# Forces, Elasticity and Newton's Laws

> CIE IGCSE Physics · 2026-2028 CIE IGCSE Physics 0625
> Source: https://www.owlsprep.com/study/cie-0625-u1-forces-elasticity-and-newton-s/

This guide covers core force concepts, Hooke’s Law for elastic springs, and Newton’s three laws of motion for CIE IGCSE Physics 0625, aligned with the 2026-2028 syllabus for both Core and Extended tiers.

**Prerequisites:** Basic unit conversions (cm to m, g to kg); Interpreting linear line graphs

## Learning objectives

- Define force and distinguish between contact and non-contact force types
- Apply $W=mg$ to calculate weight from mass, and explain the difference between mass and weight
- Sketch and interpret load-extension graphs (Core); use Hooke's Law $F=kx$ to calculate spring constant, force and extension (Extended)
- State and apply Newton's First and Third Laws of Motion to explain real-world scenarios (Core)
- Calculate resultant forces and apply Newton's Second Law $F=ma$ to solve motion problems (Extended)

## Core: Types of Force and Weight Calculation

**Force** — A push or pull that acts on an object due to interaction with another object, measured in newtons (N). Forces are either contact or non-contact.

- **Contact forces**: Act only when objects touch: friction, air resistance, normal contact force, tension
- **Non-contact forces**: Act at a distance: gravitational force, electrostatic force, magnetic force

**Mass and Weight** — Mass is the amount of matter in an object (unit: kg, scalar, constant everywhere). Weight is the gravitational force acting on an object's mass (unit: N, vector, varies with gravitational field strength). The relationship is given by $W = m \times g$.

*Example:* A 5kg object has mass 5kg on Earth and the Moon, but weight 49N on Earth and 8N on the Moon.

**Worked example:** Calculate the weight of a 7.2 kg suitcase on Earth, where $g = 9.8 N/kg$.

1. Step 1: List known values: $m = 7.2 kg$, $g = 9.8 N/kg$
2. Step 2: Substitute into the weight formula:
3. $$W = m \times g = 7.2 \times 9.8 = 70.56 N$$

> **Exam tip:** Use $g = 9.8 N/kg$ for calculations unless the question explicitly gives a different value.

## Core: Elasticity and Load-Extension Graphs

**Elastic deformation** — A change in the size or shape of an object caused by a force. A deformation is elastic if the object returns to its original size and shape once the force is removed, as happens with a spring stretched by small loads.

When you hang masses (loads) from a spring and measure how far it stretches (its extension), you can plot a load-extension graph with load on one axis and extension on the other. At Core level you need to be able to sketch, plot and interpret these graphs, but you are not required to use an equation.

> **info**
>
> Extension is the *increase* in length of the spring, not the total stretched length. Calculate it as: $x = \text{stretched length} - \text{original unstretched length}$.

**Worked example:** A student hangs increasing loads on a spring and records its extension each time. Describe qualitatively what the load-extension graph looks like and what it tells you about the spring.

1. Step 1: For small loads the graph is a straight line through the origin, showing that equal increases in load give equal increases in extension.
2. Step 2: For larger loads the line begins to curve, showing that the same increase in load now produces a larger increase in extension.
3. Step 3: Interpret the straight part as the region where the spring stretches steadily, and the curved part as the region where the spring stretches more easily and may not return to its original length.

> **Exam tip:** On a load-extension graph, a straight line through the origin means extension increases in equal steps for equal increases in load; when the line curves, equal loads no longer give equal extensions.

## Core: Newton's First and Third Laws of Motion

**Newton's First Law of Motion** — An object remains at rest, or moves at constant speed in a straight line, unless acted on by a non-zero resultant force.

*Example:* A book resting on a table stays still because upward normal contact force balances downward weight, so resultant force = 0.

**Newton's Third Law of Motion** — When two objects interact, the forces they exert on each other are equal in size, opposite in direction, and act on *different* objects.

*Example:* When you push a wall with a force of 20N, the wall pushes back on you with a 20N force in the opposite direction.

**Worked example:** Use Newton's First Law to explain why a passenger falls backward when a stationary bus suddenly accelerates forward.

1. Step 1: Before acceleration, the passenger is stationary relative to the bus.
2. Step 2: When the bus accelerates, the passenger's feet are pulled forward by friction with the bus floor, but their upper body has no resultant force acting on it immediately, so it stays stationary relative to the ground, making the passenger fall backward.

## Extended: Newton's Second Law and Resultant Forces

**Newton's Second Law of Motion** — The resultant force acting on an object equals the object's mass multiplied by its acceleration, in the direction of the resultant force. $F$ = resultant force (N), $m$ = mass (kg), $a$ = acceleration (m/s²)

*Notation:* $F_{resultant} = m \times a$

The resultant force is the single force that has the same effect as all individual forces acting on an object. Calculate it by adding forces acting in the same direction, and subtracting forces acting in opposite directions.

