Ch 6 · 1.5.2 Friction · O Level Physics 5054

Topic 6 of 17 in Motion, Forces and Energy — O Level Physics 5054 — 78 lessons.

Introduction

Rub your hands together and they warm up. A skydiver stops speeding up after a few seconds. A car on a wet road takes longer to stop. All three are about friction and drag. This lesson explains how resistive forces change motion, how terminal velocity is reached, and what affects a car's stopping distance. (Syllabus section 1.5.2.)

Key ideas — Friction

Friction is a force between surfaces that may impede (oppose) motion and produces heating. Rubbing surfaces get warm because work done against friction is transferred to thermal energy. Friction is not always unwanted: it lets tyres grip the road and lets you walk.

🧑‍🏫 Teacher's explanation

Imagine you are sliding a heavy wooden crate across a carpeted floor. You have to push hard to get it moving, and once it is moving, you have to keep pushing just to keep it at a constant speed. This is because friction is a force that always acts in the opposite direction to the motion. If you rub your hands together quickly, they feel warm; this is because the kinetic energy you provide is being converted into thermal energy due to the friction between your palms.

📖 Definitions

🔍 Worked example — Easy

A wooden block is pulled across a floor with a driving force of 15 N. The frictional force acting against the motion is 12 N. Calculate the resultant force acting on the block.

🔍 Worked example — Standard

A skydiver of mass 80 kg is falling through the air. The weight of the skydiver is W = mg. Taking g = 10 m/s^2, calculate the weight. If the skydiver has reached terminal velocity, what is the magnitude of the air resistance acting on them?

🔍 Worked example — Exam-style

A car is traveling at 20 m/s. The driver's thinking distance is 0.7 s. Calculate the thinking distance.

⚠️ Common errors

🧠 Quick check 1

🧠 Quick check 2

🎯 Exam question

Key ideas — Drag

Drag is the resistive force on an object moving through a fluid: in air it is air resistance, in water it is resistance in a liquid. Drag always acts opposite to the motion and increases as the speed increases.

🧑‍🏫 Teacher's explanation

Imagine you are running through a swimming pool. The faster you try to move, the harder the water pushes back against you. This 'pushing back' is drag. When an object moves through a fluid (like air or water), it has to push the particles of that fluid out of its way. This requires work, which results in a resistive force acting in the opposite direction to the motion.

📖 Definitions

🔍 Worked example — Easy

A skydiver of mass 70 kg falls from a plane. Initially, the only force acting is weight. Calculate the weight of the skydiver. (Use g = 10 m/s^2 for O Level).

🔍 Worked example — Standard

A ball is dropped into a thick liquid. At a certain speed, the drag force is 5.0 N acting upwards. If the weight of the ball is 5.0 N acting downwards, describe the motion of the ball.

🔍 Worked example — Exam-style

A car is traveling at 20 m/s. The driver sees an obstacle and takes 0.6 s to react. Calculate the thinking distance.

⚠️ Common errors

🧠 Quick check 1

🧠 Quick check 2

🎯 Exam question

Key ideas — Motion with a constant force and no drag

If a constant force (a driving force, or weight in free fall) acts and there is no drag, the resultant force stays the same, so the object keeps accelerating at a constant rate (F = ma).

🧑‍🏫 Teacher's explanation

Imagine you are pushing a heavy shopping trolley across a perfectly smooth ice rink. If you push with a steady, constant force and there is absolutely no friction or air resistance to push back, the trolley won't just move at a steady speed; it will actually keep getting faster and faster every single second. This is because there is a 'net' or 'resultant' force acting on it.

📖 Definitions

🔍 Worked example — Easy

A toy car is pushed with a constant driving force of 1.5 N on a frictionless surface. The mass of the car is 0.5 kg. Calculate the acceleration of the car.

🔍 Worked example — Standard

A skydiver of mass 80 kg falls from a plane. When the air resistance (drag) is 784 N, calculate the acceleration of the skydiver. Use g = 10 m/s^2 for weight calculations.

🔍 Worked example — Exam-style

A car is traveling at 20 m/s. The driver sees a red light and takes 0.75 s to react (thinking distance). The brakes then apply a constant deceleration, bringing the car to a stop in 40 m (braking distance). Calculate the total stopping distance.

⚠️ Common errors

🧠 Quick check 1

🧠 Quick check 2

🎯 Exam question

Key ideas — Motion with drag

With drag, the resultant force decreases as the object speeds up, because drag grows while the driving force (or weight) stays the same. The acceleration therefore decreases.

🧑‍🏫 Teacher's explanation

Imagine you are running through a swimming pool. At the start, you can push off the wall and accelerate easily. But as you move faster, the water pushes back against you much harder. This 'push back' is drag. Because the water's resistance increases as you speed up, your acceleration isn't constant; it actually drops until you reach a speed where the water's push is exactly equal to your swimming force. At that point, you stop accelerating and move at a steady speed.

📖 Definitions

🔍 Worked example — Easy

A skydiver of mass m = 80 kg falls from a plane. Initially, the only force acting is weight W = mg. Calculate the weight of the skydiver. (Use g = 10 m/s^2)

🔍 Worked example — Standard

A car is traveling at 20 m/s. The driver sees a red light and takes 0.6 s to react. Calculate the thinking distance.

