Potential energy

A drawn bow, a boulder perched on a cliff edge, a book resting on a high shelf — none of these are moving, yet each is loaded with a capacity to do work the instant it is released. Let the string go and the arrow leaps forward; nudge the boulder and it thunders downhill; knock the book off and it falls, gathering speed. Where was that "stored" ability to do work hiding before anything moved? Physics answers with a single idea: potential energy, the energy a system holds because of the relative positions of its parts, not because anything is in motion. This article builds that idea from the work-energy theorem you already know and pins it down with algebra you already own. We ask exactly which forces are allowed to have a potential energy at all — only the well-behaved, path-independent conservative forces, never friction — and derive the defining relation $W = -\Delta U$ that links work done to energy stored. From that one relation the two potential-energy formulas the CBSE Class 11 syllabus actually tests fall out directly: gravitational PE near Earth's surface, $U = mgh$ , and spring PE, $U = \tfrac12 kx^2$ . Along the way we settle the question every student eventually asks — if potential energy needs a reference level, is it even real? — and connect PE to its constant partner, kinetic energy, through the conservation law that makes falling objects, swinging pendulums and stretched springs all solvable with the same handful of lines. By the end you should be able to look at any configuration — a raised mass, a compressed spring, a stretched bowstring — and immediately know whether it has a definable potential energy, write down its formula, and use $W=-\Delta U$ to move fluently between work, potential energy and kinetic energy in board-level and competitive problems alike.

Key takeaways

Potential energy $U$ is energy stored due to the position or configuration of a system, not its motion — height above the ground, or a spring's stretch/compression.
Only conservative forces get a potential energy — gravity and the spring force qualify; friction does not, because its work depends on the path taken.
Defining relation: $W = -\Delta U$ — the work done by a conservative force equals the negative of the change in potential energy. Positive work by gravity means $U$ goes down.
Gravitational PE near Earth: $U = mgh$ , valid when $h$ is small enough that $g$ is essentially constant.
Spring PE: $U = \tfrac12 kx^2$ , where $x$ is the extension or compression from the natural length — always positive, whichever way the spring is deformed.
Only changes in $U$ matter. The reference level ( $h=0$ or $x=0$ ) is a free choice; physics (accelerations, speeds, forces) never depends on it.
Key idea: potential energy is position-stored energy, defined only for conservative forces through W = -ΔU, giving U = mgh for gravity and U = ½kx² for a spring, with the reference level chosen freely because only ΔU has physical meaning.

The core insight

For a conservative force the work done equals the negative change in potential energy, $W=-\Delta U$ , which gives $U=mgh$ for gravity near Earth and $U=\tfrac12kx^2$ for a spring, with only changes in $U$ — never an absolute value — carrying physical meaning.

Where it fits in the syllabus

Potential energy sits in the CBSE Class 11 unit Work, Energy and Power, placed right after kinetic energy and the work-energy theorem and just before conservation of mechanical energy, which is built directly on top of it. The board syllabus expects the general definition of PE, the conservative-versus-non-conservative distinction, the relation $W=-\Delta U$ , and the two named formulas $U=mgh$ (gravity) and $U=\tfrac12kx^2$ (spring); a formal derivation of the spring formula from $F=-kx$ by integration is also within scope since the spring force is explicitly listed in the chapter. Gravitation's more general $U=-\dfrac{GMm}{r}$ belongs to a later chapter and is only mentioned here as a forward pointer, not tested in this unit.

