Physics · Units 3 & 4

Newton's Laws and Forces

Master Newton's three laws and forces the easy way, with plain English intuition, an interactive simulation, free body diagrams, worked examples and an auto marked practice test. VCE Physics Units 3 and 4.

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Give a shopping trolley a shove and it rolls, push harder and it speeds up faster, load it with bricks and the same push barely moves it. Hidden inside that everyday moment are Newton’s three laws of motion, the rules that connect a force to the way things move. Once you can spot the forces acting on an object and add them up, you can predict exactly how it will move, every single time.

The three laws in plain English

A force is just a push or a pull. Newton noticed that forces explain everything about how objects start moving, stop moving, or change direction, and he boiled it down to three short laws.

  • First law (inertia). An object stays at rest, or keeps moving at constant velocity, unless a net force acts on it. Things do not change their motion on their own. They are lazy.
  • Second law. When there is a net force, the object accelerates. The bigger the net force the bigger the acceleration, and the heavier the object the smaller the acceleration. In symbols, Fnet=maF_{net} = ma.
  • Third law. Every action has an equal and opposite reaction. If you push on a wall, the wall pushes back on you just as hard. The two forces are the same size, point opposite ways, and crucially act on different objects.

The most important word in all three is net. You add up every force on the object as vectors first, and only the leftover net force decides what happens.

The forces you will meet

Almost every VCE question is built from the same small cast of forces. Learn to recognise these and you can draw the picture for any problem.

  • Weight, the pull of gravity, always points straight down and equals Fg=mgF_g = mg with g=9.8g = 9.8 m/s2^2.
  • Normal force NN, the push of a surface, always points at right angles out of the surface.
  • Tension TT, the pull of a rope or string, always points along the rope away from the object.
  • Friction ff, which resists sliding and acts along the surface. Its maximum size is f=μNf = \mu N, where μ\mu is the coefficient of friction.

The clearest way to keep track of all this is a free body diagram: draw the object as a dot or box and every force as an arrow pointing the way it acts.

Fg = mgNpushfriction f

The blue arrows are the normal force pushing up and the push driving the block to the right. The red arrows are the weight pulling down and the friction resisting the slide. Add up the arrows in each direction and the leftover is the net force.

Newton’s second law is the workhorse

If you remember one equation from this topic, make it Fnet=maF_{net} = ma. It is the bridge between the forces you draw and the motion you measure.

To use it, add up all the forces as vectors to get the net force, then divide by the mass to get the acceleration. The acceleration always points the same way as the net force. If the forces cancel, the net force is zero, the acceleration is zero, and you are back to the first law: rest or constant velocity.

A frictionless incline is the classic example. Gravity points straight down, but only the part of it that runs along the slope can speed the object up. That component works out to gsin⁡θg\sin\theta, and because mass cancels, the acceleration down a frictionless slope is simply a=gsin⁡θa = g\sin\theta, no matter how heavy the object is. The part of gravity pressing into the slope is gcos⁡θg\cos\theta, and that is balanced by the normal force.

See it for yourself

Pile on the pushers, change the load, and watch how the net force and the motion respond. Try balancing two equal pushes on opposite sides and notice the object does not move, then add a single extra newton to one side and watch it accelerate.

Interactive simulation, Forces and Motion: Basics Source: PhET Interactive Simulations, University of Colorado Boulder (CC BY 4.0)

How to actually solve one

Nearly every forces question follows the same short recipe.

  1. Draw a free body diagram with every force as an arrow on the object.
  2. Choose a positive direction, then add the forces as vectors to find the net force.
  3. Apply Fnet=maF_{net} = ma to find the unknown, whether that is the acceleration, the mass, or a missing force.
  4. For a surface, remember NN balances the perpendicular forces and friction is at most f=μNf = \mu N.

Watch the difference between mass and weight. Mass in kilograms goes into Fnet=maF_{net} = ma, while weight in newtons is a force you draw on the diagram using Fg=mgF_g = mg.

Lock it in with active recall

Cover the answer and say each one out loud before you flip. Rate yourself honestly — the cards you find hard come back sooner, the ones you know are spaced further out.

Active recall

Answer from memory first, then flip. Rate yourself and each card returns on a spaced schedule (1 → 3 → 7 → 16 days).

State Newton’s second law and what each symbol means.
What does Newton’s first law say about an object with no net force on it?
Why don’t action and reaction forces cancel out?
How do you find the weight of a mass, and how is it different from mass?
What is the acceleration of a block sliding down a frictionless incline at angle θ\theta?
Write the formula for the maximum friction force before sliding.
Recall · Momentum and Impulse
How is impulse related to force and time?
Recall · Work, Energy and Power
What is the work done by a force over a distance?

