Exercises — Efficiency

What we want: the fraction of the $70\text{W}$ that became useful air-motion. Why divide: by definition $\eta =P_{\text{useful}}/P_{\text{in}}$, and both numbers are already powers in the same unit (watts), so no conversion is needed. $\eta =\frac{42}{70}=0.6=60\%$ The missing $70-42=28\text{W}$ leaked out as heat — that is why fan motors feel warm (see Heat and Internal Energy).

Why use $1-\eta$: the parent identity $\eta =1-E_{\text{wasted}}/E_{\text{in}}$ rearranges to $E_{\text{wasted}}/E_{\text{in}}=1-\eta$. $\frac{E_{\text{wasted}}}{E_{\text{in}}}=1-0.25=0.75=75\%$ Three-quarters of everything you feed this machine is thrown away.

Step 1 — Find the useful energy. The useful job is raising the crate, so useful energy = gravitational PE gained (see Gravitational Potential Energy): $E_{\text{useful}}=mgh=120\times 10\times 5=6000\text{J}$ Why this is the useful part: the height gained is exactly what we paid the crane to achieve. Step 2 — Divide by input. $\eta =\frac{6000}{8000}=0.75=75\%$ The lost $2000\text{J}$ became heat in the cables and gears (Friction).

Why rearrange: we know $\eta$ and the output, but want the input. Start from $\eta =E_{\text{useful}}/E_{\text{in}}$ and flip it: $E_{\text{in}}=\frac{E_{\text{useful}}}{\eta }=\frac{9\text{kJ}}{0.6}=15\text{kJ}$ Sanity check: input ($15$) is bigger than output ($9$) — good, a real machine always needs to be fed more than it usefully returns.

Step 1 — Convert to one unit. $4\text{kW}=4000\text{W}$ (kilo means $\times 1000$). Why the power form: the question gives power, not energy. Since input and output share the same time, $t$ cancels and $\eta =P_{\text{useful}}/P_{\text{in}}$ (see Work and Power). Step 2 — Multiply. $P_{\text{useful}}=\eta P_{\text{in}}=0.80\times 4000=3200\text{W}$

(a) $\eta =\frac{30}{50}=0.6=60\%$. (b) By conservation, input splits into useful + wasted: $P_{\text{wasted}}=P_{\text{in}}-P_{\text{useful}}=50-30=20\text{kW}$ (c) Energy = power × time. Over $2$ hours: $E_{\text{wasted}}=20\text{kW}\times 2\text{h}=40\text{kWh}$ Why kWh works directly: a kilowatt-hour is literally "kilowatts multiplied by hours," so no unit gymnastics are needed.

Step 1 — Useful energy first. Only $70\%$ of the input reaches the load: $E_{\text{useful}}=\eta E_{\text{in}}=0.70\times 2100=1470\text{J}$ Step 2 — Convert that PE into a height. The useful energy is $mgh$, so solve for $h$: $h=\frac{E_{\text{useful}}}{mg}=\frac{1470}{30\times 10}=\frac{1470}{300}=4.9\text{m}$ Why we used $E_{\text{useful}}$, not $E_{\text{in}}$: the wasted joules never lifted the mass, so plugging in $2100$ would over-predict the height.

Why raw joules mislead: Y produces more useful energy ($630>150$), but it also eats far more. Efficiency normalises away size (that is the whole point of the ratio), letting us compare fairly. ${\eta }_X=\frac{150}{200}=0.75=75\%,{\eta }_Y=\frac{630}{900}=0.70=70\%$ X is more efficient, even though Y does more total work.

Why multiply: each stage's output becomes the next stage's input, so the surviving fraction of energy is multiplied stage by stage (parent note: chained efficiencies multiply). Trace $100$ units of fuel through the chain in the figure below.

${\eta }_{\text{total}}=0.40\times 0.90\times 0.85=0.306=30.6\%$ Lesson: the total ($30.6\%$) is below the smallest single stage ($40\%$). Weak links drag the whole chain down.

Step 1 — Useful energy. $E_{\text{useful}}=0.65\times 5000=3250\text{J}$. Step 2 — Set it equal to kinetic energy. All useful energy becomes motion, so $\frac{1}{2}m{v}^2=3250$. Step 3 — Solve for $v$. Why take a square root: $v$ is buried inside a square, so undoing the square (the inverse operation) frees it. $v=\sqrt{\frac{2E_{\text{useful}}}{m}}=\sqrt{\frac{2\times 3250}{10}}=\sqrt{650}\approx 25.5\text{m/s}$

(a) Going output → input, divide by $\eta$: $P_{\text{in}}=\frac{P_{\text{useful}}}{\eta }=\frac{4.5}{0.75}=6\text{kW}$ (b) Wasted = input − useful: $P_{\text{wasted}}=6-4.5=1.5\text{kW}$ Check: $1.5/6=0.25=1-\eta$. Consistent with the $1-\eta$ waste-fraction identity.

(a) PE released per second = input power. Per second, $2000\text{kg}$ falls $40\text{m}$: $P_{\text{in}}=\frac{mgh}{t}=\frac{2000\times 10\times 40}{1}=800,000\text{W}=800\text{kW}$ Why "per second" makes it a power: energy divided by the one-second interval is exactly power (Work and Power). (b) Turbine output. $P_{\text{turbine}}=0.90\times 800=720\text{kW}$ (c) After the cables (a second stage at $90\%$). $P_{\text{town}}=0.90\times 720=648\text{kW}$ ${\eta }_{\text{total}}=0.90\times 0.90=0.81=81\%$ The two identical $90\%$ stages combine to $81\%$, not $90\%$ — chaining always costs more than any single stage suggests.

Step 1 — Wasted power. $P_{\text{wasted}}=(1-\eta )P_{\text{in}}=0.40\times 500=200\text{W}$, i.e. $200\text{J}$ of heat every second. Step 2 — Heat → temperature. The energy to raise temperature is $Q=mc\Delta T$ (see Heat and Internal Energy). Rearrange for the rise per second, using $Q=200\text{J}$ in one second: $\Delta T=\frac{Q}{mc}=\frac{200}{0.2\times 500}=\frac{200}{100}=2{}^{\circ }\text{C per second}$ Why efficiency mattered: the temperature rise is driven purely by the wasted fraction — a more efficient drill would heat up more slowly.

Step 1 — Required useful energy. $E_{\text{useful}}=mgh=4\times 10\times 8=320\text{J}$. Step 2 — Compute the implied efficiency. $\eta =\frac{320}{300}\approx 1.07=107\%$ Step 3 — Judge it. $\eta >100\%$ is impossible — it would mean more useful energy came out than the total energy put in, breaking Conservation of Energy. So the claim is bogus: either the input is understated or the lift is overstated. Real machines always give $\eta <100\%$; the true limit for heat-based engines is even stricter (see Heat Engines & Carnot Efficiency).

L1 just spots "useful ÷ in." L2 rearranges it to hunt a missing quantity. L3 makes you compare fractions instead of raw joules. L4 chains machines — energy gets cut again and again, so you multiply, never average. L5 fuses efficiency with motion, heat, and the iron rule that nothing beats $100\%$. Same one formula the whole way up — only the surrounding physics grows.