Potential Energy: Gravitational & Elastic

The world's biggest batteries are still buckets of $mgh$. Pumped-storage hydroelectric plants — about 95% of all grid storage on Earth — pump water uphill when electricity is cheap and let it fall through turbines at peak demand; a single large plant banks around $10^{13}$ J, the $mgh$ of a small lake raised a few hundred metres. Gravity batteries hoisting concrete blocks in disused mine shafts run the same equation at startup scale. Meanwhile $\tfrac12 kx^2$ earns its living everywhere metal bends: vehicle suspension and crash structures, the power springs of watches, archery's compound bows (engineered force–draw curves storing over 100 J), and prosthetic running blades, which bank a sprinter's landing energy and refund it into the next stride — a spring pretending, briefly, to be a tendon.

Kinetic energy is a current account: visible, spendable, riding with the motion. Potential energy is the fixed deposit. Lift a brick to a shelf and the work you did has not vanished — it sits there, silently, for a year if need be, and the day the brick slips, gravity refunds every joule as speed. Draw a bow and your muscles' work waits in the bent limbs; release, and the arrow collects. The two storage devices of this module — height and deformation — are the only ones the mechanics syllabus needs, and they share one defining habit: the stored amount depends only on the configuration now, not on how it was reached. Haul the brick up a ladder, a ramp, or a wild spiral staircase: same shelf, same $mgh$, same refund.

One habit of language keeps you honest for the rest of physics: the energy is stored in the system, not in the object. "The ball's gravitational potential energy" is convenient shorthand, but the storage device is the ball–Earth pair — remove the Earth and no amount of height means anything. Likewise the spring's energy lives in the spring's stretched bonds, not in the block attached to it. Exams rarely punish the shorthand, but conceptual questions ("where is the energy stored?") pay the precise answer. And a warning made friendly early: the zero is yours. Measuring heights from the floor, the tabletop, or the ceiling changes every $U$ value on the page and changes no physics whatsoever, because every prediction routes through $\Delta U$. Choose the zero once, write it down, never move it mid-problem.

The deposit rule defines the account. When you lift a body at constant speed, your lifting force does work $+mgh$ while gravity does $-mgh$; the deposit is what gravity will later refund, so we define the change in gravitational potential energy as the negative of gravity's work:

where: $W_{\rm grav}$ — gravity's work over the move. Read the sign dial both ways: gravity works negatively on a rising ball (account filling, $\Delta U > 0$); positively on a falling one (account draining, $\Delta U < 0$). This single line is the master convention — every potential energy in physics, from springs to atoms, is defined by the same "minus the storing force's work." Since gravity near the surface bills by height alone ($W_{\rm grav} = -mg\,\Delta h$, the ramp theorem), the account balance is

where: $m$ — mass (kg); $g$ — $9.8$ m/s² (CBSE/ISC) or $10$ (JEE/NEET unless stated); $h$ — vertical height above your chosen zero, in metres, sign included (a cellar sits at negative $h$ and negative $U$). Read it physically: 2 kg raised 3 m banks about 60 J — and banks the same 60 J whether it went straight up or around a helical ramp, because gravity never billed the horizontal wandering. The spring's account comes straight from the variable-force module's triangle. Deforming a spring by $x$ from natural length takes work equal to the area under $F = kx$:

where: $k$ — spring constant (N/m); $x$ — deformation measured from natural length, never from an equilibrium-under-load position, never the spring's total length. The square makes compression and extension identical depositors, and it makes the account grow quadratically: between two stretches the stored energy changes by $\tfrac12 k(x_2^2 - x_1^2)$ — the difference of squares, big triangle minus small, never $\tfrac12 k(x_2 - x_1)^2$.

1 · Board level — the water-tank invoice (CBSE/ICSE pattern, 3 marks)

A pump lifts $500$ kg of water from a well $20$ m deep to a rooftop tank $10$ m above the ground. Take the ground as the zero and $g = 10$ m/s². Find (a) the water's initial and final potential energies, (b) the minimum work the pump must do.