See the recipe in action in the Worked Examples tab, then test yourself in Try It.

Worked examples

Worked Example 1Pushing a box on a frictionless floor

A 5.05.0 kg box is pushed across a frictionless floor by a horizontal force of 2020 N. Find its acceleration.

  1. 1

    The floor is frictionless and the push is horizontal, so the only horizontal force is the 2020 N push. That is the net force.

    Fnet=20 NF_{net} = 20 \text{ N}
  2. 2

    Newton's second law links the net force to the acceleration through the mass.

    a=Fnetma = \frac{F_{net}}{m}
  3. 3

    Substitute the numbers and divide.

    a=205.0=4.0 m/s2a = \frac{20}{5.0} = 4.0 \text{ m/s}^2
Answer
a=Fm=205.0=4.0 m/s2a = \dfrac{F}{m} = \dfrac{20}{5.0} = 4.0 \text{ m/s}^2
Worked Example 2A block sliding down a frictionless incline

A 2.02.0 kg block slides down a frictionless incline at 30∘30^\circ. Taking g=9.8g = 9.8 m/s2^2, find its acceleration.

  1. 1

    On a frictionless slope the only force pulling the block along the slope is the part of gravity that points down the slope.

    Fnet=mgsin⁡θF_{net} = mg\sin\theta
  2. 2

    Newton's second law gives the acceleration, and the mass cancels out, so it does not matter how heavy the block is.

    a=mgsin⁡θm=gsin⁡θa = \frac{mg\sin\theta}{m} = g\sin\theta
  3. 3

    Substitute g=9.8g = 9.8 and sin⁡30∘=0.5\sin 30^\circ = 0.5.

    a=9.8×0.5=4.9 m/s2a = 9.8 \times 0.5 = 4.9 \text{ m/s}^2
Answer
a=gsin⁡30∘=9.8×0.5=4.9 m/s2a = g\sin 30^\circ = 9.8 \times 0.5 = 4.9 \text{ m/s}^2

Practice questions

Practice test

Try it yourself

6 questions, 8 marks

Choose your answers, then submit to see your score and the full worked solutions. Multiple choice is marked for you, just like Exam 2 Section A.

Q1.A car travels along a straight road at a constant 6060 km/h. According to Newton's first law, the net force acting on the car is:

1mark
Need a hint?
Constant velocity means no change in motion. What net force does no change in motion require?

Q2.A net force of 1212 N acts on a 3.03.0 kg trolley. Using Fnet=maF_{net} = ma, the acceleration of the trolley is:

1mark
Need a hint?
Rearrange to a=Fnet/ma = F_{net}/m.

Q3.A box has a mass of 8.08.0 kg. Taking g=9.8g = 9.8 m/s2^2, its weight is closest to:

1mark
Need a hint?
Weight is Fg=mgF_g = mg, not the same as mass.

Q4.A book rests on a table. The reaction force to the weight of the book, as described by Newton's third law, is the:

1mark
Need a hint?
An action and reaction pair act on two different objects and are the same type of force.

Q5.A 4.04.0 kg block sits on a rough horizontal floor where the coefficient of friction is μ=0.25\mu = 0.25. Taking g=9.8g = 9.8 m/s2^2, find the normal force on the block and the maximum friction force before it slips. Show your working.

3marks

Work this on paper. The worked solution appears once you submit.

Q6.Jacinta and her ski (combined mass 6565 kg) are towed in a straight line by a horizontal rope, and at one instant she accelerates at 4.04.0 m/s2^2. The total resistance on her at that instant is 500500 N. Which one of the following is closest to the tension in the rope?

1mark
Need a hint?
The net force is mama. The rope must both overcome the resistance and supply the acceleration: T=ma+resistanceT = ma + \text{resistance}.

VCAA 2025 Physics Exam, Section A Q1

Frequently asked questions

What is the difference between mass and weight?
Mass is the amount of matter in an object, measured in kilograms, and it never changes. Weight is the gravitational force on that mass, measured in newtons, and it is found by multiplying the mass by g. The same astronaut has the same mass on the Moon but a smaller weight there because gravity is weaker.
Why do action and reaction forces not just cancel out?
Newton's third law forces always act on two different objects, never on the same one. The book pushes down on the table and the table pushes up on the book, but those two forces act on different bodies, so they never appear in the same net force calculation and cannot cancel.
Why does mass cancel out on a frictionless incline?
The force pulling the block down the slope is mg sin theta, but acceleration is force divided by mass, so the m on top and the m on the bottom cancel. That leaves a = g sin theta, which means a heavy block and a light block slide down the same slope with the same acceleration.