10 GCSE Physics questions about the robot vacuum cleaner that get harder as you go, from Foundation to Higher, each with a full worked solution. A robot vacuum runs on a rechargeable battery. Electric motors turn its wheels, brushes and suction fan, and infrared sensors underneath stop it falling down the stairs.
Course: Combined Science + Separate Physics | Tier: Foundation + Higher | Topics: Electricity, Energy, Forces
📌 KEY DEFINITIONS – learn them word for word
Current: the rate of flow of electrical charge. Current = charge ÷ time.
Power: the rate at which energy is transferred, or the rate at which work is done.
Resultant force: the single force that has the same effect as all the forces acting on an object. If it is zero, the object stays still or moves at a steady speed.
Equations you’ll use
| Quantity | Symbol | Unit |
|---|---|---|
| distance | s | metres (m) |
| speed | v | metres per second (m/s) |
| time | t | seconds (s) |
| potential difference | V | volts (V) |
| current | I | amps (A) |
| power | P | watts (W) |
| energy transferred | E | joules (J) |
| charge | Q | coulombs (C) |
| force | F | newtons (N) |
| mass | m | kilograms (kg) |
| acceleration | a | metres per second squared (m/s²) |
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Questions
Q1 (F) A robot vacuum moves 6.0 m across a kitchen in 20 s. Calculate its average speed.
Show solution
F: s = v t, so v = s ÷ t
I: v = 6.0 ÷ 20
F: v is already the subject
A: 0.30 m/s
Q2 (F) The robot uses infrared sensors to detect the edge of the stairs. (a) Name the group of waves that infrared belongs to. (b) Give one other everyday use of infrared.
Show answer
(a) Electromagnetic waves (the electromagnetic spectrum).
(b) Any one: TV remote controls, electric heaters, cooking (grills), thermal imaging cameras, optical fibre communication.
Q3 (F) The robot’s battery has a potential difference of 14.4 V. While cleaning, the current is 2.5 A. Calculate the power.
Show solution
F: P = V I
I: P = 14.4 × 2.5
F: P is already the subject
A: 36 W
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Q4 (F) The robot cleans for 60 minutes at a power of 36 W. Calculate the energy transferred.
Show solution
Convert first: 60 minutes = 60 × 60 = 3600 s
F: E = P t
I: E = 36 × 3600
F: E is already the subject
A: 129 600 J (about 130 kJ)
Q5 (F/H) When fully charged, the battery can supply a current of 2.6 A for 1.0 hour. Calculate the total charge the battery can deliver.
Show solution
Convert first: 1.0 hour = 60 × 60 = 3600 s
F: Q = I t
I: Q = 2.6 × 3600
F: Q is already the subject
A: 9360 C
Q6 (F/H) Use your answer to Q5. Calculate the energy stored in the fully charged 14.4 V battery. Give your answer to 2 significant figures.
Show solution
F: E = Q V
I: E = 9360 × 14.4
F: E is already the subject
A: 134 784 J = 1.3 × 105 J (2 s.f.)
Q7 (F/H) The battery delivers 9360 C of charge. The robot draws a steady current of 2.5 A. Calculate how long the robot can run, in minutes.
Show solution
F: Q = I t
I: 9360 = 2.5 × t
F: t = 9360 ÷ 2.5 = 3744 s
A: 3744 ÷ 60 = 62 minutes (to the nearest minute)
Q8 (F/H) The robot moves across a rug at a steady speed. Explain, in terms of forces, why it moves at a steady speed, and why it needs more power on the rug than on a hard floor. (4 marks)
Show answer
Marking points:
• The driving force from the wheels is equal to the resistive forces (friction and air resistance) (1)
• So the resultant force is zero, so the speed stays the same (1)
• The rug causes more friction, so a bigger driving force is needed (1)
• More work is done every second, so the motors need more power / transfer energy faster (1)
Q9 (H) The robot has a mass of 3.2 kg. It speeds up from rest to 0.30 m/s in 0.50 s. Calculate the resultant force needed.
Show solution
Step 1 – acceleration: a = Δv ÷ t = (0.30 − 0) ÷ 0.50 = 0.60 m/s²
Step 2 – force
F: F = m a
I: F = 3.2 × 0.60
F: F is already the subject
A: 1.9 N (2 s.f.)
Q10 (H) In 10 minutes, the 36 W suction motor transfers 7560 J usefully to moving air. Calculate the efficiency of the motor, and state what happens to the rest of the energy.
Show solution
Step 1 – total input energy: E = P t = 36 × (10 × 60) = 21 600 J
F: efficiency = useful output energy ÷ total input energy
I: efficiency = 7560 ÷ 21 600
A: 0.35 = 35%
The rest (14 040 J) is dissipated, mostly heating the motor and the surroundings, and as sound.
⚠️ Where students lose marks
- Forgetting to change hours into seconds before using Q = I t.
- Using minutes instead of seconds in E = P t.
- Saying a steady speed means there are no forces. There are forces – they are balanced, so the resultant force is zero.
- Calculating efficiency the wrong way up. Useful output always goes on top, so efficiency can never be more than 1 (100%).
More devices: all 40 everyday devices · Electricity · Forces
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