Exploring Mechanics / 3. Energy
Energy methods often solve a problem more directly than force-based methods.
Make a prediction first: you carry a heavy bag at a perfectly constant height across a flat room. Your arm is clearly exerting a force the whole time. Does physics count that as "doing work" on the bag?
W = F d cosθ · Net work = ΔKE (the Work-Energy Theorem)
Work only counts the part of a force that acts along the direction of motion. Carrying that bag at constant height, your upward supporting force is perpendicular to your (horizontal) walking displacement — cosθ = cos90° = 0, so despite feeling tired, you do exactly zero work on the bag. Lift it straight up instead and force and displacement point the same way — positive work. Lower it slowly against your own resistance and they point opposite ways — negative work.
W = F d cosθ is exactly the dot product F⃗·d⃗ from the Toolkit's vectors tab, applied to force and displacement specifically. That's also why work is a scalar (a single signed number, no direction of its own) even though both of its ingredients are vectors — the dot product strips away everything except "how much of one vector points along the other."
For a force that stays constant, work is just base × height — the rectangular area under a flat force-distance line. Switch to a spring and the force grows as you stretch it further (F = kx), so the area becomes a triangle instead of a rectangle, and the work formula changes shape right along with it.
A triangle's area is ½ × base × height. Here the base is the stretch x and the height is the force at that final point, kx — so work stretching a spring from 0 to x is W = ½x(kx) = ½kx². The factor of ½ isn't arbitrary; it's simply what "area of a triangle" requires once the force itself is changing linearly instead of staying constant.
Pull a box across a rough floor. Your applied force does positive work; friction does negative work (and turns that energy into heat). The net work — whatever's left after friction takes its share — is exactly how much kinetic energy the box gains.
Rotate the angle between the force and the direction of motion and watch the sign of the work flip right at 90°.
Make a prediction first: a car crashes at 60 km/h instead of 30 km/h — twice the speed. Is the energy that has to go somewhere in that crash twice as much, or something else entirely?
KE = ½mv²
Kinetic energy grows with the square of speed, not speed itself. Double the speed and you don't double the kinetic energy — you quadruple it. That's the real answer to the prediction above: a crash at 60 km/h carries four times the energy of one at 30 km/h, which is exactly why speed limits matter so much more than they'd seem to at first glance.
Start an object from rest and apply a constant force F over a distance d. The work-energy theorem says that work becomes its kinetic energy: KE = Fd = (ma)d. From kinematics, v² = 2ad, so ad = v²/2. Substituting: KE = m(v²/2) = ½mv². The formula isn't a separate law — it falls straight out of F = ma and the definition of work.
Drag the speed slider and watch the marker race up the curve far faster than the slider moves — that steepening curve is the v² relationship.
A moving object strikes a stationary one. Toggle between a perfectly-inelastic collision (they stick together) and an elastic one (they bounce apart) and watch how much kinetic energy survives — the full momentum-based derivation behind these numbers is coming in Momentum → Collisions; here, the focus is purely on where the energy goes.
"Elastic" is defined as a collision where kinetic energy is conserved — it's not a coincidence, it's the definition. Real collisions (car crashes, a dropped ball of clay) are almost always somewhere between perfectly elastic and perfectly inelastic; the "lost" energy was never destroyed, it converted into heat, sound, and permanent deformation of the colliding objects.
Make a prediction first: a ball rolls back and forth in a smooth, frictionless bowl. Does it ever rise higher than the height it was released from? Does it ever stop?
Total mechanical energy = KE + PE = constant, as long as no dissipative force (like friction or air resistance) is removing energy from the system.
The ball in the prediction above trades kinetic energy for gravitational potential energy and back, forever, in an idealized frictionless bowl — it always returns to exactly its starting height, never higher (that would create energy from nothing) and never lower forever (nothing is removing energy). Add real friction and the bowl slowly "wins," converting mechanical energy into heat until the ball settles at the bottom.
Release the ball from a chosen height and watch the three bars — kinetic, potential, and (if friction is on) thermal — trade height while their total stays fixed.
Compress a spring and release a ball up a frictionless ramp. All of the spring's stored elastic energy becomes kinetic energy the moment it leaves the spring, then converts again into gravitational potential energy as the ball coasts uphill.
This is a three-step conversion: elastic PE (½kx²) → all kinetic energy (½mv²) the instant the ball leaves the spring → all gravitational PE (mgh) at the highest point, where it's momentarily at rest again. Since energy is conserved throughout, all three quantities are equal: ½kx² = ½mv² = mgh, which is exactly how v and h above were solved for without ever tracking the motion in between.
A pendulum released from an angle swaps potential energy for kinetic energy on the way down, reaching its fastest point exactly at the bottom of the swing.
Make a prediction first: a braking car's speed drops from 25 m/s to 0. Its kinetic energy clearly disappears — so where, specifically, did it go?
In an isolated system (one with no energy crossing its boundary), total energy is exactly conserved — it only ever changes form. "Lost" energy is never actually destroyed; it's converted into a form that's easy to overlook, usually heat, sound, or permanent deformation.
