Two Roads to the Same Answer
A car of mass $1000\ \text{kg}$ brakes to rest under a constant braking force of $4500\ \text{N}$. The kinematics route uses $F = ma$ and then $v^{2} = u^{2} + 2as$; the energy route writes $\dfrac{1}{2}mv^{2} = Fd$ and is done in one line. Both are correct — and they must always agree. Move the speed slider and watch both distances stay in step.
energy route $d = \dfrac{mv^{2}}{2F}$ = 36.0 m · kinematics route $d = \dfrac{v^{2}}{2a}$ = 36.0 m
Two different toolkits, one distance — choosing the quicker route is the skill of this lesson.
Try each in your head before clicking an answer — this is a test, not a reveal. See how many you get first try. The clock starts on your first answer.
R1. At speed v with constant power P, the driving force is:
R2. A ball is dropped from rest and falls 20 m with no air resistance. Its speed just before landing is about:
R3. The work done against a resistance force R acting over a path of length L is:
| The question asks for... | Reach for |
|---|---|
| a speed or distance, with no time mentioned | energy: $W = \Delta E_{k}$, $mgh$, the audit |
| a force or acceleration at an instant | $F = \dfrac{P}{v}$ then $F - R = ma$ |
| time, or a constant-acceleration journey | the SUVAT formulae with $a$ from $F = ma$ |
| how fast work is being delivered | $P = \dfrac{W}{t}$ or $P = Fv$ |
Most P4 questions are chosen so that one route is much shorter than the others — the exam rewards finding it.
A $1200\ \text{kg}$ car with a constant $18\ \text{kW}$ is doing $15\ \text{m s}^{-1}$ against $700\ \text{N}$: first $F = \dfrac{18\,000}{15} = 1200\ \text{N}$; then $a = \dfrac{1200 - 700}{1200} = 0.417\ \text{m s}^{-2}$. If the stage lasts $6\ \text{s}$, the SUVAT formula adds $v = 15 + 0.417 \times 6 = 17.5\ \text{m s}^{-1}$ — tools stacked, each feeding the next.
Braking from $v$ under a constant force $F$: the energy route gives $d = \dfrac{mv^{2}}{2F}$; the kinematics route uses $a = \dfrac{F}{m}$ and $d = \dfrac{v^{2}}{2a} = \dfrac{mv^{2}}{2F}$. Identical — because both encode the same physics.
In an exam, if time allows, run the second route over the key step: if the answers disagree, one of them has a slip, and the audit will find it.
Groups of 4: each question stem on the left has one best tool on the right. Tap a left card, then its partner on the right — the colour-coded line and badges mark your answer, and cards lock once linked. Draw all six lines, then press Check to score them at once. Any wrong link can be rethought after a Reset.
Work alone up the ladder. Every stage comes from a different part of the week — read it, choose the tool, then answer. A wrong first choice locks, shows a hint and still lets you continue; first-try stages score on the HUD. The final stage is the top-speed pattern you met in Lesson 3.
A winch pulls a crate of mass $1500\ \text{kg}$ up a slope at $20^{\circ}$ to the horizontal, the wire running parallel to the slope. The crate takes $50\ \text{s}$ to move $25\ \text{m}$ up the slope at a constant speed while the winch supplies $6\ \text{kW}$. (i) How much work is done by the tension in the wire in the $50\ \text{s}$? (ii) Calculate the resistance to the motion up the slope. (iii) Show that the coefficient of friction between the crate and the slope is $0.5$ correct to one decimal place. (iv) The winch breaks down and the crate is pushed up the slope by a constant force of $16\,000\ \text{N}$ inclined at $15^{\circ}$ to the slope. Calculate the distance travelled as the crate speeds up from rest to $2.5\ \text{m s}^{-1}$, taking the coefficient of friction as exactly $0.5$.
Part (i) is pure power: $W = Pt$. For part (ii), the tension's work is spent on the climb $mgL\sin 20^{\circ}$ plus the resistance work $RL$. For part (iv), resolve the push along the slope, then subtract both friction $\mu R_{N}$ and the weight component before using $v^{2} = 2as$.
$W = Pt = 6000 \times 50 = 300\,000\ \text{J} = 300\ \text{kJ}$.
Climb: $1500 \times 9.8 \times 25 \times \sin 20^{\circ} = 125\,692\ \text{J}$. So $300\,000 - 125\,692 = 174\,308\ \text{J}$ went against resistance over $25\ \text{m}$: $R = \dfrac{174\,308}{25} = 6972\ \text{N}$ (3 s.f.).
Normal reaction $= 1500 \times 9.8 \times \cos 20^{\circ} = 13\,814\ \text{N}$. Then $\mu = \dfrac{6972}{13\,814} = 0.505$, which is $0.5$ correct to one decimal place.
Along the slope: push $= 16\,000\cos 15^{\circ} = 15\,455\ \text{N}$; friction $= 0.5 \times 13\,814 = 6907\ \text{N}$; weight component $= 1500 \times 9.8 \times \sin 20^{\circ} = 5028\ \text{N}$. Resultant $= 15\,455 - 6907 - 5028 = 3520\ \text{N}$, so $a = \dfrac{3520}{1500} = 2.35\ \text{m s}^{-2}$. From rest to $2.5\ \text{m s}^{-1}$: $s = \dfrac{v^{2}}{2a} = \dfrac{6.25}{4.70} = 1.33\ \text{m}$.
