Topic 07 | Physical sciences

Forces, balanced & unbalanced

Year 7 (Levels 7-8 band): contact and non-contact forces, free-body diagrams, balanced vs unbalanced forces, and how net force changes an object's motion.

40-60 min Printable practice Answer key Challenge included
How to use this page

Read the explanation, work through the examples, then complete the core practice before printing.

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What you will learn

Worked example 0 Real-world example: why a book on a desk is still

A textbook sits on a table. Gravity pulls it down with a force of 1010 N. What holds it up?

  1. The table pushes back on the book with an equal upward force — called the normal force — of 1010 N.
  2. The two forces are equal in size and opposite in direction.
  3. Net force =1010=0= 10 - 10 = 0 N.
  4. With zero net force, motion does not change — the book stays still.

Key idea: “at rest” does not mean “no forces” — it means the forces balance out to zero. Whenever the net force is zero, motion is unchanged.

1. What is a force?

A force is a push or a pull. It can:

Force is measured in newtons (N). On Earth, an object of mass 11 kg weighs about 1010 N. Forces are drawn as arrows: the length shows size, the direction shows which way the force pushes.

2. Types of force

Contact forces require touching:

Non-contact forces act across space:

Worked example 1 Classifying forces in a real situation

A gymnast swings on a rope. List the forces on the gymnast.

  • Gravity — downward (non-contact).
  • Tension in the rope — along the rope toward where it is tied (contact).
  • Air resistance — small, opposite to motion (contact).

Key idea: nearly every real situation involves several forces at once. Always ask “is there a surface touching the object? Is there a rope? Is gravity acting?“

3. Free-body diagrams and net force

A free-body diagram shows the object as a dot or box with every force drawn as an arrow from it.

Weight (10 N)Normal (10 N)Weight (20 N)Normal (20 N)Push (30 N)Friction (10 N)
Free-body diagrams. Left: book at rest on a table (balanced). Right: box being dragged across a floor with friction (unbalanced forward).

The net force is the single force that has the same effect as all the individual forces combined.

Net force (one dimension)

Forces along a line

Take one direction as positive (e.g. right). Add forces in that direction; subtract forces in the opposite direction.

Fnet=F1+F2+FoppositeF_{\text{net}} = F_1 + F_2 + \ldots - F_{\text{opposite}}
Worked example 2 Adding forces along one direction

A box is pulled right with 3030 N; friction pulls left with 1010 N. Find the net force.

Fnet=3010=20 N to the right.F_{\text{net}} = 30 - 10 = 20 \text{ N to the right}.

Net force is not zero, so the box accelerates to the right.

Worked example 3 Tug-of-war

Team A pulls right with 250250 N. Team B pulls left with 220220 N. What is the net force and which team moves?

Fnet=250220=30 N to the right.F_{\text{net}} = 250 - 220 = 30 \text{ N to the right}.

The rope (and everyone holding it) accelerates to the right — Team A wins.

4. Balanced vs unbalanced forces

Worked example 4 Cyclist at constant speed

A cyclist pedals at steady 2020 km/h on flat ground. Air resistance + friction together add up to 4040 N backwards. What forward force is the cyclist producing?

Steady speed → balanced forces → net force is zero.

Therefore the forward force from pedalling = 4040 N.

Key idea: constant speed in a straight line means the forces are balanced, even though the cyclist is working hard.

5. Mass, weight and acceleration

Weight

Weight from mass
W=mgW = m g

where gg is the gravitational field strength. On Earth g10g \approx 10 N/kg.

Worked example 5 Mass vs weight

Your mass is 5050 kg. Find your weight on Earth and on the Moon (where g1.6g \approx 1.6 N/kg).

  • Earth: W=50×10=500W = 50 \times 10 = 500 N.
  • Moon: W=50×1.6=80W = 50 \times 1.6 = 80 N.

Mass stays at 5050 kg in both places — only the pull of gravity changes.

Newton’s second law (conceptually): the bigger the unbalanced force, the bigger the acceleration. The bigger the mass, the smaller the acceleration for the same force.

Fnet=m×aF_{\text{net}} = m \times a

Practice: Year 7

Fluency

Tier 1: recall and identify

    1. What is a force? Give its unit.
    2. Give two contact forces and two non-contact forces.
    3. Define “net force”.
    4. What is the difference between mass and weight?
    5. A box is pushed right with 2525 N and friction pulls left with 1010 N. Find the net force.
    6. A bag weighs 200200 N on Earth. What is its mass?
    7. Explain what a free-body diagram shows.
    8. Two students pull on a rope; each pulls 5050 N in opposite directions. What is the net force? Describe the motion.
    9. An object in space has no air resistance. If a small thruster pushes it for one second and then stops, what happens next?
    10. A person has a mass of 6060 kg. What is their weight on Earth? (g=10g = 10 N/kg.)
Reasoning

Tier 2: explain and reason

    1. Explain why a book on a table stays still even though gravity pulls it down.
    2. A car travels at a constant 8080 km/h in a straight line. Are the forces on it balanced or unbalanced? Justify.
    3. A parachutist falls at terminal velocity. Draw a labelled free-body diagram and describe the net force.
    4. Why does it take more force to push a heavy box than a light one to the same acceleration?
    5. Explain why your mass is the same on the Moon but your weight is less.
    6. A satellite orbiting Earth moves at constant speed in a curve. Is the net force zero? Justify.
Problem solving

