Work power and efficiency

Work is the transfer of energy performed by a force as a body moves: $W = F s \cos\theta$, where $\theta$ is the angle between the force and the displacement; the unit of work is the joule (J). A free-body diagram shows that only the component of force along the displacement does work, while forces perpendicular to the displacement do no work. Power tells how fast work is done: $P = W/t$, measured in watts (W). Efficiency compares useful work with the work supplied: $\eta = \frac{W_{\text{useful}}}{W_{\text{in}}},$ and for real devices always $\eta < 1$ because of friction, heating and other losses. Conservation of energy gives the balance of the supplied energy: $W_{\text{in}} = W_{\text{useful}} + W_{\text{loss}}.$

A box is pushed along a horizontal floor by a constant horizontal force $F = 30\ \text{N}$ over a distance $s = 4.0\ \text{m}$; the displacement takes $t = 5.0\ \text{s}$. Find the work done by this force and the average power.

A sled is pulled by a rope with force $T = 50\ \text{N}$ at $37^\circ$ above the horizontal, with $\cos 37^\circ = 0.80$. The sled moves uniformly along the horizontal for $s = 12\ \text{m}$ in $t = 8.0\ \text{s}$; the friction force is $f = 40\ \text{N}$. Calculate the work of the tension, the work of friction, the work of gravity and of the normal force, the net work, and the average power of the tension.

A hoist motor lifts a load of mass $m = 20\ \text{kg}$ uniformly to a height $h = 6.0\ \text{m}$ in $t = 10\ \text{s}$. The average input power of the motor is $P_{\text{in}} = 150\ \text{W}$. Taking $g = 9.8\ \text{m/s}^2$, find the useful work, the useful power, the efficiency of the hoist and the energy lost.

Safe self-study measurement (small masses only, no bending or sudden movements).

1. Take a small book and measure its mass with a kitchen scale; compute the weight force $F = m g$ with $g = 9.8\ \text{m/s}^2$.
2. Measure the height $h$ (for example $0.50\ \text{m}$) with a ruler. Lift the book slowly to this height in time $t_1$, then to the same height in a shorter time $t_2$ (time it with a phone stopwatch).
3. The work is the same in both cases: $W = m g h$. Compare the powers $P_1 = W/t_1$ and $P_2 = W/t_2$ and explain why the second is larger although $W$ is unchanged.
4. If lifting an object is inconvenient or medically inadvisable, replace steps 1–3 with a simulation: in PhET "Energy Skate Park: Basics" turn on friction and compare the mechanical energy at the start and at the end, taking $\eta = E_{\text{end}}/E_{\text{start}}$.
5. Write a conclusion: into which forms the lost energy turned in your experiment or in the simulation.

Show that the supplied energy splits into useful work and losses, and that efficiency is the share of the useful part in the supplied energy.

An energy-flow diagram (Sankey diagram): a wide arrow $W_{\text{in}}$ enters from the left and splits into an arrow $W_{\text{useful}}$ (upward) and an arrow $W_{\text{loss}}$ (downward), with arrow widths proportional to energy. On the right the relations are written as $W_{\text{in}} = W_{\text{useful}} + W_{\text{loss}}, \qquad \eta = \frac{W_{\text{useful}}}{W_{\text{in}}} \cdot 100\%, \qquad P = \frac{W}{t}.$

A cyclist with a bicycle of total mass $80\ \mathrm{kg}$ climbs a hill of height $120\ \mathrm{m}$ in $15\ \mathrm{min}=900\ \mathrm{s}$. The useful work against gravity is $W=mgh=80\cdot9.8\cdot120\approx94\,000\ \mathrm{J}=94\ \mathrm{kJ}$. The average useful power is $P=W/t=94\,000/900\approx105\ \mathrm{W}$. With a body efficiency of about $22\%$, the expended chemical power is $105/0.22\approx480\ \mathrm{W}$.

The scenario exercises the work formula $W=Fs\cos\theta$ for gravity, the average power formula $P=W/t$, and the efficiency definition $\eta=W_{\text{useful}}/W_{\text{input}}$; on a climb the useful work is $mgh$, while excess energy becomes heating of muscles.

A car of mass $1200\ \mathrm{kg}$ accelerates from $0$ to $100\ \mathrm{km/h}=27.8\ \mathrm{m/s}$ in $10\ \mathrm{s}$. For constant acceleration, the average driving force is $F=ma=1200\cdot(27.8/10)\approx3.3\times10^3\ \mathrm{N}$, the distance is $s=\frac12 vt\approx139\ \mathrm{m}$, and the useful work is $W=Fs\approx463\ \mathrm{kJ}$. The average useful power is $P=W/t\approx46\ \mathrm{kW}$. With an engine efficiency of $25\%$, the fuel's chemical power is $46/0.25=185\ \mathrm{kW}$.

The scenario uses work done by a constant force over a displacement $W=Fs$, average power $P=W/t$, and engine efficiency $\eta=W_{\text{useful}}/E_{\text{fuel}}$; useful mechanical energy is only $25\%$ of the fuel energy, and the rest becomes heat.

A charger draws $10\ \mathrm{W}$ from a wall outlet for $2\ \mathrm{h}=7200\ \mathrm{s}$; the input electrical energy is $W_{\text{input}}=Pt=10\cdot7200=72\,000\ \mathrm{J}=72\ \mathrm{kJ}$. The battery stores $15\ \mathrm{Wh}=54\,000\ \mathrm{J}=54\ \mathrm{kJ}$. The charging efficiency is $\eta=54/72=0.75=75\%$, and $18\ \mathrm{kJ}$ becomes heating of the phone and charger.

Here energy is transferred by the work of electric current, power gives the rate of transfer, and efficiency compares the useful stored energy with the total input energy.