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Newton's second law

Key idea: Balanced and unbalanced forces, two-dimensional free-body diagrams, F = ma, friction and action–reaction pairs.

  • SEC G3 Combined Science Physics component 2027
Syllabus and review details

Free-body diagrams may involve forces in at most two dimensions. This Science lesson excludes graphical three-force equilibrium, inertia as a separate extension and terminal-velocity treatment.

Retrieval check
Before starting, name the body you are analysing and distinguish velocity from acceleration. The answer depends on the selected body and the change in velocity, not on motion alone.

Build the force model

Use resultant force = mass × acceleration only after choosing a direction and combining the real interaction forces. State force in newtons, mass in kilograms and acceleration in metres per second squared.

Worked reasoning

Modelled example 1

Model the decisive reasoning

Core

Problem

A 5.0 kg crate has 24 N right and 9.0 N left; its vertical forces balance. Calculate its acceleration.
Study the worked solution

  1. Select and apply the model

    Method

    Form the horizontal resultant, then use acceleration = resultant force divided by mass.

    Reason

    The force in Fnet = ma is the vector sum, not the 24 N applied force alone.

    Working

    Fnet = 24 - 9.0 = 15 N right; a = 15/5.0 = 3.0 m/s² right.

Guided practice 2

Complete a guided force model

About 5 min

Problem

A 4.0 kg trolley experiences 19 N right, 5.0 N left and another 2.0 N left. Find its acceleration.

Write your reasoning before opening the solution

Hints

Hint 1: choose the decisive model
Combine all three horizontal forces with signs before dividing by mass.
View solution step by step

  1. Compare your model with the physical interactions

    Method

    The resultant is 12 N right and the acceleration is 3.0 m/s² right.

    Reason

    Both leftward forces oppose the rightward force.

    Working

    Fnet = 19 - 5.0 - 2.0 = 12 N; a = 12/4.0 = 3.0 m/s².

Misconception contrast

Write ma as another force arrow on the free-body diagram.
ma is the value of the resultant of the real interaction forces. It is not an additional interaction force.

Changed-context transfer

Challenge 3

Transfer the force model

Minimal support

Problem

A 6.0 kg object accelerates 2.5 m/s² left while one force of 22 N acts right. Determine the required leftward force.

Commit to a model before using the hint

Hints

Hint 1: identify what changed
Take right as positive; the acceleration and resultant are negative.
View solution step by step

  1. Apply the model in the new context

    Method

    The resultant must be 15 N left, so the leftward force is 37 N.

    Reason

    A rightward 22 N and leftward 37 N combine to -15 N.

    Working

    Fnet = ma = 6.0(-2.5) = -15 N; 22 - Fleft = -15, so Fleft = 37 N.

Independent check

A 3.0 kg drone has 42 N upward thrust, 29 N weight downward and 4.0 N drag downward. Select the model, calculate its acceleration and state the direction.

Check the force-model reasoning

Use Newton's second law after forming the vertical resultant: 42 - 29 - 4.0 = 9.0 N upward, so a = 9.0/3.0 = 3.0 m/s² upward.

Start the Dynamics topic check. Use its feedback to practise the smallest weak step.

Continue with the next resource in this course.

Course and syllabus information
Course
SEC G3 Combined Science Physics component
Edition
SEC G3 Combined Science Physics component 2027