Work, energy and power

The **energy transferred** when a **force moves something through a distance**. Unit: the **joule (J)**.

$W = Fs\cos\theta$ — force × distance × the cosine of the angle between them. (Given in the data booklet.)

The **angle between the force and the direction of motion**. If the force is along the motion, θ = 0 and cos 0 = 1, so W = Fs.

**Zero** — cos 90° = 0, so W = 0. (e.g. the normal force on a block sliding along a floor.)

The **work done** by the force.

The **joule (J)** — the same unit as all forms of energy.

**Zero** — no movement means no distance, so no work, however hard you push.

The work becomes kinetic energy: set **W = ½mv²** and solve for **v**.

The energy of a moving object: $E_k = \tfrac{1}{2}mv^{2}$ (m = mass, v = speed). Given in the data booklet.

W = Fs = 9.0 × 4.0 = **36 J** (which equals the kinetic energy gained).

The energy an object has because it is **moving**. It depends on the mass and the **speed squared**. Unit: the joule (J).

$E_k = \tfrac{1}{2}mv^{2} = \dfrac{p^{2}}{2m}$ — use ½mv² with the speed, or p²/2m with the momentum.

It becomes **four times** as big, because the speed is squared (2² = 4).

The **net work done** on an object equals its **change in kinetic energy**: W_{net} = ΔE_k.

Friction does **negative work**, removing kinetic energy. The object stops when all its E_k is used up.

Set **friction force × distance = E_k**, then **distance = E_k ÷ friction force**.

The **joule (J)** — the same unit as work and all other forms of energy.

When you're **given the momentum** p (= mv) instead of the speed — both forms give the same energy.

E_k = ½ × 3.0 × 4.0² = ½ × 3.0 × 16 = 24 J.

distance = E_k ÷ friction = 90 ÷ 18 = 5.0 m.

Kinetic energy ½mv² is a **scalar** (no direction) measured in joules; momentum mv is a **vector** measured in kg m s⁻¹.

The energy an object has **because of its height** in a gravitational field. It increases when the object is raised.

The energy an object has **because of its motion**. The faster it moves, the more KE it has.

$\Delta E_p = mg\Delta h$ — mass × gravitational field strength × change in height. (Given in the data booklet.)

$E_k = \tfrac{1}{2}mv^{2}$ — half × mass × speed². (Given in the data booklet.)

With no air resistance, **PE + KE stays constant**: the PE lost equals the KE gained.

$mg\Delta h = \tfrac{1}{2}mv^{2}$ — set the PE lost equal to the KE gained.

**All PE, no KE** — it is at maximum height and not yet moving.

**All KE, no PE** (taking the bottom as the reference height) — all the PE has converted to KE.

**No** — in mgΔh = ½mv² the mass cancels, so heavy and light objects reach the same speed (no air resistance).

**A quarter** — KE gained = PE lost, so the fraction of height fallen = the fraction now KE.

**Half-way down** — there it has lost half its PE, which has become KE, so PE = KE.

The **joule (J)**. PE and KE are both measured in joules.

The energy **stored in a spring** (or springy material) when it is **stretched or squashed**. Unit: the joule (J).

$E_H = \tfrac{1}{2}k\,\Delta x^{2}$ — half × spring constant × extension squared. (Also written E_p = ½kx².)

How **stiff** a spring is — the force needed per metre of stretch. Unit: N m⁻¹ (newtons per metre).

The **extension or compression** — how far the spring is stretched or squashed from its natural length, in metres.

It becomes **four times** as big, because the extension is squared (2² = 4).

By conservation of energy: **E_H = kinetic energy before − kinetic energy of the combined motion**.

From **conservation of momentum**: total momentum before = (combined mass) × common speed.

The **joule (J)** — the same unit as all other forms of energy.

E_H = ½ × 300 × 0.020² = ½ × 300 × 0.0004 = 0.060 J.

Power = energy ÷ time = 0.60 ÷ 0.015 = 40 W.

Elastic PE (½kΔx²) is stored by **stretching/squashing** a spring; gravitational PE (mgΔh) is stored by **lifting** a mass to a height. Both are in joules.

The **rate of energy transfer** — the energy transferred (or work done) each **second**. Unit: the watt (W).

**1 watt = 1 joule per second** (1 W = 1 J s⁻¹).

$P = \dfrac{\Delta W}{\Delta t} = Fv$ — energy ÷ time, or force × speed.

**P = Fv**, where F is the **resistive (drag) force** — it equals the driving force at constant speed.

**Average** = total energy ÷ total time (ΔW/Δt). **Instantaneous** = the power at one instant, using Fv with the speed right now.

The fraction of the energy put in that comes out as **useful** energy: η = useful out ÷ total in (× 100 for a %). It has no unit.

**No** — you can never get more useful energy out than you put in; some is always wasted (mostly as heat).

Mostly to **thermal energy (heat)**, plus some sound — energy spread out and no longer useful.

P = ΔW/Δt = 600 ÷ 5.0 = 120 W.

P = Fv = 400 × 30 = 12 000 W = 12 kW.

At constant speed P = Fv = cv², so c = P ÷ v². Its SI unit is kg s⁻¹.

Energy cannot be created or destroyed — it is only **transferred** from one store to another. The total amount stays the same.

It has been transferred to a **less useful** store — almost always **thermal energy (heat)** — that spreads out and can't easily be reused. It is NOT destroyed.

The **useful fraction** of the energy (or power) supplied: η = useful output ÷ total input. It has no unit and is often given as a %.

$\eta = \dfrac{\text{useful out}}{\text{total in}}$ — useful energy (or power) out ÷ total energy (or power) in.

An arrow diagram showing how the input energy splits into useful and wasted branches; the **width** of each arrow shows the amount of energy.

The useful and wasted branches **add up to the input arrow** — energy is conserved, so nothing is missing.

**Thermal energy (heat)** — and sometimes **sound** in moving parts. It spreads into the surroundings.

No — the useful output can never be larger than the total input, so efficiency is always between **0 and 1** (0–100%).

wasted = total energy in − useful energy out.

η = useful ÷ total = 350 ÷ 500 = 0.70 = 70%.

Wasted = 60 − 9 = 51 J, transferred as thermal energy (heat).

Because energy is **conserved** — it isn't lost, only **transferred** to a less useful store (heat). The total is unchanged.