Efficiency: Worked examples — Physics deep dive

This is the problem-drill child of Efficiency. The parent built the idea; here we hunt down every kind of question efficiency can throw at you and solve one of each — including the weird ones (zero output, 100% ideal, chained losses, "which is useful?" traps, and an exam twist). Nothing new is assumed: if a symbol appears, it was earned in the parent or is re-earned here.

Recall The two formulas we lean on (from the parent)

Efficiency is the useful fraction of what went in: $η = \frac{E _{useful out}}{E _{total in}} = \frac{P _{useful out}}{P _{in}} = 1 - \frac{E _{wasted}}{E _{in}}$ Here $E$ = energy in joules ( $J$ ), $P$ = power in watts ( $W$ ), $η$ (Greek letter "eta") = a pure number from $0$ to $1$ . Multiply by $100$ for a percentage (\%).

The scenario matrix

Before solving, let's list every case class a question can be. Each worked example below is tagged with the cell it fills.

Cell	Case class	What makes it tricky	Example
C1	Standard energy input/output	Pick the useful energy	Ex 1
C2	Power form ( $P$ , not $E$ )	Same-unit division	Ex 2
C3	Reverse solve (find input/output/waste)	Rearrange the formula	Ex 3
C4	Zero useful output (degenerate)	$η = 0$ , machine does no wanted work	Ex 4
C5	Ideal / limiting ( $η \to 1$ )	Frictionless bound; why real $η < 1$	Ex 5
C6	Chained machines (series)	Efficiencies multiply	Ex 6
C7	"Which output is useful?" trap	Total ≠ useful	Ex 7
C8	Real-world word problem + unit clash	kJ vs J, per-second rates	Ex 8
C9	Exam twist (efficiency hidden inside a bigger quantity)	Solve for a non- $η$ unknown	Ex 9

We'll walk them top to bottom. Ex 1 also carries the master figure.

C1 — Standard energy input/output

Worked example Ex 1 — Crane lifts a crate

A crane consumes $8000 J$ of electrical energy to lift a $40 kg$ crate through a height of $15 m$ . Take $g = 10 m/s^{2}$ . Find $η$ .

Forecast: Guess — will $η$ beat $80%$ , or come in lower? (pause)

Step 1 — Identify the useful output. The wanted job is raising the crate, so useful energy = gain in Gravitational Potential Energy: $E_{useful} = m g h = 40 \times 10 \times 15 = 6000 J .$ Why this step? Efficiency's numerator is only the energy that did the intended task — here, lifting.

Step 2 — Apply the definition. $η = \frac{E _{useful}}{E _{in}} = \frac{6000}{8000} = 0.75 = 75%.$ Why this step? Input is the full electrical energy supplied; the definition does the rest.

Verify: Waste = $8000 - 6000 = 2000 J$ (heat in motor/cables). Check $1 - 2000/8000 = 0.75$ ✓. Answer under $80%$ , so the forecast "lower" was right.

Look at the figure: the burnt-orange bar is the full $8000 J$ input; the teal segment (75%) is what climbed into PE; the plum tail (25%) leaked as heat — the "leaky bucket" from the parent, drawn to scale.

C2 — Power form

C3 — Reverse solve

C4 — Zero useful output (degenerate)

Worked example Ex 4 — Pushing against a locked wall

You supply $250 J$ of muscular energy shoving a wall that does not move. What is the efficiency of this "lifting" attempt?

Forecast: Feels like effort should count for something — but does it?

Step 1 — Useful output. Useful work = force $\times$ distance the load moves. The wall's displacement is $0$ , so: $E_{useful} = 0 J .$ Why this step? Work needs motion (see Work and Power). No displacement ⇒ no useful energy delivered, however tired you feel.

Step 2 — Efficiency. $η = \frac{0}{250} = 0 = 0%.$ Why this step? Zero in the numerator forces $η = 0$ — the lower limit of the $0$ -to- $1$ range.

Verify: All $250 J$ became heat in your muscles (Heat and Internal Energy): waste $= 250 - 0 = 250 J$ , and $1 - 250/250 = 0$ ✓. This is the degenerate floor of the matrix — efficiency's minimum.

C5 — Ideal / limiting case

Worked example Ex 5 — The frictionless dream

A frictionless, resistance-free ramp system lifts a $2 kg$ mass by $3 m$ using exactly $m g h$ of input. What is $η$ ? Then: a real version loses $12 J$ to Friction. New $η$ ?

Forecast: Ideal case — will $η$ hit exactly $100%$ , or just approach it?

