FORCESAND PRESSURE

KS3 Physics · Lessons 1–7
⬇ Student Booklet ⬇ Teacher Booklet

Key Terms — Forces and Pressure

All Terms

01
Force
DefinitionA push, a pull or a twist, arising from the interaction between two objects. Measured in newtons (N).
02
Contact force
DefinitionA force that acts only while two objects are physically touching — friction, drag and the reaction force are all contact forces.
03
Friction
DefinitionA contact force caused by two surfaces moving over each other. It always acts in the opposite direction to the movement.
04
Contact points
DefinitionThe few tiny points where two surfaces actually touch. Surfaces that look smooth are covered in microscopic peaks and hollows.
05
Lubrication
DefinitionAdding oil or grease between two surfaces so they slide past each other more easily, reducing friction.
06
Thermal energy store
DefinitionFriction transfers energy to the thermal energy store of the surfaces. This is why rubbing your hands warms them and why brake blocks wear away and get hot.
07
Drag force
DefinitionThe resistance force on an object moving through a fluid. It increases with speed and with the area facing the direction of motion.
08
Air resistance
DefinitionThe drag force acting on an object moving through air.
09
Water resistance
DefinitionThe drag force acting on an object moving through water.
10
Streamlined
DefinitionA smooth, tapered shape that lets air or water flow around it easily instead of being churned up behind it, so the drag force is reduced.
11
Thrust
DefinitionThe forward driving force produced by an engine.
12
Resultant force
DefinitionThe single force you get by adding together all the forces acting on an object. A resultant force changes the object's motion.
13
Balanced forces
DefinitionForces that are equal in size and opposite in direction, so the resultant force is zero and the object stays still or keeps moving at a steady speed.
14
Newton's first law
DefinitionThe speed of an object stays the same unless a resultant force acts on it.
15
Newton's second law
DefinitionThe larger the resultant force, the larger the acceleration.
16
Newton's third law
DefinitionIf object A pushes on object B, then B pushes back on A with an equal force in the opposite direction.
17
Terminal velocity
DefinitionThe maximum steady speed of a falling object, reached when the drag force has grown until it balances the weight, so the resultant force is zero.
18
Weight
DefinitionThe downward force on an object due to gravity, measured in newtons.

Lesson 1 · Friction and Drag

Do Now

Q1
What is a force, and what unit is it measured in?
Model AnswerA push, a pull or a twist. Measured in newtons (N).
Q2
What does it mean if the forces on an object are balanced?
Model AnswerThe resultant force is zero, so the object stays still or keeps moving at the same speed.
Q3
Name one contact force and one non-contact force.
Model AnswerContact: friction / air resistance / reaction force. Non-contact: weight (gravity) / magnetism.

Part 1 · Friction

Read the passage, then answer the questions.

Friction is a contact force caused by two surfaces moving over each other. Surfaces that look and feel smooth are not really smooth at all. If you look at wood, glass or plastic under a microscope you see peaks and hollows.

Fig 1.1 — A piece of wood looks and feels perfectly smooth (left). Under a scanning electron microscope the same surface is a mass of peaks and hollows (right). The scale bar on the magnified image is 20 micrometres.Fig 1.1 — A piece of wood looks and feels perfectly smooth (left). Under a scanning electron microscope the same surface is a mass of peaks and hollows (right). The scale bar on the magnified image is 20 micrometres.
Fig 1.1 — A piece of wood looks and feels perfectly smooth (left). Under a scanning electron microscope the same surface is a mass of peaks and hollows (right). The scale bar on the magnified image is 20 micrometres.

When two surfaces slide past one another these tiny peaks catch on each other, and this is what produces friction. Friction always acts in the opposite direction to the movement, so it makes objects harder to move and slows moving objects down. Rough surfaces produce more friction than smooth surfaces. Friction also transfers energy to the thermal energy store of the surfaces, which is why rubbing your hands together warms them up, and why brake blocks wear away.

Fig 1.2 — The block and the surface look perfectly smooth, but magnified they touch at only a few contact points.
Fig 1.2 — The block and the surface look perfectly smooth, but magnified they touch at only a few contact points.

Friction is often useful. You need friction between your shoes and the ground in order to walk, and the brakes on a bicycle or a car only work because of friction. At other times friction is a nuisance: it wears away tyres and moving parts, and it wastes energy.