The physical picture

Start from something you already trust: the work-energy theorem, $W_{net}=\Delta K$ . Now separate the total work into work done by conservative forces ( $W_c$ ) and everything else. For a conservative force, work done moving between two points does not depend on the path — only on where you start and end — and the work done around any closed loop is exactly zero. That path-independence is precisely what lets us assign a number, potential energy $U$ , to each position, so that the work done in going from one point to another is simply the drop in that number:

W_c = -\Delta U = U_i - U_f

where: $W_c$ is the work done by the conservative force, $U_i$ and $U_f$ are the potential energies at the start and end points. Read it physically: when a conservative force does positive work (gravity pulling something down, a stretched spring snapping back), the configuration's stored energy falls by exactly that amount — the energy has changed form, from potential to kinetic, not vanished. Friction fails this test. Push a block across a table by a short direct path and by a long winding path of the same start and end points, and friction does more negative work on the longer path — the work depends on the path, not just the endpoints. So friction cannot be given a potential energy; its lost mechanical energy simply becomes heat, tracked separately, never "stored" for later recovery. Applying $W_c=-\Delta U$ to gravity near Earth's surface, where the force $mg$ is essentially constant over ordinary heights, gives the familiar

U = mgh

where: $m$ is mass, $g\approx 9.8\ \text{m/s}^2$ , and $h$ is height measured from a chosen reference level ( $h=0$ can be the floor, the table, or anywhere convenient). Applying the same idea to a spring, where the restoring force is $F=-kx$ (Hooke's law, $k$ the spring/force constant, $x$ the displacement from the natural length), gives

U = \tfrac12 kx^2

where: $x$ is the extension or compression from natural length. Because $x$ is squared, $U$ is positive whether the spring is stretched or compressed — both states have work "loaded into" the spring, ready to be released.

Figure

A hill and a spring side by side showing potential energy converting to kinetic energy, a ball rolling down from height h gaining speed while a compressed spring of displacement x pushes a block away, both governed by the same relation W equals minus delta U

Deriving the spring potential energy

Derivation APotential energy stored in a stretched or compressed spring

The goal is to find the work needed to stretch (or compress) an ideal spring from its natural length to a displacement $x$ , and hence the potential energy stored, starting only from Hooke's law $F=-kx$ for the restoring force and the definition of work done by a variable force as the area under an $F$ – $x$ graph — no result beyond algebra and the geometry of a triangle is assumed.

Step 1. Write the applied force needed to stretch the spring quasi-statically (slowly, so kinetic energy stays negligible). To hold the spring at extension $x$ against its own restoring force $F_{\text{spring}}=-kx$ , an external agent must apply $F_{\text{ext}}=+kx$ — equal and opposite to the spring's pull, growing in direct proportion to $x$ .

Step 2. Because $F_{\text{ext}}$ is not constant, the work done in stretching the spring from $0$ to $x$ is the area under the $F_{\text{ext}}$ -versus- $x$ graph, not simply force times distance. That graph is a straight line through the origin with slope $k$ , so the area from $0$ to $x$ is a right triangle of base $x$ and height $kx$ .

Step 3. Compute the triangle's area: $W = \tfrac12 \times \text{base} \times \text{height} = \tfrac12 \times x \times kx = \tfrac12 kx^2$ . This is the work done by the external agent on the spring, and since the spring started and ended at rest, all of it is stored — none went into kinetic energy.

Step 4. Identify this stored work as the potential energy. By definition, the work done on the system in a slow, reversible change equals the increase in its potential energy, so

U(x) = \tfrac12 kx^2

Notice $x$ appears squared, so compressing the spring by $x$ stores exactly as much energy as stretching it by the same $x$ — the sign of the displacement does not matter, only its magnitude.

Worked examples

Board level — lifting a book onto a shelf

A book of mass $1.2\ \text{kg}$ is lifted from a table to a shelf $1.5\ \text{m}$ above the table top. Find the increase in its gravitational potential energy, taking the table top as the reference level. Use $g=9.8\ \text{m/s}^2$ .

Step 1. Identify the quantities: $m=1.2\ \text{kg}$ , $h=1.5\ \text{m}$ , reference $h=0$ at the table.

Step 2. Apply $U=mgh$ : $U = 1.2 \times 9.8 \times 1.5$ .

