A force of 20\,N acts on a 4\,kg mass on a smooth surface. Find the acceleration. [2 marks]

Find the weight of a 7\,kg mass, taking g = 9.8\,m s^-2. [1 mark]

CCEA CCEA 2018-style practice: A box of mass 12\,kg rests on a smooth horizontal floor. A horizontal force of 30\,N acts on it. Find the acceleration, and state the magnitude of the normal reaction. Take g = 9.8\,m s^-2.

Horizontally, the only force is the applied 30\,N, so by Newton's second law: F = ma 30 = 12a a = 2.5\,m s^-2. Vertically the box does not accelerate, so the normal reaction balances the weight: R = mg = 12 × 9.8 = 117.6\,N. Markers reward F = ma for the horizontal acceleration, recognising vertical equilibrium, and the normal reaction equal to the weight.

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How do Newton's laws relate the forces on a body to its acceleration?

Forces as vectors, modelling assumptions, Newton's three laws, the equation $F = ma$ , weight, normal reaction, tension and the motion of connected particles.

A CCEA A-Level Mathematics answer on forces as vectors, the standard modelling assumptions, Newton's three laws, the equation F = ma, weight, normal reaction and tension, and the motion of connected particles such as a mass on a string over a pulley.

Generated by Claude Opus 4.813 min answerUpdated 2026-06-16

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

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What this dot point is asking

CCEA wants you to treat forces as vectors, apply the standard modelling assumptions, state and use Newton's three laws, work with the equation $F = ma$ , identify the common forces (weight, normal reaction, tension), and analyse the motion of connected particles such as masses joined by a string over a pulley. This is the heart of the mechanics half of AS 2.

The answer

Forces and modelling assumptions

Newton's three laws

Weight, reaction and tension

The weight of a body is $W = mg$ , acting vertically down. The normal reaction $R$ is the push of a surface, perpendicular to it. Tension $T$ is the pull of a string, along the string and away from the body. To solve a problem, draw all the forces, then apply $F = ma$ in each direction, taking the direction of motion as positive.

Connected particles

For two particles joined by a light inextensible string over a smooth pulley, both share the same acceleration magnitude, and the tension is the same throughout the string. Write an equation of motion ( $F = ma$ ) for each particle separately, then add or solve them simultaneously to find the acceleration and the tension.

Worked example: a mass pulling another along a table

Find the acceleration of a connected system

A $4\,\text{kg}$ block on a smooth horizontal table is connected by a light inextensible string over a smooth pulley at the table's edge to a hanging $6\,\text{kg}$ mass. Find the acceleration and the tension. Take $g = 9.8\,\text{m s}^{-2}$ .

step Equation for the hanging mass

The $6\,\text{kg}$ mass falls, so taking down as positive:

6g - T = 6a.

step Equation for the block on the table

The $4\,\text{kg}$ block accelerates horizontally, pulled by the tension:

T = 4a.

step Solve simultaneously

Substituting $T = 4a$ into the first equation:

6g - 4a = 6a \Rightarrow 6g = 10a \Rightarrow a = \frac{6 \times 9.8}{10} = 5.88\,\text{m s}^{-2}.

Then

T = 4a = 23.52\,\text{N}

. The block and the hanging mass share the same acceleration because the string is inextensible.

Examples in context

Example 1. A lift accelerating upwards. A person of mass $m$ in a lift accelerating up at $a$ feels a floor reaction $R = m(g + a)$ , larger than their weight, which is why you feel heavier as a lift starts to rise. Applying $F = ma$ vertically explains the everyday sensation.

Example 2. Towing a trailer. A car towing a trailer is a connected-particle problem: the tow bar tension is found by writing $F = ma$ for the trailer alone, while the engine force drives the whole combined mass. The same method as the pulley problem applies on the flat.

Try this

Q1. A force of $20\,\text{N}$ acts on a $4\,\text{kg}$ mass on a smooth surface. Find the acceleration. [2 marks]

Cue. $a = F/m = 20/4 = 5\,\text{m s}^{-2}$ .

Q2. State Newton's first law. [1 mark]

Cue. A body stays at rest or moves at constant velocity unless acted on by a resultant force.

Q3. Find the weight of a $7\,\text{kg}$ mass, taking $g = 9.8\,\text{m s}^{-2}$ . [1 mark]

Cue. $W = mg = 7 \times 9.8 = 68.6\,\text{N}$ .

Exam-style practice questions

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

CCEA 20207 marksTwo particles of mass

5\,\text{kg}

and

3\,\text{kg}

are connected by a light inextensible string over a smooth pulley. The system is released from rest. Taking

g = 9.8\,\text{m s}^{-2}

, find the acceleration of the system and the tension in the string.

Show worked answer →

Let the acceleration be $a$ and the tension $T$ . The heavier mass falls and the lighter mass rises.

For the $5\,\text{kg}$ mass (falling): $5g - T = 5a$ .

For the $3\,\text{kg}$ mass (rising): $T - 3g = 3a$ .

Adding the equations eliminates $T$ : $5g - 3g = 8a$ , so $2g = 8a$ and $a = \frac{2 \times 9.8}{8} = 2.45\,\text{m s}^{-2}$ .

Substitute back: $T = 3a + 3g = 3(2.45) + 3(9.8) = 7.35 + 29.4 = 36.75\,\text{N}$ .

Markers reward an equation of motion for each particle, adding to eliminate $T$ , the acceleration, and the tension.

CCEA 20185 marksA box of mass

12\,\text{kg}

rests on a smooth horizontal floor. A horizontal force of

30\,\text{N}

acts on it. Find the acceleration, and state the magnitude of the normal reaction. Take

g = 9.8\,\text{m s}^{-2}

Show worked answer →

Horizontally, the only force is the applied $30\,\text{N}$ , so by Newton's second law:

$F = ma \Rightarrow 30 = 12a \Rightarrow a = 2.5\,\text{m s}^{-2}.$

Vertically the box does not accelerate, so the normal reaction balances the weight:

$R = mg = 12 \times 9.8 = 117.6\,\text{N}.$

Markers reward $F = ma$ for the horizontal acceleration, recognising vertical equilibrium, and the normal reaction equal to the weight.

Related dot points

Sources & how we know this

CCEA GCE Mathematics specification — CCEA (2018)