function U(\vec{r}) is defined by:

U(\vec{r}) - U(\vec{r}_0) = -\int_{\vec{r}_0}^{\vec{r}} \vec{F} \cdot d\vec{r}

where \vec{r}_0 is the reference point where U = 0.

Why this works: a conservative force has the property that the work it does depends only on the endpoints, not the path. This means the integral above gives the same result regardless of which path you take from \vec{r}_0 to \vec{r}. That path-independence is what makes U a well-defined function of position.

For gravity near the surface: \vec{F} = -mg\hat{y}, integrate from y_0 = 0 to y = h, and you get U = mgh.

For a spring: F = -kx, integrate from x_0 = 0 to x, and you get U = \frac{1}{2}kx^2.

For Newtonian gravity at arbitrary distances: \vec{F} = -\frac{GMm}{r^2}\hat{r}, integrate from r_0 = \infty to r, and you get:

U(r) = -\frac{GMm}{r}

Why the negative sign and why the reference at infinity: setting U = 0 at r = \infty is the standard convention for gravitational problems. As the object falls from infinity toward Earth, gravity does positive work, and PE decreases from 0 toward -\infty. The total energy of a bound orbit is negative — meaning you would need to add energy to free the satellite.

The force–PE connection: F = -dU/dx

There is a powerful inverse relationship: if you know the potential energy as a function of position, you can find the force without doing any new physics:

F(x) = -\frac{dU}{dx}

Why: this is the fundamental theorem of calculus applied to the definition \Delta U = -\int F \, dx. Differentiating both sides with respect to x gives dU/dx = -F, or F = -dU/dx.

Check it against the two formulas you know:

  • Gravitational: U = mgy, so F = -d(mgy)/dy = -mg. Correct — gravity points downward.
  • Elastic: U = \frac{1}{2}kx^2, so F = -d(\frac{1}{2}kx^2)/dx = -kx. Correct — the spring force is -kx.

This relationship is the foundation of energy diagrams, where you plot U(x) and read off the force from the slope. Minima of U correspond to stable equilibrium; maxima to unstable equilibrium; and inflection points to neutral equilibrium.

Why non-conservative forces don't have PE

Friction does work that depends on the path — slide a book in a straight line across a table and friction does less work than if you push it in a zigzag covering the same start-to-end distance. Because the work depends on the path, you cannot define a potential energy function for friction. The integral -\int \vec{F} \cdot d\vec{r} gives different answers for different paths, so U would not be a unique function of position.

This is precisely why energy is "lost" to friction — it cannot be stored as PE. The kinetic energy that friction removes goes into thermal energy (heat), which is not recoverable as mechanical energy in the same way PE is. Conservation of mechanical energy (\text{KE} + \text{PE} = \text{constant}) holds only when all forces are conservative. When friction is present, the total energy is still conserved (first law of thermodynamics), but mechanical energy decreases.

Where this leads next