Work, energy and power - TCE Physics (Tasmania)

Syllabus dot point

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

TCEPhysicsUnit 3: Gravity and Electromagnetism9 min readUpdated 2 Jun 2026

Quick answer

Work is energy transferred by a force acting over a distance, $W = Fs\cos\theta$ . It changes kinetic energy ( $E_k = \tfrac12 mv^2$ ) or potential energy ( $E_p = mgh$ ). In a closed system total energy is conserved, and power is the rate of doing work, $P = \dfrac{W}{t} = Fv$ .

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

Work-energy theorem

The net work done by all forces on an object equals the change in its kinetic energy. If the net work is positive the object speeds up, and if it is negative the object slows down.

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.

Worked example: speed from energy conservation

Equate potential and kinetic energy

A $2.0\ \text{kg}$ ball is released from rest at a height of $1.8\ \text{m}$ . Ignoring air resistance, all the potential energy becomes kinetic energy:

mgh = \tfrac{1}{2}mv^2.

Solve for the speed

The mass cancels, so

v = \sqrt{2gh} = \sqrt{2(9.8)(1.8)} = \sqrt{35.3} = 5.9\ \text{m s}^{-1}.

The ball lands at about

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

The work-energy method skips the time

Using energy conservation, you can find a final speed from a height or distance without ever knowing the time taken or the acceleration. This makes energy methods faster than the equations of motion when time is not asked for.

Including the angle when force and motion are parallel

Students sometimes drop the $\cos\theta$ or apply it incorrectly. When the force acts along the direction of motion, $\theta = 0$ and $\cos\theta = 1$ , so $W = Fs$ . The $\cos\theta$ only reduces the work when the force acts at an angle to the displacement, and gives zero when they are perpendicular.

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 questions

Questions in the style of TASC 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.

Practice4 marks

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.

Practice4 marks

A 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 $\tfrac12 m v^2$ .

Practice2 marks

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.

Practise this

Exam-style questions 3 on this page
Flashcards Physics deck

Sources & how we know this

Physics Level 4 (PHY415115) course document (2015 onwards) — TASC