Step 1. (a) Initial: the water sits $20$ m below the zero: $U_i = mgh_i = 500 \times 10 \times (-20) = \boxed{-1.0 \times 10^5\ \mathrm{J}}$. Final: $U_f = 500 \times 10 \times 10 = \boxed{+5.0 \times 10^4\ \mathrm{J}}$.

Step 2. (b) Minimum work (no speed gained, no friction losses) is the deposit: $W = \Delta U = U_f - U_i = 5.0\times 10^4 - (-1.0\times10^5) = \boxed{1.5 \times 10^5\ \mathrm{J}}$.

Step 3. Audit with the total lift: the water rose $30$ m in all, and $mg \times 30 = 1.5\times10^5$ J ✓ — the negative initial balance was doing honest work, not decoration.

2 · Board level — the pre-stretched spring (ICSE/ISC pattern)

A spring of constant $k = 400$ N/m is already stretched $5.0$ cm. How much additional work is needed to stretch it to $10.0$ cm? Compare with the work that took it from natural length to $5.0$ cm.

Step 1. First leg, 0 → 5 cm: $U_1 = \tfrac12 \times 400 \times (0.05)^2 = \boxed{0.50\ \mathrm{J}}$.

Step 2. Second leg, 5 → 10 cm — difference of squares, never the square of the difference: $\Delta U = \tfrac12 \times 400 \times \left(0.10^2 - 0.05^2\right) = 200 \times 0.0075 = \boxed{1.5\ \mathrm{J}}$.

Step 3. Compare: the second 5 cm costs $3\times$ the first — the quadratic account in action. (The wrong formula $\tfrac12 k(0.05)^2$ returns 0.50 J again and forfeits the comparison mark.)

3 · Competitive level — the movable zero, audited (JEE-Main pattern, ~2.5 minutes)

A $0.50$ kg ball is thrown from a $20$ m rooftop and lands on the ground. Student A puts the zero at the ground; student B at the rooftop. For EACH student, write the ball's initial and final potential energies and the change $\Delta U$. ($g = 10$ m/s².)

Step 1. Student A (zero at ground): $U_i = 0.5 \times 10 \times 20 = +100$ J; $U_f = 0$. Change: $\Delta U = \boxed{-100\ \mathrm{J}}$.

Step 2. Student B (zero at rooftop): $U_i = 0$; $U_f = 0.5 \times 10 \times (-20) = -100$ J. Change: $\Delta U = \boxed{-100\ \mathrm{J}}$.

Step 3. The balances disagree everywhere; the change agrees exactly — and the change is all that gravity's refund (the ball's kinetic-energy gain) ever reads. Both students predict the same landing speed.

Step 4. MCQ armour: any option that lets a physical outcome (speed, height reached, tension) depend on where the zero sits is wrong by structure. Timing note: ~2.5 min, almost all of it careful sign work. Write $h$ WITH its sign relative to your zero and the table fills itself. The designed error is giving B's $\Delta U$ as $+100$ J by subtracting in the wrong order — it is always final minus initial.

4 · Competitive level — same force versus same stretch (JEE-Main/NEET conceptual, ~3 minutes)

Two springs have constants $k_1 = 100$ N/m and $k_2 = 400$ N/m. (a) Each is stretched by the same force $F = 20$ N. Which stores more energy, and how much each? (b) Each is stretched by the same extension $x = 10$ cm. Which stores more now?

Step 1. (a) Same force: each spring settles at $x = F/k$, so $x_1 = 0.20$ m, $x_2 = 0.05$ m. Energy: $U = \tfrac12 kx^2 = \tfrac12 k (F/k)^2 = \dfrac{F^2}{2k}$ — inverse in $k$. $U_1 = \dfrac{400}{200} = \boxed{2.0\ \mathrm{J}}$, $U_2 = \dfrac{400}{800} = \boxed{0.5\ \mathrm{J}}$. The soft spring stores more: it yields further, and distance is where work lives.

