What power means physically
Work and energy say nothing about time. Carrying a load up four floors involves the same work whether you take ten seconds or ten minutes — the displacement against gravity is identical. NCERT introduces power to capture the missing dimension: "We say a person is physically fit if he not only climbs four floors of a building but climbs them fast." Power is the time rate at which work is done or energy is transferred.
The same idea recurs across the syllabus. A man may take hours to load a truck with cement bags; a machine does it in minutes (NIOS §6.3). Both transfer the same energy, but the machine is more powerful because it delivers that energy in less time. Power, then, is not "how much" but "how fast."
Average power
The average power of a force is the ratio of the total work $W$ it does to the total time $t$ taken (NCERT Eq. before 5.20; NIOS Eq. 6.14):
$$P_{\text{av}} = \frac{W}{t}$$
This is the value to use whenever a problem gives you a total energy transfer and a total duration — a pump filling a tank in fifteen minutes, a dieter lifting a mass a thousand times, water falling onto a turbine at a steady rate. Average power smooths over any moment-to-moment variation and reports the overall rate.
Instantaneous power and F·v
When the rate of working changes from instant to instant, the average no longer describes any single moment. The instantaneous power is the limiting value of the average power as the time interval approaches zero (NCERT Eq. 5.20; NIOS Eq. 6.15):
$$P = \frac{dW}{dt}$$
This form has a powerful rewrite. The small work done by a force $\vec F$ over a small displacement $d\vec r$ is $dW = \vec F\cdot d\vec r$. Dividing by $dt$ and recognising $d\vec r/dt$ as the instantaneous velocity (NCERT Eq. 5.21):
$$P = \vec F\cdot\frac{d\vec r}{dt} = \vec F\cdot\vec v$$
where $\vec v$ is the instantaneous velocity when the force is $\vec F$. Writing it out with the angle $\theta$ between force and velocity, $P = Fv\cos\theta$. Three consequences follow at once, and they are exactly what NEET tests.
- Constant velocity does not mean zero power. A lift rising at steady speed has zero net force, but the motor still delivers $P=\vec F\cdot\vec v$ to balance gravity and friction. The "minimum power of the motor" is the recurring NEET ask.
- Power can be negative. Friction and braking forces have $\theta > 90^\circ$, so $\cos\theta < 0$ and they remove energy. Positive power feeds kinetic energy; negative power drains it.
- A time-varying force gives a time-varying power. If $\vec F$ and $\vec v$ both depend on $t$, compute $\vec v(t)$ first, then dot with $\vec F(t)$ — the NEET 2016 vector-force item below is exactly this.
Scalar nature, watt and horsepower
Power, like work and energy, is a scalar quantity (NCERT §5.10). The dot product $\vec F\cdot\vec v$ multiplies two vectors but returns a single number with no direction. Its dimensional formula is $[\text{ML}^2\text{T}^{-3}]$, derived as force times distance over time.
The SI unit is the watt (W), defined as one joule per second, named after James Watt of steam-engine fame. An agent has a power of $1\,\text{W}$ if it does one joule of work each second. Larger multiples — the kilowatt ($10^3\,\text{W}$) and megawatt ($10^6\,\text{W}$) — rate everyday appliances and power stations. The older British unit, the horsepower, survives on engines:
$$1\ \text{hp} = 746\ \text{W}$$
| Quantity | Symbol / formula | SI unit | Notes |
|---|---|---|---|
| Average power | P_av = W / t | watt (W) | Total work over total time |
| Instantaneous power | P = dW/dt = F·v | watt (W) | Scalar; equals $Fv\cos\theta$ |
| Watt | 1 W = 1 J s⁻¹ | — | SI unit; dimensions $[\text{ML}^2\text{T}^{-3}]$ |
| Horsepower | 1 hp = 746 W | — | Older unit; rates vehicles |
| Kilowatt-hour | 1 kWh = 3.6 × 10⁶ J | joule (J) | A unit of ENERGY, not power |
The kilowatt-hour trap
The electricity bill carries energy consumption in units of kWh, and the unit's name — kilo-watt-hour — invites the single most exploited confusion in this chapter. NCERT pre-empts it in one line: "Note that kWh is a unit of energy and not of power." A kilowatt is power; multiply it by an hour of time and you obtain energy.
NCERT's own demonstration: a 100-watt bulb left on for 10 hours uses
$$100\,\text{W} \times 10\,\text{h} = 1000\,\text{W h} = 1\,\text{kWh} = 10^3\,\text{W}\times 3600\,\text{s} = 3.6\times10^{6}\,\text{J}.$$
The same logic strips other distractors. The dieter who lifts a mass a thousand times (NCERT Exercise 5.22) does a fixed quantity of work — that is energy. Divide by the time and only then do you have power. Keep the two quantities apart before you reach for any formula.
Worked example 1 — minimum power of a lift
An elevator with a maximum load of 1800 kg (elevator + passengers) is moving up with a constant speed of $2\ \text{m s}^{-1}$. The frictional force opposing the motion is 4000 N. Determine the minimum power delivered by the motor in watts and in horsepower. Take $g = 10\ \text{m s}^{-2}$.
