HookAt the bottom of a bungee jump your speed is zero — the force is not
The bungee platform on the Macau Tower sits \(233\ \text{m}\) above the ground — the highest commercial jump in the world. You step off, free-fall for several seconds, and then the cord begins to stretch. Here is the strange part: the most violent moment of the whole jump is the instant you are not moving at all. At the lowest point your speed is momentarily zero, yet the cord is hauling upward on you with three to four times your own weight. The kinetic energy you built up falling has not vanished — every joule of it is now stored in the stretched cord, about to throw you back up the way you came.
That is this section in one picture. Work is how energy moves between stores; kinetic, gravitational potential and elastic potential energy are the stores themselves; conservation of energy is the audit that says the ledger must balance; and power is how fast the transfers happen. AQA's Further Mechanics option leans on energy methods because they answer questions that Newton's second law makes miserable — a bungee cord's tension changes continuously as it stretches, so \(F=ma\) gives a differential equation, while the energy ledger gives one line of algebra.
ModelWork — only the component along the motion counts
The work done by a constant force \(F\) whose line of action makes angle \(\theta\) with the direction of motion is \(W=Fd\cos\theta\), where \(d\) is the distance moved. The \(\cos\theta\) is doing real physics: only the component of the force along the motion transfers energy. A sledge hauled \(12\ \text{m}\) by a rope at \(30^\circ\) above the horizontal under a tension of \(40\ \text{N}\) receives \(W=40\times 12\times\cos 30^\circ\approx 416\ \text{J}\) — not \(480\ \text{J}\), because part of the pull is wasted lifting rather than dragging. The unit is the joule: one newton acting through one metre.
Two special cases carry most of the marks. A force perpendicular to the motion does no work at all (\(\cos 90^\circ=0\)) — the normal reaction on a sliding block and the tension in a conical pendulum's string transfer no energy, which is why they never appear in an energy equation. And a force opposing the motion, like friction, has \(\theta=180^\circ\), so it does negative work; we usually flip the sign and speak of work done against friction, \(W=Fd\).
Gravity has one more property worth noticing now: the work it does depends only on the vertical drop, \(W=mgh\), whatever path the body takes. A ball lowered straight down \(2\ \text{m}\) and a ball rolled down a \(10\ \text{m}\) helter-skelter losing \(2\ \text{m}\) of height receive exactly the same work from gravity. That path-blindness is the reason a 'gravitational potential energy' store can exist at all.
ModelThe energy stores — kinetic and gravitational potential
Kinetic energy is \(\text{KE}=\frac{1}{2}mv^2\), and it is not an arbitrary formula. For a constant force acting along the motion, \(W=Fs=mas\), and the constant-acceleration result \(v^2=u^2+2as\) gives \(as=\frac{1}{2}(v^2-u^2)\); substituting, \[W=\tfrac{1}{2}mv^2-\tfrac{1}{2}mu^2.\] Work done equals the change in \(\frac{1}{2}mv^2\) — that is the work–energy principle, and it is why \(\frac{1}{2}mv^2\) deserves the name 'energy of motion'.
Gravitational potential energy is \(\text{GPE}=mgh\), measured from a datum level you choose and must state. Only changes in GPE ever matter, so put the datum wherever it kills the most terms — usually the lowest point of the motion. When gravity is the only force doing work, mechanical energy is conserved: \(\frac{1}{2}mv^2+mgh\) has the same value at every instant. When resistance acts, the ledger still balances, but some energy leaves through friction: \[\text{KE}_\text{start}+\text{GPE}_\text{start}=\text{KE}_\text{end}+\text{GPE}_\text{end}+\text{work done against resistance}.\]
The strategic reason examiners love energy methods: they are path-blind. Newton's second law needs the slope angle at every point of a curved track; the energy equation needs only the endpoints.
A \(55\ \text{kg}\) skier starts from rest and descends a winding run that drops \(40\ \text{m}\) vertically over \(180\ \text{m}\) of slope, against a constant resistance of \(90\ \text{N}\). Find her speed at the bottom.
No suvat equation can touch this — the slope curves, so the acceleration keeps changing. The energy ledger does not care. GPE released: \(mgh=55\times 9.8\times 40=21{,}560\ \text{J}\). Work done against resistance: \(90\times 180=16{,}200\ \text{J}\). What remains becomes kinetic energy: \(\tfrac{1}{2}mv^2=21{,}560-16{,}200=5{,}360\ \text{J}\), so \(v^2=\dfrac{2\times 5{,}360}{55}\approx 195\) and \(v\approx 14.0\ \text{m s}^{-1}\). One line of accounting replaced an impossible kinematics problem — and notice the resistance term used the \(180\ \text{m}\) path length while gravity used the \(40\ \text{m}\) drop. Mixing those two distances is the single most common error in this topic.
