Work, energy and power

Work is done when a force moves its point of application through a displacement. It measures the energy transferred by that force. For a constant force: $W = Fs\cos\theta$ where $\theta$ is the angle between the force and the displacement. Work is a scalar, measured in joules (J); $1\,\text{J} = 1\,\text{N m}$.

The role of $\cos\theta$ — and where marks are won or lost:

• If the force is along the displacement ($\theta = 0°$): $W = Fs$ (maximum work).
• If the force is perpendicular to the displacement ($\theta = 90°$): $W = 0$ (no work done). This is why the centripetal force does no work, and why carrying a bag horizontally at constant height does no work against gravity.
• If the force opposes motion ($\theta = 180°$, e.g. friction): $W = -Fs$ (negative work — energy is removed from the object).

Work from a graph: the area under a force–displacement graph equals the work done — useful when the force varies (e.g. stretching a spring).

Energy is the capacity to do work; it comes in many forms and is measured in joules.

Kinetic energy — energy of motion: $E_k = \tfrac{1}{2}mv^2 = \frac{p^2}{2m}$ (The second form, using momentum $p = mv$, is handy in collision problems.) Note $E_k \propto v^2$ — doubling the speed quadruples the kinetic energy (and roughly the braking distance).

Gravitational potential energy — energy due to position in a gravitational field. Near Earth's surface, the change as height changes by $\Delta h$ is: $\Delta E_p = mg\Delta h$ Only changes in height matter for SL mechanics; you may set any convenient level as the zero of PE.

Elastic potential energy — energy stored in a stretched or compressed spring (or elastic material) obeying Hooke's law ($F = k\Delta x$): $E_e = \tfrac{1}{2}k\Delta x^2$ This equals the area under the force–extension graph (a triangle, hence the $\tfrac{1}{2}$).

The principle of conservation of energy states that energy cannot be created or destroyed, only transferred from one form to another; the total energy of an isolated system is constant.

In mechanics this usually means the interchange of kinetic and potential energy, plus any energy dissipated (usually as heat by friction or drag): $E_{k,\text{initial}} + E_{p,\text{initial}} = E_{k,\text{final}} + E_{p,\text{final}} + E_{\text{dissipated}}$

Why this is so powerful: it lets you solve problems without knowing the time or the force, and it works even when the acceleration is not uniform (so suvat would fail).

The classic frictionless case — a falling or sliding object with no resistance: $\tfrac{1}{2}mv^2 = mg\Delta h \quad\Rightarrow\quad v = \sqrt{2g\Delta h}$ Notice the mass cancels — all objects reach the same speed after falling the same height (ignoring air resistance).

When friction is present: the "lost" mechanical energy has not vanished — it has become thermal energy (heat) in the surfaces and surrounding air. Include it as a work-done-against-friction term ($W_f = f \times d$) to keep the energy books balanced.

Power is the rate at which work is done or energy is transferred: $P = \frac{W}{t} = \frac{\Delta E}{t}$ The SI unit is the watt (W); $1\,\text{W} = 1\,\text{J s}^{-1}$.

The very useful form $P = Fv$: for an object moving at speed $v$ against (or driven by) a constant force $F$, $P = Fv$ This is derived from $P = W/t = Fs/t = Fv$. It is the go-to equation for vehicles and motors:

• At top (constant) speed, the driving force equals the total resistive force, so the useful output power $P = F_{\text{resist}}\,v_{\max}$.
• To find the maximum speed of a car of known engine power against a known resistive force, use $v_{\max} = P / F_{\text{resist}}$.

Distinguish energy and power clearly: energy (J) is the total transferred; power (W) is how fast it is transferred. A 2 kW kettle and a 100 W bulb can deliver the same energy — the kettle just does it 20× faster.

Understanding where $E_k = \tfrac{1}{2}mv^2$ and $P = Fv$ come from means you can rebuild them under pressure and apply them to novel situations — a hallmark of grade-9 candidates.

1. The work–energy theorem: deriving $E_k = \tfrac{1}{2}mv^2$. Consider a constant resultant force $F$ acting on a mass $m$ (initially at rest, $u = 0$) over a displacement $s$, giving it a final speed $v$.

The work done is $W = Fs$. Using Newton's second law $F = ma$ and the suvat equation $v^2 = u^2 + 2as$ (with $u = 0$, so $s = \dfrac{v^2}{2a}$): $W = Fs = (ma)\left(\frac{v^2}{2a}\right) = \tfrac{1}{2}mv^2$ The acceleration cancels, leaving the work done = the kinetic energy gained. This is the work–energy theorem: the net work done on an object equals its change in kinetic energy, $W_{\text{net}} = \Delta E_k = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mu^2.$

2. Deriving $P = Fv$. Power is work per unit time. For a constant force $F$ moving an object a distance $s$ in time $t$: $P = \frac{W}{t} = \frac{Fs}{t} = F\left(\frac{s}{t}\right) = Fv$ because $v = s/t$ is the speed. This is why a vehicle's engine power, its driving force and its speed are locked together — and why, at top speed (where driving force = resistance), $P = F_{\text{resist}}\,v_{\max}$.

Micro-example (work–energy theorem in action): a 1000 kg car speeds up from 10 m s⁻¹ to 20 m s⁻¹. The net work done on it is $W_{\text{net}} = \tfrac{1}{2}(1000)(20^2 - 10^2) = 500 \times 300 = 1.5 \times 10^5\,\text{J},$ without ever needing the distance or the time — the theorem gets you straight from the speeds to the energy.

No real device transfers all its input energy usefully — some is always wasted (usually as heat). Efficiency measures how good a device is: $\text{efficiency} = \frac{\text{useful output energy}}{\text{total input energy}} = \frac{\text{useful output power}}{\text{total input power}}$ It is a ratio between 0 and 1, often written as a percentage ($\times 100\%$). It can never exceed 100% (that would violate conservation of energy).

Sankey diagrams picture the energy flow. Rules:

• The width of each arrow is proportional to the amount of energy it represents.
• The total input enters from the left; useful output continues straight; wasted energy (usually heat) branches off downwards.
• Because energy is conserved, the input width equals the sum of the output widths.