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How do forces change the way an object moves, and what does Newton's second law tell us?

State Newton's laws of motion and use the relationship F = ma for a single resultant force

State Newton's three laws of motion, find the resultant of balanced and unbalanced forces, and use F = ma with simple N(A)-Level numbers to link force, mass and acceleration.

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
  4. Try this

What this dot point is asking

SEAB wants you to state Newton's three laws of motion, to find the resultant of two or more forces, and to use the relationship F=maF = ma for the resultant force on an object. The big idea is that a force is a push or a pull, and it is the resultant (overall) force that changes how an object moves.

The answer

What a force does

A force is a push or a pull, measured in newtons (N\text{N}). A force can change an object's speed, its direction, or its shape. When several forces act, what matters is the resultant force: the single force that has the same effect as all of them together.

If forces act along the same line, you add forces in the same direction and subtract forces in opposite directions.

Newton's first law

Newton's first law says that an object stays at rest, or keeps moving at a constant velocity in a straight line, unless a resultant force acts on it.

So if the resultant force is zero, the motion does not change: a still object stays still, and a moving object keeps the same speed and direction. This is why a car at a steady speed in a straight line has a zero resultant force, with the driving force balancing friction and air resistance.

Newton's second law

Newton's second law links the resultant force to the acceleration:

F=maF = ma

where FF is the resultant force in newtons, mm is the mass in kilograms, and aa is the acceleration in m s2\text{m s}^{-2}. A bigger resultant force gives a bigger acceleration; a bigger mass gives a smaller acceleration for the same force.

You can rearrange the formula: a=Fma = \dfrac{F}{m} to find acceleration, or m=Fam = \dfrac{F}{a} to find mass.

Newton's third law

Newton's third law says that if object A pushes on object B, then object B pushes back on A with an equal and opposite force. Forces always come in pairs that act on two different objects.

Examples: a swimmer pushes water backwards and the water pushes the swimmer forwards; a rocket pushes gas downwards and the gas pushes the rocket upwards.

Examples in context

Example 1. A lift starting upward. A lift cable must pull up with more force than the weight of the lift to make it accelerate upward. The resultant force (cable tension minus weight) is what appears in F=maF = ma. Once the lift moves at a steady speed, the tension equals the weight and the resultant is zero.

Example 2. A car braking. When a driver brakes, friction from the road acts backwards on the tyres. This backward resultant force gives a negative acceleration (deceleration), slowing the car. A heavier car needs a bigger braking force for the same deceleration, which is why loaded lorries take longer to stop.

Try this

  • Cue. A 3.0 kg3.0\ \text{kg} mass has a resultant force of 9.0 N9.0\ \text{N} acting on it. Find the acceleration. [2 marks] a=Fm=9.03.0=3.0 m s2a = \dfrac{F}{m} = \dfrac{9.0}{3.0} = 3.0\ \text{m s}^{-2}.

  • Cue. State Newton's third law and give one everyday example. [2 marks] If A pushes B, then B pushes A with an equal and opposite force; for example, a gun recoils backward as it pushes the bullet forward.

  • Cue. A force of 30 N30\ \text{N} acts forwards on a box and friction of 10 N10\ \text{N} acts backwards. The box has mass 4.0 kg4.0\ \text{kg}. Find the acceleration. [3 marks] Resultant =3010=20 N= 30 - 10 = 20\ \text{N}; a=204.0=5.0 m s2a = \dfrac{20}{4.0} = 5.0\ \text{m s}^{-2}.

Exam-style practice questions

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

Original4 marksA box of mass 5.0 kg5.0\ \text{kg} is pushed along the floor with a force of 20 N20\ \text{N}. Friction on the box is 5.0 N5.0\ \text{N}. (a) Find the resultant force. (b) Calculate the acceleration. (c) State the unit of acceleration.
Show worked answer →

(a) The push and friction act in opposite directions, so the resultant force =205.0=15 N= 20 - 5.0 = 15\ \text{N} in the direction of the push.

(b) Using F=maF = ma, rearrange to a=Fm=155.0=3.0 m s2a = \dfrac{F}{m} = \dfrac{15}{5.0} = 3.0\ \text{m s}^{-2}.

(c) The unit of acceleration is metres per second squared (m s2\text{m s}^{-2}).

What markers reward: subtracting friction to get the resultant force, using F=maF = ma with the resultant (not the applied) force, and the correct unit.

Original3 marks(a) State Newton's first law of motion. (b) A car travels at a steady speed in a straight line. What can you say about the resultant force on it? (c) State one example of Newton's third law.
Show worked answer →

(a) An object stays at rest, or keeps moving at a constant velocity in a straight line, unless a resultant force acts on it.

(b) The resultant force is zero, because the velocity is constant (steady speed in a straight line). The driving force balances the friction and air resistance.

(c) Any matched pair, for example: a swimmer pushes the water backwards, and the water pushes the swimmer forwards with an equal and opposite force.

What markers reward: the first law stated correctly, recognising constant velocity means zero resultant force, and a clear equal-and-opposite pair for the third law.

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