HSC Exam Practice

Sample response. Work is the energy transferred when a force causes a displacement in the direction of the force. Quantitatively, W = Fs cosθ, where θ is the angle between the force and the displacement. A force does zero work when it acts perpendicular to the displacement (θ = 90°, cos90° = 0), even if both the force and displacement are non-zero. Examples include the normal force on a horizontally moving object and centripetal force on a object in circular motion.

Marking notes. 1 mark for definition of work including reference to displacement in the direction of force; 1 mark for W = Fs cosθ with θ defined; 1 mark for correctly identifying θ = 90° as the zero-work condition with at least one example.

Sample response. The work-energy theorem states that the net work done on an object equals the change in its kinetic energy: W_net = ΔKE = KE_final − KE_initial. It is essential to use net work (the sum of work done by all forces) because multiple forces act simultaneously. For example, if friction acts against the applied force, friction does negative work that removes energy from the system. Using only the applied force work would overestimate ΔKE and give a final speed that is too high.

Marking notes. 1 mark for correctly stating W_net = ΔKE; 1 mark for explaining that net work = sum of work by all forces; 1 mark for a clear example showing what happens if only applied force is used (e.g. overestimate of speed, friction doing negative work).

Sample response. Conservation of mechanical energy applies only when no non-conservative forces (such as friction or air resistance) do work on the system — only conservative forces (gravity, ideal springs) act. In this case, KE₁ + U₁ = KE₂ + U₂ at every point. Example where it applies: a ball rolling on a frictionless slope — no friction, so all PE converts to KE. Example where it does not apply: a box sliding down a rough ramp (μ_k > 0) — friction is a non-conservative force that converts mechanical energy to thermal energy, reducing the total mechanical energy.

Marking notes. 1 mark for stating condition (no friction/non-conservative forces, only conservative forces do work); 1 mark for the conservation equation; 1 mark for a valid apply-example with justification; 1 mark for a valid does-not-apply example with justification referencing friction/non-conservative forces.

Sample response. KE = ½mv²; the v² term arises from kinematics: from Newton’s 2nd Law and the kinematic equation v² = u² + 2as, the work done equals force times displacement, which gives F × s = ½mv². So KE is proportional to v², not v. Going from 10 to 20 m/s: KE_i = ½m(10)² = 50m; KE_f = ½m(20)² = 200m. Ratio = 200m/50m = 4. Kinetic energy quadruples when speed doubles. This means crash severity increases dramatically with speed — stopping a car at 20 m/s requires four times the work compared to 10 m/s.

Marking notes. 1 mark for stating KE ∝ v² (not v); 1 mark for a correct explanation or derivation showing why (even qualitative: squaring the speed term); 1 mark for calculation showing ratio = 4; 1 mark for clear conclusion: speed doubles, KE quadruples.

Sample response. The student is incorrect because weight (mg) acts vertically downward and does zero work on a horizontally moving car. On a flat road at constant velocity, Newton’s First Law requires F_drive = F_resistance (net horizontal force = 0). Therefore the correct approach is to find the driving force by equating it to the horizontal resistance force stated in the problem, then use P = F_resistance × v. Using mg as the driving force confuses a vertical force with a horizontal one and grossly overestimates engine power (a typical car would appear to need hundreds of kW rather than tens).

Marking notes. 1 mark for identifying the error (weight is vertical, does no work horizontally); 1 mark for stating the correct approach (Newton 1: F_drive = F_resistance); 1 mark for identifying F_resistance as the correct force to use in P = Fv.

Sample response. The hill is frictionless, so only gravity does work — mechanical energy is conserved. Apply KE₁ + U₁ = KE₂ + U₂: taking ground level as reference, 0 + mgh = ½mv² + 0. Masses cancel: v = √(2gh) = √(2 × 9.8 × 12) = √235.2 = 15.3 m/s. Conservation of mechanical energy is justified because no friction acts on the hill.

Marking notes. 1 mark for correctly selecting conservation of mechanical energy and justifying (no friction on hill); 1 mark for v = √(2gh) with correct substitution; 1 mark for 15.3 m/s.

Sample response. At constant cruise speed, net force = 0. P = F_resistance × v_cruise. v_cruise = P/F = 240/60 = 4 m/s. The cruise speed (4 m/s) is much lower than the hill-bottom speed (15.3 m/s) because the 240 W the cyclist can produce is only enough to overcome the 60 N resistance at 4 m/s. To maintain 15.3 m/s the cyclist would need P = 60 × 15.3 = 918 W — nearly four times their output. So the cyclist decelerates after the hill bottom until reaching the speed at which their power output exactly balances the resistive force.

Marking notes. 1 mark for v = P/F = 240/60 = 4 m/s; 1 mark for explaining the energy/power argument (240 W insufficient to maintain 15.3 m/s; 918 W would be needed); 1 mark for explicit conclusion (decelerates until power output balances resistance).

Sample response. t = s/v = 600/4 = 150 s.

Marking notes. 1 mark for 150 s (accept carry-forward from (b)).

Sample response.

Formula choice and justification: Friction (μ_k = 0.2) is present. Friction is a non-conservative force that converts mechanical energy to thermal energy, so E_mech is not conserved. Conservation of mechanical energy must not be used. The correct approach is W_net = ΔKE, summing work done by all forces (gravity and friction). [1 mark — must state why, not just which]

Slope geometry: sinθ = 18/60 = 0.300; cosθ = √(1 − 0.09) = 0.954. [1 mark]

Normal force: F_N = mg cosθ = 800 × 9.8 × 0.954 = 7479 N. [1 mark]

Friction force and work: f_k = μ_k × F_N = 0.2 × 7479 = 1496 N. W_friction = −f_k × s = −1496 × 60 = −89 760 J. (Negative because friction opposes motion.) [1 mark]

Work done by gravity: W_gravity = mgh = 800 × 9.8 × 18 = 141 120 J. [1 mark]

Apply W_net = ΔKE: W_net = 141 120 + (−89 760) = 51 360 J. Starting from rest: ½mv² = 51 360. v = √(2 × 51 360/800) = √128.4 = 11.3 m/s. (Compare: if frictionless, v = √(2gh) = √352.8 = 18.8 m/s — friction reduces speed significantly.) [1 mark]

Percentage PE lost to friction: % = |W_friction| / W_gravity × 100 = 89 760/141 120 × 100 = 63.6%. [1 mark]

Energy efficiency comment: Only 36.4% of the initial gravitational PE was converted to kinetic energy; 63.6% was dissipated as heat by friction. This is very inefficient for an engineered system. In practice, reducing μ_k (better lubrication, smoother surfaces) or reducing the slope angle (longer, shallower path) would improve efficiency by decreasing the normal force and thus friction work. [1 mark]

Energy transformation summary: Gravitational PE → KE (useful) + thermal energy (friction loss). Total E_mech decreases by 89 760 J, all of which becomes thermal energy in the surfaces. [1 mark — for coherent energy transformation narrative]

Marking criteria (9 marks): 1 = formula choice justified (W_net = ΔKE because friction non-conservative); 1 = sinθ = 0.3, cosθ = 0.954; 1 = F_N = 7479 N; 1 = W_friction = −89 760 J (sign required); 1 = W_gravity = 141 120 J; 1 = v = 11.3 m/s; 1 = % lost = 63.6%; 1 = evaluative comment on efficiency with physics reasoning; 1 = coherent energy transformation narrative (gravitational PE → KE + thermal).