Most houses in an Indian town have a water tank on the roof. An electric pump lifts water into it from a sump at the ground. The shopkeeper who sells pumps asks two questions: how high is the tank, and how quickly do you want it filled? A bigger pump does not lift the water any higher. It lifts it faster.
Work per second
Two students climb four floors of a building, \(12\,\text{m}\) in all. Each has a mass of \(50\,\text{kg}\), so each does \(50 \times 10 \times 12 = 6000\,\text{J}\) of work against gravity. One runs up in \(20\,\text{s}\); the other walks up in \(60\,\text{s}\). The work is the same, but the runner does it three times as fast. The rate of doing work is called power. Over a time \(\Delta t\), the average power is the work divided by the time: \(6000/20 = 300\,\text{J/s}\) for the runner and \(100\,\text{J/s}\) for the walker.
Power is the rate at which work is done, or at which energy is transferred. The average power over a time \(\Delta t\) is \(P_{\text{avg}} = W/\Delta t\). The instantaneous power is \(P = dW/dt\). Power is a scalar. Its SI unit is the watt: \(1\,\text{W} = 1\,\text{J/s}\).
The unit of power, one joule per second, is called the watt, W, after the Scottish engineer James Watt. Engines are often rated in kilowatts, and sometimes in horsepower: \(1\,\text{hp} \approx 746\,\text{W}\). The runner above works at \(300\,\text{W}\), about \(0.4\,\text{hp}\).
Power at an instant
Power can change from moment to moment. A sprinter works hardest in the first seconds of a race, while speeding up, and less once running steadily. The average over the whole race hides this. The power at one instant is the limit of the average power as the time interval shrinks: \(P = dW/dt\). On a graph of work against time, the average power is the slope of a chord. The instantaneous power is the slope of the tangent. It is the same idea as average and instantaneous velocity.
Back from power to work
If power is the rate of doing work, then work is power added up over time: \(W = \int P\,dt\). On a graph of power against time, the work is the area under the graph, just as it was for force against position. A meter that records a motor's power every second can give the total work without anyone knowing a force or a distance.
Energy sold as electricity is measured the same way. The unit on an electricity bill is the kilowatt hour: the work done in one hour at a power of one kilowatt. That is \(1000\,\text{W} \times 3600\,\text{s} = 3.6 \times 10^6\,\text{J}\). An air conditioner of \(1.5\,\text{kW}\) that runs for \(8\) hours uses \(12\,\text{kWh}\), which the bill calls \(12\) units.
Two workers lift identical sacks of cement to the same height. The first takes half the time the second takes. Which statement is correct?
A \(2\,\text{kW}\) geyser runs for \(30\) minutes. How much energy does it use, in megajoules?
A sprinter's average power over a \(100\,\text{m}\) race is \(400\,\text{W}\). What can you say about her power two seconds after the start?
Force times velocity
A force \(\vec F\) acts on a body that moves with velocity \(\vec v\). How fast does the force do work? Over a tiny time \(dt\), the body moves through \(d\vec r = \vec v\,dt\).
\[ dW = \vec F\cdot d\vec r \]
Why. the work of the force over a tiny displacement, during which the force is constant.
\[ P = \dfrac{dW}{dt} = \vec F\cdot\dfrac{d\vec r}{dt} \]
Why. divide by the tiny time \(dt\). Power is the rate of doing work.
\[ P = \vec F\cdot\vec v = Fv\cos\theta \]
Read it. the rate of change of position is the velocity. \(\theta\) is the angle between the force and the velocity.
Why a hill costs more power
A car keeps a steady speed \(v\). On a level road, the engine's driving force \(F\) balances the resistance \(R\) of the air and the road, so \(F = R\) and \(P = Rv\). On a slope of angle \(\theta\), the driving force must also balance the component \(mg\sin\theta\) of the weight down the slope. So \(F = R + mg\sin\theta\), and at the same speed the power is larger.
A slope of "1 in 20" rises \(1\,\text{m}\) for every \(20\,\text{m}\) along the road, so \(\sin\theta = 1/20\). For the car in Fig 6.1, \(mg\sin\theta = 10\,000/20 = 500\,\text{N}\), which doubles the driving force and the power. An engine that can give only \(10\,\text{kW}\) cannot hold \(20\,\text{m/s}\) on that slope. With \(1000\,\text{N}\) needed, \(P = Fv\) allows only \(10\,000/1000 = 10\,\text{m/s}\). This is why loaded trucks crawl up the ghats.
A car cruises at a steady \(20\,\text{m/s}\) on a level road against a total resistance of \(600\,\text{N}\). What power does the engine deliver, in kilowatts?
