Phase 2 Consolidation — Energy Mastery
In the 2014 Formula 1 season, Mercedes W05 engineers calculated that each lap their Kinetic Energy Recovery System (KERS) stored 2 MJ of braking energy — recovered by decelerating from 320 km/h to 80 km/h entering chicanes. This single figure blends $\Delta KE = \frac{1}{2}m\Delta v^2$, the work-energy theorem, and $P = \Delta E/\Delta t$ simultaneously. This consolidation lesson teaches you to recognise and sequence exactly these six formulae under exam conditions.
Practise this lesson
Phase 2 consolidation worksheets.
Rate your confidence with each of the six Phase 2 energy formulae (1 = need help, 5 = confident). Identify your weakest area to focus on in this lesson.
Which formula would you use FIRST to find the speed of a ball at the bottom of a frictionless ramp if you know its mass and starting height?
Core Content
A skateboard rolls down a 3 m ramp (frictionless), hits flat ground, then coasts to a stop in 8 m against 12 N of friction. You need to find the speed at the bottom, the force the ground exerts to stop it, and the power output if it stops in 4 s. This single scenario requires three different Phase 2 formulae applied in sequence — and knowing which formula fits each stage is the entire skill this lesson develops.
| # | Formula | When to use | Key condition |
|---|---|---|---|
| 1 | $W = Fs\cos\theta$ | Force at angle acts over displacement | θ = angle between force and motion |
| 2 | $KE = \frac{1}{2}mv^2$ | Find KE from mass + speed | v is speed (scalar) |
| 3 | $W_{net} = \Delta KE$ | Net work done changes speed | Use when friction acts |
| 4 | $\Delta U = mg\Delta h$ | Height changes (up or down) | $\Delta h$ positive upward |
| 5 | $KE_1 + U_1 = KE_2 + U_2$ | No friction — energy conserved | Frictionless only |
| 6a | $P = \Delta E/\Delta t$ | Average power over time | Known total energy and time |
| 6b | $P = Fv$ | Instantaneous power at speed v | Force and velocity known |
Phase 2 formula hierarchy: $W_{net} = \Delta KE$ (F3) is the most general and always works; conservation $KE_1+U_1=KE_2+U_2$ (F5) is a special case of F3 when friction is zero; $P = Fv$ applies at constant speed; $P = \Delta E/\Delta t$ applies whenever total energy change and time are known.
Pause — copy the highlighted formula hierarchy into your book before moving on.
A car brakes from 20 m/s to rest over 40 m. Which formula gives the braking force most directly?
We just saw the full Phase 2 formula table and when to use each formula. That raises a question: how does that selection process look in practice when a multi-step problem is worked through? This card answers it → two worked examples demonstrate the decision sequence and correct formula chaining.
A 900 kg car brakes from 25 m/s to rest. The braking force is 6750 N. Find the stopping distance.
A 70 kg ski jumper leaves a 40 m high ramp from rest and reaches the base. Find (a) speed at base, (b) average power if the descent takes 4 s.
Formula selection guide: height change without friction → Conservation F5; force + distance + friction → Work-energy theorem F3; energy + time → $P = \Delta E/\Delta t$; force + speed (constant) → $P = Fv$; braking force problems → $W_{net} = \Delta KE$, substitute $W = Fs\cos180°$.
Pause — copy the highlighted selection guide into your book before moving on.
True or false: $W_{net} = \Delta KE$ can be used to find stopping distance when a braking force is known, even if there is no height change.
A 10 kg ball falls 5 m and then another 5 m. Its KE after the second 5 m compared to after the first 5 m is:
Activities
For each scenario below, state which formula(e) you would use and in what order, then solve:
- A 5 kg box pushed 8 m along the floor by a 30 N force at 20° below horizontal, with 8 N friction. Find the final speed from rest.
- A 400 kg roller-coaster car drops 60 m (frictionless), then drives at constant speed against 1200 N drag. Find the power needed.
A student writes: "The car descends 20 m. KE = mgh = 1500 × 9.8 × 20 = 294 000 J. Speed = KE/m = 294 000/1500 = 196 m/s." Identify ALL errors in this working.
ApplyBand 4(3 marks) 1. A 1000 kg car brakes from 20 m/s to rest. Friction is 5000 N. Calculate stopping distance using the work-energy theorem.
AnalyseBand 5(3 marks) 2. A waterfall drops 80 m. 1000 kg of water falls every second. (a) Calculate the GPE released per second. (b) If this converts entirely to electrical power, what is the output power in kW?
EvaluateBand 6(4 marks) 3. A 2 kg ball is projected horizontally at 15 m/s from the top of a 10 m cliff (ignore air resistance). Find: (a) KE at launch, (b) GPE lost in falling 10 m, (c) total KE at ground level, (d) speed at ground level. ($g = 9.8$ m/s²)
Show all answers
Short Answer — Model Answers
Q1 (3 marks): $\Delta KE = 0 - \frac{1}{2}(1000)(20^2) = -200\,000\text{ J}$. $W_{net} = -F \times s = -5000s$. $-5000s = -200\,000$; $s = 40\text{ m}$.
Q2 (3 marks): (a) $\Delta U/s = mgh/t = 1000 \times 9.8 \times 80 / 1 = 784\,000\text{ J/s} = 784\text{ kW}$. (b) If 100% efficiency: $P = 784\text{ kW}$. (In practice, real hydroelectric plants achieve ~90% efficiency, giving ~706 kW.)
Q3 (4 marks): (a) $KE_{launch} = \frac{1}{2}(2)(15^2) = 225\text{ J}$. (b) $\Delta U = mgh = 2 \times 9.8 \times 10 = 196\text{ J}$ lost (converted to KE). (c) $KE_{total} = 225 + 196 = 421\text{ J}$. (d) $v = \sqrt{2KE/m} = \sqrt{2 \times 421/2} = \sqrt{421} \approx 20.5\text{ m/s}$. Note: This can also be found by Pythagoras using horizontal (15 m/s) and vertical (=$\sqrt{2gh} = 14.0\text{ m/s}$) components: $v = \sqrt{15^2 + 14.0^2} = \sqrt{225+196} = \sqrt{421} \approx 20.5\text{ m/s}$ ✓
The 2014 Mercedes W05 KERS example requires exactly the formula sequence this lesson teaches: $\Delta KE = \frac{1}{2}m(v_f^2 - v_i^2)$ first, then $W_{net} = \Delta KE$ to confirm work done, then $P = \Delta E/\Delta t$ to find charging power. The most common student error — dividing energy by mass to get speed instead of using $v = \sqrt{2KE/m}$ — would give nonsense answers for the F1 engineers. The formula selection table above prevents exactly this mistake.