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How do work, kinetic energy and power describe the transfer and rate of transfer of energy in mechanical systems?

Calculate work done by a force, relate net work to change in kinetic energy via the work-energy theorem, and define power as the rate of doing work.

Defining mechanical work including the angle factor, the work-energy theorem linking net work to change in kinetic energy, and power as the rate of energy transfer, with worked examples.

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  1. What this dot point is asking
  2. Work done by a force
  3. The work-energy theorem
  4. Power
  5. Gravitational potential energy
  6. How SACE assesses this

What this dot point is asking

You need to calculate work done by a force (including when the force is at an angle to the motion), use the work-energy theorem to connect net work and change in kinetic energy, and define and calculate power.

Work done by a force

Work is done only when a force has a component along the displacement.

The cosθ\cos\theta factor is important:

  • Force along the motion (θ=0\theta = 0): W=FsW = Fs (maximum positive work).
  • Force perpendicular to the motion (θ=90\theta = 90^\circ): W=0W = 0. A centripetal force, or the normal force on a sliding box, does no work.
  • Force opposing the motion (θ=180\theta = 180^\circ): W=FsW = -Fs (negative work, removes energy, e.g. friction).

The work-energy theorem

The total (net) work done on an object equals its change in kinetic energy.

Kinetic energy itself is Ek=12mv2E_k = \tfrac{1}{2}mv^2. The theorem follows directly from F=maF=ma combined with v2=u2+2asv^2 = u^2 + 2as. The key word is net: you must add the work done by every force, including the negative work done by friction or drag.

Power

Power is how quickly work is done, the rate of energy transfer.

A more powerful engine does the same work in less time, or more work in the same time. At a steady speed against a constant resistive force, the driving power equals FvFv because the driving force just balances the resistance.

Gravitational potential energy

Lifting a mass through a height hh stores gravitational potential energy Ep=mghE_p = mgh. The work you do against gravity equals the energy stored. In energy problems, total mechanical energy (Ek+EpE_k + E_p) is conserved when only gravity does work; friction or drag does negative work and converts mechanical energy to heat. This conservation lets you find the speed at the bottom of a ramp or the height a thrown ball reaches without using forces at all.

How SACE assesses this

SACE Stage 2 work-energy questions frequently combine several ideas in one item: compute the work done by a force (watching for the cosθ\cos\theta factor), apply the work-energy theorem to find a final speed, and then find the average power. The most common error is forgetting that the work-energy theorem uses the net work, so any friction or drag (which does negative work) must be subtracted before equating to ΔEk\Delta E_k. Set out each force's work separately, sum them for the net work, then solve for the unknown. For power, decide whether the question wants average power (P=W/tP = W/t) or instantaneous power (P=FvP = Fv), and quote the correct form before substituting.

Exam-style practice questions

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

SACE 20244 marksA 1500 kg1500\ \text{kg} car accelerates from rest along a flat road under a constant driving force of 4500 N4500\ \text{N} over a distance of 80.0 m80.0\ \text{m} (ignore resistance). (a) Calculate the work done by the driving force. (b) Use the work-energy theorem to find the final speed. (c) If the car takes 5.4 s5.4\ \text{s}, calculate the average power.
Show worked answer →

(a) The force is along the motion, so W=Fscos0=(4500)(80.0)=3.6×105 JW = Fs\cos 0^\circ = (4500)(80.0) = 3.6\times10^5\ \text{J}. (1 mark)

(b) The car starts at rest, so Wnet=12mv2W_\text{net} = \tfrac{1}{2}mv^2:

3.6×105=12(1500)v2v2=480v=21.9 m s1.3.6\times10^5 = \tfrac{1}{2}(1500)v^2 \Rightarrow v^2 = 480 \Rightarrow v = 21.9\ \text{m s}^{-1}.
(2 marks)

(c) P=Wt=3.6×1055.4=6.7×104 WP = \dfrac{W}{t} = \dfrac{3.6\times10^5}{5.4} = 6.7\times10^4\ \text{W} (about 67 kW67\ \text{kW}). (1 mark) Markers reward the cosθ\cos\theta in the work, the work-energy theorem for vv, and the rate definition for PP.

SACE 20223 marksA worker pulls a 40.0 kg40.0\ \text{kg} crate 6.0 m6.0\ \text{m} across a level floor using a rope at 30.030.0^\circ above the horizontal with a force of 120 N120\ \text{N}. Friction does 450 J-450\ \text{J} of work on the crate. (a) Calculate the work done by the rope. (b) Use the work-energy theorem to find the crate's final speed if it started from rest.
Show worked answer →

(a) Only the component of the rope force along the motion does work:

Wrope=Fscosθ=(120)(6.0)cos30.0=624 J.W_\text{rope} = Fs\cos\theta = (120)(6.0)\cos 30.0^\circ = 624\ \text{J}.
(1 mark)

(b) Net work is the sum of all works: Wnet=624+(450)=174 JW_\text{net} = 624 + (-450) = 174\ \text{J}.

Wnet=12mv2174=12(40.0)v2v2=8.70v=2.95 m s1.W_\text{net} = \tfrac{1}{2}mv^2 \Rightarrow 174 = \tfrac{1}{2}(40.0)v^2 \Rightarrow v^2 = 8.70 \Rightarrow v = 2.95\ \text{m s}^{-1}.
(2 marks) Markers reward including the negative friction work in the net work, not just the rope work.

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