Mechanics: forces and Newton's laws (Former VCE Specialist Mathematics study design)

Syllabus dot point

“Newton's laws of motion, the resultant of forces acting on a particle, the resolution of forces into components, the relationship $\mathbf{F} = m\mathbf{a}$ , and the analysis of equilibrium and of motion under constant forces including weight, normal reaction and friction”

6 min readUpdated 23 Sep 2026

Quick answer

Newton's first law: a particle stays at rest or moves at constant velocity unless acted on by a non-zero resultant force. Newton's second law: the resultant force equals mass times acceleration, $\mathbf{F} = m\mathbf{a}$ . Newton's third law: forces between two bodies are equal and opposite. To analyse a particle, draw all forces, resolve them into perpendicular components (often horizontal and vertical, or along and perpendicular to an incline), and apply $\mathbf{F} = m\mathbf{a}$ in each direction. In equilibrium the acceleration is zero, so the components of the resultant are each zero. Weight is $mg$ downward; friction opposes motion with magnitude up to $\mu N$ .

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What this dot point is asking
Newton's laws
Resolving forces
Equilibrium and motion
Examples in context
Try this
Exam-style questions

What this dot point is asking

Former study design. Dynamics (Newton's laws, momentum and connected particles) was part of the previous VCE Specialist Mathematics study design and last appeared on the 2022 examinations. The current study design (from 2023) has no dynamics in Units 3 and 4: forces appear only in Units 1 and 2 (vectors applied to statics and motion under a constant force), and the Unit 4 calculus is kinematics (rectilinear motion and vector calculus). Keep this page for Units 1 and 2 vectors, for Physics, or for older resources.

VCAA wants you to apply Newton's laws to a particle, find the resultant of several forces, resolve forces into perpendicular components, use $\mathbf{F} = m\mathbf{a}$ to relate net force and acceleration, and analyse both equilibrium (zero acceleration) and motion under constant forces such as weight, normal reaction and friction. This connects the vector and calculus work to physical situations.

Newton's laws

First law. A particle continues at rest or in uniform straight-line motion unless a non-zero resultant force acts. This defines the natural state and identifies acceleration as the signature of a net force.

Second law. The resultant force on a particle equals its mass times its acceleration:

\mathbf{F}_{\text{net}} = m\mathbf{a}.

This is a vector equation, so it holds component by component.

Third law. When body A exerts a force on body B, body B exerts an equal and opposite force on A. These act on different bodies, so they do not cancel on a single particle.

Resolving forces

To use the second law, find the resultant of all forces. The most reliable method is to resolve each force into perpendicular components, sum the components in each direction, and apply $\mathbf{F} = m\mathbf{a}$ separately along each axis. Choose axes to simplify the problem: horizontal and vertical for free motion, or parallel and perpendicular to a surface for motion on an incline.

Common forces:

Weight: magnitude $mg$ , directed vertically downward.
Normal reaction $N$ : the push of a surface, perpendicular to it.
Friction: parallel to the surface, opposing relative motion, with magnitude up to $\mu N$ where $\mu$ is the coefficient of friction. When sliding, friction equals $\mu N$ .

Equilibrium and motion

In equilibrium the acceleration is zero, so the resultant force is zero and each component balances. For motion under constant forces the acceleration is constant, and you can find the resultant, divide by the mass to get $\mathbf{a}$ , and then use constant-acceleration kinematics for velocity and displacement.

Worked example

A block on a rough incline

A block of mass $5\ \text{kg}$ rests on a rough plane inclined at $30^\circ$ to the horizontal. The coefficient of friction is $\mu = 0.2$ . Find the acceleration of the block as it slides down. Take $g = 9.8\ \text{m/s}^2$ .

Resolve along and perpendicular to the incline. The weight $mg = 5\times 9.8 = 49\ \text{N}$ splits into:

component down the slope: $mg\sin 30^\circ = 49 \times 0.5 = 24.5\ \text{N}$ ;
component into the slope: $mg\cos 30^\circ = 49 \times \frac{\sqrt{3}}{2} \approx 49 \times 0.8660 = 42.4\ \text{N}$ .

Normal reaction. Perpendicular to the slope there is no acceleration, so $N = mg\cos 30^\circ \approx 42.4\ \text{N}$ .

Friction. The block slides down, so friction acts up the slope with magnitude

\mu N = 0.2 \times 42.4 = 8.49\ \text{N}.