You can reduce friction by lubricating a surface with oil or grease so the surfaces slide past each other more easily, or by polishing a surface so it is smoother. You can increase friction by using a rougher surface, such as the tread on a tyre or the studs on a boot.

Key idea Friction is a contact force that acts in the opposite direction to movement. Rough surfaces produce more friction than smooth surfaces.

Questions — Friction

Q1
State what causes friction. (1 mark)
Model AnswerTwo surfaces moving over each other (the rough peaks on the surfaces catching).
Q2
State the direction in which friction acts. (1 mark)
Model AnswerIn the opposite direction to the movement.
Q3
State whether friction is a contact or a non-contact force. (1 mark)
Model AnswerA contact force.
Q4
Describe why a rough surface produces more friction than a smooth surface. (2 marks)
Model AnswerA rough surface has larger peaks and hollows, so more of them catch on the other surface, producing a larger frictional force.
Q5
Give two examples of friction being useful. (2 marks)
Model AnswerAny two of: walking (grip between shoe and ground); braking on a bike or in a car; holding a pencil; a nail staying in wood.
Q6
State two ways of reducing the friction between two surfaces. (2 marks)
Model AnswerLubricate with oil or grease; polish or smooth the surfaces.
Q7
A cyclist's brake blocks wear away over time. Explain why. (2 marks)
Model AnswerThe brake blocks rub against the wheel rim, so there is friction between them. Friction wears material away from the softer brake block each time the brakes are used.
Q8
A student uses a newton meter to pull a wooden block across four different surfaces, recording the force needed to make the block start moving on each one. (3 marks)(a) State the independent variable.(b) State the dependent variable.(c) State one control variable.
Model Answer(a) The type of surface.(b) The force needed to make the block start moving.(c) The block used / the mass on the block / the speed of pulling.

Part 2 · Drag Forces

Read the passage and study Fig 1.3 and Fig 1.4, then answer the questions.

Any object moving through air or water has a resistance force acting on it that slows it down. This is called a drag force. Air resistance and water resistance are both drag forces. They happen because the object has to push the particles of air or water out of the way as it moves.

Drag always acts in the opposite direction to the motion, in the same way that friction does. The faster an object moves, the greater the drag force on it. The larger the surface area facing the direction of travel, the greater the drag force as well.

Fig 1.3 — The four forces acting on a moving car.
Fig 1.3 — The four forces acting on a moving car.
Fig 1.4 — Water resistance is the drag force on a submarine.
Fig 1.4 — Water resistance is the drag force on a submarine.
Fig 1.5 — A blunt shape has to push more air aside than a streamlined shape.
Fig 1.5 — A blunt shape has to push more air aside than a streamlined shape.

Engineers reduce drag by making vehicles streamlined. A streamlined shape is smooth and tapered so that air flows around it easily instead of being churned up behind it. Racing cyclists crouch low and wear smooth helmets, and lorries have curved deflectors on their roofs, all for the same reason. A peregrine falcon tucks its wings in when it dives, which makes it more streamlined so it can reach a much higher speed.

Sometimes a large drag force is wanted. A parachute has a very large surface area so that the drag force on it is big enough to slow a falling skydiver down safely.

Key idea Drag is the resistance force on an object moving through a fluid. It increases with speed and with the area facing the direction of motion.

Questions — Drag Forces

Q9
State the name given to the drag force acting on an object moving through air. (1 mark)
Model AnswerAir resistance.
Q10
Describe what causes a drag force. (2 marks)
Model AnswerThe object has to push the particles of the air or water out of its way, and these particles push back on the object in the opposite direction to its motion.
Q11
State what happens to the drag force on a car as its speed increases. (1 mark)
Model AnswerThe drag force increases.
Q12
Describe what is meant by a streamlined shape. (2 marks)
Model AnswerA smooth, tapered shape that lets air or water flow around it easily, so the drag force on it is reduced.
Q13
Give two ways a racing cyclist reduces the drag force acting on them. (2 marks)
Model AnswerAny two of: crouching low over the handlebars; wearing tight-fitting clothing; wearing a smooth streamlined helmet; riding closely behind another rider.
Q14
Explain why a peregrine falcon tucks its wings into its body when it dives. (2 marks)
Model AnswerTucking the wings in reduces the area facing the direction of motion and makes the bird more streamlined. This reduces the drag force, so the falcon reaches a higher speed.
Q15
Use Fig 1.3 to answer the following. (3 marks)(a) State whether the vertical forces on the car are balanced. Give a reason.(b) Calculate the resultant horizontal force on the car.(c) State what will happen to the speed of the car.
Model Answer(a) Balanced — the reaction force and the weight are both 12 000 N and act in opposite directions.(b) 10 000 − 6 000 = 4000 N forwards.(c) It will increase — the car accelerates forwards.
Q16
A dragster releases a parachute at the end of a race. (3 marks)(a) State what happens to the area facing the direction of motion.(b) State what happens to the drag force on the car.(c) State what happens to the speed of the car.
Model Answer(a) It increases greatly.(b) The drag force increases greatly.(c) The car slows down (decelerates).