**Worked example:** A 3kg cart is pushed forward with a force of 15N, and experiences a friction force of 6N acting backward. Calculate the acceleration of the cart.

1. Step 1: Calculate resultant force: $F = 15 N - 6 N = 9 N$ forward
2. Step 2: Rearrange Newton's Second Law to solve for acceleration:
3. $$a = \frac{F}{m} = \frac{9}{3} = 3 m/s^2 \text{ forward}$$

> **Exam tip:** Always state the direction of resultant force and acceleration in Extended responses to earn full marks.

## Extended: Hooke's Law and the Spring Constant

**Hooke's Law** — The extension of an elastic object is directly proportional to the force applied, provided the limit of proportionality is not exceeded. $F$ = applied force (N), $k$ = spring constant (N/m), $x$ = extension (m).

*Notation:* $F = kx$

*Example:* A spring with $k = 100 N/m$ extends 0.01 m when 1 N of force is applied.

**Spring constant** — The force per unit extension of a spring, measured in newtons per metre (N/m). A stiffer spring has a larger spring constant.

*Notation:* $k = \frac{F}{x}$

> **info**
>
> The **limit of proportionality** is the point beyond which extension is no longer proportional to the applied force. $F = kx$ is only valid for loads below this limit (the straight-line part of the load-extension graph).

**Worked example:** A spring has an original length of 0.1 m. When a 6 N weight is hung from it, it stretches to a total length of 0.13 m. Calculate its spring constant, assuming the limit of proportionality is not reached.

1. Step 1: Calculate extension: $x = 0.13 m - 0.1 m = 0.03 m$
2. Step 2: Rearrange Hooke's Law to solve for $k$:
3. $$k = \frac{F}{x} = \frac{6}{0.03} = 200 N/m$$

> **Exam tip:** If a force-extension graph curves upward at higher forces, the limit of proportionality has been passed, so $F=kx$ no longer applies for those values.

## Common pitfalls

- **Wrong:** Using total spring length instead of extension in Hooke's Law calculations
  - Why it fails: Hooke's Law uses the increase in length of the spring, not its total stretched length, so using total length gives an incorrect value for $x$.
  - Correct: Subtract the original unstretched length of the spring from its stretched length to calculate extension before using $F=kx$.
- **Wrong:** Confusing mass and weight, using kg as the unit for weight
  - Why it fails: Mass is measured in kg, while weight is a force measured in N. Using the wrong unit leads to lost marks and incorrect force calculations.
  - Correct: Use $W=mg$ to convert mass in kg to weight in N whenever force calculations are required.
- **Wrong:** Stating Newton's Third Law force pairs cancel each other out
  - Why it fails: Third law pairs act on two different interacting objects, so they never cancel. Cancelling balanced forces act on the same object.
  - Correct: When identifying Third Law pairs, confirm each force acts on a separate object, e.g. force of Earth on you, force of you on Earth.
- **Wrong:** Applying Hooke's Law beyond the limit of proportionality
  - Why it fails: Once the limit of proportionality is exceeded, extension is no longer proportional to applied force, so $F=kx$ is no longer valid.
  - Correct: Only use $F=kx$ if the question confirms the limit of proportionality is not reached, or if the force-extension graph is linear.
- **Wrong:** Assuming a positive resultant force acts on an object moving at constant speed
  - Why it fails: Newton's First Law states constant speed means there is no resultant force acting on the object, not a positive force.
  - Correct: If an object is stationary or moving at constant velocity, all forces acting on it are balanced, so resultant force = 0.

## Cheatsheet

| Concept | Formula | Key Rule | Tier |
| --- | --- | --- | --- |
| Weight | $W = mg$ | Use $g=9.8 N/kg$ unless stated otherwise | Core |
| Hooke's Law | $F = kx$ | $x$ = extension, not total spring length | Extended |
| Newton's First Law | - | Zero resultant force = stationary / constant speed | Core |
| Newton's Third Law | - | Equal, opposite forces acting on different objects | Core |
| Newton's Second Law | $F=ma$ | $F$ = resultant force, state direction of acceleration | Extended |

## What's next

Now that you have mastered forces, elasticity and Newton's laws, you are ready to apply these concepts to more advanced motion and energy topics in CIE IGCSE Physics 0625. Next, you will learn to analyse force-extension graphs to identify the limit of proportionality, and calculate work done by forces as part of the energy unit. You will also use Newton's laws to interpret motion graphs and solve real-world scenarios involving friction and air resistance, which are frequently tested in both Extended MCQ and theory papers. Practice structured response questions to build your ability to explain laws and calculations clearly, as this is where most marks are awarded in theory exams.

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