Course: Combined Science + Separate Physics | Tier: Foundation + Higher | Linked: Kinetic energy questions · Work done questions · Stopping distances notes
✏️ Work on paper. Write out every step before you open the answer.
📌 KEY DEFINITIONS – learn them word for word
Work done: work is done when a force moves an object through a distance. Work done = energy transferred. Unit: joules (J).
Braking: friction between the brakes and the wheel does work. This transfers energy from the kinetic store of the vehicle to the thermal store of the brakes, so the brakes get hot.
To stop completely, the work done by the brakes must equal all the kinetic energy of the vehicle.
🧮 Three steps every time
Step 1 – Kinetic energy.
Kinetic energy = 0.5 × mass × speed × speed (Ek = ½ m v²)
Square the speed first, then multiply.
Step 2 – Work done by the brakes = kinetic energy.
To stop, the brakes must take away all the kinetic energy.
Step 3 – Braking force.
Work done = force × distance (W = F s)
So braking force = work done ÷ braking distance
Units: mass in kg, speed in m/s, distance in m, energy in J, force in N.
Level 1: Kinetic energy given (Step 3 only)
Q1 (F) 🛴 Electric scooter. A scooter and rider have 1500 J of kinetic energy. The rider brakes and stops in 5 m. Calculate the braking force.
Show answer
Work done by brakes = 1500 J
Braking force = 1500 ÷ 5 = 300 N
Q2 (F) 🚗 Car. A car has 200 000 J of kinetic energy. It brakes and stops in 40 m. Calculate the braking force.
Show answer
Work done by brakes = 200 000 J
Braking force = 200 000 ÷ 40 = 5000 N
Level 2: Work out the kinetic energy first
Q3 (F) 🚲 Cyclist. A cyclist and bike have a mass of 80 kg and travel at 5 m/s. The cyclist brakes and stops in 10 m. Calculate the braking force.
Show answer
Step 1: Kinetic energy = 0.5 × 80 × 5 × 5 = 0.5 × 80 × 25 = 1000 J
Step 2: Work done by brakes = 1000 J
Step 3: Braking force = 1000 ÷ 10 = 100 N
Q4 (F) 🚙 Car in town. A car of mass 1000 kg travels at 10 m/s. It brakes and stops in 25 m. Calculate the braking force.
Show answer
Step 1: Kinetic energy = 0.5 × 1000 × 10 × 10 = 50 000 J
Step 2: Work done by brakes = 50 000 J
Step 3: Braking force = 50 000 ÷ 25 = 2000 N
Q5 (F) 🚗 Car on a main road. A car of mass 1200 kg travels at 20 m/s. Its braking distance is 40 m. Calculate the braking force.
Show answer
Step 1: Kinetic energy = 0.5 × 1200 × 20 × 20 = 0.5 × 1200 × 400 = 240 000 J
Step 2: Work done by brakes = 240 000 J
Step 3: Braking force = 240 000 ÷ 40 = 6000 N
Q6 (F/H) 🚚 Lorry. A lorry of mass 10 000 kg travels at 15 m/s. Its braking distance is 75 m. Calculate the braking force.
Show answer
Step 1: Kinetic energy = 0.5 × 10 000 × 15 × 15 = 0.5 × 10 000 × 225 = 1 125 000 J
Step 2: Work done by brakes = 1 125 000 J
Step 3: Braking force = 1 125 000 ÷ 75 = 15 000 N
Level 3: Find the braking distance
Q7 (F/H) 🚗 Small car. A car of mass 800 kg travels at 10 m/s. The brakes give a braking force of 4000 N. Calculate the braking distance.
Hint: braking distance = work done ÷ braking force.
Show answer
Step 1: Kinetic energy = 0.5 × 800 × 10 × 10 = 40 000 J
Step 2: Work done by brakes = 40 000 J
Step 3: Braking distance = 40 000 ÷ 4000 = 10 m
Q8 (F/H) ⚠️ Double the speed. A car of mass 1000 kg has a braking force of 5000 N.
(a) Calculate its braking distance at 10 m/s.
(b) Calculate its braking distance at 20 m/s.
(c) The speed doubled. What happened to the braking distance?
Show answer
(a) Kinetic energy = 0.5 × 1000 × 10 × 10 = 50 000 J. Distance = 50 000 ÷ 5000 = 10 m
(b) Kinetic energy = 0.5 × 1000 × 20 × 20 = 200 000 J. Distance = 200 000 ÷ 5000 = 40 m
(c) Doubling the speed made the braking distance 4 times bigger. This is because the speed is squared in the kinetic energy equation, so doubling the speed gives 4 times the kinetic energy.
Level 4: Challenge (H)
Q9 (H) 🛣️ Convert first. A car of mass 1.5 tonnes travels at 72 km/h. It stops in a braking distance of 50 m. Calculate the braking force.
Hint: 1 tonne = 1000 kg. To change km/h to m/s, divide by 3.6.
Show answer
Convert: 1.5 tonnes = 1500 kg. 72 km/h ÷ 3.6 = 20 m/s
Step 1: Kinetic energy = 0.5 × 1500 × 20 × 20 = 300 000 J
Step 2: Work done by brakes = 300 000 J
Step 3: Braking force = 300 000 ÷ 50 = 6000 N
Q10 (H) 🔥 Where does the energy go? Use the car from Q9.
(a) How much energy is transferred to the thermal store of the brakes?
(b) Use F = ma to calculate the deceleration of the car.
(c) Suggest why a very large deceleration is dangerous.
Show answer
(a) All the kinetic energy: 300 000 J (the brakes get very hot).
(b) Deceleration = force ÷ mass = 6000 ÷ 1500 = 4 m/s²
(c) The brakes can overheat, and the driver can lose control of the vehicle.
⚠️ Where students lose marks
- Forgetting to square the speed. 20 m/s squared is 20 × 20 = 400, not 40.
- Forgetting the 0.5 (the ½) in the kinetic energy equation.
- Using the stopping distance instead of the braking distance. Only the braking distance counts – the thinking distance happens before the brakes are on.
- Saying the energy is “lost”. Say it is transferred to the thermal store of the brakes.
Next: Stopping distances notes · F = ma: find the resultant force first · Kinetic energy questions