Part 3 · Terminal Velocity

Read the passage and study the graph, then answer the questions.

We use Newton's laws to describe how forces change motion. Newton's first law says the speed of an object stays the same unless a resultant force acts on it. Newton's second law says the larger the resultant force, the larger the acceleration. Newton's third law says if object A pushes on object B, then B pushes back on A with an equal force in the opposite direction.

When a skydiver jumps from an aeroplane, the only force acting at first is their weight, so there is a large resultant force downwards and they accelerate. As they speed up, the drag force on them increases. The resultant force downwards gets smaller, so the acceleration gets smaller too.

Eventually the drag force becomes equal to the weight. The forces are now balanced, the resultant force is zero and the acceleration is zero, so the skydiver falls at a steady speed. This maximum steady speed is called the terminal velocity.

When the parachute opens, the surface area increases sharply and so does the drag force. There is now a resultant force upwards, so the skydiver decelerates. As they slow down the drag force falls again until it equals the weight, and they reach a second, much slower terminal velocity — slow enough to land safely.

Fig 1.6 — Speed–time graph for a skydiver, showing two terminal velocities.
Fig 1.6 — Speed–time graph for a skydiver, showing two terminal velocities.
Key idea An object reaches terminal velocity when the drag force on it has grown until it balances the weight, so the resultant force is zero.

Questions — Terminal Velocity

Q17
State what happens to the drag force on a falling object as its speed increases. (1 mark)
Model AnswerThe drag force increases.
Q18
State Newton's first law. (1 mark)
Model AnswerThe speed of an object stays the same unless a resultant force acts on it.
Q19
A skydiver has just jumped from the aeroplane. State the only force acting on them at that moment. (1 mark)
Model AnswerTheir weight.
Q20
Explain why the skydiver's acceleration decreases as they fall. (3 marks)
Model AnswerAs the skydiver speeds up the drag force increases. The weight stays the same, so the resultant force downwards gets smaller. By Newton's second law a smaller resultant force gives a smaller acceleration.
Q21
Explain how a skydiver reaches terminal velocity. (3 marks)
Model AnswerThe drag force increases as the skydiver speeds up until it is equal to the weight. The forces are then balanced and the resultant force is zero, so the skydiver stops accelerating and falls at a constant maximum speed.
Q22
State and explain what happens to the skydiver's speed immediately after the parachute opens. (3 marks)
Model AnswerThe speed decreases. The parachute greatly increases the area facing the motion, so the drag force becomes larger than the weight and there is a resultant force upwards, which decelerates the skydiver.
Q23
Use the speed–time graph in Fig 1.6 to answer the following. (3 marks)(a) Mark with an A a region where the skydiver is accelerating.(b) Mark with a T the two regions of terminal velocity.(c) Mark with a P the moment the parachute opens.
Model Answer(a) Any point on the rising part of the curve, between 0 s and about 13 s.(b) The flat section at 55 m/s (about 13–20 s) and the flat section at 8 m/s (after about 30 s).(c) At 20 s, where the graph turns sharply downwards.

