Newton's Laws of Motion — Interactive Simulator

Newton's three laws of motion are the foundation of classical mechanics. Published in 1687 in the Principia Mathematica, they describe how objects behave when forces act on them — and they remain the most useful framework for understanding everyday motion, from cars braking to rockets launching.

The three laws work together: Law 1 defines what "no force" looks like, Law 2 quantifies how force produces acceleration, and Law 3 ensures forces always come in pairs.

An object in motion stays in motion unless acted on by an external force. Friction is an external force — without it, the object never stops.

10 m/s
0.30

The Three Laws

Law 1 — The Law of Inertia

An object at rest stays at rest, and an object in motion stays in motion at constant velocity, unless acted upon by a net external force.

Inertia is resistance to change in motion. A heavier object has more inertia. Without friction, a sliding block would never stop — there is no force to change its velocity.

Law 2 — F = ma

The net force on an object equals its mass times its acceleration: F = ma.

QuantitySymbolSI Unit
Net forceFNewton (N)
Massmkilogram (kg)
Accelerationam/s²

This law is quantitative: doubling the force doubles the acceleration. Doubling the mass halves the acceleration for the same force.

Law 3 — Action and Reaction

For every action there is an equal and opposite reaction.

The forces in Law 3 act on different objects. When you push a wall, the wall pushes back on you with the same force. When a rocket expels gas downward, the gas pushes the rocket upward.


Worked Examples

Worked Example

Example 1 — Braking car

A car of mass 1200 kg brakes with a net force of 6000 N. What is its deceleration?

Using F = ma: a=Fm=60001200=5 m/s2a = \frac{F}{m} = \frac{6000}{1200} = 5 \text{ m/s}^2

The car decelerates at 5 m/s² (i.e. loses 5 m/s of speed every second).

If it was travelling at 20 m/s, it takes: t=va=205=4 st = \frac{v}{a} = \frac{20}{5} = 4 \text{ s}

to come to a complete stop.

Worked Example

Example 2 — Rocket thrust and reaction

A rocket engine expels exhaust gas at 3000 m/s and burns 10 kg of fuel per second. What thrust force does the rocket experience?

By Newton's 3rd Law, the thrust equals the reaction force to expelling the exhaust:

F=m˙×ve=10×3000=30,000 NF = \dot{m} \times v_e = 10 \times 3000 = 30{,}000 \text{ N}

The rocket experiences 30 kN of upward thrust, while the exhaust experiences the same force downward.


Frequently Asked Questions

Explore more simulations

Every concept on PhysicStuff has an interactive visualisation — no login, no setup required.