Force the motor must overcome. The elevator moves at constant speed, so the upward force equals the total downward force — weight plus friction:
$$F = mg + F_f = (1800\times 10) + 4000 = 22000\ \text{N}.$$
Apply $P = \vec F\cdot\vec v$. Force and velocity are both upward, so $\cos\theta = 1$:
$$P = Fv = 22000 \times 2 = 44000\ \text{W}.$$
Convert to horsepower using $1\,\text{hp} = 746\,\text{W}$: $P = 44000/746 \approx 59\ \text{hp}$.
Answer: $44000\ \text{W} \approx 59\ \text{hp}$. Constant speed does not make the motor power zero — it is the power needed to cancel gravity and friction.
The lift problem rests on $W = \vec F\cdot\vec d$ — revisit work — definition and calculation if the dot product feels shaky.
Worked example 2 — power of a water pump
A pump on the ground floor can pump up water to fill a tank of volume $30\ \text{m}^3$ in 15 min. The tank is 40 m above the ground, and the efficiency of the pump is 30%. How much electric power is consumed by the pump? Take $g = 10\ \text{m s}^{-2}$ and density of water $= 10^3\ \text{kg m}^{-3}$.
Mass of water raised. $m = \rho V = 10^3 \times 30 = 3\times10^4\ \text{kg}$.
Useful work done against gravity. $W = mgh = 3\times10^4 \times 10 \times 40 = 1.2\times10^{7}\ \text{J}$.
Useful (output) power. Time $t = 15\ \text{min} = 900\ \text{s}$, so by average power $P_{\text{out}} = W/t = 1.2\times10^{7}/900 \approx 1.33\times10^{4}\ \text{W}$.
Account for efficiency. Only 30% of the electrical input emerges as useful work, so the input power is
$$P_{\text{in}} = \frac{P_{\text{out}}}{0.30} = \frac{1.33\times10^{4}}{0.30} \approx 4.4\times10^{4}\ \text{W} \approx 44\ \text{kW}.$$
Answer: the pump consumes about 44 kW. Note the order: compute useful power first, then divide by efficiency to get input — never the reverse.
Worked example 3 — electrical power from a windmill
The blades of a windmill sweep out a circle of area $A$. Wind flows at velocity $v$ perpendicular to the circle. (a) Find the mass of air passing through in time $t$. (b) Find the kinetic energy of that air. (c) If the windmill converts 25% of the wind's energy into electrical energy, with $A = 30\ \text{m}^2$, $v = 36\ \text{km/h}$ and air density $\rho = 1.2\ \text{kg m}^{-3}$, find the electrical power produced.
(a) Mass of air in time $t$. In time $t$ the air sweeps a cylinder of length $vt$ and cross-section $A$, so $m = \rho \times (\text{volume}) = \rho A v t$.
(b) Kinetic energy of that air. $K = \tfrac{1}{2}mv^2 = \tfrac{1}{2}(\rho A v t)v^2 = \tfrac{1}{2}\rho A v^3 t$.
Power available in the wind. Dividing the kinetic energy by $t$ (average power), $P_{\text{wind}} = \tfrac{1}{2}\rho A v^3$.
(c) Convert units and apply 25%. $v = 36\ \text{km/h} = 10\ \text{m s}^{-1}$, so
$$P_{\text{wind}} = \tfrac{1}{2}\times 1.2 \times 30 \times (10)^3 = 1.8\times10^{4}\ \text{W}.$$
Electrical power $= 0.25 \times 1.8\times10^{4} = 4.5\times10^{3}\ \text{W} = 4.5\ \text{kW}$.
Answer: about 4.5 kW. The crucial step is that wind power scales as $v^3$ — doubling the wind speed multiplies the available power eightfold.
These three examples cover the entire NEET vocabulary of power problems: a steady-speed motor solved by $\vec F\cdot\vec v$, a fixed energy transfer divided by time and corrected for efficiency, and a continuous flow whose power follows from a mass-flow rate. Recognise which of the three a question belongs to and the formula chooses itself.
Power in one breath
- Power is the time rate of doing work or transferring energy — "how fast," not "how much."
- Average power $P_{\text{av}} = W/t$; instantaneous power $P = dW/dt = \vec F\cdot\vec v = Fv\cos\theta$.
- Power is a scalar, dimensions $[\text{ML}^2\text{T}^{-3}]$, SI unit the watt $= 1\ \text{J s}^{-1}$. $1\ \text{hp} = 746\ \text{W}$.
- The kilowatt-hour is a unit of energy: $1\ \text{kWh} = 3.6\times10^{6}\ \text{J}$.
- Constant speed gives zero net power but a non-zero motor power; power is negative when force opposes velocity.
- Efficiency: input $= $ output $/\,\eta$. Wind/water power follows from the mass-flow rate; wind power $\propto v^3$.