The braking car's kinetic energy becomes heat in the brake pads and tires (and a little sound) — track your hand along a car's wheel well right after a hard stop and you can feel it directly. Widget 1 tracks exactly this kind of accounting, second by second, for two different scenarios.
Pick a scenario and watch three running totals — kinetic, potential, and heat — evolve over time. However they individually change, their sum never does.
The heat curve isn't computed from some new law — it's just whatever's missing: Heat(t) = E₀ − KE(t) − PE(t). That's the entire content of "energy is conserved": once you know how kinetic and potential energy changed, the heat term is forced, not independently measured. It's the same accounting trick used to find rope tension in the last section — solve for everything else, and the remaining unknown is whatever balances the books.
Make a prediction first: two people each climb the same flight of stairs, lifting the same body weight the same height. One sprints up in 5 seconds; the other strolls up in 30 seconds. Did they do the same amount of work? Did they produce the same power?
P = W/t = Fv · Efficiency = Pout/Pin
Both stair-climbers in the prediction above do exactly the same work (same weight, same height gained) — work doesn't care about time at all. But the sprinter produces six times the power, because power is work divided by how long it took. This is the entire reason a small motor can still lift a heavy load (given enough time) while a large motor is needed to lift it quickly.
P = W/t works for a single event with a clear start and end. For continuous motion — a car cruising, a cyclist pedaling — it's often more useful to write P = Fv: since W = Fd and v = d/t, dividing both sides by t gives W/t = F(d/t) = Fv. Same physics, whichever form fits the situation better.
Estimate your own power output climbing a real staircase, and see how it stacks up against a lightbulb and a horsepower.
Cycling power has to fight three things at once: gravity on a climb, rolling resistance from the tires, and air drag — and drag alone grows with the cube of speed, which is why going twice as fast on the flat takes vastly more than twice the effort.
Air-drag force already grows with v² (more air hit per second and each hit is harder). Power is force times speed, so drag power is v² × v = v³. That cubic growth is why elite cyclists spend enormous effort shaving off small amounts of drag (aerodynamic helmets, tucked positions) — at high speed, drag dominates the whole power budget.
No real machine is 100% efficient — friction, heat, and vibration always claim some of the input power. A motor lifting a crane's load, a winch, or a bicycle's chain drive all follow the same accounting.
Make a prediction first: a coaster car crests the top of a hill. Do you feel heavier, lighter, or the same as standing still — and does that feeling have anything to do with your actual weight changing?
A coaster (ignoring track friction and air resistance) is a pure energy-conservation machine: ½mv² + mgh = constant along the whole ride. But what you feel isn't your speed or height — it's the track's normal force on you, your apparent weight, which can differ sharply from mg.
Drag the probe along the track. Speed depends only on how far below the very first hill the probe currently is — nothing else about the track's shape matters.
At the very top of the first hill the car is (approximately) at rest, so its total energy is entirely mgh1. At any later point, ½mv² + mgh = mgh1, so v² = 2g(h1 − h). For v² to stay non-negative, h can never exceed h1 — which is exactly why no later hill on a real coaster (without an extra motor) is ever built taller than the first one.
This is the part of coaster design with real safety stakes: taking a dip or crest too fast, at too tight a radius, can push apparent weight to levels that cause tunnel vision, blackout, or (in documented historical cases of poorly-designed rides) fatal injury. Real coasters are engineered to keep momentary G-forces within a well-studied safe envelope.
At a hilltop, both gravity and the track's normal force point toward the center of the curve (downward), so Newton's Second Law along that direction reads mg − N = mv²/r, giving N = mg − mv²/r — less than your ordinary weight, and it can reach exactly zero (true weightlessness, momentarily) or even go negative, meaning the track would need to actively hold you down (this is exactly a vertical-circle problem, developed fully in Circular Motion → Vertical Circles). At a dip, the normal force points up while gravity points down, both relative to a center of curvature that's now above the car, giving N − mg = mv²/r, so N = mg + mv²/r — always more than ordinary weight.
A car needs a minimum speed at the very top of a loop just to keep contact with the track at all (N ≥ 0) — which sets a minimum height the first hill must have, purely from energy conservation. The circular-motion side of this derivation belongs to Circular Motion → Vertical Circles; here, the focus is just the energy bookkeeping that gets the car there.
At minimum speed, gravity alone supplies the centripetal force at the loop's top: mg = mvtop²/r, so vtop² = gr. Energy conservation from the first hill (height h, at rest) to the loop's top (height 2r, speed vtop) gives mgh = mg(2r) + ½mvtop² = 2mgr + ½mgr, so h = 2.5r — the hill must be built two and a half times taller than the loop itself, regardless of the car's mass.
Make a prediction first: a full coaster profile — several hills of decreasing height, with real friction — is released from the very top. Does it necessarily make it all the way to the end, or could it get stuck partway?
This combines every idea from this section into one continuous ride: kinetic energy, potential energy, and (once friction is turned up) heat, tracked bar by bar as the car actually moves along the whole track — not just at one isolated point.
Watch a car released from the highest hill. Add friction to see mechanical energy turn into heat — and find where the car stalls.