Mechanics

Work and Energy

Force times distance — and the theorem that says work done equals change in kinetic energy

Work is the energy transferred when a force moves an object — W = F·d (force times distance, dot product). The work-energy theorem says the net work done on an object equals its change in kinetic energy — W_net = ΔKE = ½mv² − ½mv₀². Together with conservation of energy, these are the most powerful tools for solving mechanics problems without integrating forces over time.

  • WorkW = F·d cos θ (scalar; dot product)
  • Units1 J = 1 N·m = 1 kg·m²/s²
  • Work-energy theoremW_net = ΔKE = ½mv² − ½mv₀²
  • Kinetic energyKE = ½mv²
  • PowerP = dW/dt = F·v
  • SignPositive (force aids motion), negative (opposes)

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Work

Work is the energy transferred to an object by a force acting over a distance.

W = F · d = |F| · |d| · cos θ

where θ is the angle between the force and displacement vectors. The dot product means only the component of force ALONG the motion does work.

Units: 1 joule (J) = 1 newton-meter (N·m) = 1 kg·m²/s².

Work can be:

  • Positive when force aids motion (F and d in same direction; θ < 90°).
  • Negative when force opposes motion (θ > 90°). Friction does negative work.
  • Zero when force is perpendicular to motion (θ = 90°). Centripetal forces, normal forces.

Kinetic energy

Kinetic energy is the energy of motion:

KE = ½mv²

It's a scalar — direction doesn't matter. Always non-negative. A 1 kg object at 1 m/s has 0.5 J of KE; at 10 m/s, 50 J (100×). Doubling speed quadruples KE.

ObjectMassSpeedKE
Pitched fastball0.145 kg40 m/s116 J
Cyclist80 kg10 m/s4,000 J
Car at 60 mph1500 kg27 m/s~547,000 J
Bullet (5.56 mm)0.004 kg1000 m/s2,000 J
Asteroid (Chicxulub)~10¹⁵ kg~20 km/s~2 × 10²³ J
Earth orbiting Sun6 × 10²⁴ kg30 km/s~2.7 × 10³³ J

The work-energy theorem

The net work done on an object equals its change in kinetic energy.

W_net = ΔKE = ½mv² − ½mv₀²

This is one of the most useful equations in physics. It connects forces (left side) directly to speeds (right side), without needing time as an intermediary.

Derivation

Start with F = ma. Then:

W = ∫F·dx = ∫ma·dx = ∫m(dv/dt)·dx = ∫m·v·dv = ½mv² − ½mv₀²

(Using a·dx = (dv/dt)·dx = v·dv.)

Why it's powerful

  • Bypasses time entirely — only positions and forces matter.
  • Works for any path, not just straight lines.
  • Scalar equation — no vector decomposition.
  • Direct way to find speeds from energy considerations.

Worked examples

Block sliding down a frictionless ramp

A block of mass m on a frictionless ramp of height h, starting from rest. Find speed at the bottom.

Only gravity does work. W_gravity = mg·h (force is mg down, vertical drop is h). By work-energy:

mg·h = ½mv² − 0
v = √(2g·h)

This works for ANY ramp shape (loop, curve), as long as it's frictionless — work depends only on the height drop.

Stopping a car

A 1500 kg car going 30 m/s brakes to a stop with constant friction force F. The braking distance is d.

W = -F·d = ΔKE = 0 − ½(1500)(30)² = -675,000 J

If F = 6000 N (typical max for tire on dry asphalt), d = 675000/6000 = 112.5 m.

Doubling speed (60 m/s) quadruples stopping distance (450 m). Why brakes lose effectiveness at high speeds — KE grows as v².

Power

Power is the rate of doing work:

P = dW/dt

For constant force, P = F·v. Units: watts (W) = J/s. 1 hp ≈ 746 W.

SourcePower
Human (sustained)~75-100 W
Olympic cyclist (sprint)~2,000 W
Car engine (typical)~150,000 W (200 hp)
Saturn V rocket~150 GW
Sun (total output)~3.8 × 10²⁶ W
USA average power consumption~3,000 GW (electricity ~450 GW)

Air resistance scales as v², so power needed to overcome drag scales as v³. A car going 100 mph needs 8× the power as 50 mph just to fight drag — explains why fuel economy crashes at highway speeds.

JavaScript — work-energy calculations

// Work done by constant force over distance d at angle theta
function work(force, distance, angleRad = 0) {
  return force * distance * Math.cos(angleRad);
}

// Kinetic energy
function kineticEnergy(mass, velocity) {
  return 0.5 * mass * velocity * velocity;
}

// Work-energy theorem: solve for final velocity given net work
function finalVelocity(mass, vInitial, netWork) {
  const KEInitial = kineticEnergy(mass, vInitial);
  const KEFinal = KEInitial + netWork;
  if (KEFinal < 0) return null;  // not enough energy
  // KE = ½mv²  →  v = √(2KE/m)
  const v = Math.sqrt(2 * KEFinal / mass);
  return v * Math.sign(vInitial); // preserve direction
}

// Stopping distance: how far does a car go with friction force f?
function stoppingDistance(mass, vInitial, frictionForce) {
  const KEInitial = kineticEnergy(mass, vInitial);
  // KEInitial = f · d  →  d = KE/f
  return KEInitial / frictionForce;
}

console.log(stoppingDistance(1500, 30, 6000));  // 112.5 m (60 mph stop)
console.log(stoppingDistance(1500, 60, 6000));  // 450 m (120 mph stop, 4× distance)