MechanismHooke's law, the Further Maths way: T = λx/l
At A-level Maths a spring obeys \(T=kx\). Further Maths upgrades this to \[T=\frac{\lambda x}{l},\] where \(l\) is the natural length, \(x\) the extension beyond it, and \(\lambda\) the modulus of elasticity. Look at what \(\lambda\) is: set \(x=l\) and you get \(T=\lambda\). So \(\lambda\) is a force, in newtons — the tension needed to double the string's length. It is not a stiffness; the stiffness is \(k=\lambda/l\).
Why bother? Because \(\lambda\) belongs to the material and cross-section, while \(k\) depends on how long a piece you cut. Halve a bungee cord and each half is twice as stiff — same \(\lambda\), half the \(l\). That is exactly the bookkeeping a manufacturer needs, and it is why AQA questions hand you \(\lambda\) and expect you to keep the natural length in view at all times. A cord of natural length \(25\ \text{m}\) and modulus \(1{,}470\ \text{N}\) stretched to \(30\ \text{m}\) has \(x=5\ \text{m}\), so \(T=\dfrac{1470\times 5}{25}=294\ \text{N}\).
One modelling trap: a string can only pull. If the 'extension' would be negative — the ends closer than the natural length — the string is slack and \(T=0\); the formula switches off. A spring can also be compressed, in which case \(x\) is the compression and the force pushes outward.
MechanismVariable forces: work as an integral — and where λx²/2l comes from
\(W=Fd\) assumes the force is the same over the whole distance. A stretching cord breaks that assumption — the tension grows as \(x\) grows. The fix is the same one calculus always offers: over a sliver \(\delta x\) the force is effectively constant, the sliver of work is \(F\,\delta x\), and summing gives \[W=\int F\,dx.\] Graphically, work is the area under a force–displacement graph, exactly as displacement is the area under velocity–time.
Apply that to Hooke's law and the elastic energy formula falls out in two lines. Stretching a string from natural length to extension \(x\), the tension when the extension is \(t\) is \(\lambda t/l\), so the work stored is \[\text{EPE}=\int_0^x \frac{\lambda t}{l}\,dt=\frac{\lambda}{l}\left[\frac{t^2}{2}\right]_0^x=\frac{\lambda x^2}{2l}.\] You can see the same result without calculus: the tension–extension graph is a straight line from \(0\) to \(\lambda x/l\), and the triangular area under it is \(\frac{1}{2}\times x\times\frac{\lambda x}{l}\). AQA has asked for this derivation outright — it is a reasoning (AO2) mark, not background reading.
A trolley of mass \(2\ \text{kg}\) is released from rest on a smooth horizontal track at \(x=2\ \text{m}\) from a fixed magnet that repels it with force \(F=\dfrac{48}{x^2}\ \text{N}\). Find its speed as it passes \(x=6\ \text{m}\).
The force varies with position, so integrate: \[W=\int_2^6 \frac{48}{x^2}\,dx=\left[-\frac{48}{x}\right]_2^6=\left(-8\right)-\left(-24\right)=16\ \text{J}.\] By the work–energy principle all of it becomes kinetic energy: \(\tfrac{1}{2}\times 2\times v^2=16\), so \(v=4\ \text{m s}^{-1}\). Notice the shape of the method: identify that \(F\) depends on \(x\), write \(W=\int F\,dx\) with limits taken from the journey, then hand the result to the energy ledger. Trying \(W=Fd\) with the starting force would give \(12\times 4=48\ \text{J}\) — three times too big.
CaseThe bungee ledger — GPE to KE to EPE
Now run the whole jump. Three stores trade with each other: while the cord is slack, GPE becomes KE (free fall); once the cord goes taut, EPE starts to claim its share; at the lowest point KE is zero and the ledger reads GPE lost = EPE stored.
The subtlety examiners test: maximum speed is not at the lowest point. The jumper keeps accelerating downward until the cord's tension has grown to equal their weight — \(\frac{\lambda x}{l}=mg\), at extension \(x=\frac{mgl}{\lambda}\). Below that, tension exceeds weight and the jumper decelerates while still moving down. Speed peaks where acceleration is zero; force peaks where speed is zero. Keeping those two 'maximum' points separate is the difference between a method mark and a full solution.
A jumper of mass \(60\ \text{kg}\) steps off the anchor point of a bungee cord of natural length \(25\ \text{m}\) and modulus of elasticity \(1{,}470\ \text{N}\). Find the total drop to the lowest point and the cord tension there.
Let the extension at the lowest point be \(x\), so the drop is \(d=25+x\). All GPE has become EPE: \[mg(25+x)=\frac{\lambda x^2}{2l}\quad\Rightarrow\quad 588(25+x)=\frac{1470x^2}{50}=29.4x^2.\] Dividing by \(29.4\): \(20(25+x)=x^2\), i.e. \(x^2-20x-500=0\), so \(x=\dfrac{20+\sqrt{400+2000}}{2}=10+10\sqrt{6}\approx 34.5\ \text{m}\) (the negative root is meaningless). Total drop: \(d\approx 59.5\ \text{m}\). Tension there: \(T=\dfrac{1470\times 34.5}{25}\approx 2{,}030\ \text{N}\) — about \(3.4\) times the jumper's \(588\ \text{N}\) weight, giving an upward acceleration of roughly \(2.4g\) at the very moment the jumper is stationary.