Exam-style question

Q1
A lorry is travelling along a motorway at a steady speed. The driver fits a curved deflector to the roof of the cab. (6 marks)
Fig 1.7 — A lorry fitted with a roof deflector.
Fig 1.7 — A lorry fitted with a roof deflector.
(a) The lorry travels at a steady speed. State what this tells you about the forces acting on the lorry. (1 mark)(b) Explain how the deflector reduces the amount of fuel the lorry uses. (3 marks)(c) The lorry driver brakes hard. Explain, in terms of friction, why the brake discs become hot. (2 marks)
Mark Scheme(a) The forces are balanced — the resultant force on the lorry is zero.(b) The deflector makes the lorry more streamlined so air flows around it more easily. This reduces the drag force on the lorry. A smaller forward force (thrust) is then needed to keep the lorry moving at the same steady speed, so less fuel is used.(c) There is friction between the brake pads and the discs as they rub together. Friction transfers energy to the thermal energy store of the discs, so their temperature rises.

Exit Ticket

Q1
Name the two contact forces met in this lesson that oppose motion.
Model AnswerFriction and drag.
Q2
State one way of increasing friction and one way of decreasing it.
Model AnswerIncrease: use a rougher surface (tread, studs). Decrease: lubricate with oil or grease.
Q3
Write a sentence defining terminal velocity.
Model AnswerThe maximum steady speed reached when the drag force on a falling object balances its weight.

Lesson 2 · Squashing and Stretching

Do Now

Q1
What is a contact force?
Model AnswerA force that acts only when two objects are physically touching.
Q2
What happens to an object when the forces on it are unbalanced?
Model AnswerIt accelerates — it speeds up, slows down or changes direction.
Q3
How are the particles arranged in a solid?
Model AnswerIn a fixed, regular pattern, touching their neighbours and held by strong bonds.

Part 1 · Deformation and the Reaction Force

Read the passage, then answer the questions.

When you apply a force to an object you can change its shape. Changing the shape of an object is called deformation. You can compress an object, which means squashing it so it gets shorter, or you can stretch it, which means making it longer without breaking it.

The particles in a solid are held together by strong bonds. These bonds behave like tiny springs. When you stand on the floor, your weight pushes the particles closer together and the bonds are compressed. Squashed bonds push back, and this outward push is what we call the reaction force. It is the reason you do not fall through the chair you are sitting on.

When an object is stretched instead, the bonds are pulled further apart and they pull back. This pulling force in a stretched object is called tension. A bungee cord stretches as the jumper falls, and the tension in the cord eventually pulls the jumper back upwards.

Fig 2.1 — Bonds behave like springs. The floor is squashed under your feet and pushes back; a bungee cord is stretched and pulls back.
Fig 2.1 — Bonds behave like springs. The floor is squashed under your feet and pushes back; a bungee cord is stretched and pulls back.
Key idea Deformation is a change of shape. A compressed surface pushes back (reaction force) and a stretched object pulls back (tension).

Both the reaction force and tension are contact forces — they exist only while the two objects are touching.

Questions — Deformation and the Reaction Force

Q1
State what is meant by deformation. (1 mark)
Model AnswerA change in the shape of an object.
Q2
State the difference between compressing and stretching an object. (2 marks)
Model AnswerCompressing squashes the object so it becomes shorter; stretching pulls it so it becomes longer without breaking.
Q3
State the name of the force in a stretched bungee cord. (1 mark)
Model AnswerTension.
Q4
Explain, in terms of particles and bonds, why you do not fall through your chair. (3 marks)
Model AnswerYour weight pushes down on the chair, compressing the bonds between its particles. The compressed bonds push back like springs, producing an upward reaction force that is equal to your weight, so the forces are balanced.
Q5
Describe what happens to a tennis ball when it hits the ground. (2 marks)
Model AnswerThe force from the ground compresses (squashes) the ball, deforming it. The ball then pushes back and returns to its original shape, which pushes it off the ground.
Q6
A footballer heads a ball. (2 marks)(a) State which object or objects are deformed.(b) Name the force the ball exerts back on the player's head.
Model Answer(a) Both the ball and the player's head.(b) The reaction force.
Q7
State whether the reaction force is a contact or a non-contact force. Give a reason for your answer. (2 marks)
Model AnswerA contact force, because it only acts while the two objects are touching each other.

Part 2 · Hooke's Law and the Elastic Limit

Read the passage and study Fig 2.2 and Fig 2.3, then answer the questions.

When a force is applied to a spring, the spring stretches. The increase in length is called the extension. To find the extension you subtract the original length of the spring from its new length.

If you hang one mass on a spring you get an extension. If you hang two identical masses on the same spring you double the force, and the extension doubles as well. This relationship is called Hooke's law: the extension of a spring is directly proportional to the force applied to it.