Step 1 — Ideal output = input. With no losses, every input joule becomes PE: $E_{useful} = m g h = 2 \times 10 \times 3 = 60 J, E_{in} = 60 J .$ $η_{ideal} = \frac{60}{60} = 1 = 100%.$ Why this step? $η = 1$ is the theoretical ceiling — reachable only when $E_{wasted} = 0$ . This is a limit, never a real machine.

Step 2 — Add real friction. Now input must cover PE plus the $12 J$ friction loss: $E_{in} = 60 + 12 = 72 J, η = \frac{60}{72} = 0.8 \overline{3} \approx 83.3%.$ Why this step? Friction shows why real $η < 1$ : extra input is needed to still deliver the same useful $60 J$ .

Verify: $60/72 = 5/6 \approx 0.8333$ ✓, and $1 - 12/72 = 1 - 1/6 = 5/6$ ✓. The ceiling ( $100%$ ) and a realistic value both confirmed.

C6 — Chained machines (series)

Worked example Ex 6 — Generator then motor

A diesel generator is $35%$ efficient. Its electrical output feeds a motor that is $80%$ efficient. Find the overall efficiency, given $10000 J$ of diesel energy in.

Forecast: Will overall $η$ be above or below the smaller stage ( $35%$ )?

Step 1 — Multiply the stage efficiencies. The motor's input is the generator's output: $η_{total} = η_{1} \times η_{2} = 0.35 \times 0.80 = 0.28 = 28%.$ Why this step? Fractions of fractions multiply — $80%$ of the $35%$ that survived stage one.

Step 2 — Trace the joules to confirm. $10000 \times 0.35 3500 J \times 0.80 2800 J useful .$ $η_{total} = \frac{2800}{10 000} = 0.28 ✓$ Why this step? Following the actual energy proves the multiply rule and that the result ( $28%$ ) is below even the smallest stage ( $35%$ ).

Verify: $0.35 \times 0.80 = 0.28$ and $2800/10000 = 0.28$ ✓. Forecast "below $35%$ " correct.

C7 — "Which output is useful?" trap

C8 — Real-world word problem + unit clash

C9 — Exam twist (efficiency hidden inside)

Worked example Ex 9 — Solve for the height, not

$η$ A $70%$ efficient hoist uses $4200 J$ of electrical energy to raise a $10 kg$ load. To what height does it lift the load? ( $g = 10 m/s^{2}$ .)

Forecast: They give you $η$ but ask for $h$ — will $h$ be bigger or smaller than the "ideal" no-loss height?

Step 1 — Find the useful energy first. Efficiency tells us how much input became PE: $E_{useful} = η \times E_{in} = 0.70 \times 4200 = 2940 J .$ Why this step? The hoist only delivers $70%$ of its input as lifting energy; the rest is waste.

Step 2 — Convert useful energy to height. Useful energy is PE $= m g h$ , so: $h = \frac{E _{useful}}{m g} = \frac{2940}{10 \times 10} = \frac{2940}{100} = 29.4 m .$ Why this step? Rearranging $m g h = E_{useful}$ isolates $h$ . Efficiency is now inside the calculation, not the answer.

Verify: Check backward: PE gained $= 10 \times 10 \times 29.4 = 2940 J$ ; $η = 2940/4200 = 0.70$ ✓. Ideal (100%) height would be $4200/100 = 42 m$ , so $29.4 m$ being smaller is correct — losses steal height.

Cross-check: every matrix cell hit

Recall Coverage tick-list

Which cell forces $η = 0$ ? ::: C4 — zero useful output (locked wall). Which cell is the ceiling $η = 1$ ? ::: C5 — the ideal frictionless case. Which cell requires multiplying efficiencies? ::: C6 — machines in series. Which cell is the "total ≠ useful" trap? ::: C7 — the light bulb. Which cell needs unit conversion before dividing? ::: C8 — kettle (kW·s → kJ). Which cell asks for a non- $η$ unknown? ::: C9 — solving for height $h$ .

Connections

Efficiency — parent topic; this page drills its every case.
Conservation of Energy — every "waste = in − useful" step is conservation.
Work and Power — the $E = P t$ conversion in Ex 2 and Ex 8.
Gravitational Potential Energy — useful output as $m g h$ in Ex 1, 5, 9.
Heat and Internal Energy — where the wasted joules go.
Friction — the loss added in Ex 5.
Heat Engines & Carnot Efficiency — the thermodynamic ceiling behind Ex 5's $η < 1$ .