Grade 12 Physical Sciences: Newton's laws & gravitation

a) Static friction is the force that opposes the tendency of motion of a stationary object relative to the surface it is on. ✓✓

b) Free-body diagram of the box (horizontal forces only): tension $T$ pointing towards the pulley ✓ and static friction $f_s$ pointing in the opposite direction ✓. (Both drawn from a dot/box and labelled.)

c) On the point of sliding the box is in equilibrium, so $T=f_s^{max}$ and, for the hanging bucket, $T=m_{hang}g$.
$f_s^{max}={\mu }_{s}N$ with $N=m_{box}g=6\times 9,8=58,8$ N ✓
${\mu }_s=\frac{f_s^{max}}{N}=\frac{2,7\times 9,8}{6\times 9,8}$ ✓✓ $=0,45$ ✓

d) Kinetic friction: $f_k={\mu }_{k}N=0,38\times 6\times 9,8=22,34$ N ✓
Apply Newton's Second Law to the system:
$F_{net}=ma$ ✓
$5\times 9,8-22,34=(6+5)a$ ✓✓
$a=\frac{26,66}{11}=2,42$ m·s⁻² ✓ (formula ✓ substitution ✓ answer + unit ✓)

e) INCREASES ✓ — the added brick increases the normal force, so the maximum static friction $f_s^{max}={\mu }_{s}N$ increases, and a greater hanging weight (tension) is needed to start the box sliding. ✓

Modelled on Nov 2024 Paper 1, Question 2 — same skills, new scenario.

B. Forces in equilibrium means the resultant force is zero, so by Newton's first law the acceleration is zero and the velocity (magnitude AND direction) stays constant. A is wrong: zero net force means zero acceleration in every direction. C is wrong: a decreasing speed requires a non-zero net force opposing the motion. D is wrong: equilibrium means the net force is exactly zero.

Modelled on Nov 2024 Paper 1, Question 1.1 — same skill, new scenario.

a) Newton's First Law: An object continues in its state of rest or of uniform (constant) velocity unless a net (resultant) force acts on it. ✓✓

b) Free-body diagram of the bag: weight $w$ (or $F_g$) vertically downwards ✓; normal force $N$ vertically upwards ✓; tension $T=40$ N along the rope, at angle θ above the horizontal ✓; static friction $f_s$ horizontally, opposite to the horizontal component of the tension ✓.

c) $T\sin \theta =24$ N
$\sin \theta =\frac{24}{40}$ ✓
$\theta =36,87°$ ✓

d) Horizontal component: $T_x=\sqrt{40^2-24^2}=32$ N (or $40\cos 36,87°=32$ N) ✓
Vertical equilibrium: $N=mg-T_y=12\times 9,8-24=93,6$ N ✓✓
On the point of sliding: $f_s^{max}=T_x=32$ N and $f_s^{max}={\mu }_{s}N$ ✓
${\mu }_s=\frac{32}{93,6}=0,34$ ✓

e) DECREASES ✓ While the bag remains at rest it is in equilibrium, so the friction is equal in magnitude to the horizontal component of the tension, $T\cos \theta$. ✓ When θ increases, $\cos \theta$ decreases, so the horizontal component decreases. ✓ The static friction adjusts to balance this smaller component, so the friction decreases. ✓

Modelled on Nov 2025 Paper 1, Question 2 — same skills, new scenario.

B. Newton's Second Law, $F_{net}=ma$, means the net force always points in the direction of the acceleration. A fails whenever an object slows down — the net force then points opposite to the velocity. C is false: a moving object has zero net force only if its velocity is constant, not merely because it is moving. D is false: net force is tied to acceleration, not speed — a very fast object can have zero net force. Modelled on Nov 2025 Paper 1, Question 1.1 — same skills, new scenario.