HSC Exam Practice

Sample response. Power is the rate of energy transfer — the amount of energy transferred (or work done) per unit time. Its SI unit is the watt (W), where 1 W = 1 J/s. Power differs from work: work is the total energy transferred by a force through a displacement (measured in joules) and does not depend on time; power measures how quickly that work is done and is measured in watts.

Marking notes. 1 mark for a correct definition of power (rate of energy transfer); 1 mark for stating the SI unit as watt (W) = 1 J/s; 1 mark for correctly distinguishing work (total energy, joules, not a rate) from power (rate, watts, time-dependent).

Sample response. Formula 1: P = ΔE/Δt (average power). Applies when total energy transferred and total time are known; gives the average power over that interval. Formula 2: P = Fv cosθ (instantaneous power). Applies when force and velocity are known at a specific instant; for horizontal force and motion, θ = 0° so P = Fv. Formula 3: P = mgh/Δt (power for vertical lifting). Applies only when an object is being lifted purely vertically; it accounts only for work done against gravity.

Marking notes. 2 marks per formula: 1 for the correct formula stated, 1 for the correct condition. Accept P = W/Δt for formula 1. For formula 2 accept θ = 0 stated or implied (horizontal motion). For formula 3 — must state ‘vertical lifting only’ or equivalent.

Sample response. The student’s error is using the weight (mg) as the driving force. The weight acts vertically downward; on a flat road it is balanced by the normal force and plays no role in the horizontal power calculation [1]. At constant speed on a flat road, Newton’s First Law gives F_net = 0, meaning the horizontal driving force equals the total horizontal resistance force, not the weight [1]. The correct approach: identify constant v → apply Newton 1: F_drive = F_resistance → substitute into P = F_drive × v [1].

Marking notes. 1 mark for identifying the error (weight is not the driving force on a flat road); 1 mark for invoking Newton 1 correctly (F_drive = F_resistance at constant v); 1 mark for stating the correct approach (Newton 1 first, then P = F_drive × v).

Sample response. Average power is the total energy transferred divided by the total time interval: P_av = ΔE/Δt. Instantaneous power is the power at a specific moment: P = Fv cosθ. They differ when the power output varies over time [1]. An example where they differ significantly: a car accelerating from rest to 100 km/h. As speed increases, instantaneous power P = Fv increases continuously from zero; the average power over the whole interval is less than the final instantaneous power [1 — correct example with explanation]. Accept any other valid example (e.g. sprinter starting a race; a lift starting to move) [1 — third mark for clear statement of when they are equal: at constant power output over an interval, P_av = P_{instantaneous}; or accept correct example alone for full marks].

Marking notes. 1 mark for correct definition of average power; 1 mark for correct definition of instantaneous power; 1 mark for a valid example with a clear explanation of why the values differ.

Sample response. The instantaneous power output increases continuously as the car accelerates [1]. Using P = Fv: at rest, v = 0 so P = 0 W. As v increases to 30 m/s, P = 2000 × 30 = 60 000 W = 60 kW at the final speed [1]. Power is not constant during acceleration — it increases in direct proportion to v because F is constant. The power increases from 0 W to 60 kW over the acceleration phase [1].

Marking notes. 1 mark for stating power increases during acceleration; 1 mark for correctly calculating power at v = 30 m/s (P = 60 000 W = 60 kW); 1 mark for explaining that P increases in direct proportion to v because F is constant (or equivalent reasoning).

Sample response. Both motors do the same work (W = mgh) because they lift the same load the same height [1]. The 10 000 W motor does this work 100 times faster than the 100 W motor — it is described as “more powerful” because power = work/time; same work in less time means higher power. The 100 W motor can do the same job; it just takes 100 times longer [1].

Marking notes. 1 mark for stating that both do the same work (W = mgh); 1 mark for explaining that the 10 000 W motor is more powerful because it does the work in less time (power = work/time).

Sample response (a). At 12 m/s, upright: P = 200 × 12 = 2400 W; crouched: P = 140 × 12 = 1680 W (both readable from the graph) [1]. Power saving = 2400 − 1680 = 720 W [1]. This represents a 30% reduction in power requirement for the same speed.

Sample response (b). Both lines are straight because P = Fv: with F constant for each posture, P varies linearly with v; doubling v doubles P — a direct proportionality gives a straight line through the origin [1]. The crouched line has a smaller gradient because the drag force (F = 140 N) is smaller than the upright drag (F = 200 N); gradient = F in P = Fv, so lower F gives a less steep line [1 — gradient = F; 1 — explains why crouched gradient is smaller in terms of drag force reduction].

