A 1200\ kg car experiences a resultant forward force of 3000\ N. Find its acceleration. [2 marks]

WJEC Eduqas Eduqas 2019-style practice: A block of mass 5.0\ kg rests on a frictionless plane inclined at 25^ to the horizontal. Calculate the component of the weight acting down the slope and the normal contact force. Take g = 9.81\ m s^-2.

Weight W = mg = 5.0 × 9.81 = 49.1\ N. Component down the slope: Wsin 25^ = 49.1 × 0.423 = 20.8\ N, about 21\ N. Normal contact force balances the perpendicular component: N = Wcos 25^ = 49.1 × 0.906 = 44.5\ N, about 45\ N. Markers reward resolving the weight parallel and perpendicular to the slope, the down-slope component about 21\ N, and the normal force about 45\ N.

EnglandPhysicsSyllabus dot point

How do Newton's three laws connect force, mass and motion, and how do we analyse forces in equilibrium?

Dynamics: Newton's three laws of motion, free-body diagrams, weight and the normal force, resolving forces on inclined planes, moments and the conditions for equilibrium, and terminal velocity.

A focused answer to the Eduqas A-Level Physics Component 1 dynamics content, covering Newton's three laws of motion, drawing free-body diagrams, weight and the normal force, resolving forces on inclined planes, moments and couples, the conditions for equilibrium, and terminal velocity.

Generated by Claude Opus 4.813 min answerUpdated 2026-06-02

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

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

Quick answer

Newton's first law: an object stays at rest or in uniform motion unless acted on by a resultant force. Newton's second law: the resultant force equals the rate of change of momentum, $F = ma$ for constant mass. Newton's third law: forces act in pairs that are equal, opposite and of the same type, acting on different bodies. Weight is $W = mg$ ; the normal contact force acts perpendicular to a surface. On a slope, resolve the weight into $mg\sin\theta$ down the slope and $mg\cos\theta$ into it. The moment of a force is $Fd$ (force times perpendicular distance); a couple is two equal antiparallel forces. For equilibrium the resultant force is zero and the resultant moment about any point is zero. At terminal velocity the resultant force is zero, so the object moves at constant speed.

Jump to a section

What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

Eduqas wants you to state and apply Newton's three laws, draw free-body force diagrams, treat weight as $W = mg$ and the normal contact force, resolve forces on an inclined plane, use moments and couples, apply the two conditions for equilibrium, and explain terminal velocity in terms of balanced forces.

The answer

Newton's three laws

A common Eduqas error is to treat a Newton's third law pair as the cause of equilibrium. The pair acts on different objects (for example the Earth pulls a book down, the book pulls the Earth up), whereas equilibrium of the book balances the weight against the normal force, which act on the same object.

Free-body diagrams, weight and normal force

Resolving forces on an inclined plane

Moments, couples and equilibrium

Terminal velocity

Connected masses over a pulley

A $3.0\ \text{kg}$ mass hangs from a light string over a frictionless pulley, connected to a $2.0\ \text{kg}$ mass on a horizontal frictionless table. Find the acceleration of the system and the tension in the string. Take $g = 9.81\ \text{m s}^{-2}$ .

step 1: Resultant force on the system

Only the hanging weight drives the motion: $F = m_1 g = 3.0 \times 9.81 = 29.4\ \text{N}$ .

step 2: Acceleration of the whole system

The string is inextensible, so both masses share one acceleration. Total mass $= 3.0 + 2.0 = 5.0\ \text{kg}$ , so $a = \dfrac{F}{m_\text{total}} = \dfrac{29.4}{5.0} = 5.9\ \text{m s}^{-2}$ .

step 3: Tension from the table mass

Apply $F = ma$ to the $2.0\ \text{kg}$ mass, on which the only horizontal force is the tension: $T = 2.0 \times 5.9 = 11.8\ \text{N}$ , about $12\ \text{N}$ .

Examples in context

Free-body analysis underpins vehicle design (braking forces and grip), structural engineering (beams, bridges and the moments at their supports), and rocketry (Newton's third law thrust). Terminal velocity governs parachute and skydiving safety, the settling of dust and the design of viscous dampers. Moments explain why a long spanner loosens a tight bolt and how a crane balances its load against a counterweight.

Try this

Q1. State Newton's second law in its most general form. [1 mark]

Cue. The resultant force equals the rate of change of momentum.

Q2. A $1200\ \text{kg}$ car experiences a resultant forward force of $3000\ \text{N}$ . Find its acceleration. [2 marks]

Cue. $a = \frac{F}{m} = \frac{3000}{1200} = 2.5\ \text{m s}^{-2}$ .

Q3. State the two conditions for an object to be in equilibrium. [2 marks]

Cue. The resultant force is zero and the resultant moment about any point is zero.

Exam-style practice questions

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

Eduqas 20194 marksA block of mass

5.0\ \text{kg}

rests on a frictionless plane inclined at

25^\circ

to the horizontal. Calculate the component of the weight acting down the slope and the normal contact force. Take

g = 9.81\ \text{m s}^{-2}

Show worked answer →

Weight $W = mg = 5.0 \times 9.81 = 49.1\ \text{N}$ .

Component down the slope: $W\sin 25^\circ = 49.1 \times 0.423 = 20.8\ \text{N}$ , about $21\ \text{N}$ .

Normal contact force balances the perpendicular component: $N = W\cos 25^\circ = 49.1 \times 0.906 = 44.5\ \text{N}$ , about $45\ \text{N}$ .

Markers reward resolving the weight parallel and perpendicular to the slope, the down-slope component about $21\ \text{N}$ , and the normal force about $45\ \text{N}$ .

Eduqas 20213 marksA uniform beam of weight

120\ \text{N}

and length

4.0\ \text{m}

is pivoted at one end. A vertical force is applied at the far end to hold the beam horizontal. Calculate the size of this force, using moments about the pivot.

Show worked answer →

The weight acts at the centre of the uniform beam, $2.0\ \text{m}$ from the pivot. Taking moments about the pivot, the anticlockwise moment of the applied force $F$ at $4.0\ \text{m}$ must balance the clockwise moment of the weight at $2.0\ \text{m}$ .

$F \times 4.0 = 120 \times 2.0$ , so $F = \dfrac{240}{4.0} = 60\ \text{N}$ .

Markers reward taking moments about the pivot, placing the weight at the centre of the uniform beam, and the applied force $60\ \text{N}$ .

Related dot points

Sources & how we know this

Eduqas GCE AS/A Level Physics specification (A720QS) — WJEC Eduqas (2015)