Apply $F = ma$ along the slope (taking down-slope positive).

mg\sin 30^\circ - \mu N = ma,

24.5 - 8.49 = 5a,

16.01 = 5a \quad\Rightarrow\quad a = 3.20\ \text{m/s}^2.

The block accelerates down the slope at about $3.20\ \text{m/s}^2$ . The friction force is smaller than the driving component, so a positive down-slope acceleration is consistent with the block sliding down.

Common exam traps

Using $mg$ as the normal reaction on an incline: On a slope $N = mg\cos\theta$ , not $mg$ . Resolve perpendicular to the surface.
Wrong direction for friction: Friction opposes motion (or impending motion). For a block sliding down, friction acts up the slope.
Treating action and reaction as cancelling: Newton's third-law pairs act on different bodies, so they never cancel on the same particle.
Forgetting to resolve: Apply $\mathbf{F} = m\mathbf{a}$ along each chosen axis separately; do not add perpendicular force magnitudes directly.

Examples in context

Example 1. A $2\ \text{kg}$ mass hanging in equilibrium on a string has string tension $T = mg = 2\times 9.8 = 19.6\ \text{N}$ .

Example 2. A horizontal force of $10\ \text{N}$ on a $4\ \text{kg}$ block on a smooth surface gives $a = \frac{10}{4} = 2.5\ \text{m/s}^2$ .

Try this

Q1. State Newton's second law as an equation. [1 mark]

Cue. $\mathbf{F}_{\text{net}} = m\mathbf{a}$ .

Q2. A $3\ \text{kg}$ object has a resultant force of $12\ \text{N}$ . Find its acceleration. [1 mark]

Cue. $a = \frac{12}{3} = 4\ \text{m/s}^2$ .

Q3. Resolve the weight of a $10\ \text{kg}$ block on a $20^\circ$ incline into components along the slope. [2 marks]

Cue. Down-slope $mg\sin 20^\circ$ , into-slope $mg\cos 20^\circ$ , with $mg = 98\ \text{N}$ .

Exam-style questions

Questions in the style of VCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice (former study design)1 mark

A tourist in a hot air balloon, which is rising vertically at 2.5 m s^-1, accidentally drops a phone over the side when the phone is 80 metres above the ground. Assuming air resistance is negligible, how long in seconds, correct to two decimal places, does it take for the phone to hit the ground? A. 2.86 B. 2.98 C. 3.79 D. 4.04 E. 4.30

Show worked answer →

The answer is E.

The only force on the phone is its weight, so by F = ma the acceleration is g = 9.8 m s^-2 downward (constant force, constant acceleration). When released, the phone shares the balloon's upward velocity of 2.5 m s^-1.

Take up as positive with origin at the ground. Then y = 80 + 2.5t - 4.9t^2, with initial velocity +2.5 and acceleration -9.8.

Set y = 0 (ground): 4.9t^2 - 2.5t - 80 = 0. Solve with the quadratic formula: t = (2.5 + sqrt(2.5^2 + 4 . 4.9 . 80)) / (2 . 4.9) = (2.5 + sqrt(1574.25)) / 9.8.

This gives t = (2.5 + 39.68) / 9.8 = 4.30 s (taking the positive root), which is option E.

Practice (former study design)1 mark

From an open window, a person projects a ball vertically up using an outstretched arm. The point of projection is 50 m above the ground and its velocity of projection is 20 m s^-1. The time, in seconds, it takes for the ball to reach the tray of a truck that is 1 m above the ground directly below the point of projection is closest to A. 1.72 B. 5.80 C. 5.83 D. 1.75

Show worked answer →

The answer is B.

The ball moves under its weight alone, so the acceleration is the constant g = 9.8 m s^-2 downward (from F = ma). The initial velocity is +20 m s^-1 (up) and the launch height is 50 m.

Take up as positive with the ground as origin: y = 50 + 20t - 4.9t^2.

The truck tray is at y = 1, so 50 + 20t - 4.9t^2 = 1, that is 4.9t^2 - 20t - 49 = 0.

Solve: t = (20 + sqrt(20^2 + 4 . 4.9 . 49)) / (2 . 4.9) = (20 + sqrt(1360.4)) / 9.8 = (20 + 36.88) / 9.8 = 5.80 s (positive root), which is option B.