You are viewing a free preview of this lesson.
Subscribe to unlock all 10 lessons in this course and every other course on LearningBro.
This lesson brings together all the key concepts from the AQA GCSE Physics Forces topic (4.5) to help you prepare for the exam. It includes worked examples, common question types, mark scheme guidance, and revision strategies. Use this lesson to consolidate your understanding and identify any areas that need further revision.
The following equations are essential for the Forces topic. Some are on the AQA equation sheet, but you should know them all by heart.
| Equation | Meaning | On Equation Sheet? |
|---|---|---|
| W = m x g | Weight = mass x gravitational field strength | Yes |
| F = m x a | Force = mass x acceleration | Yes |
| F = k x e | Force = spring constant x extension (Hooke's law) | Yes |
| M = F x d | Moment = force x perpendicular distance from pivot | Yes |
| s = d / t | Speed = distance / time | No (must memorise) |
| a = (v - u) / t | Acceleration = change in velocity / time | No (must memorise) |
| E = 0.5 x k x e^2 | Elastic potential energy = 0.5 x spring constant x extension^2 | Yes |
Exam Tip: At the start of the exam, write out the key equations on the question paper. This saves time during the exam and reduces the chance of making mistakes. Even though some equations are on the sheet, having them ready to use quickly is an advantage.
Use this checklist to ensure you have revised every part of the AQA GCSE Physics Forces topic (4.5):
| Topic | Key Points | Revised? |
|---|---|---|
| Scalar and vector | Scalars have magnitude only; vectors have magnitude and direction | |
| Contact forces | Friction, air resistance, normal contact force, tension, upthrust | |
| Non-contact forces | Gravitational, electrostatic, magnetic | |
| Weight | W = m x g; weight is a force (N), mass is in kg | |
| Resultant forces | Add forces in same direction, subtract in opposite directions | |
| Balanced/unbalanced | Balanced = constant velocity; unbalanced = acceleration | |
| Newton's first law | Object stays at rest or constant velocity unless resultant force acts | |
| Newton's second law | F = m x a; acceleration proportional to force, inversely proportional to mass | |
| Newton's third law | Equal and opposite forces on different objects, same type | |
| Hooke's law | F = k x e; limit of proportionality | |
| Elastic/inelastic deformation | Elastic: returns to original shape; inelastic: permanently deformed | |
| Moments [H] | M = F x d; principle of moments | |
| Levers and gears [H] | Force multipliers; gear ratios | |
| Speed and velocity | s = d / t; velocity = speed + direction | |
| d-t graphs | Gradient = speed | |
| v-t graphs | Gradient = acceleration; area = distance | |
| Stopping distance | Thinking distance + braking distance | |
| Required practical: springs | F = k x e; method, variables, graph | |
| Required practical: acceleration | F = m x a; method, variables, graph |
Q: Calculate the weight of a 65 kg person on Earth (g = 9.8 N/kg).
A: W = m x g W = 65 x 9.8 W = 637 N (2 marks)
Q: A car has a driving force of 5000 N and friction forces totalling 2000 N. Calculate the resultant force and state its direction.
A: Resultant force = 5000 - 2000 = 3000 N (1 mark) Direction: forwards (in the direction of the driving force) (1 mark)
Q: A resultant force of 900 N acts on a car of mass 1200 kg. Calculate the acceleration.
A: F = m x a (1 mark) 900 = 1200 x a a = 900 / 1200 a = 0.75 m/s^2 (2 marks — 1 for rearranging, 1 for correct answer with unit)
Q: A spring has a spring constant of 40 N/m. Calculate the force needed to extend the spring by 0.15 m.
A: F = k x e (1 mark) F = 40 x 0.15 F = 6.0 N (2 marks — 1 for substitution, 1 for correct answer with unit)
Q: A spanner is used to tighten a nut. A force of 25 N is applied at a perpendicular distance of 0.3 m from the pivot. Calculate the moment.
A: M = F x d (1 mark) M = 25 x 0.3 M = 7.5 Nm (2 marks — 1 for substitution, 1 for correct answer with unit)
Q: A cyclist travels 450 m in 30 seconds. Calculate the average speed.
A: s = d / t (1 mark) s = 450 / 30 s = 15 m/s (2 marks)
Exam Tip: Always write the equation first, then substitute the values with their units, then calculate. Even if you get the final number wrong, you can still pick up marks for the equation and substitution. Never skip these steps.
Model Answer:
When the skydiver first jumps from the plane, the only significant force acting is their weight (W = m x g), which acts vertically downwards. Air resistance is very small because the skydiver's speed is low. The resultant force is large and acts downwards, so the skydiver accelerates downwards (Newton's second law: F = m x a).
As the skydiver's speed increases, the air resistance (drag) increases because air resistance depends on speed. The resultant downward force is now weight minus air resistance, which is smaller than the weight alone. The skydiver still accelerates, but the acceleration decreases because the resultant force is decreasing.
Eventually, the air resistance becomes equal to the weight. The resultant force is now zero. By Newton's first law, the skydiver continues at a constant velocity — this is called terminal velocity.
When the parachute opens, the surface area increases dramatically, causing a sudden large increase in air resistance. Air resistance is now greater than the weight, so the resultant force acts upwards. The skydiver decelerates (slows down).
As the skydiver slows down, air resistance decreases again. A new, lower terminal velocity is reached when air resistance once again equals weight. The skydiver descends at this lower constant speed until landing.
graph TD
subgraph "Skydiver Forces Sequence"
A["Jump: Weight >> Air resistance"] -->|"Accelerates"| B["Falling: Weight > Air resistance"]
B -->|"Acceleration decreases"| C["Terminal velocity: Weight = Air resistance"]
C -->|"Parachute opens"| D["Air resistance >> Weight"]
D -->|"Decelerates"| E["New terminal velocity: Weight = Air resistance"]
end
style A fill:#e74c3c,color:#fff
style B fill:#e67e22,color:#fff
style C fill:#27ae60,color:#fff
style D fill:#2980b9,color:#fff
style E fill:#27ae60,color:#fff
Exam Tip: For 6-mark questions, the examiner is looking for a logical sequence of points. Use connective phrases like "as a result," "therefore," "this causes," and "because" to show the chain of cause and effect. Always refer to specific forces by name and use the correct physics terminology.
Understanding command words is essential for answering questions correctly:
Subscribe to continue reading
Get full access to this lesson and all 10 lessons in this course.