Step 3. Compute: $U = 17.64\ \text{J}$ . Answer: the book's potential energy increases by $17.64\ \text{J}$ relative to the table, equal to the (slow-lift) work done against gravity, assuming the book starts and ends essentially at rest.

Competitive level — a falling ball, then a spring catch at the bottom

A ball of mass $0.5\ \text{kg}$ is released from rest $10\ \text{m}$ above the ground and falls freely. (a) Find the work done by gravity as it falls from $10\ \text{m}$ to $4\ \text{m}$ above the ground, using $W=-\Delta U$ . (b) The ball then lands on a vertical spring of force constant $k=2000\ \text{N/m}$ fixed to the ground; using energy conservation, estimate the maximum compression $x$ of the spring, ignoring air resistance and treating the ball's height loss during compression as approximately $x$ itself (take $g=9.8\ \text{m/s}^2$ ).

Step 1. Potential energies relative to the ground for part (a): $U_i = mgh_i = 0.5\times9.8\times10=49\ \text{J}$ , $U_f=mgh_f=0.5\times9.8\times4=19.6\ \text{J}$ , so $\Delta U = 19.6-49=-29.4\ \text{J}$ .

Step 2. Apply $W=-\Delta U$ : $W = -(-29.4) = 29.4\ \text{J}$ . Gravity does $29.4\ \text{J}$ of positive work as the ball falls from $10\ \text{m}$ to $4\ \text{m}$ , matching the kinetic energy gained over that stretch.

Step 3. For part (b), treat the whole fall of $10\ \text{m}$ (plus the small extra drop $x$ during compression, folded in as part of the total height) as converting entirely into spring PE at maximum compression, when the ball is momentarily at rest: $mg(10+x) = \tfrac12 kx^2$ . Since $x$ will be small compared with $10\ \text{m}$ , first approximate $mg(10) \approx \tfrac12 kx^2$ : $0.5\times9.8\times10 = 49\ \text{J} = \tfrac12 \times 2000 \times x^2 = 1000x^2$ .

Step 4. Solve: $x^2 = 49/1000 = 0.049$ , so $x \approx 0.221\ \text{m}$ . Answer: the ball falls with gravity doing $29.4\ \text{J}$ of work over the $10\ \text{m}\to4\ \text{m}$ stretch, and the spring compresses by roughly $22\ \text{cm}$ at the moment the ball momentarily stops. Timing note: under three minutes. Part (a) is a direct two-line substitution into $W=-\Delta U$ ; part (b) is the same idea run in reverse — total mechanical energy converts entirely to spring PE at the instant of maximum compression (zero kinetic energy there). The one shortcut worth remembering: since $x\ll10\ \text{m}$ here, dropping the extra $mgx$ term on the left keeps the algebra to a single quadratic in $x^2$ rather than a full quadratic in $x$ — always check afterward that the neglected term really is small, which $0.221\ \text{m}$ against $10\ \text{m}$ confirms.

Common traps

Trying to assign a potential energy to friction or any non-conservative force — PE is defined only for conservative forces; friction's effect is tracked as work converted to heat, never "stored."
Dropping the minus sign in $W=-\Delta U$ — writing $W=\Delta U$ reverses the physics: positive work by gravity must decrease $U$ , not increase it.
Treating $U$ as an absolute quantity with one fixed value — only changes in $U$ are physically meaningful; the zero of height (or of spring extension) is a free choice that never affects forces, accelerations or speeds.
Forgetting that spring PE depends on $x^2$ , so a compression and an equal-magnitude extension store identical energy — the sign of $x$ does not matter, only the squared magnitude.
Assuming "potential energy" always means gravitational PE — the concept is general; each conservative force gets its own formula (gravity: $mgh$ ; spring: $\tfrac12kx^2$ ).

