Work, energy and power: VCE Physics Unit 2 Year 11

Syllabus dot point

“Work $W = Fd \cos\theta$ , kinetic energy $\frac{1}{2}mv^2$ , gravitational potential energy $mgh$ , elastic potential energy $\frac{1}{2}kx^2$ , conservation of mechanical energy, and power $P = W/t = Fv$ ”

10 min readUpdated 21 May 2026

Quick answer

Work $W = Fd\cos\theta$ done by a force changes an object's kinetic energy ( $W_{\text{net}} = \Delta KE$ ); the total mechanical energy ( $KE + PE$ ) is conserved in systems with only conservative forces (gravity, springs), while friction dissipates energy as heat; power is the rate of energy transfer, $P = W/t = Fv$ for an object moving at velocity $v$ under a parallel force.

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What this dot point is asking
Work
Kinetic energy
Potential energy
Conservation of mechanical energy
Power
Energy types and conversions
Examples in context
Try this
Exam-style questions

What this dot point is asking

VCAA wants you to apply the concepts of work, energy, and power to mechanical systems, use conservation of energy in problems involving multiple energy types, and analyse situations with and without friction.

Work

Work done by a constant force $F$ acting over a displacement $d$ :

W = F d \cos\theta

where $\theta$ is the angle between force and displacement.

$\theta = 0$ (force along motion): $W = F d$ (positive).
$\theta = 90$ degrees (perpendicular): $W = 0$ . The force does no work.
$\theta = 180$ degrees (force opposite motion): $W = -F d$ (negative; the force takes energy away).

Units: joule (J) = N m.

Kinetic energy

The energy of motion:

KE = \frac{1}{2} m v^2

The work-kinetic energy theorem: the net work on an object equals its change in kinetic energy.

W_{\text{net}} = \Delta KE

Potential energy

Gravitational PE. $PE_g = mgh$ where $h$ is height above a reference point. The reference is arbitrary; only changes in $PE$ matter.

Elastic PE. A spring with spring constant $k$ stretched or compressed by $x$ from equilibrium: $PE_e = \frac{1}{2} k x^2$ .

Conservation of mechanical energy

In an isolated system with only conservative forces (gravity, springs), mechanical energy ( $KE + PE$ ) is conserved:

KE_i + PE_i = KE_f + PE_f

In practice, friction, air resistance and similar forces dissipate energy as heat:

KE_i + PE_i = KE_f + PE_f + E_{\text{lost}}

where $E_{\text{lost}}$ is the work done by friction etc.

Power

Rate of doing work or transferring energy:

P = \frac{W}{t} = \frac{\Delta E}{t}

Units: watt (W) = J/s.

For an object moving at velocity $v$ with a force $F$ applied:

P = F v

(For force parallel to motion. More generally $P = \vec{F} \cdot \vec{v}$ .)

Worked example

Falling ball. A 0.5 kg ball is dropped from 10 m. Final speed (ignoring air resistance)?

Conservation: $mgh = \frac{1}{2} m v^2$ . $v = \sqrt{2gh} = \sqrt{196} = 14$ m/s.

Spring launch. A 0.2 kg ball is launched by a spring with $k = 100$ N/m compressed by 0.1 m. Maximum speed?

Spring PE: $\frac{1}{2} k x^2 = \frac{1}{2} \times 100 \times 0.01 = 0.5$ J.

Equal to KE at maximum speed: $\frac{1}{2} m v^2 = 0.5$ , so $v^2 = 5$ , $v \approx 2.24$ m/s.

Power of a car. A 1500 kg car accelerates from 0 to 27 m/s in 5 s. Average power?

$\Delta KE = \frac{1}{2}(1500)(27^2 - 0) = 546,750$ J.

Power = $\Delta KE / t = 546,750 / 5 \approx 109$ kW.

Energy types and conversions

Mechanical (KE + PE) is one form. Others include:

Thermal (heat).
Chemical (in fuels).
Nuclear.
Electromagnetic (radiation).

Energy can convert between forms. In a falling object with air resistance: gravitational PE -> KE + heat (friction with air).