Sample response (c). The statement is inconsistent [1]. Work = F × d. Over 1 km (1000 m) at constant speed: W_upright = 200 × 1000 = 200 000 J = 200 kJ; W_crouched = 140 × 1000 = 140 000 J = 140 kJ [1]. The upright cyclist does 43% more work over the same distance. They are at the same speed but experience different resistance forces, so different amounts of work are done by each. “Same work over 1 km” is only true if the forces are the same [1].

Marking notes. Part (a): 1 mark for correct power values from graph or formula; 1 mark for correct power saving (720 W). Part (b): 1 mark for explaining linearity via P = Fv with constant F; 1 mark for gradient = F and larger F gives steeper line; 1 mark for applying this to explain why crouched line is shallower. Part (c): 1 mark for identifying the statement as inconsistent; 1 mark for correct work calculation for each posture; 1 mark for explaining that different forces mean different work over the same distance.

Sample response (a). P_mech = F × v, where F = mg = 400 × 9.8 = 3920 N (weight lifted at constant speed, so F_drive = mg). P_mech = 3920 × 0.8 = 3136 W ≈ 3.14 kW [1 formula correctly identified; 1 correct calculation with units].

Alternative: P = mgh/t; in 1 s, h = v × t = 0.8 m. P = (400 × 9.8 × 0.8)/1 = 3136 W. Accept either method.

Sample response (b). Efficiency = P_output/P_input; 0.75 = 3136/P_input; P_input = 3136/0.75 = 4181 W ≈ 4.18 kW [1].

Sample response (c). Efficiency cannot reach 100% because no real motor is perfectly frictionless — friction in bearings, windings, and pulleys converts some electrical energy to heat [1]. The ‘lost’ energy is dissipated as thermal energy (heat) in the motor windings and mechanical components, and some is converted to sound [1].

Marking notes. (a) 1 mark for formula (P = Fv or P = mgh/t with constant speed); 1 mark for correct calculation (3136 W). (b) 1 mark for correct P_input = 4181 W from efficiency formula. (c) 1 mark for a correct physical reason (friction/heat); 1 mark for identifying the destination of ‘lost’ energy (thermal energy / heat).

Sample response. Power, force and velocity are linked by the relationship P = Fv for a vehicle maintaining constant speed on a flat road. At constant velocity, Newton’s First Law gives F_net = 0, so the driving force exactly equals the total resistance force (air drag and rolling resistance). This means the engine must supply P = F_resistance × v watts to maintain any given speed. Since P is directly proportional to v when resistance is constant, the power required increases linearly with speed: doubling the cruise speed requires double the power. For example, on the Hume Highway connecting Sydney and Melbourne, trucks and cars travelling at 110 km/h (30.6 m/s) experience a significantly higher resistance than at 80 km/h; if resistance at 110 km/h is approximately 1200 N for a passenger car, P = 1200 × 30.6 = 36 720 W ≈ 37 kW, whereas at 22 m/s (80 km/h) with the same resistance P = 1200 × 22 = 26 400 W = 26.4 kW — demonstrating why fuel efficiency drops at highway speeds. In reality, air resistance increases approximately with v², so the power–speed relationship becomes steeper than linear at high speeds; this makes high-speed driving disproportionately expensive in energy terms. When a vehicle operates below its maximum engine power at cruise speed, the difference is ‘reserve power’. If the driver uses the full engine power (say 120 kW) at a speed where only 37 kW is needed, the new driving force is F = P_max/v = 120 000/30.6 = 3922 N, far exceeding the resistance of 1200 N. The net force (3922 − 1200 = 2722 N) causes acceleration (Newton’s Second Law: F_net = ma). The car speeds up — it is no longer at constant velocity — until either the driver reduces power or resistance increases enough (at higher speed) to restore equilibrium. Exceeding cruise power is therefore the mechanism for overtaking and hill climbing; it breaks the F_drive = F_resist condition and produces a net forward force. Understanding this P–v relationship is essential for Australian road freight engineers designing fuel-efficient interstate trucks that spend thousands of kilometres at steady highway speeds.

Marking criteria (7 marks). 1 = correctly states and derives P = Fv for constant speed via Newton 1 (F_drive = F_resistance). 1 = analyses how power changes with speed (linear proportionality when F constant; steeper at high speed when drag ∝ v² — accept either). 1 = uses a numerical example or calculation to support the analysis. 1 = correctly explains Newton’s First Law as the reasoning step for constant velocity (F_drive = F_resist). 1 = correctly describes the physical consequence of exceeding cruise power (net force forward → acceleration via Newton 2; vehicle speeds up). 1 = uses a specific named Australian transport or engineering context appropriately (Hume Highway, Pacific Highway, East–West Link tunnels, Snowy Hydro vehicles, mining haul trucks, etc.). 1 = reaches an explicit, evidence-based synthesis or conclusion (e.g. P–v relationship explains why high-speed travel is energy-intensive and why reserve power is essential for overtaking and gradients).