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Learning objectives
Core idea
Key formulae
Worked examples
Explore the graph
Common mistakes
Exam tip
Guided practice
Practice questions
Challenge
Summary
Quick quiz
Same topic, different board
Past paper questions

Learn7357Paper 2 · Pure and MechanicsForces and Newton's lawsNewton’s laws and F = ma

Forces and Newton's laws · Subtopic 3 of 8

Newton's Laws and F = ma

Resultant force equals mass times acceleration. It links the forces on an object to how its motion changes, and with suvat it lets you predict exactly where the object will be.

Cambridge 9709 Paper 4Edexcel 9MA0 Paper 3AQA 7357 Paper 2≈ 35 minRevision 5 · updated 2026-10-27

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Section 1 of 14

Learning objectives

State Newton's three laws of motion.
Use $F=ma$ , where $F$ is the resultant force in the direction of motion.
Combine $F=ma$ with the suvat formulae.
Apply $F=ma$ when a force acts at an angle and there is friction.

Pause and think

A car's engine provides $3000$ N of driving force, but the car moves at a steady $25\ \text{m s}^{-1}$ . What can you say about the resistance forces?

Section 2 of 14

Core idea

Start herePush harder, go faster

Push an empty trolley and a full one with the same force: the empty one accelerates more. Push the empty one twice as hard: twice the acceleration. Newton summarised it in three words: $F=ma$ .

The big idea

Resultant force $=ma$ (Newton's 2nd law). No resultant ⇒ constant velocity (1st law). Forces come in equal-and-opposite pairs on different bodies (3rd law).

1. The laws

Acceleration produced by a force, for different masses

mass m (kg) = 1.5

Double the force: double the acceleration. Double the mass: half the acceleration.

1st: an object stays at rest or at constant velocity unless a resultant force acts.
2nd: $F_{\text{resultant}}=ma$ .
3rd: A on B equals and opposes B on A.

A 4 kg box: 30 N pull, 10 N resistance

Vertically:

R=40

N, balanced. Horizontally: resultant

30-10=20

N, so

a=\dfrac{20}{4}=5

m/s².

2. Applying F = ma

Solving any forces problem

Drawing diagram…

The same routine solves equilibrium, friction, slopes, connected particles and lifts.

Worked example1200 kg car: 3000 N driving, 600 N resistance

Find $a$ , then speed and distance after 5 s from rest.

Goal: Values.
Strategy: Resultant, then suvat.

$2400=1200a\Rightarrow a=2$
$F=ma$ .
$v=10,\ s=25$
suvat.

Answer

2

m s

^{-2}

;

10

m/s;

25

✓ Check

v^2=2as=100

✓.

Worked exampleTowing

Car 1200 kg tows a 400 kg trailer; driving force 2000 N; resistances 300 N (car) and 100 N (trailer). Find $a$ and the tow-bar tension.

Goal: $a$ , $T$ .
Strategy: Whole system, then trailer alone.

$2000-400=1600a\Rightarrow a=1$
System: $T$ is internal.
$T-100=400(1)\Rightarrow T=500$
Trailer.

Answer

1

m s

^{-2}

;

500

✓ CheckCar alone:

2000-300-500=1200

✓.

Common misconceptionUsing one force instead of the resultant

F

F=ma

is the resultant.

Pause and think

2 kg accelerates at 3 m s⁻² against 4 N resistance. Driving force?

Summary

Resultant $=ma$ .
System or part.

Newton: Force giving 1 kg an acceleration of 1 m s $^{-2}$ .

HistoryThe laws in Latin. Newton published his three laws in Latin in the Principia in 1687. His second law actually says that the change of motion (what we call momentum) is proportional to the force applied. The tidy form

F=ma

was written down later, by mathematicians such as Leonhard Euler in the 1750s.

Fun factWhy 1 N is what it is. The newton is defined by the second law:

1

N is the force that gives

1

kg an acceleration of

1\ \text{m s}^{-2}

. The unit was officially named after Newton in 1948, more than 200 years after his death.

Section 3 of 14

Key formulae

Newton's second law

F_{\text{resultant}}=ma

Driving force D against resistance R

D-R=ma

Force P at angle θ above the horizontal, rough floor

P\cos\theta-\mu(mg-P\sin\theta)=ma

TrickForwards minus backwards. Write every

F=ma

equation as (forces pointing forwards)

-

(forces pointing backwards)

=ma

. If you get a negative

a

, the object is decelerating; you have not made a mistake.

Section 4 of 14

Worked examples

Example 1 · Driving force0 / 2 steps

A car of mass

900

kg accelerates uniformly from rest to

15\ \text{m s}^{-1}

10

s against a resistance of

400

N. Find the driving force.

Example 2 · Angled pull with friction0 / 3 steps

A box of mass

8

kg is pulled across a rough floor (

\mu=0.25

) by a force of

40

N at

20^\circ

above the horizontal. Find its acceleration.

Example 3 · Braking0 / 2 steps

A car of mass

1500

kg travelling at

20\ \text{m s}^{-1}

brakes and stops in

50

m. Find the braking force, assumed constant.

Section 5 of 14

Explore the graph

Velocity under a constant resultant force

Loading the interactive graph…

Try this

A car starts from rest with driving force $f$ , resistance $r$ and mass $m$ . The line is its velocity against time.
The gradient is $a=\dfrac{f-r}{m}$ . Double $m$ : what happens to the gradient?
Make $f=r$ . What does Newton's first law predict, and what does the line do?

Constant resultant force means constant acceleration: a straight-line velocity–time graph.