Step 2. (b) Same stretch: $U = \tfrac12 kx^2$ directly — linear in $k$. $U_1 = \tfrac12 \times 100 \times 0.01 = \boxed{0.5\ \mathrm{J}}$, $U_2 = \tfrac12 \times 400 \times 0.01 = \boxed{2.0\ \mathrm{J}}$. Now the stiff spring wins.

Step 3. The inversion is the entire question, and it is pure formula hygiene: at fixed $F$ use $F^2/2k$; at fixed $x$ use $\tfrac12 kx^2$. Identify which variable the problem holds equal before comparing. Timing note: ~3 min. The NEET phrasing is often "work done in stretching" rather than "energy stored" — same number. Options bait the unconverted with the other case's winner; note the elegant symmetry that the winners swap and the winning margins ($4\times$) match.

Path length for height. $h$ in $mgh$ is the vertical rise, never the ramp length. A 5 m ramp to a 3 m shelf banks $mg \times 3$. If an incline appears, multiply by $\sin\theta$ once — and only once.

Stretch measured from the wrong place. The $x$ in $\tfrac12 kx^2$ runs from natural length. A spring hanging with a load at equilibrium is already deformed; further motion questions must track total deformation from natural length, not from the loaded rest point. (When energy methods and this trap collide, the mass-dropped-on-spring archetype settles it — ahead in this unit.)

The difference-of-squares evasion. Between stretches: $\tfrac12 k(x_2^2 - x_1^2)$. Writing $\tfrac12 k(x_2-x_1)^2$ undercharges — it forgets the spring was already fighting. Big triangle minus small triangle, every time.

Moving the zero mid-problem. Choose the reference level once, declare it, and keep it. The classic self-sabotage: measuring the initial $U$ from the ground and the final $U$ from the table. All balances, one zero.

Signs into $\Delta U = -W_{\rm grav}$. Gravity does positive work on the way down and the account drains. If your rising ball shows decreasing $U$ or your falling ball shows increasing $U$, a sign died. Cross-check with common sense: higher = more banked, always.

"Energy stored in the mass." For conceptual/assertion-reason questions: gravitational PE belongs to the ball–Earth system; spring PE lives in the spring. Shorthand is fine in calculations; the concept question wants the system answer.

mgh beyond its passport. $U = mgh$ assumes uniform $g$ — fine for buildings, wells, mountains; wrong for satellites and escape problems, where the gravitation chapter's $-GMm/r$ takes over. If the height rivals Earth's radius, $mgh$ has expired.

Which $g$. CBSE/ISC: $9.8$ m/s² unless told; JEE-Main/NEET: almost always $10$. State your choice.

Texts. NCERT Physics Class 11, Work, Energy and Power chapter (potential energy of a body and of a spring) · HC Verma, Concepts of Physics Vol. 1, Ch. 8 · Halliday–Resnick–Walker Ch. 8 (the potential-energy treatment this module downshifts).

Provenance. All results are standard. A verifier should recompute: the beam's 30 kJ ↔ 17 m/s round trip; the ramp check (60 J both routes); the spring ladder 0.04 / 0.12 / 0.16 J; the water tank ($-10^5$, $+5\times10^4$, $1.5\times10^5$ J); the rooftop double audit ($\Delta U = -100$ J for both zeros); the spring pair ($2.0$ vs $0.5$ J swapping winners between fixed-$F$ and fixed-$x$).

Prerequisites. Work Done by a Force · Work by a Variable Force & F–x Graphs · Spring Force & Hooke's Law. Leads to. Conservative & Non-Conservative Forces · Conservation of Mechanical Energy & Efficiency · ARCHETYPE: Block Meets Spring · Conservative Forces & Potential Energy (UG deep dive).