Fig 2.2 — Doubling the force doubles the extension. Image: Svjo, CC BY-SA 3.0, via Wikimedia Commons.
Fig 2.2 — Doubling the force doubles the extension. Image: Svjo, CC BY-SA 3.0, via Wikimedia Commons.
Fig 2.3 — Force–extension graph for a spring.
Fig 2.3 — Force–extension graph for a spring.

If you plot a graph of force against extension for a spring obeying Hooke's law you get a straight line that passes through the origin. A straight line through the origin is what directly proportional looks like on a graph.

If you keep adding force, you eventually reach a point where the spring is stretched so far that it no longer returns to its original length when the load is removed. This point is called the elastic limit. Past the elastic limit the graph is no longer a straight line, and the spring is permanently deformed.

Key idea Hooke's law: extension is directly proportional to the force applied — but only up to the elastic limit.

Questions — Hooke's Law and the Elastic Limit

Q8
State what is meant by the extension of a spring. (1 mark)
Model AnswerThe increase in length of the spring when a force is applied (new length − original length).
Q9
State Hooke's law. (1 mark)
Model AnswerThe extension of a spring is directly proportional to the force applied to it.
Q10
A spring is 5 cm long. When a force is applied it becomes 8 cm long. Calculate the extension of the spring. (2 marks)
Model Answer8 − 5 = 3 cm.
Q11
A spring extends by 3 cm when a force of 2 N is applied. State the extension when a force of 4 N is applied. (1 mark)
Model Answer6 cm.
Q12
Describe the shape of a force–extension graph for a spring obeying Hooke's law. (2 marks)
Model AnswerA straight line that passes through the origin.
Q13
State what is meant by the elastic limit. (2 marks)
Model AnswerThe point beyond which a spring no longer returns to its original length when the load is removed; the spring is permanently deformed.
Q14
Describe how the graph in Fig 2.3 changes beyond the elastic limit. (2 marks)
Model AnswerThe line is no longer straight — it curves, so the extension is no longer directly proportional to the force.

Part 3 · The Spring Constant

Read the passage, then work through each worked example and the questions that follow it.

Hooke's law can be written as an equation. The force applied to a spring is equal to the spring constant multiplied by the extension.

The spring constant, k, tells you how stiff a spring is. It is the force needed to extend the spring by one metre, so it is measured in newtons per metre (N/m). A spring with a high spring constant is stiff: it needs a large force to stretch it. A spring with a low spring constant is easy to stretch.

Force = spring constant × extension     F = k e

F is measured in newtons (N), k in newtons per metre (N/m) and e in metres (m).

Worked Example — Calculating a Force
V
A spring has a spring constant of 4 N/m. It is extended by 0.1 m. Calculate the force applied to the spring.
Variables k = 4 N/m · e = 0.1 m · F = F
E
Equation F = k e
S
Substitute F = 4 × 0.1
S
Solve F = 0.4
U
Units F = 0.4 N

Questions — set out your working using VESSU

Questions — Calculating a Force

Q15
A spring has a spring constant of 20 N/m and is extended by 0.05 m. Calculate the force applied. (2 marks)
Model AnswerF = k e = 20 × 0.05 = 1 N.
Q16
A rubber band has a spring constant of 10 N/m and is extended by 0.2 m. Calculate the force applied. (2 marks)
Model AnswerF = k e = 10 × 0.2 = 2 N.
Q17
A spring has a spring constant of 50 N/m and is extended by 0.1 m. Calculate the force applied. (2 marks)
Model AnswerF = k e = 50 × 0.1 = 5 N.
Q18
A spring has a spring constant of 8 N/m and is extended by 0.25 m. Calculate the force applied. (2 marks)
Model AnswerF = k e = 8 × 0.25 = 2 N.
Q19
A spring has a spring constant of 120 N/m and is extended by 0.05 m. Calculate the force applied. (2 marks)
Model AnswerF = k e = 120 × 0.05 = 6 N.
Q20
A spring has a spring constant of 15 N/m and is extended by 0.6 m. Calculate the force applied. (2 marks)
Model AnswerF = k e = 15 × 0.6 = 9 N.
Q21
A spring has a spring constant of 250 N/m and is extended by 0.02 m. Calculate the force applied. (2 marks)
Model AnswerF = k e = 250 × 0.02 = 5 N.
Q22
A bungee cord has a spring constant of 3.5 N/m and is extended by 2 m. Calculate the force applied. (2 marks)
Model AnswerF = k e = 3.5 × 2 = 7 N.
Worked Example — Calculating an Extension
V
A rubber band has a spring constant of 60 N/m and is pulled with a force of 15 N. Calculate the extension of the rubber band.
Variables F = 15 N · k = 60 N/m · e = e
E
Equation F = k e
S
Substitute 15 = 60 × e
S
Solve e = 1560 = 0.25
U
Units e = 0.25 m