In the real worldLifting a rocket. At lift-off, a Saturn V moon rocket had a mass of about

2.9

million kg and its engines gave a thrust of about

34

million N. Its weight was about

29

million N, so the resultant force was only about

5

million N: an initial acceleration of under

2\ \text{m s}^{-2}

. It got faster as it burned fuel and became lighter.

Section 6 of 14

Common mistakes

Incorrect

F=ma

with

F

as the driving force only

Correct

F

is the resultant

Subtract resistances and friction first.

IncorrectMass and weight used interchangeably

CorrectWeight

=mg

newtons

Mass is in kg; only weight is a force.

Incorrect

R=mg

when the pull is angled

Correct

R=mg-P\sin\theta

Resolve perpendicular to the motion first.

IncorrectConstant velocity needs a resultant force

CorrectResultant force is zero

Newton's first law.

the car is moving, so there must be a resultant force

Newton

is it accelerating?

…constant speed. Forces balanced. Sorry, Isaac.

Section 7 of 14

Exam tip

Exam tipF = ma is your bridge to suvat. Most Paper 4 dynamics questions are "forces →

a

→ suvat" or "suvat →

a

→ forces". Decide which way you are going, find

a

in the middle, and say which formula you are using at each end.

Section 8 of 14

Guided practice

SA-9709-P4-NL-G011 markL1 FoundationCalculatorF = ma

A resultant force of

10

N acts on a

4

kg mass. Find its acceleration.

Salam Academy Original Practice · Proposed mark scheme

Section 9 of 14

Practice questions

SA-9709-P4-NL1-0012 marksL1 FoundationCalculatorF = ma

A particle of mass

3

kg is pushed along a horizontal surface by a force of

20

N against a resistance of

8

N. Find its acceleration.

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P4-NL-0012 marksL1 FoundationCalculatorF = ma

A car of mass

1200

kg has a driving force of

3000

N and experiences a resistance of

600

N. Find its acceleration.

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P4-NL1-0023 marksL2 StandardCalculatorF = ma

A car of mass

900

kg accelerates uniformly from rest to

15\ \text{m s}^{-1}

10

s. The resistance to motion is

400

N. Find the driving force.

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P4-NL1-0063 marksL2 StandardCalculatorF = ma

A car of mass

1500

kg travelling at

20\ \text{m s}^{-1}

brakes with a constant force and stops in

50

m. Find the braking force.

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P4-NL1-0033 marksL2 StandardCalculatorAngled forces

A box of mass

5

kg on a smooth horizontal floor is pulled by a force of

20

N at

30^\circ

above the horizontal. Find its acceleration (3 s.f.) and the normal reaction.

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P4-NL1-0045 marksL3 ExamCalculatorAngled forces

A box of mass

8

kg is pulled across a rough horizontal floor, with

\mu=0.25

, by a force of

40

N at

20^\circ

above the horizontal. Find its acceleration, correct to 3 significant figures.

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P4-NL1-0074 marksL3 ExamCalculatorAngled forces

A sledge of mass

10

kg is pulled across rough horizontal ground, with

\mu=0.2

, by a rope at

15^\circ

above the horizontal. It moves at constant speed. Find the tension in the rope, correct to 3 significant figures.

Salam Academy Original Practice · Proposed mark scheme

More questions on every skill in this subtopic are in the question bank under Practise.

Section 10 of 14

Challenge

SA-9709-P4-NL-0094 marksL3 ExamCalculatorF = ma

0.5

kg particle starts from rest on a horizontal surface and a horizontal force of

2

N acts for

3

s. The force is then removed and a constant resistance of

1

N brings the particle to rest. Find the greatest speed and the time taken to stop after the force is removed.

Salam Academy Original Practice · Proposed mark scheme

Section 11 of 14

Summary

First law: no resultant force, no change in velocity.
Second law: resultant force $=ma$ .
Third law: equal and opposite forces on two different bodies.
Find $R$ by resolving perpendicular to the motion, then friction $=\mu R$ if sliding.
Use $a$ to connect forces with suvat.

Section 12 of 14

Quick quiz

Q1 · SA-9709-P4-NL-Q011 markL1 FoundationCalculatorF = ma

F=ma

F

Salam Academy Original Practice · Proposed mark scheme

Q2 · SA-9709-P4-NL-Q031 markL1 FoundationCalculatorF = ma

Newton's third law says that

Salam Academy Original Practice · Proposed mark scheme

Q3 · SA-9709-P4-NL1-0081 markL1 FoundationCalculatorF = ma

A particle moves in a straight line at constant velocity. The resultant force on it is

Salam Academy Original Practice · Proposed mark scheme

Q4 · SA-9709-P4-NL1-0051 markL1 FoundationCalculatorF = ma

A resultant force of

12

N gives a particle an acceleration of

1.5\ \text{m s}^{-2}

. Find its mass.

Salam Academy Original Practice · Proposed mark scheme

Section 13 of 14

Same topic, different board

Specification: 7357 A-level Mathematics
Where: Section R, Forces and Newton’s laws
Expected: Newton’s three laws; F = ma in one and two dimensions; resolving forces, including friction.
Calculator: Permitted on all papers. AQA uses g = 9.8 m s⁻².
Typical style: Often asks you to criticise the model, for example the effect of air resistance.
Open in: 7357 · Paper 2

Section 14 of 14

Past paper questions

1 question (6 marks) from real exams test this subtopic. Try each one under timed conditions, then mark it with the official mark scheme.

9709/42 · Oct/Nov 2022Question 46 marks
Engine power from Newton’s second law, then distance using work done.
Question paper, page 5 Mark scheme

All 9709 past paper questions by topic

Salam Academy original content · Content status: reviewed and published. Syllabus mapping: AQA 7357, First teaching 2017 (official specification).