Why work out the net force first?

The F in F = ma is the sum of all forces, not just the push.

Why does the scale read higher in a lift going up?

The support force must exceed the weight to give an upward net force.

Why is the tension the same at both ends of a light string?

A light string has negligible mass, so its tension is equal throughout.

S4 Physics: Forces and Newton Laws Practice Paper 2

2026-11-01

Tutor Prep Guide · Secondary Physics

Paper two warms up with 24 N on 8 kg, weight W = mg and the first law, then works net force against friction, four seconds from rest, a gram to kilogram conversion and a lift scale, before a two-block tension and a sliding stone, ending with the inertia mix-up.

Quick answer　3 m/s^2, 20 N, rest or steady, 1.6 m/s^2, 2 m/s^2 with 8 m/s and 16 m, 5 m/s^2, 575 N, 2 m/s^2 with 12 N, 3.75 m/s^2 with 18.75 N, and more inertia.

01　Paper details (check answers afterwards)

Item	Detail
Questions	10 questions
Suggested time	25 minutes
Difficulty mix	Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1
Chapter	Chapter: S4 Physics: Forces and Newton Laws – Tutor Lesson Plan
Level	S4

02　Questions (10 items)

FoundationA 24 N net force acts on an 8 kg box. Find the acceleration.
FoundationWhat is the weight of a 2 kg object on the Earth surface (g = 10 m/s^2)?
FoundationAccording to Newton first law, how does an object move when the net force is zero?
FoundationA 12 N push acts on a 5 kg crate while friction is 4 N. Find the acceleration.
IntermediateA 6 N net force acts for 4 s on a 3 kg object starting from rest. Find the acceleration, the speed after 4 s and the displacement.
IntermediateA 4 N net force acts on an 800 g ball. Find the acceleration.
IntermediateA 50 kg person stands on a scale in a lift accelerating upwards at 1.5 m/s^2 (g = 10). What does the scale read?
ChallengeBlocks of 4 kg and 6 kg joined by a light string rest on a smooth table. A 20 N horizontal force pulls the 4 kg block. Find the acceleration and the tension.
ChallengeA 5 kg stone slides at 15 m/s and stops in 4 s. Find the deceleration and the friction force.
Error redoA student says a larger mass means less inertia. What is wrong?

03　Answers and one-line notes

#	Answer	One-line note
1	3 m/s^2 (24 / 8)	From F = ma, acceleration is the net force divided by the mass.
2	20 N	Weight is W = mg, measured in newtons.
3	It stays at rest or moves at steady speed in a straight line	No net force means no acceleration, so the motion does not change.
4	1.6 m/s^2 (net 8 N / 5 kg)	First take the net force 12 – 4 = 8 N, then divide by the mass.
5	2 m/s^2, 8 m/s and 16 m	a is 2, v is at = 8 and s is 1/2 a t^2 = 16.
6	5 m/s^2 (4 / 0.8)	Convert the mass to kilograms first: 800 g is 0.8 kg.
7	575 N	The support force must balance the weight and provide the upward acceleration: 50 x 11.5.
8	2 m/s^2 and tension 12 N	For the whole system a is 20 / 10 = 2; the string pulls only the 6 kg block, so T is 6 x 2 = 12 N.
9	3.75 m/s^2 and 18.75 N	The deceleration is 15 / 4 = 3.75, so the force is 5 x 3.75.
10	A larger mass means more inertia, so the motion is harder to change	Inertia is the resistance to a change of motion, and it grows with mass.

04　What to do after marking

After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
Repeat the same item two days later; move to the next paper only when both attempts are correct.
Parents only need the answer and note columns; there is no need to explain every item.

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