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NotesPhysics HLTopic 1.2Newton's laws of motion (F = ma)
Back to Physics HL Topics
1.2.26 min read

Newton's laws of motion (F = ma) (Physics HL)

IB Physics • Unit 1

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Contents

  • Newton's three laws
  • Net force = ma
  • Exam-style question
The big idea: Push an empty shopping trolley and it leaps forward. Load it with bricks and the same push barely moves it.

That's the heart of Newton's laws: the net force on an object (all the forces added up) and its mass decide how it speeds up — and with no net force, it just keeps going.

1st law — no net force, no change

With zero net force, an object stays still or keeps moving at constant velocity. Motion doesn't need a force — only a change in motion does.

2nd law — net force makes it accelerate

A net force gives an acceleration in the same direction: F = ma. Bigger force → bigger acceleration; bigger mass → smaller acceleration.

3rd law — forces come in pairs

If A pushes B, then B pushes A back equally hard, the opposite way. The two forces act on different objects, so they never cancel on one body.

Spot which law you need: Constant velocity or at rest? → 1st law: net force = 0.

Speeding up, slowing down or turning? → 2nd law: net force = ma.

Two objects pushing on each other? → 3rd law: an equal, opposite pair.

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Newton's second law links the net force on an object to its acceleration. The same law can be written using momentum (momentum = mass × velocity): the net force equals how fast the momentum changes.

Given in the data booklet — Newton's second law. F is the NET force on the object.
net (resultant) force (N)
mass (kg)
acceleration (m s⁻²)
change in momentum (kg m s⁻¹)
time interval (s)
It's the NET force: The F in F = ma is the net force — every force on the object added together (direction matters).

Always find the net force first, then divide by the mass to get the acceleration.

Free-body diagram: the horizontal NET force is the pull minus friction; weight and normal cancel vertically.

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IB-style questionCalculate[2 marks]

A 4.0 kg trolley is pulled forward by a 30 N force while a 6.0 N friction force acts backward. Find its acceleration.

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How this is tested — most forces questions are: draw a free-body diagram, then apply F = ma. It comes up two ways:

Paper 1A

  • Quick net-force calculations.
  • A block driven by an angled force; the acceleration of a single body.

Paper 2

  • Connected systems — two masses on a string, an elevator cable, stacked blocks.
  • Apply F = ma to one body to find a tension or contact force.
The classic trap: Plugging in one force instead of the net force — or forgetting that connected objects share the same acceleration.
Connected bodies share an acceleration: When two objects are joined (by a string, or stacked so they move together), they have the same acceleration.

To find a connecting force (a string tension, or the friction between stacked blocks), cover everything except one object and apply F = ma to it.

The trick is asking what is still touching it: a force applied to the other block never acts on this one, so usually only the string (or the friction) is left — and that force must be what accelerates it.
Set-upWhole systemOne body alone
Two masses on a stringa = (driving force) ÷ (total mass)tension = (that body's mass) × a
Elevator going up/downnet force = T − mga = (T − mg) ÷ m
Stacked blocksa = F ÷ (total mass)friction on top = (top mass) × a

Tap each set-up to see the same two moves play out with real numbers.

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IB-style questionDetermine[3 marks]

Two blocks are joined by a light string on a smooth floor: a 3.0 kg block in front, a 5.0 kg block behind. A 24 N force pulls the front block. (a) Find the acceleration of the pair. (b) Find the tension in the string.

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Newton's first law of motion. [2 marks]

Related Physics HL Topics

Continue learning with these related topics from the same unit:

1.1.1Velocity and displacement
1.1.2Acceleration
1.1.3Displacement from a velocity–time graph
1.1.4The suvat equations
View all Physics HL topics

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1.2.1Free-body diagrams, equilibrium & resolving forces
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Friction (static & dynamic)1.2.3

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