SACE Board SACE 2024-style practice: A 3.0\ kg block on a frictionless bench is connected by a light string over a frictionless pulley to a hanging 2.0\ kg mass. (a) Calculate the acceleration of the system. (b) Calculate the tension in the string. Use g = 9.80\ m s^-2.

Treat the two masses as one system to find the acceleration, then isolate one mass for the tension. (a) The only external force driving the system is the weight of the hanging mass, W = (2.0)(9.80) = 19.6\ N. The total mass is 3.0 + 2.0 = 5.0\ kg. $a = F_netm_total = 19.65.0 = 3.92\ m s^-2.$ (2 marks) (b) Isolate the 3.0\ kg block on the bench: the only horizontal force is the tension, so T = m a = (3.0)(3.92) = 11.8\ N. (2 marks) Markers reward the whole-system step for a and isolating one mass for T.

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How do Newton's three laws relate the forces on an object to its motion?

Apply Newton's three laws of motion, drawing free-body diagrams and resolving forces to determine the acceleration of objects and the forces in interacting systems.

Newton's three laws of motion, free-body diagrams, resolving forces on inclines, connected systems, and using net force to find acceleration, with worked examples.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

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

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What this dot point is asking
The three laws
Free-body diagrams
Forces on an inclined plane
Connected systems
Equilibrium

What this dot point is asking

You need to state and apply Newton's three laws, draw free-body diagrams, resolve forces (including on inclined planes), and calculate acceleration from the net force on an object or connected system.

Block on a frictionless incline

Acceleration down a 25 degree slope

A $4.0\ \text{kg}$ block slides down a frictionless incline at $25^\circ$ . Find its acceleration and the normal force. Use $g = 9.8\ \text{m s}^{-2}$ .

Along the slope: $F_{\text{net}} = mg\sin\theta = 4.0 \times 9.8 \times \sin 25^\circ = 16.6\ \text{N}$ .

So $a = g\sin\theta = 9.8 \times \sin 25^\circ = 4.1\ \text{m s}^{-2}$ down the slope.

Perpendicular: $N = mg\cos\theta = 4.0 \times 9.8 \times \cos 25^\circ = 35.5\ \text{N}$ .

Note the mass cancels in the acceleration but not in the forces.

The three laws

The second law is the workhorse. "Net force" means the vector sum of all forces acting on the object. If the net force is zero the object is in equilibrium (first law); if not, it accelerates in the direction of the net force, with $a = F_\text{net}/m$ .

Free-body diagrams

A free-body diagram shows one object as a point (or simple shape) with every force on it drawn as an arrow from that point. Typical forces are:

Weight $W = mg$ , always vertically down.
Normal force $N$ , perpendicular to the surface.
Tension $T$ , along a rope or cable, pulling away from the object.
Friction $f$ , parallel to the surface, opposing relative motion.
Applied force $F$ , whatever pushes or pulls.

Once drawn, resolve the forces into perpendicular axes (usually horizontal and vertical, or along and across an incline) and apply $\vec{F}_{\text{net}} = m\vec{a}$ to each axis separately.

Forces on an inclined plane

For an object on a frictionless incline at angle $\theta$ , choose axes along and perpendicular to the slope. The weight resolves into:

W_{\parallel} = mg\sin\theta\ (\text{down the slope}) \qquad W_{\perp} = mg\cos\theta.

Perpendicular to the slope there is no acceleration, so

N = mg\cos\theta

. Along the slope,

mg\sin\theta

produces an acceleration

a = g\sin\theta

(frictionless). If friction acts, subtract

f

from

mg\sin\theta

before dividing by the mass.

Connected systems

For objects joined by a rope (e.g. a trolley pulled by a hanging mass over a pulley), treat the whole system to find the common acceleration, then isolate one object to find the tension. For the whole system:

a = \frac{\text{net external force}}{\text{total mass}}.

Then apply Newton's second law to a single object to get the internal tension. The rope tension is an internal force for the system but an external force for each block, which is why isolating one block reveals it.

Equilibrium

When $\vec{F}_{\text{net}} = 0$ the object is in equilibrium (at rest or constant velocity). Then the components of all forces balance in each direction. This is how you find unknown tensions in a hanging sign or the friction holding a box still on a slope. Set the sum of components to zero along each axis and solve.

Exam-style practice questions

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

SACE 20244 marksA

3.0\ \text{kg}

block on a frictionless bench is connected by a light string over a frictionless pulley to a hanging

2.0\ \text{kg}

mass. (a) Calculate the acceleration of the system. (b) Calculate the tension in the string. Use

g = 9.80\ \text{m s}^{-2}

Show worked answer →

Treat the two masses as one system to find the acceleration, then isolate one mass for the tension.

(a) The only external force driving the system is the weight of the hanging mass, $W = (2.0)(9.80) = 19.6\ \text{N}$ . The total mass is $3.0 + 2.0 = 5.0\ \text{kg}$ .

a = \dfrac{F_\text{net}}{m_\text{total}} = \dfrac{19.6}{5.0} = 3.92\ \text{m s}^{-2}.

(2 marks)

(b) Isolate the $3.0\ \text{kg}$ block on the bench: the only horizontal force is the tension, so $T = m a = (3.0)(3.92) = 11.8\ \text{N}$ . (2 marks) Markers reward the whole-system step for $a$ and isolating one mass for $T$ .

SACE 20233 marksA

4.0\ \text{kg}

block is held on a frictionless incline at

25.0^\circ

to the horizontal, then released. (a) Calculate the acceleration of the block down the slope. (b) Calculate the normal force from the surface on the block. Use

g = 9.80\ \text{m s}^{-2}

Show worked answer →

Choose axes along and perpendicular to the slope and resolve the weight $W = mg$ .

(a) Along the slope the net force is $mg\sin\theta$ , so $a = g\sin\theta = (9.80)\sin 25.0^\circ = 4.14\ \text{m s}^{-2}$ down the slope. The mass cancels. (2 marks)

(b) Perpendicular to the slope there is no acceleration, so $N = mg\cos\theta = (4.0)(9.80)\cos 25.0^\circ = 35.5\ \text{N}$ . (1 mark) Markers reward resolving the weight into the two slope components and recognising $N = mg\cos\theta$ , not $mg$ .