Unit 2: Linear motion and waves

QLDPhysicsSyllabus dot point

Topic 1: Linear motion and force

Recall, describe and apply Newton's three laws of motion, including the use of free-body diagrams to identify forces acting on an object and solve problems involving weight, normal force, friction and tension

A focused answer to the QCE Physics Unit 2 dot point on Newton's three laws and force analysis. States each law, walks through free-body diagrams for the standard QCAA problem types (level surface with friction, inclined plane, connected bodies, hanging tension), and works a force-on-an-incline example that recurs in IA1 stimulus and EA Paper 2.

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What this dot point is asking

QCAA expects you to state and apply Newton's three laws of motion, and to construct free-body diagrams to analyse the forces on a single body. The standard problem types are level-surface motion with friction, motion on an inclined plane, connected bodies (pulleys and trains of carts), and tension in a hanging string or cable.

Newton's three laws

First law (inertia). An object continues in a state of rest or uniform motion in a straight line unless a net external force acts on it. A net force of zero means constant velocity (which includes being at rest).

Second law. The net force on an object equals its mass times acceleration:

Fnet=ma\vec{F}_{\text{net}} = m \vec{a}

Force has SI unit newton (N): 11 N is the force that accelerates a 11 kg mass at 11 m s2^{-2}. Force is a vector. Add forces vectorially.

Third law. For every action there is an equal and opposite reaction. If body A exerts a force on body B, then body B exerts a force of equal magnitude and opposite direction on body A. The forces act on different bodies and never on the same body, which is why a third-law pair does not "cancel out" the way two opposing forces on one body do.

Free-body diagrams

Isolate one object and draw every external force acting on it as an arrow from the object's centre. Standard forces:

  • Weight (W=mgW = mg). Straight down. Always present unless explicitly in deep space.
  • Normal force (NN). Perpendicular to the contact surface, pointing away from the surface.
  • Friction (ff). Parallel to the surface, opposing relative motion or relative motion tendency. Kinetic: fk=μkNf_k = \mu_k N. Static: fsμsNf_s \le \mu_s N.
  • Tension (TT). Along a string or cable, pulling away from the body.
  • Applied force. As stated.

Sum forces vectorially in two perpendicular directions and apply F=maF = ma in each.

Inclined planes

On a frictionless ramp at angle θ\theta:

  • Weight component along the slope (down the slope): W=mgsinθW_\parallel = mg \sin\theta.
  • Weight component perpendicular to slope: W=mgcosθW_\perp = mg \cos\theta.
  • Normal force balances WW_\perp: N=mgcosθN = mg \cos\theta.
  • If friction acts up the slope (object sliding down): Fnet=mgsinθμkmgcosθF_{\text{net}} = mg \sin\theta - \mu_k mg \cos\theta.

Choose the xx-axis along the slope and the yy-axis perpendicular. This eliminates the need to resolve the normal force or friction.

Connected bodies

For two masses joined by a light, inextensible string over a frictionless pulley, both masses have the same magnitude of acceleration and the tension is the same throughout the string. Write F=maF = ma for each mass, then solve simultaneously.

Worked example

A 2.02.0 kg block sits on a frictionless ramp inclined at 30°30°. Find the block's acceleration down the ramp. Use g=9.8g = 9.8 m s2^{-2}.

F=mgsinθ=(2.0)(9.8)sin30°=9.8F_\parallel = mg \sin\theta = (2.0)(9.8)\sin 30° = 9.8 N down the slope.

a=F/m=9.8/2.0=4.9a = F_\parallel / m = 9.8 / 2.0 = 4.9 m s2^{-2} down the slope.

(Equivalent: a=gsinθ=9.8sin30°=4.9a = g \sin\theta = 9.8 \sin 30° = 4.9 m s2^{-2}.)

Common traps

Confusing action-reaction with balanced forces. A book on a table has weight (down) and normal force (up). These are not a third-law pair because both act on the book. The third-law pair to the book's weight is the book's gravitational pull on the Earth.

Forgetting to resolve weight on an incline. Weight does not act along the slope unless the slope is vertical.

Treating static friction as μsN\mu_s N at all times. Static friction is only equal to μsN\mu_s N at the threshold of slipping. Before slipping, it equals whatever force it needs to to keep the object stationary, up to μsN\mu_s N.

Using F=maF = ma without summing all forces. Always tally the net force first.

In one sentence

Newton's first law says zero net force gives constant velocity, the second law Fnet=maF_{\text{net}} = ma relates net force to acceleration, and the third law pairs equal and opposite forces on different bodies; a free-body diagram showing weight, normal force, friction and tension lets you sum forces and solve for the acceleration of a single body.

Past exam questions, worked

Real questions from past QCAA papers on this dot point, with our answer explainer.

Year 11 SAC5 marksA $5.0$ kg block is pulled along a horizontal surface by a horizontal force of $30$ N. The coefficient of kinetic friction is $\mu_k = 0.30$. Using $g = 9.8$ m s$^{-2}$, calculate (a) the friction force and (b) the acceleration of the block.
Show worked answer →

(a) Friction force. Normal force on a horizontal surface equals weight.

N=mg=(5.0)(9.8)=49N = mg = (5.0)(9.8) = 49 N.

Kinetic friction: fk=μkN=(0.30)(49)=14.7f_k = \mu_k N = (0.30)(49) = 14.7 N, opposing motion.

(b) Acceleration. Net horizontal force divided by mass.

Fnet=Fappliedfk=3014.7=15.3F_{\text{net}} = F_{\text{applied}} - f_k = 30 - 14.7 = 15.3 N.

a=Fnet/m=15.3/5.0=3.06a = F_{\text{net}} / m = 15.3 / 5.0 = 3.06 m s2^{-2}.

Markers reward an explicit free-body diagram, separate horizontal and vertical force balances, and the substitution into F=maF = ma rather than guessing.

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