Tier 3: apply to a novel context

    1. A rocket engine provides 50005000 N upward. The rocket’s weight is 40004000 N. Find the net force and describe what happens.
    2. A 33 kg trolley is pushed with a net force of 1212 N. Find its acceleration using F=maF = ma.
    3. A skydiver initially accelerates downward. Describe the forces at (a) the moment of jumping, (b) mid-fall as speed rises, (c) at terminal velocity.
    4. Two dogs pull on a lead attached to a post. Dog A pulls north with 8080 N; Dog B pulls east with 6060 N. Describe qualitatively the direction the post would be pulled (assume the post is only loosely set).

Challenge

Reasoning

Harder reasoning

    1. A car of mass 12001200 kg accelerates from 00 to 2020 m/s in 1010 seconds. Find its acceleration and the net force required. How is this force generated?
    2. Explain using forces why a person in a lift feels momentarily “heavier” when the lift starts going up and “lighter” when it starts going down.
    3. A rock with weight 5050 N on Earth is taken to a planet where gravity is 3×3\times Earth’s. State its mass, weight on that planet, and one everyday consequence for the astronaut.
    4. Two skaters push off from each other on smooth ice: a 4040 kg skater and a 6060 kg skater. Predict which moves faster and justify using forces and mass.
Answers

Answer key

Attempt the practice first. When you're ready to check, expand the answers below.

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Year 7 answers

Fluency

Tier 1: recall and identify

    1. A push or pull, measured in newtons (N).
    2. Contact: friction, normal, tension, applied, air resistance. Non-contact: gravity, magnetic, electrostatic.
    3. The single force that has the same effect as all the individual forces on an object combined.
    4. Mass is how much matter is in an object (kg) — it does not change. Weight is the gravitational force on that mass (N) — it depends on gravity.
    5. 2510=1525 - 10 = 15 N to the right.
    6. m=W/g=200/10=20m = W/g = 200/10 = 20 kg.
    7. A diagram showing every force acting on an object as an arrow from a single point, with correct direction and relative size.
    8. Net force =5050=0= 50 - 50 = 0 N. The rope does not accelerate; it stays at rest (or moves at constant speed if already moving).
    9. It continues at constant speed in a straight line forever (no friction, no other forces).
    10. W=60×10=600W = 60 \times 10 = 600 N.
Reasoning

Tier 2: explain and reason

    1. Gravity pulls the book down; the table pushes it up with an equal normal force. The two forces cancel, giving zero net force — so the book’s motion does not change.
    2. Balanced. Constant velocity in a straight line means zero net force; the driving force equals air resistance + rolling friction.
    3. Forces: gravity (weight) downward, air resistance equal and upward. Net force =0= 0, so the skydiver falls at constant (terminal) speed.
    4. From F=maF = ma, for the same acceleration, a larger mass requires a proportionally larger force.
    5. Mass is the amount of matter — unchanged by location. Gravity on the Moon is about 1/61/6 of Earth’s, so the gravitational force (weight) is 1/61/6 — but the amount of matter is identical.
    6. No. Although speed is constant, direction is changing (circular motion). A change in direction means the velocity changes, which requires a net force — gravity, pulling toward Earth.
Reasoning

Tier 3: apply to a novel context

    1. Fnet=50004000=1000F_{\text{net}} = 5000 - 4000 = 1000 N upward. The rocket accelerates upward.
    2. a=F/m=12/3=4a = F/m = 12/3 = 4 m/s2^2.
    3. (a) On jumping: gravity much greater than air resistance → large net force down → accelerates down. (b) As speed grows, air resistance grows → net downward force shrinks → acceleration decreases. (c) At terminal velocity: air resistance = weight → net force zero → constant speed.
    4. The post is pulled at an angle between north and east. The pull is stronger toward north (80 N > 60 N), so the direction is closer to north than to east — roughly north-north-east.
Reasoning

Challenge

    1. a=Δv/Δt=20/10=2a = \Delta v/\Delta t = 20/10 = 2 m/s2^2. F=ma=1200×2=2400F = ma = 1200 \times 2 = 2400 N. This force is generated by friction between the drive wheels and the road (the engine spins the wheels; the ground pushes the car forward).
    2. When a lift accelerates upward, the floor must push harder than gravity to both support the person and accelerate them upward — so the normal force is greater than weight, making you feel heavier. Going down: floor pushes less than gravity briefly, so you feel lighter.
    3. Mass: W/g=50/10=5W/g = 50/10 = 5 kg, unchanged. On the new planet: W=5×30=150W = 5 \times 30 = 150 N. The astronaut feels three times heavier — walking, lifting and standing all become very tiring.
    4. By Newton’s third law the forces are equal and opposite. By F=maF = ma the skater with less mass (4040 kg) has greater acceleration and moves faster (6060 kg skater moves slower). Ratio of speeds: 60:40=3:260:40 = 3:2.

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