Maximum speed happens earlier, where \(T=mg\): \(x=\dfrac{mgl}{\lambda}=\dfrac{588\times 25}{1470}=10\ \text{m}\), a depth of \(35\ \text{m}\). The ledger there: \(\tfrac{1}{2}\times 60\times v^2=588\times 35-\dfrac{1470\times 10^2}{50}=20{,}580-2{,}940=17{,}640\ \text{J}\), so \(v^2=588\) and \(v=14\sqrt{3}\approx 24.2\ \text{m s}^{-1}\) — about \(87\ \text{km h}^{-1}\), reached with \(24.5\ \text{m}\) of falling still to go.
ModelPower — the rate of working, P = Fv
Power is the rate of doing work, measured in watts (\(1\ \text{W}=1\ \text{J s}^{-1}\)). For a driving force \(F\) along the motion, \(W=Fs\) gives, on differentiating with respect to time, \[P=\frac{dW}{dt}=F\frac{ds}{dt}=Fv.\] The formula is exam shorthand for a physical trade-off: at fixed engine power, the faster you go, the less driving force is available — which is why a car accelerates hard at \(20\ \text{m s}^{-1}\) and barely at all at \(40\ \text{m s}^{-1}\).
The phrase 'maximum speed' is a coded instruction. At top speed the acceleration is zero, so the driving force exactly balances the total resistance (plus any weight component on a hill). Resolve, set \(F\) equal to the resistance, then apply \(P=Fv\). The \(F\) in the power formula is always the driving force the engine provides — never the resultant, which is zero at precisely the moment these questions are set.
A car of mass \(1{,}200\ \text{kg}\) has a maximum engine power of \(60\ \text{kW}\) and moves against a constant resistance of \(2{,}400\ \text{N}\). (a) Find its maximum speed on the flat. (b) Find its maximum steady speed up a slope inclined at \(\theta\) where \(\sin\theta=\frac{1}{20}\). (c) Find its acceleration at \(15\ \text{m s}^{-1}\) on the flat, at full power.
(a) At top speed the driving force equals the resistance, \(2{,}400\ \text{N}\), so \(v=\dfrac{P}{F}=\dfrac{60{,}000}{2{,}400}=25\ \text{m s}^{-1}\).
(b) Uphill the engine also fights the weight component \(mg\sin\theta=1200\times 9.8\times\frac{1}{20}=588\ \text{N}\). Required force: \(2{,}400+588=2{,}988\ \text{N}\), so \(v=\dfrac{60{,}000}{2{,}988}\approx 20.1\ \text{m s}^{-1}\).
(c) At \(15\ \text{m s}^{-1}\) the engine can supply \(F=\dfrac{60{,}000}{15}=4{,}000\ \text{N}\). The resultant is \(4{,}000-2{,}400=1{,}600\ \text{N}\), so \(a=\dfrac{1{,}600}{1{,}200}\approx 1.33\ \text{m s}^{-2}\). The same engine that cannot accelerate at all at \(25\ \text{m s}^{-1}\) delivers a brisk push at \(15\ \text{m s}^{-1}\) — \(P=Fv\) is the whole story of why.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
Open every energy question the same way: state the datum for GPE, then write the whole ledger in one line — initial KE + GPE (+ EPE) = final KE + GPE (+ EPE) + work done against resistance. AQA's mechanics mark schemes give the method mark for an energy equation with all terms present and lose you both M and A marks for one missing store; the term candidates drop most often is the EPE of a still-stretched string at the 'end' position. Keep the two distances distinct: gravity acts through the vertical drop, resistance through the path length.
With elastic strings, define the extension explicitly — \(x=\text{current length}-l\) — on your first line, and check whether the string is taut at each position; a slack string contributes zero EPE and zero tension, and AQA regularly builds a part (b) around spotting exactly that. Remember Further Maths weights AO1 (standard techniques) at roughly 60%, but the ~20% of AO2 reasoning marks hide in derivations: be ready to derive \(\text{EPE}=\frac{\lambda x^2}{2l}\) by integrating the tension, and to justify \(P=Fv\) from \(W=Fs\), because 'show that' versions of both have appeared.
Translate trigger phrases before calculating: 'maximum speed' or 'constant speed' means acceleration zero, so resolve along the slope, set driving force = resistance + \(mg\sin\theta\), then use \(P=Fv\) with the driving force, never the resultant. Take \(g=9.8\ \text{m s}^{-2}\) as AQA instructs, give final answers to three significant figures unless told otherwise, and carry exact values (like \(x=10+10\sqrt{6}\)) through the working — rounding early is the classic accuracy-mark leak in multi-stage energy problems.