// Power required to overcome drag at velocity v
function dragPower(dragCoefficient, area, density, v) {
  // Drag force = ½ρCd·A·v². Power = F·v
  const dragForce = 0.5 * density * dragCoefficient * area * v * v;
  return dragForce * v;
}

// Typical car: Cd ~0.3, A ~2.2 m², ρ_air ~1.2 kg/m³
console.log(`30 m/s drag power: ${dragPower(0.3, 2.2, 1.2, 30).toFixed(0)} W`);
console.log(`60 m/s drag power: ${dragPower(0.3, 2.2, 1.2, 60).toFixed(0)} W`); // ~8× more

// Tsiolkovsky rocket equation Δv from energy + momentum
function rocketDeltaV(mInitial, mFinal, vExhaust) {
  return vExhaust * Math.log(mInitial / mFinal);
}

Where work-energy shows up

  • Trajectory and speed problems. Anytime you need final speed but not detailed motion. Roller coasters, projectiles, energy ramps.
  • Spring problems. Springs store potential energy ½kx². Compress, release, calculate launch speed via energy conservation.
  • Pendulum analysis. Convert between KE and PE as pendulum swings; max speed at bottom from drop height.
  • Engineering — efficiency. Useful work / total energy input. Engines, motors, power generation.
  • Sports physics. Pole vault (KE → PE), baseball bat (work done by bat = KE imparted to ball).
  • Astrodynamics. Gravitational potential and KE balance for orbit determination, escape velocity.
  • Particle physics. Accelerators do work on charged particles to give them KE — measured in eV.

Common mistakes

  • Forgetting cos θ in work formula. If force is at an angle, only the component along motion does work. Pushing a cart at 45° gets you cos 45° = 0.71× the horizontal force.
  • Treating ½mv² as "kinetic energy at this point" when v is direction-sensitive. KE is scalar — it doesn't care about direction. But final KE depends on net work, which depends on path-integrated forces.
  • Using work-energy when forces depend on time, not position. If F is purely a function of t (e.g., a time-varying push), it's easier to use F = ma over time. Work-energy excels when F depends on position (springs, gravity).
  • Forgetting that gravity does no work in horizontal motion. A book pushed across a level table has gravity acting (vertical) and motion (horizontal). cos 90° = 0 → gravity does no work on it.
  • Sign errors. Friction does negative work. If you forget the sign, you'll get the wrong final speed.
  • Confusing power with work. A 100 W bulb uses 100 J every second. Power × time = energy. Watt-hour is a unit of energy (= 3600 J).

Frequently asked questions

Why is work F·d cos θ and not just F·d?

Only the component of force ALONG the direction of motion does work. If you push a cart at 45°, only the horizontal component (F cos 45°) contributes to forward motion. The vertical component (F sin 45°) does no work — it's perpendicular to motion. So W = F·d·cos θ where θ is the angle between force and displacement. Mathematically, this is the dot product F⃗·d⃗.

When is work zero?

When force is perpendicular to motion (θ = 90°, cos 90° = 0). Examples — gravity does no work on a satellite in circular orbit (force is centripetal, perpendicular to velocity); normal force does no work on objects sliding on a surface; magnetic force does no work on charged particles (always perpendicular to velocity). Counterintuitive but powerful.

How does work-energy compare to F = ma?

F = ma is per instant — needs time integration to find changes. Work-energy uses position changes — much easier when forces depend on position (springs, gravity), not time. KE is a scalar — no vector decomposition needed. Trade-off — work-energy gives only kinetic energy, not detailed motion (timing, exact path). Use F = ma for trajectories; use W = ΔKE for "what's the speed at point B?"

Why ½mv² and not just mv²?

From integration. Starting from F = ma, multiply both sides by v (velocity)— Fv = mav. Note that v·a = d/dt(½v²). So Fv·dt = ½m·d(v²). Integrating — ∫F·v·dt = ½m·v². Or via dW = F·dx = m·a·dx = m·v·dv. Integrating ∫m·v·dv = ½mv². The factor of ½ falls out of the calculus.

What's the difference between work done BY a force vs ON an object?

Same thing usually. Work done BY a force equals work done ON the object that the force acts on. For example, if you push a box and it moves 5 m with 50 N, you do 250 J of work, and the box receives 250 J. In multi-object systems, identify which object you're computing work FOR — pay attention to direction (whose displacement, whose force).

How does power relate to work?

Power is the rate of doing work — P = dW/dt. For constant force, P = F·v (force times velocity). A 100 hp engine (≈ 75 kW) at 30 m/s produces force F = P/v = 75000/30 = 2500 N. Cars at high speeds need disproportionately more power because air resistance grows with v² — and overcoming it requires power proportional to v³.

What about non-conservative forces like friction?

Friction does negative work — W_friction = -f·d (force opposes motion). The lost kinetic energy becomes heat (thermal energy). For non-conservative forces, the work-energy theorem still holds — net work = ΔKE — but mechanical energy isn't conserved. Friction's work goes into the molecular kinetic energy (heat) of the surfaces.