Questions — set out your working using VESSU

Questions — Calculating an Extension

Q23
A spring has a spring constant of 200 N/m and is stretched with a force of 15 N. Calculate the extension. (2 marks)
Model Answere = F ÷ k = 15 ÷ 200 = 0.075 m.
Q24
A spring has a spring constant of 50 N/m and a force of 20 N is applied to it. Calculate the extension. (2 marks)
Model Answere = F ÷ k = 20 ÷ 50 = 0.4 m.
Q25
A force of 6 N is applied to a spring with a spring constant of 30 N/m. Calculate the extension. (2 marks)
Model Answere = F ÷ k = 6 ÷ 30 = 0.2 m.
Q26
A force of 12 N is applied to a spring with a spring constant of 40 N/m. Calculate the extension. (2 marks)
Model Answere = F ÷ k = 12 ÷ 40 = 0.3 m.
Q27
A force of 2.5 N is applied to a spring with a spring constant of 25 N/m. Calculate the extension. (2 marks)
Model Answere = F ÷ k = 2.5 ÷ 25 = 0.1 m.
Q28
A force of 45 N is applied to a spring with a spring constant of 90 N/m. Calculate the extension. (2 marks)
Model Answere = F ÷ k = 45 ÷ 90 = 0.5 m.
Q29
A force of 9 N is applied to a spring with a spring constant of 300 N/m. Calculate the extension. (2 marks)
Model Answere = F ÷ k = 9 ÷ 300 = 0.03 m.
Q30
A force of 1.8 N is applied to a spring with a spring constant of 12 N/m. Calculate the extension. (2 marks)
Model Answere = F ÷ k = 1.8 ÷ 12 = 0.15 m.

Forces are not always given to you in newtons. A large force may be written in kilonewtons and a very small one in millinewtons. Before you substitute a force into F = k e you must always convert it into newtons first.

Complete the multiplier column, then use it in the questions below.

Unit prefix conversion
PrefixSymbolUnitMultiply the number by
megaMMN1 000 000
kilokkN1000
millimmN0.001
microµµN0.000 001
Worked Example — Spring Constant from a Force in Kilonewtons
V
A steel spring is stretched by a force of 2.4 kN. The extension is 0.6 m. Calculate the spring constant of the spring.
Variables F = 2.4 kN = 2.4 × 1000 = 2400 N · e = 0.6 m · k = k
E
Equation F = k e
S
Substitute 2400 = k × 0.6
S
Solve k = 24000.6 = 4000
U
Units k = 4000 N/m

Questions — convert each force into newtons first, then use VESSU

Questions — Converting Units First

Q31
A force of 3 kN extends a spring by 0.5 m. Calculate the spring constant. Give the unit. (3 marks)
Model Answer3 kN = 3000 N. k = F ÷ e = 3000 ÷ 0.5 = 6000 N/m.
Q32
A force of 1.5 kN extends a spring by 0.25 m. Calculate the spring constant. (2 marks)
Model Answer1.5 kN = 1500 N. k = 1500 ÷ 0.25 = 6000 N/m.
Q33
A force of 250 mN extends a delicate spring by 0.05 m. Calculate the spring constant. (2 marks)
Model Answer250 mN = 0.25 N. k = 0.25 ÷ 0.05 = 5 N/m.
Q34
A force of 800 mN extends a spring by 0.4 m. Calculate the spring constant. (2 marks)
Model Answer800 mN = 0.8 N. k = 0.8 ÷ 0.4 = 2 N/m.
Q35
A force of 60 mN extends a spring by 0.02 m. Calculate the spring constant. (2 marks)
Model Answer60 mN = 0.06 N. k = 0.06 ÷ 0.02 = 3 N/m.
Q36
A bridge cable is stretched by a force of 0.5 MN and extends by 2 m. Calculate the spring constant of the cable. (3 marks)
Model Answer0.5 MN = 500 000 N. k = 500 000 ÷ 2 = 250 000 N/m.
Q37
The same force is applied to Spring X, which has a high spring constant, and Spring Y, which has a low spring constant. State and explain which spring has the larger extension. (2 marks)
Model AnswerSpring Y. It has the lower spring constant, so it is less stiff and extends more for the same force.

