Skip to main content
ScotlandPhysicsSyllabus dot point

How do forces change the motion of an object, and what is the link between force, mass and acceleration?

Newton's laws: balanced and unbalanced forces, Newton's first and second laws including F equals ma, the difference between mass and weight, and friction and free-body force diagrams.

An SQA National 5 Physics answer on Newton's laws, covering balanced and unbalanced forces, Newton's first and second laws including F equals ma, the difference between mass and weight using W equals mg, terminal velocity, and how to find a resultant force from a free-body diagram.

Generated by Claude Opus 4.810 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this key area is asking
  2. Balanced forces and Newton's first law
  3. Unbalanced forces and Newton's second law
  4. Mass and weight
  5. Friction and Newton's third law
  6. Try this

What this key area is asking

The SQA wants you to explain balanced and unbalanced forces, state and use Newton's first and second laws (including F=maF = ma), find a resultant force from a force diagram, and tell the difference between mass and weight using W=mgW = mg.

Balanced forces and Newton's first law

This is often tested with terminal velocity. A skydiver speeds up at first because weight is bigger than air resistance, but air resistance grows with speed until it balances weight. At that point the forces are balanced, the resultant is zero, and the skydiver falls at a constant terminal velocity.

Unbalanced forces and Newton's second law

A larger force gives a larger acceleration for the same mass, and a larger mass gives a smaller acceleration for the same force. To use F=maF = ma you must first find the resultant force by combining all the forces, taking one direction as positive.

Mass and weight

The gravitational field strength gg has the same value as the acceleration due to gravity, 9.8 m s29.8 \text{ m s}^{-2}, because for a freely falling object the only force is weight, so F=maF = ma becomes mg=mamg = ma, giving a=ga = g.

Friction and Newton's third law

Friction is a contact force that always acts to oppose the motion (or tendency to move) between two surfaces. It can be useful (grip for tyres and shoes) or a nuisance (wasting energy as heat). The SQA also expects you to know Newton's third law in its simple form: for every action force there is an equal and opposite reaction force. When you push on a wall, the wall pushes back on you with an equal force.

Try this

Q1. State Newton's first law. [1 mark]

  • Cue. If the forces on an object are balanced it stays at rest or moves at constant velocity in a straight line.

Q2. A box of mass 5.0 kg5.0 \text{ kg} experiences a resultant force of 15 N15 \text{ N}. Calculate its acceleration. [2 marks]

  • Cue. a=F/m=15/5.0=3.0 m s2a = F/m = 15/5.0 = 3.0 \text{ m s}^{-2}.

Q3. Calculate the weight of a 60 kg60 \text{ kg} person on Earth, taking g=9.8 N kg1g = 9.8 \text{ N kg}^{-1}. [2 marks]

  • Cue. W=mg=60×9.8=588 NW = mg = 60 \times 9.8 = 588 \text{ N}.

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SQA N5 style3 marksA car of mass 1200 kg accelerates at 2.5 m s per second per second. Calculate the unbalanced force acting on the car.
Show worked answer →

Use Newton's second law, which links unbalanced force, mass and acceleration.

Relationship: F=maF = ma.

Substitution: F=1200×2.5=3000 NF = 1200 \times 2.5 = 3000 \text{ N}.

Markers reward selecting F=maF = ma, correct substitution of the mass and acceleration, and a final answer with the unit newton (N\text{N}).

SQA N5 style3 marksAn astronaut has a mass of 80 kg. Calculate the astronaut's weight on Earth (g = 9.8 N per kg) and on the Moon (g = 1.6 N per kg).
Show worked answer →

Use the relationship between weight, mass and gravitational field strength.

Relationship: W=mgW = mg.

On Earth: W=80×9.8=784 NW = 80 \times 9.8 = 784 \text{ N}.

On the Moon: W=80×1.6=128 NW = 80 \times 1.6 = 128 \text{ N}.

The mass stays at 80 kg80 \text{ kg} everywhere; only the weight changes because gg is different. Markers reward W=mgW = mg, both substitutions, and the units in newtons.

Related dot points

Sources & how we know this