IB Physics SLTopic A.3 â Work, Energy & PowerPaper 1 & 2Efficiency~6 min read
Efficiency
No real device converts energy perfectly. Efficiency puts a number on exactly how good a system is at turning its energy input into the output you actually wanted.
ð What you need to know
Efficiency is the ratio of useful energy or power output to total energy or power input
It’s calculated using η = EoutâEin = PoutâPin
Multiplying by 100% turns the ratio into a percentage
Efficiency has no units â it’s a ratio of two quantities measured in the same units
What counts as “useful” versus “wasted” energy depends entirely on the purpose of the system
What Does Efficiency Measure?
Efficiency describes how successfully a system transfers energy toward its intended purpose. A highly efficient system sends most of its input energy where you want it to go; a low-efficiency system loses most of it along the way, typically as heat.
Whether a particular energy transfer counts as “useful” or “wasted” depends on what the system is for. In a hair dryer, thermal energy delivered to the air stream is useful â but in a laptop’s cooling fan, thermal energy escaping to the surroundings is exactly what you want, making that same kind of transfer the useful one in that context.
Efficiency equationη = EoutâEin = PoutâPin
Where η (the Greek letter “eta”) is efficiency, E is energy in joules, and P is power in watts. Multiply the result by 100% to express it as a percentage rather than a decimal.
Efficiency is simply the fraction of the input bar taken up by the useful-output section
Quick recap: η = useful output ÷ total input, as a fraction or a percentage. It has no units, since both quantities are measured the same way.
ð§ Recipe: Working With Efficiency
Identify the input and useful output â decide exactly which quantities are relevant to the system described
Choose energy or power â use whichever pair of quantities the question actually provides
Apply the equation â divide useful output by total input
Convert if needed â multiply by 100% if the answer is required as a percentage rather than a ratio
WE 1
A conveyor motor has an efficiency of 42%. It lifts a 12 kg crate through a height of 3.5 m in 4.0 s. Calculate the total input power required by the motor.
Step 1 â Find the useful power outputUseful power = mgh ÷ t = (12 à 9.8 à 3.5) ÷ 4.0 â 103 WStep 2 â Rearrange the efficiency equation for input powerP_in = P_out ÷ η = 103 ÷ 0.42â 245 W (3 s.f.)
WE 2
A small pump uses the kinetic energy of flowing water to lift a portion of that water. The supply pipe holds 550 kg of water, and a valve shuts once the water’s speed reaches 2.8 m sâ»Â¹, converting that kinetic energy to lift some of the water by 9.0 m. The pump’s efficiency is 25%. Calculate the mass of water that can be lifted.
Step 1 â Find the kinetic energy of the water in the pipeE_k = Âœmv² = Âœ à 550 à 2.8² â 2160 JStep 2 â Find the useful energy, using the pump’s efficiencyE_useful = 0.25 à 2160 â 539 JStep 3 â Convert useful energy into a lifted massm = E_useful ÷ (gh) = 539 ÷ (9.8 à 9.0)â 6.1 kg
ð¡ Top tips
Efficiency can be given as a ratio (0 to 1) or a percentage (0% to 100%) â check which format a question wants before you answer
Remember efficiency has no units, since it’s the same quantity divided by itself
Multiplying an intermediate energy by an efficiency only converts that specific store â not automatically the final quantity you’re solving for
Real systems always have η < 1 (or < 100%) â a value of 100% or more signals an error somewhere
â Common mistakes
Forgetting to convert a percentage back into a decimal (e.g. 42% â 0.42) before using it in a calculation
Applying an efficiency to the wrong energy store partway through a multi-step problem
Reporting efficiency with units attached, when it should always be a plain ratio or percentage
Assuming “useful” energy is fixed, rather than checking what the system is actually meant to achieve
Up next: Energy Density â where we compare how much energy different fuels pack into the same volume.
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