Part 4 · Planning the Spring Practical

Read the apparatus list and method, study Fig 2.4, then complete the risk assessment and answer the questions.

Fig 2.4 — The apparatus for the spring practical.
Fig 2.4 — The apparatus for the spring practical.

Practical — Investigating the extension of a spring

Apparatus: clamp stand and boss, spring, 100 g slotted masses and hanger, pointer, metre rule, safety goggles, cushioned tray

  1. Set up the apparatus as shown in Fig 2.4: hang the spring from the clamp, fix the pointer to the bottom of the mass hanger, and stand the metre rule vertically beside it.
  2. Read the position of the pointer against the ruler. This is your starting reading.
  3. Add one 100 g mass (a weight of about 1 N) and read the new pointer position.
  4. Subtract the starting reading from the new reading to find the extension.
  5. Repeat, adding one mass at a time, up to five masses. Take each reading three times and find a mean.

Risk assessment — complete the two right-hand columns before you start.

HazardRisk — what could happenControl — how you will reduce the risk
Falling massesThey land on a foot, or bounce off the benchStand a cushioned tray under the masses; keep feet clear
An over-stretched springIt flies off the clamp and hits someone in the eyeWear safety goggles; never add more than five masses
A top-heavy clamp standThe stand topples and falls off the benchWeight or clamp the base; work away from the bench edge

Questions — Planning the Practical

Q38
State the independent variable in this investigation. (1 mark)
Model AnswerThe force applied to the spring (the number of masses).
Q39
State the dependent variable. (1 mark)
Model AnswerThe extension of the spring.
Q40
State one control variable. (1 mark)
Model AnswerThe spring used (or the point on the ruler the pointer is read against).
Q41
Explain why each reading is taken three times and a mean calculated. (2 marks)
Model AnswerRepeating and taking a mean reduces the effect of random errors, which makes the result more reliable.
Q42
Explain why the metre rule must be held vertically. (2 marks)
Model AnswerIf the rule is tilted the measured length is longer than the true vertical length, so every reading would be too large.
Q43
Explain why safety goggles are needed for this practical. (2 marks)
Model AnswerA spring stretched past its elastic limit can spring off the clamp with a lot of energy and could damage an eye.
Q44
A student suggests using a 1 kg mass on its own instead of five 100 g masses. Explain why this would spoil the investigation. (2 marks)
Model AnswerYou would only get one reading, so you could not see how the extension changes as the force increases and could not plot a graph.

Exit Ticket

Q1
Give one word for a change in the shape of an object.
Model AnswerDeformation.
Q2
Complete: the extension of a spring is directly ______ to the force applied.
Model Answerproportional.
Q3
What does a spring constant of 30 N/m mean?
Model AnswerA force of 30 N is needed to extend the spring by one metre.

Lesson 3 · Practical: Investigating a Spring

Do Now

Q1
What is meant by the extension of a spring?
Model AnswerThe increase in length when a force is applied (new length − original length).
Q2
State Hooke's law.
Model AnswerThe extension of a spring is directly proportional to the force applied to it.
Q3
What is the weight, in newtons, of a 100 g mass?
Model AnswerAbout 1 N.

Part 1 · Apparatus and Method

Set up the apparatus as shown in Fig 3.1 and follow the method carefully.

Fig 3.1 — The apparatus for the spring practical.
Fig 3.1 — The apparatus for the spring practical.