🔍 Worked example — Exam-style

A stone falls through thick oil. At a certain moment, the weight of the stone is 5.0 N downwards and the drag force is 2.0 N upwards. Calculate the resultant force acting on the stone.

⚠️ Common errors

🧠 Quick check 1

🧠 Quick check 2

🎯 Exam question

Key ideas — Terminal velocity

Follow a skydiver:

🧑‍🏫 Teacher's explanation

Imagine you are sliding a heavy box across a carpet. You push hard, and it moves. But as you push, you feel the carpet rubbing against the bottom of the box, making it harder to move. This 'rubbing' is friction. Friction always acts in the opposite direction to motion, and because it works against the movement, it converts kinetic energy into heat, making the surfaces warmer.

📖 Definitions

🔍 Worked example — Easy

A skydiver has a weight of 800 N. When they reach terminal velocity, the air resistance is 800 N. Calculate the resultant force acting on the skydiver.

🔍 Worked example — Standard

A ball falls through a thick liquid. Its weight is 5.0 N. At a certain speed, the drag force is 2.0 N. Calculate the resultant force.

🔍 Worked example — Exam-style

A car is traveling at 20 m/s. The driver sees a red light and takes 0.7 s to react. Calculate the thinking distance.

⚠️ Common errors

🧠 Quick check 1

🧠 Quick check 2

🎯 Exam question

(a) Explain why the skydiver's acceleration decreases as they fall faster. [2] (b) Describe the state of motion of the skydiver when they reach terminal velocity. [1] [3 marks]

Key ideas — Stopping a vehicle

between seeing the hazard and applying the brakes.

🧑‍🏫 Teacher's explanation

Imagine you are playing a game of 'Statues'. You see a friend move, and there is a tiny delay between your brain seeing the movement and your muscles actually jumping. That delay is like 'thinking distance'. Once you actually jump, the distance you travel before landing is like 'braking distance'. Together, they make up your total movement.

📖 Definitions

🔍 Worked example — Easy

A car is travelling at 20 m/s. The driver's reaction time is 0.5 s. Calculate the thinking distance.

🔍 Worked example — Standard

A car travels at 15 m/s. The thinking distance is 5 m and the braking distance is 12 m. Calculate the total stopping distance.

🔍 Worked example — Exam-style

A car is moving at 30 m/s. The driver sees a hazard and takes 0.7 s to react. The car then brakes to a stop over a distance of 45 m. Calculate the total stopping distance and state one factor that would increase the braking distance.

⚠️ Common errors

🧠 Quick check 1

🧠 Quick check 2

🎯 Exam question

(a) Calculate the total stopping distance. [2] (b) State two factors that would increase the braking distance. [2] [4 marks]

Key ideas — Factors affecting stopping distance

🧑‍🏫 Teacher's explanation

When you are driving, your total stopping distance is the sum of two very different processes. First, there is the 'thinking distance', which is the distance your car travels while your brain is processing the need to brake. Think of this like the time it takes for a signal to travel from your eyes to your foot; if you are distracted or tired, that signal takes longer to arrive.

📖 Definitions

🔍 Worked example — Easy

A car is travelling at 20 m/s. The driver's reaction time is 0.5 s. Calculate the thinking distance.

🔍 Worked example — Standard

A car travelling at 15 m/s has a braking distance of 12 m. Calculate the total stopping distance.

⚠️ Common errors

🧠 Quick check 1

🧠 Quick check 2

🎯 Exam question

(a) Calculate the thinking distance. (b) Calculate the total stopping distance. (c) State one factor that would increase the braking distance. [4 marks]

Worked example

Example 1: thinking distance. A driver travelling at 20 m/s has a reaction time of 0.70 s.

🔍 Worked example — Easy

A car is travelling at 25 m/s. The driver takes 0.65 s to react to a red light. Calculate the thinking distance.

🔍 Worked example — Standard

A skydiver is falling through the air. Initially, the only force acting on them is their weight (W). As they speed up, air resistance (D) increases. Describe the change in acceleration and explain why the skydiver eventually reaches terminal velocity.

🔍 Worked example — Exam-style

A car travelling at 30 m/s has a thinking distance of 12 m and a braking distance of 45 m. Calculate the total stopping distance.

⚠️ Common errors

Common mistakes

Practice

Answers: (1) Friction between the brake pads and the wheel does work, transferring kinetic energy into thermal energy. (2) Air resistance increases with speed until it equals the driving force; the resultant force is then zero, so there is no further acceleration. (3) 25 × 0.60 = 15\ m. (4) Braking distance: wet or icy road, worn tyres, greater load, higher speed (any two). Thinking distance only: tiredness, alcohol or drugs. (5) Air resistance becomes greater than weight, so there is an upward resultant force and she decelerates; as she slows, air resistance falls until it equals her weight again and she falls at a new, lower terminal velocity.

Ch 6 · 1.5.2 Friction · O Level Physics 5054

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