A truck of mass \(10\,000\,\text{kg}\) has an engine of greatest power \(60\,\text{kW}\). The resistance to its motion is \(2000\,\text{N}\). It climbs a slope with \(\sin\theta = 0.02\). Take \(g = 10\,\text{m/s}^2\). Find its greatest steady speed on the slope, in m/s.
The engine at a steady speed
A car at a steady speed on a level road has no change in kinetic energy. By the work-energy theorem, the net work on it is zero. That does not mean the engine is idle. It means the engine does work exactly as fast as the resistance removes it. That energy becomes heat in the air and the tyres. A cyclist riding at a steady \(5\,\text{m/s}\) against a resistance of \(20\,\text{N}\) works at \(20 \times 5 = 100\,\text{W}\).
The same rule gives an engine's top speed. As the speed rises, \(P = Fv\) leaves less force for the same power, until the force only just balances the resistance. A car whose engine gives at most \(30\,\text{kW}\), against a resistance of \(1200\,\text{N}\), can reach at most \(30\,000/1200 = 25\,\text{m/s}\), or \(90\,\text{km/h}\). While it is still speeding up, the engine supplies both the resistance and the growing kinetic energy: \(P = (R + ma)v\).
Air drag is not constant. At highway speeds it grows roughly as \(v^2\). Then the power to hold a speed, \(P = Rv\), grows roughly as \(v^3\). Driving at twice the speed needs about eight times the power, and burns fuel much faster per kilometre. This is the physics behind the advice that a moderate speed saves fuel.
The power of a pump
A pump does two jobs on the water it moves. It lifts the water through a height \(h\), and it may send it out of a pipe at a speed \(v\). If the pump moves a mass \(\Delta m\) of water in a time \(\Delta t\), the work on that water is \(\Delta m\,gh + \tfrac12 \Delta m\,v^2\). Dividing by the time gives the useful power, or output:
\[ P_{\text{out}} = \frac{\Delta m}{\Delta t}\left(gh + \tfrac12 v^2\right) \]
No real pump turns all its input power into this output. The ratio is the efficiency, \(\eta = P_{\text{out}}/P_{\text{in}}\), always less than one. The rest becomes heat in the motor and the pipes.
A pump fills a \(1000\) litre tank on a roof, \(15\,\text{m}\) above the sump, in \(25\) minutes. The water leaves the pipe slowly, so its speed can be neglected. The pump's efficiency is \(40\,\%\). Take \(g = 10\,\text{m/s}^2\). Find the power the pump draws.
\(1000\) litres of water have a mass of \(1000\,\text{kg}\). The time is \(25 \times 60 = 1500\,\text{s}\).
\(P_{\text{out}} = mgh/\Delta t\), and \(P_{\text{in}} = P_{\text{out}}/\eta\).
\(P_{\text{out}} = 1000 \times 10 \times 15/1500 = 100\,\text{W}\). Then \(P_{\text{in}} = 100/0.4 = 250\,\text{W}\).
The pump draws \(250\,\text{W}\), a common rating for a small domestic pump. To fill the tank in half the time, it would need twice the power.
Now return to the shopkeeper's two questions. The height of the tank fixes the work on each litre, \(gh\) per kilogram. The time you allow fixes how fast that work must be done. A bigger pump does the same work on the water, only faster. Power is about the rate, never about how high the water can go.
A pump lifts \(2\,\text{kg}\) of water every second through \(10\,\text{m}\) and delivers it at \(10\,\text{m/s}\). Take \(g = 10\,\text{m/s}^2\). Find the useful power of the pump, in watts.
A \(60\,\text{kg}\) student runs up one floor of stairs, \(3\,\text{m}\) high, in about \(3\,\text{s}\). Estimate her power, in watts. Take \(g = 10\,\text{m/s}^2\).
James Watt improved the steam engine in Scotland in the 1760s and 1770s. To sell his engines to mill and mine owners, he had to say how many horses one engine could replace. Watt watched horses turning mill wheels and estimated what a horse could do in a minute: \(33\,000\) foot-pounds of work. He called that rate one horsepower, about \(746\,\text{W}\). The unit of power in SI, the watt, was later named after him.
- Confusing work and power.Work is the energy transferred; power is how fast it is transferred. The same work done faster is more power, not more work.
- Dividing force by speed to get power.Power is force times velocity: \(P = Fv\cos\theta\). Check the units: \(\text{N} \times \text{m/s} = \text{J/s} = \text{W}\).
- Thinking that an engine at a steady speed does no work.The net work is zero, but the engine works at the rate \(Rv\) against the resistance.
- Forgetting the delivery speed in a pump problem.A pump that sends water out fast must also supply \(\tfrac12 v^2\) per kilogram.