Practical — Investigating the extension of a spring

Apparatus: clamp stand and boss, spring, 100 g slotted masses and hanger, pointer, metre rule, safety goggles, cushioned tray

  1. Set up the apparatus as shown in Fig 3.1: hang the spring from the clamp, fix the pointer to the bottom of the mass hanger, and stand the metre rule vertically beside it.
  2. Read the position of the pointer against the ruler and record it. This is your starting reading.
  3. Add one 100 g mass (a weight of about 1 N) and read the new pointer position.
  4. Subtract the starting reading from the new reading to find the extension.
  5. Repeat, adding one mass at a time, up to five masses. Take each reading three times and find a mean.
  6. Remove the masses one at a time and check the spring returns to its original length.
Risk assessment Wear safety goggles throughout. Keep the cushioned tray under the masses and your feet clear of the bench. Do not add more than five masses.

Part 2 · Results Table

Record your measurements in the table as you go. Fill in every column.

A good results table is drawn before you start collecting data. Every column needs a heading and a unit, and the independent variable goes in the first column. Write down every reading as you take it rather than trying to remember it.

The mean is the average of your three readings. The extension is the mean reading minus your starting reading, so the extension in the top row will always be zero.

Force (N)Reading 1 (cm)Reading 2 (cm)Reading 3 (cm)Mean (cm)Extension (cm)
0
1
2
3
4
5

If one of your three readings is very different from the other two, it is an anomalous result — leave it out when you calculate the mean.

Questions — Results Table

Q1
Highlight any anomalies on your results table. (2 marks)
Model AnswerAny reading that does not fit the pattern of the other two readings for that force should be highlighted.
Q2
Calculate the mean for each force step. (3 marks)
Model AnswerAdd the readings for each force and divide by how many you used, leaving out any highlighted anomaly. Record each value in the Mean column.

Part 3 · Drawing a Graph

Graph paper for plotting your results.

Use the checklist to make sure your graph is drawn properly. Tick each box once you have checked it, then swap booklets with a partner and check theirs.

Graph Checklist

  • Force is on the y-axis (up the side) and extension is on the x-axis (across the bottom).
  • Both axes are labelled with the quantity and its unit.
  • The graph has a title written in the space above the grid.
  • The scale uses at least half of the grid in each direction.
  • Each square is worth a sensible amount — 1, 2, 5 or 10, never 3 or 7.
  • Every point is plotted with a small, neat cross.
  • A single straight line of best fit is drawn with a ruler.
  • There are roughly as many points above the line as below it.
  • The points are not joined dot-to-dot.

Part 4 · Conclusion and Evaluation

Use your graph to answer the questions and write your conclusion.

Key idea A conclusion answers the question. An evaluation judges how much you can trust the answer.

A conclusion says what your results show, and uses your data to back it up. If your line of best fit is straight and passes through the origin, the extension is directly proportional to the force and the spring is obeying Hooke's law.

Conclusion Questions

Q3
Describe the pattern shown by your graph. (2 marks)
Model AnswerAs the force increases the extension increases by the same amount each time — a straight line through the origin.
Q4
State what your graph tells you about the relationship between force and extension. (2 marks)
Model AnswerThe extension is directly proportional to the force applied.
Q5
Name the law that describes this relationship. (1 mark)
Model AnswerHooke's law.
Q6
Use your graph to find the extension produced by a force of 2.5 N. (2 marks)
Model AnswerRead across from 2.5 N on the force axis to the line of best fit, then down to the extension axis. Any answer consistent with the student's own graph.
Q7
Explain how you could use your results to calculate the spring constant of your spring. (3 marks)
Model AnswerPick a point on the line of best fit and read off the force and the extension. Convert the extension into metres, then use k = F ÷ e.

An evaluation is different: it comments on how good the results are. Think about whether the points lie close to the line, whether there were any anomalous results, and what you would change if you did the practical again.

Evaluation Questions

Q8
State whether your points lay close to your line of best fit, and explain what this tells you about your results. (2 marks)
Model AnswerPoints lying close to the line show the results are precise, with little random error. Points scattered away from the line show larger random errors.
Q9
State whether you had any anomalous results, and describe what you did about them. (2 marks)
Model AnswerAny anomaly should have been highlighted, re-measured if possible, and left out of the mean for that force.
Q10
Suggest one thing that could have made your readings less accurate. (2 marks)
Model AnswerAny sensible source of error, e.g. reading the ruler at the wrong eye level; the pointer being too thick to read precisely; the rule not held vertically.
Q11
Suggest one improvement you would make if you repeated this practical. (2 marks)
Model AnswerAny sensible improvement, e.g. use a thinner pointer so the reading is more precise; take more than three readings; use smaller mass steps.
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