§-Syllabus dot point

TASPhysicsSyllabus dot point

How is energy transferred when a force moves an object?

Apply work, kinetic and potential energy, conservation of energy and power to mechanical systems.

Work done by a force, kinetic and gravitational potential energy, the work-energy theorem, conservation of mechanical energy, and power as the rate of doing work.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

This dot point treats motion through the lens of energy, which is often quicker than forces because energy is a scalar and you can ignore direction.

Work done by a force

Work is done when a force moves its point of application through a distance. Only the component of force along the displacement does work:

W = Fs\cos\theta

where $\theta$ is the angle between the force and the displacement. A force perpendicular to motion does no work, which is why the centripetal force and the magnetic force on a charge do no work. Work is measured in joules and can be negative when the force opposes motion, as friction does.

Kinetic and potential energy

A moving mass carries kinetic energy:

E_k = \tfrac{1}{2}mv^2

An object raised against gravity stores gravitational potential energy relative to a chosen reference height:

E_p = mgh

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

W_\text{net} = \Delta E_k

The work-energy theorem also gives a clean way to handle friction and braking. The work done against a constant friction force $f$ over a distance $d$ is $fd$ , and setting this equal to the initial kinetic energy gives the stopping distance $d = \dfrac{mv^2}{2f}$ . Because the kinetic energy depends on the square of the speed, doubling the speed quadruples the stopping distance, a result of direct relevance to road safety.

Conservation of energy

Energy cannot be created or destroyed, only transferred or transformed. In a system with no friction, mechanical energy is conserved, so the sum of kinetic and potential energy stays constant:

E_k + E_p = \text{constant}

A dropped ball converts potential energy into kinetic energy; a pendulum swaps the two back and forth. When friction is present, some mechanical energy is transformed into thermal energy, but the total energy of the system plus surroundings is still conserved.

Elastic potential energy and energy chains

Energy is often stored elastically before becoming kinetic. A stretched spring or drawn bow stores elastic potential energy equal to the area under its force-extension graph; for a spring obeying Hooke's law this is $E = \tfrac12 k x^2$ . Many TCE problems are energy chains: chemical or elastic energy becomes kinetic energy, then potential energy, with a stated efficiency telling you what fraction is usefully transferred. Track the energy from store to store, multiplying by the efficiency at each transfer.

Power

Power is the rate at which work is done or energy is transferred:

P = \frac{W}{t}

For a force moving an object at speed $v$ , this becomes $P = Fv$ . Power is measured in watts, where one watt is one joule per second.

In the exam, decide whether an energy method or a force method is faster. If the question gives heights, speeds or distances and asks for another speed, use conservation of energy. If it asks for time or acceleration, the equations of motion may be needed. Always set a clear reference height for potential energy.

Exam-style practice questions

Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

TCE 20224 marksA

50\,000\ \text{kg}

truck climbs a hill of inclination

20^\circ

at a constant

2\ \text{m s}^{-1}

. Calculate the potential energy gained per second, and the work done per second against a resistance force equal to one tenth of the weight.

Show worked answer →

Energy gained per second is a power. The truck travels $2\ \text{m}$ along the slope each second, rising $(2\sin 20^\circ)$ vertically each second.

P_\text{PE} = mg \times (\text{vertical rise per second}) = (50\,000)(9.81)(2)(\sin 20^\circ) = 3.36 \times 10^5\ \text{W}.

Resistance force $= 0.1mg = (0.1)(50\,000)(9.81) = 4.905 \times 10^4\ \text{N}$ .
Work per second against resistance $= (4.905\times10^4)(2) = 9.81 \times 10^4\ \text{W}$ .

So the truck gains about $3.4 \times 10^5\ \text{J}$ of potential energy and does about $9.8 \times 10^4\ \text{J}$ of work against resistance each second. Markers want both quantities as energy per second using the $2\ \text{m}$ travelled each second.

TCE 20234 marksA bow stores about

64\ \text{J}

of elastic energy at full draw, read from the area under its force-extension graph. If

60\%

of this energy is transferred to a

20\ \text{g}

arrow, show that the arrow leaves at about

62\ \text{m s}^{-1}

Show worked answer →

The stored elastic energy is the area under the force-extension graph, here $64\ \text{J}$ .

Energy transferred to the arrow as kinetic energy $= 0.60 \times 64 = 38.4\ \text{J}$ .

Set this equal to the arrow's kinetic energy and solve for speed:

E_k = \tfrac12 m v^2, \quad 38.4 = \tfrac12(0.020)v^2, \quad v^2 = \frac{38.4}{0.010} = 3840, \quad v = 62\ \text{m s}^{-1}.

The arrow leaves at about $62\ \text{m s}^{-1}$ , matching the printed result. Markers reward reading the graph area as stored energy and equating the transferred fraction to $\tfrac12 m v^2$ .

TCE 20232 marksA

5\ \text{g}

air-gun pellet is fired at

100\ \text{m s}^{-1}

into a

100\ \text{g}

plasticine lump suspended as a pendulum. After they move off together, calculate the loss in kinetic energy during the collision.

Show worked answer →

First find the common speed after the (perfectly inelastic) collision using conservation of momentum:

mv = (m + M)V, \quad (0.005)(100) = (0.105)V, \quad V = \frac{0.5}{0.105} = 4.76\ \text{m s}^{-1}.

$E_k$ before $= \tfrac12(0.005)(100)^2 = 25.0\ \text{J}$ .
$E_k$ after $= \tfrac12(0.105)(4.76)^2 = 1.19\ \text{J}$ .

Loss in kinetic energy $= 25.0 - 1.19 = 23.8\ \text{J}$ .

Most of the pellet's kinetic energy is lost (converted to heat and deformation) because the collision is inelastic. Markers want the post-collision speed from momentum, then the kinetic energy difference.