Common trapPotential energy belongs to the conservative force, not to the object alone — and it always needs a stated reference

Two mistakes account for most of the marks lost on this topic. First, students often speak of "the potential energy of the ball" as though it were an intrinsic property of the ball, forgetting it is really a property of the system (ball plus Earth, or block plus spring) associated with a specific conservative force and a chosen zero. Second, and more damaging in calculations, is silently dropping the sign in $W=-\Delta U$ or mixing reference levels mid-problem — for instance taking $h=0$ at the ground for the initial position but at the table for the final one. Always fix one reference level for the whole problem and state it explicitly; the physically meaningful quantity is always a difference, $\Delta U$ , never a bare value of $U$ .

You should be able to…

State why potential energy can be defined only for conservative forces, and explain why friction is excluded.
Write and apply $W=-\Delta U$ to relate work done by a conservative force to the change in potential energy, with the correct sign.
Derive $U=\tfrac12kx^2$ for a spring from Hooke's law and the area-under-the-graph definition of work.
Apply $U=mgh$ for gravitational PE near Earth's surface, choosing and stating a consistent reference level.
Combine gravitational and spring potential energy in a single energy-conservation problem (falling object landing on a spring).
Explain why only changes in potential energy, never its absolute value, carry physical meaning.

Ask the tutor: Want to see why the reference level for h truly cannot change any physical answer — try re-solving the book example with the floor as h = 0 instead of the table, and ask the AI tutor to check your working.

Why can't friction have a potential energy of its own?

Potential energy only makes sense for a force whose work between two points does not depend on the path taken — that is what lets us assign a single number to each position. Friction fails this immediately: push a block along a short straight path and friction removes a certain amount of energy as heat; push it the long way round between the same two points and friction removes more, because it acts over a longer distance. Since the "loss" depends on the route, there is no single well-defined $U$ you could assign to each position that friction's work would just equal the difference of. The energy friction removes is real, but it leaves the mechanical system as heat and cannot be recovered as motion later — so it is tracked separately as non-conservative work, not folded into a potential energy.

If I can choose the reference level for h freely, doesn't that make potential energy kind of made up?

The value of $U$ at any one point is indeed a free choice — that part is bookkeeping, not physics. But every physical prediction you can actually test — how fast something is moving after falling a certain height, how hard a spring pushes back, how much kinetic energy is gained or lost — depends only on the change $\Delta U$ between two points, and that change comes out identical no matter which reference you picked, because both $U_i$ and $U_f$ shift by the same constant. Try the book-lifting example with the reference at the floor instead of the table top: the individual $U$ values change, but $\Delta U = 17.64\ \text{J}$ stays exactly the same. That invariance is precisely what makes the freedom to choose $h=0$ harmless rather than arbitrary in a worrying sense.

Is the U = mgh formula always correct, or only sometimes?

$U=mgh$ is an approximation that holds when the height change $h$ is small enough that the gravitational force $mg$ can be treated as constant over that range — true for essentially every Class 11 mechanics problem, from a dropped ball to a book on a shelf. The fully general gravitational potential energy, valid for large distances (satellites, planetary motion), is $U=-\dfrac{GMm}{r}$ , which reduces to $mgh$ (plus a constant) only for small $h$ near Earth's surface. That general form belongs to the Gravitation chapter later in the syllabus — for this chapter, $U=mgh$ is exactly the tool you are expected to use.

Why does spring potential energy use one-half k x squared instead of just k x, like the force does?

Because the spring's restoring force $F=kx$ is not constant as you stretch it — it grows as $x$ grows, starting at zero when the spring is unstretched. Work done by a varying force is the area under the force-versus-displacement graph, not simply force times distance, and for a straight line through the origin with slope $k$ , that area up to displacement $x$ is the area of a triangle: half the base times the height, giving $\tfrac12 \times x \times (kx) = \tfrac12 kx^2$ . If you mistakenly used $F\times x = kx\times x = kx^2$ , you would be assuming the force was $kx$ throughout the whole stretch, which overcounts the work by exactly a factor of two — the derivation above shows precisely where that factor of one-half comes from.