Newton’s 2nd Law — Edexcel A-level Mathematics (Applied)

SPEC 8.02 • FORCES AND NEWTON'S LAWS

Newton's Second Law

Understanding F = ma and solving acceleration problems

Newton's Second Law

Law

The resultant force acting on an object is equal to the rate of change of momentum of that object.

For constant mass, this simplifies to: Resultant force = mass × acceleration

Key Insight

A resultant force causes an acceleration — the object speeds up, slows down, or changes direction.

Link to First Law

Newton's First Law is a special case: when resultant force = 0, acceleration = 0 (constant velocity or at rest).

The Equation

Formula

\sum F = ma

The sum of all forces (resultant) equals mass times acceleration

SymbolQuantityUnit

\sum F

Resultant force (sum of all forces)Newtons (N)

m

MassKilograms (kg)

a

AccelerationMetres per second squared (m s⁻²)

Important

\sum F

means the vector sum of all forces acting on the object — not just one individual force.

Direction Matters

Force and acceleration are vectors — they have direction. The acceleration is always in the same direction as the resultant force.

Sign Convention

Choose a positive direction. Forces in that direction are positive; forces in the opposite direction are negative.

Example: If right is positive, a force of 50 N to the left is written as −50 N.

01Worked Example

A resultant force of 24 N acts on an object of mass 8 kg. Find the acceleration of the object.

1

Write the equation:

\sum F = ma

2

Substitute known values:

24 = 8 \times a

3

Solve for

a

:

a = \frac{24}{8} = 3

Answer

a = 3 m s^{- 2}

02Worked Example

A car of mass 1200 kg accelerates at 2.5 m s⁻². Calculate the resultant force acting on the car.

1

Write the equation:

\sum F = ma

2

Substitute:

\sum F = 1200 \times 2.5

3

Calculate:

\sum F = 3000

Answer

\sum F = 3000 N

Multiple Forces

When several forces act on an object, find the resultant force first, then apply

\sum F = ma

.

Method

Draw a force diagram showing all forces

Choose a positive direction

Find resultant:

\sum F = F_{1} + F_{2} + ...

(using signs)

Apply

\sum F = ma

to find acceleration

03Worked Example

A car of mass 1000 kg has a driving force of 4000 N and experiences resistance of 1000 N. Find the acceleration.

1

Take forward (→) as positive

2

Resultant force:

\sum F = 4000 - 1000 = 3000 N

3

Apply

\sum F = ma

:

3000 = 1000 \times a

4

Solve:

a = \frac{3000}{1000} = 3

Answer

a = 3 m s^{- 2}

Vertical Motion

When an object is lifted or lowered, both tension and weight act on it. Use

g = 9.8 m s^{- 2}

.

Direction Rule

Take the direction of acceleration as positive. The equation becomes: (Forces in direction of motion) − (Forces opposing motion) = ma

04Worked Example

A crate of mass 50 kg is lifted by a crane with acceleration 2 m s⁻² upward. Find the tension in the cable. Take g = 9.8 m s⁻².

1

Take upward as positive (direction of acceleration)

2

Forces: Tension T upward, Weight = 50 × 9.8 = 490 N downward

3

Apply Newton's Second Law:

T - 490 = 50 \times 2

4

Solve:

T - 490 = 100

, so

T = 590

Answer

T = 590 N

Vectors and F = ma

When forces are given as vectors, apply Newton's Second Law to the resultant vector.

Vector Form

\sum F = m a

Add all force vectors first, then divide by mass to find acceleration vector.

Method

Add all force vectors:

\sum F = F_{1} + F_{2} + ...

Use

a = \frac{\sum F}{m}

to find acceleration vector

Remember

Add i components together and j components together separately.

05Worked Example

A particle of mass 4 kg is acted on by forces $F_{1} = (6 i + 8 j)$ N and $F_{2} = (2 i - 4 j)$ N. Find the acceleration.

1

Find resultant force:

\sum F = (6 i + 8 j) + (2 i - 4 j) = (8 i + 4 j)

N

2

Apply

\sum F = m a

:

(8 i + 4 j) = 4 a

3

Solve:

a = \frac{1}{4} (8 i + 4 j) = (2 i + j)

Answer

a = (2 i + j) m s^{- 2}

Practice

Question 1: A car of mass 1500 kg has a driving force of 5000 N and experiences resistance of 2000 N. Find the acceleration.

Resultant force:

\sum F = 5000 - 2000 = 3000 N

Apply ΣF = ma:

3000 = 1500 \times a

Solve:

a = \frac{3000}{1500} = 2 m s^{- 2}

Practice

Question 2: A person of mass 70 kg stands in a lift accelerating upward at 1.5 m s⁻². Find the normal reaction from the floor on the person. Take g = 9.8 m s⁻².

Forces: Normal reaction R upward, Weight = 70 × 9.8 = 686 N downward

Take up as positive:

R - 686 = 70 \times 1.5

Solve:

R - 686 = 105

, so

R = 791 N

Summary

Key Concepts

ConceptFormula / Description

Newton's Second Law

\sum F = ma

where ΣF is resultant force

Multiple ForcesFind resultant first, then apply ΣF = ma

Vertical Motion

T - m g = ma

(up) or

m g - T = ma

(down)

Vector Form

\sum F = m a

; add vectors then divide by mass

Exam Strategy

Always draw a force diagram first

State your positive direction clearly

Use

g = 9.8 m s^{- 2}

unless told otherwise

For vectors, add components separately