Define work, kinetic energy and gravitational potential energy, and apply the principle of conservation of mechanical energy to one-dimensional problems including those with friction

What this dot point is asking

QCAA wants you to apply work-energy principles to one-dimensional problems. The dot point combines four ideas: work done by a force, kinetic energy, gravitational potential energy, and the conservation of mechanical energy (with friction as the energy-loss term when present).

Work

Work done by a constant force is:

where $F$ is the force, $s$ is the displacement, and $\theta$ is the angle between the force and the displacement. SI unit: joule (J = N m).

• $\theta = 0°$: maximum positive work ($W = Fs$).
• $\theta = 90°$: zero work (centripetal force does no work on an object in uniform circular motion).
• $\theta = 180°$: negative work ($W = -Fs$, e.g. friction opposing motion).

For a variable force, work is the area under the force-displacement graph.

Kinetic energy

Kinetic energy is the energy of motion:

It is a scalar with SI unit joule.

Work-energy theorem

The net work done on an object equals the change in its kinetic energy:

This connects forces (which do work) to motion (which has kinetic energy).

Gravitational potential energy

Near the Earth's surface, gravitational potential energy is:

where $h$ is the height above an agreed reference level. Only differences in $PE_g$ matter; the reference level is your choice.

Conservation of mechanical energy

In the absence of friction and air resistance:

Mechanical energy converts between kinetic and potential but the total stays the same. A roller coaster at the top of a hill has maximum $PE$ and minimum $KE$; at the bottom, the reverse.

With friction (or any non-conservative force), the energy equation becomes:

where $E_{\text{lost}}$ is the energy dissipated as heat, sound or deformation. For a constant friction force $f$ acting over a distance $d$, $E_{\text{lost}} = f d$.

Examples in context

Example 1. Sunshine Coast tidal study models a $200 \text{ kg}$ instrument package winched up $25 \text{ m}$ from the seabed. Work against gravity is $W = mgh = 49000 \text{ J}$; the same package falling back releases this as kinetic energy (in vacuum). In water, drag friction dissipates roughly $30$ per cent of the descent's mechanical energy. The work-energy theorem $W_{net} = \Delta KE$ is the QCAA Unit 2 calculation, and energy conservation including friction is the next layer for the EA.

Example 2. A Cairns light-rail tram's regenerative braking from $14 \text{ m s}^{-1}$ at $45000 \text{ kg}$ would convert $\tfrac{1}{2}mv^2 = 4.41 \text{ MJ}$ to electrical energy if $100$ per cent efficient. In practice $60$ per cent ($2.65 \text{ MJ}$) is recovered to the supply and $40$ per cent dissipates as friction-brake heat. QCAA EA Unit 2 thematic items couple the work-energy theorem to such a Queensland-context engineering choice.

Try this

Q1. Define work and state the conditions for positive, negative and zero work. [2 marks]

• Cue. $W = Fs\cos\theta$; positive if force and displacement aligned, negative if opposed, zero if perpendicular.

Q2. A $1000 \text{ kg}$ car climbs from rest to $20 \text{ m s}^{-1}$ while rising $5.0 \text{ m}$. Calculate the change in mechanical energy. [3 marks]

• Cue. $\Delta KE = \tfrac{1}{2} \times 1000 \times 400 = 2.0 \times 10^5 \text{ J}$; $\Delta PE = mgh = 4.9 \times 10^4 \text{ J}$; total $= 2.49 \times 10^5 \text{ J}$.

Q3. A Cairns light-rail tram ($45000 \text{ kg}$) at $14 \text{ m s}^{-1}$ brakes regeneratively to rest. (a) Calculate the initial kinetic energy. (b) If recovery efficiency is $60$ per cent, determine the energy returned to the supply and dissipated. (c) Explain how the work-energy theorem governs the friction-brake fraction. [3+3+2 marks; ISMG: Analysis and interpretation, Evaluation]

• Cue. (a) $4.41 \text{ MJ}$; (b) $2.65 \text{ MJ}$ recovered, $1.76 \text{ MJ}$ dissipated; (c) work done by friction force equals KE removed by friction.