The UK Highway Code gives the braking distance for a car travelling at $22\ \text{m s}^{-1}$ (50 mph) to be $38\ \text{m}$. A car of mass $1300\ \text{kg}$ is brought to rest in just this distance, and the only resistance comes from the brakes. (i) Find the work done by the brakes. (ii) Find the average force exerted by the brakes. (iii) What happened to the kinetic energy of the car?
Route 1 (energy): the brakes remove all the kinetic energy, and $W = Fd$ then gives the force. Route 2 (kinematics): find the deceleration from $v^{2} = u^{2} + 2as$ with $v = 0$, then use $F = ma$. Both routes must land on the same numbers.
(i) The brakes must remove the whole kinetic energy: $\dfrac{1}{2} \times 1300 \times 22^{2} = 314\,600\ \text{J} \approx 315\ \text{kJ}$. (ii) $F = \dfrac{314\,600}{38} = 8280\ \text{N}$ (3 s.f.).
With $v = 0$: $0 = 22^{2} + 2a(38)$, so $a = -6.37\ \text{m s}^{-2}$. Then $F = ma = 1300 \times 6.37 = 8280\ \text{N}$ — the same force, and the same work $Fd = 8280 \times 38 = 314\,600\ \text{J}$.
(iii) The kinetic energy became internal (heat) energy in the brake discs and pads, with a little sound — exactly the transfer that this week's audits record as "work done against resistance".
Answer, then check. Options are shuffled every load. The clock starts on your first answer.
Q1. A question gives a car's power and asks for its acceleration at a given speed. The most direct first step is:
Q2. A 800 kg car climbs 20 m while travelling 200 m against a resistance of 500 N at constant speed. The engine's work is:
Q3. A 1000 kg car with constant power 16 kW travels at 20 m/s against a resistance of 600 N. Its acceleration is:
Q4. A crate slides down a rough slope and you must find its speed at the bottom — no time is given. The quickest route is:
Work each on paper, then reveal the solution to check your working.
A weightlifter takes $2\ \text{s}$ to lift $120\ \text{kg}$ from the floor to a position $2\ \text{m}$ above it, where the weight is held stationary. (i) Calculate the work done by the weightlifter. (ii) Calculate the average power developed. The lift is performed in stages: the weight is raised $0.8\ \text{m}$ in $0.5\ \text{s}$, held stationary for $1\ \text{s}$, then raised to the final position in another $0.5\ \text{s}$. (iii) Find the average power developed during each stage.
Reveal Solution
(i) $120 \times 9.8 \times 2 = 2352\ \text{J}$. (ii) $\dfrac{2352}{2} = 1176\ \text{W}$. (iii) First stage: $\dfrac{120 \times 9.8 \times 0.8}{0.5} = 1882\ \text{W}$. The hold does no work (no displacement), so its power is $0\ \text{W}$. Final stage: $\dfrac{120 \times 9.8 \times 1.2}{0.5} = 2822\ \text{W}$ — the average power hides two very different efforts.
A cyclist, travelling at a constant speed of $8\ \text{m s}^{-1}$ along a level road, experiences a total resistance of $70\ \text{N}$. (i) Find the power the cyclist is producing. (ii) Find the work done by the cyclist in 5 minutes under these conditions.
Reveal Solution
(i) At constant speed the cyclist's driving force equals the resistance: $P = Fv = 70 \times 8 = 560\ \text{W}$. (ii) $W = Pt = 560 \times 300 = 168\,000\ \text{J} = 168\ \text{kJ}$.
The power output from the engine of a car of mass $50\ \text{kg}$ travelling along level ground at a constant speed of $33\ \text{m s}^{-1}$ is $23\,200\ \text{W}$. (i) Find the total resistance on the car under these conditions. (ii) You were given one piece of unnecessary information. Which is it?
Reveal Solution
(i) At constant speed the driving force balances the resistance: $R = \dfrac{P}{v} = \dfrac{23\,200}{33} = 703\ \text{N}$ (3 s.f.). (ii) The mass: at constant speed the resistance balances the driving force whatever the mass, so the car's mass plays no part in the calculation.
A stone of mass $0.2\ \text{kg}$ is dropped from the top of a building $80\ \text{m}$ high. After $t$ seconds it has fallen $x$ metres and has speed $v\ \text{m s}^{-1}$. (This question uses the model $g = 10\ \text{m s}^{-2}$, as part (iii) confirms.) (i) What is the gravitational potential energy of the stone relative to ground level when it is at the top of the building? (ii) What is its potential energy $t$ seconds later? (iii) Show that $v^{2} = 20x$ and state the range of values of $t$ for which it is true. (iv) Find the speed of the stone when it is half-way to the ground. (v) At what height will the stone have half its final speed?
Reveal Solution
(i) $0.2 \times 10 \times 80 = 160\ \text{J}$. (ii) $0.2 \times 10 \times (80 - x) = 160 - 2x$ joules. (iii) With $g = 10$: $v = 10t$ and $x = 5t^{2}$, so $v^{2} = 100t^{2} = 20 \times 5t^{2} = 20x$. It holds while the stone is in the air: $0 \leq t \leq 4\ \text{s}$ (the fall takes $t = \sqrt{\dfrac{160}{10}} = 4\ \text{s}$). (iv) Half-way means $x = 40$, so $v = \sqrt{20 \times 40} = \sqrt{800} = 28.3\ \text{m s}^{-1}$. (v) The final speed is $\sqrt{20 \times 80} = 40\ \text{m s}^{-1}$; half of it is $20\ \text{m s}^{-1}$, so $x = \dfrac{20^{2}}{20} = 20\ \text{m}$ fallen — the stone is $60\ \text{m}$ above the ground.
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