In a car: chemical PE in fuel -> KE of car + heat (mostly) + sound + light.

Conservation of energy is one of the fundamental laws of physics. In any isolated system, the total energy is constant; only the form changes.

Common errors

Forgetting cosine in work formula: $W = F d \cos\theta$ . If force is perpendicular to motion, no work is done.
Confusing PE and KE: PE is positional (height for gravity, displacement for spring). KE is kinetic ( $\frac{1}{2} m v^2$ ).
Mixing $W$ as work and $W$ as weight: Context matters. Work is energy; weight is a force.
Power = work/time, not force: Power is the rate at which work is done.
Friction in conservation problems: When friction is present, mechanical energy is not conserved; account for the energy loss separately.

Examples in context

Example 1. Snowy 2.0 pumping cycle energy accounting. Snowy 2.0 raises water $700$ m vertically from Talbingo to Tantangara reservoir using off-peak grid electricity. Lifting $1$ kg by $700$ m requires $W = mgh = 1 \times 9.8 \times 700 = 6860$ J of work. With turbine and pump efficiency of about $80\%$ each, round-trip efficiency is $0.8 \times 0.8 = 64\%$ , so $10$ MWh in delivers $6.4$ MWh out. At a rated pumping power of $2200$ MW, lifting capacity is roughly $2200 \times 10^6 / 6860 = 3.2 \times 10^8$ kg per second, or $320$ tonnes of water per second uphill, demonstrating the massive energy flows in pumped hydro.

Example 2. MCG floodlight tower elevator power. The MCG floodlight tower elevator lifts a $1200$ kg cabin (including passengers) at constant $1.5$ m s $^{-1}$ up the $75$ m tower. Power required to lift against gravity is $P = Fv = mgv = 1200 \times 9.8 \times 1.5 = 1.76 \times 10^4$ W. Energy per trip is $W = mgh = 1200 \times 9.8 \times 75 = 8.82 \times 10^5$ J, or about $0.25$ kWh. Adding friction and electrical losses gives a real power draw of roughly $25$ kW. Power scales linearly with velocity at constant load, which is why high-rise elevators use variable-speed drives matched to passenger demand.

Try this

Q1. Define work and state the unit of mechanical power. [2 marks]

Cue. Work = force times displacement in the direction of force; power in watts (J s $^{-1}$ ).

Q2. A $1000$ kg car accelerates from rest to $20$ m s $^{-1}$ over $80$ m. Calculate (a) the kinetic energy gained, and (b) the average power delivered if it takes $8.0$ s. [4 marks]

Cue. (a) $\tfrac{1}{2} \times 1000 \times 400 = 2.0 \times 10^5$ J. (b) $P = W/t = 2.5 \times 10^4$ W.

Q3. Refer to Snowy 2.0. (a) Calculate the energy required to lift $1$ kg of water $700$ m. (b) Determine the round-trip efficiency given turbine and pump efficiencies of $80\%$ each. (c) Explain why pumped hydro is treated as energy storage rather than generation. [2+2+2 marks]

Cue. (a) $6860$ J. (b) $64\%$ . (c) Net energy is consumed; the plant time-shifts grid energy rather than producing new energy.

Exam-style questions

Questions in the style of VCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Year 11 SAC-style4 marks

50

kg skier descends a

10

m vertical drop on a frictionless slope. (a) Find the skier's speed at the bottom. (b) If friction does

2000

J of work, find the new speed at the bottom.

Show worked answer →

(a) Frictionless. Conservation of mechanical energy: $mgh = \frac{1}{2} m v^2$ .

$v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 10} = \sqrt{196} = 14$ m s $^{-1}$ .

(b) With friction. Energy at top = energy at bottom + energy lost to friction.

$mgh = \frac{1}{2} m v^2 + W_{\text{friction}}$

$50 \times 9.8 \times 10 = \frac{1}{2} \times 50 \times v^2 + 2000$

$4900 = 25 v^2 + 2000$

$v^2 = 2900/25 = 116$

$v = \sqrt{116} \approx 10.8$ m s $^{-1}$ .

Markers reward the conservation-of-energy equation in each case and the friction-energy-loss accounting in (b).