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How are work, energy and power defined and applied to mechanical systems?

Work W=FdcosθW = Fd \cos\theta, kinetic energy 12mv2\frac{1}{2}mv^2, gravitational potential energy mghmgh, elastic potential energy 12kx2\frac{1}{2}kx^2, conservation of mechanical energy, and power P=W/t=FvP = W/t = Fv

A focused answer to the VCE Physics Unit 2 key knowledge point on work, energy and power. Work done by a force, kinetic and gravitational potential energy, conservation of mechanical energy in conservative systems, friction and energy loss, and power P=W/t=FvP = W/t = Fv.

Generated by Claude Opus 4.810 min answer

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

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  1. What this dot point is asking
  2. Work
  3. Kinetic energy
  4. Potential energy
  5. Conservation of mechanical energy
  6. Power
  7. Energy types and conversions
  8. Examples in context
  9. Try this

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 FF acting over a displacement dd:

W=FdcosθW = F d \cos\theta

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

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

Units: joule (J) = N m.

Kinetic energy

The energy of motion:

KE=12mv2KE = \frac{1}{2} m v^2

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

Wnet=ΔKEW_{\text{net}} = \Delta KE

Potential energy

Gravitational PE. PEg=mghPE_g = mgh where hh is height above a reference point. The reference is arbitrary; only changes in PEPE matter.

Elastic PE. A spring with spring constant kk stretched or compressed by xx from equilibrium: PEe=12kx2PE_e = \frac{1}{2} k x^2.

Conservation of mechanical energy

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

KEi+PEi=KEf+PEfKE_i + PE_i = KE_f + PE_f

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

KEi+PEi=KEf+PEf+ElostKE_i + PE_i = KE_f + PE_f + E_{\text{lost}}

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

Power

Rate of doing work or transferring energy:

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

Units: watt (W) = J/s.

For an object moving at velocity vv with a force FF applied:

P=FvP = F v

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

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.

Examples in context

Example 1. Snowy 2.0 pumping cycle energy accounting. Snowy 2.0 raises water 700700 m vertically from Talbingo to Tantangara reservoir using off-peak grid electricity. Lifting 11 kg by 700700 m requires W=mgh=1×9.8×700=6860W = mgh = 1 \times 9.8 \times 700 = 6860 J of work. With turbine and pump efficiency of about 80%80\% each, round-trip efficiency is 0.8×0.8=64%0.8 \times 0.8 = 64\%, so 1010 MWh in delivers 6.46.4 MWh out. At a rated pumping power of 22002200 MW, lifting capacity is roughly 2200×106/6860=3.2×1082200 \times 10^6 / 6860 = 3.2 \times 10^8 kg per second, or 320320 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 12001200 kg cabin (including passengers) at constant 1.51.5 m s1^{-1} up the 7575 m tower. Power required to lift against gravity is P=Fv=mgv=1200×9.8×1.5=1.76×104P = Fv = mgv = 1200 \times 9.8 \times 1.5 = 1.76 \times 10^4 W. Energy per trip is W=mgh=1200×9.8×75=8.82×105W = mgh = 1200 \times 9.8 \times 75 = 8.82 \times 10^5 J, or about 0.250.25 kWh. Adding friction and electrical losses gives a real power draw of roughly 2525 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 s1^{-1}).

Q2. A 10001000 kg car accelerates from rest to 2020 m s1^{-1} over 8080 m. Calculate (a) the kinetic energy gained, and (b) the average power delivered if it takes 8.08.0 s. [4 marks]

  • Cue. (a) 12×1000×400=2.0×105\tfrac{1}{2} \times 1000 \times 400 = 2.0 \times 10^5 J. (b) P=W/t=2.5×104P = W/t = 2.5 \times 10^4 W.

Q3. Refer to Snowy 2.0. (a) Calculate the energy required to lift 11 kg of water 700700 m. (b) Determine the round-trip efficiency given turbine and pump efficiencies of 80%80\% each. (c) Explain why pumped hydro is treated as energy storage rather than generation. [2+2+2 marks]

  • Cue. (a) 68606860 J. (b) 64%64\%. (c) Net energy is consumed; the plant time-shifts grid energy rather than producing new energy.

Exam-style practice questions

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

Year 11 SAC4 marksA 5050 kg skier descends a 1010 m vertical drop on a frictionless slope. (a) Find the skier's speed at the bottom. (b) If friction does 20002000 J of work, find the new speed at the bottom.
Show worked answer →

(a) Frictionless. Conservation of mechanical energy: mgh=12mv2mgh = \frac{1}{2} m v^2.

v=2gh=2×9.8×10=196=14v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 10} = \sqrt{196} = 14 m s1^{-1}.

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

mgh=12mv2+Wfrictionmgh = \frac{1}{2} m v^2 + W_{\text{friction}}

50×9.8×10=12×50×v2+200050 \times 9.8 \times 10 = \frac{1}{2} \times 50 \times v^2 + 2000

4900=25v2+20004900 = 25 v^2 + 2000

v2=2900/25=116v^2 = 2900/25 = 116

v=11610.8v = \sqrt{116} \approx 10.8 m s1^{-1}.

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

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