Power — The Rate of Doing Work (NCERT 5.10)
1. Same Work, Different Hurry
Carry 20 sacks up a staircase in an afternoon, or hire a crane that does it in two minutes: the work is identical — same weight, same height, same mgh — but nobody calls the two performances equal. What the afternoon and the two minutes differ in is power: the rate at which work gets done. NCERT 5.10 defines it as
P = W / t (average), P = dW/dt (instantaneous)
with the SI unit watt (1 W = 1 J s⁻¹) and its everyday multiples kW, MW. Power is the chapter's pricing formula: work is the commodity, power is the exchange rate against time. And because machines are bought and sold by their power ratings — 100 W bulbs, 5 hp pumps, 100 kW cars — this shortest section of NCERT generates a disproportionate number of easy marks: conversions, P = Fv one-liners, and electricity-bill arithmetic.
2. Complete Theory: From dW/dt to P = Fv, and the Unit Zoo
The instantaneous-power derivation. In a small time dt a force F moves its point of application by ds = v dt, doing dW = F·ds. Divide by dt: P = dW/dt = F·v = F v cos θ, with θ the angle between force and velocity. Read it twice, because it encodes two exam sentences: power is force times speed — a stalled engine (v = 0) delivers zero power no matter how hard it pushes; and only the force component along the motion counts — the normal force on level cruising bills nothing at any speed. At constant power, F and v trade along a hyperbola: more pull, less pace.
The unit zoo — where the marks live. SI: the watt and its prefixes (1 kW = 10³ W, 1 MW = 10⁶ W). Engineering: 1 hp = 746 W (horsepower, still printed on Indian pump and motor nameplates). Domestic: 1 kWh = 3.6 × 10⁶ J — the kilowatt-hour is the energy a 1 kW device delivers in one hour (1000 W × 3600 s), which makes it a unit of energy, sold by electricity boards as "one unit". The kWh–power confusion is a standing NTA MCQ: the unit on your bill is energy; the rating on the appliance is power. A 2 kW heater running 5 hours burns 10 kWh = 3.6 × 10⁷ J.
Two standard problem templates. Template one — lifting: a pump raising mass m of water (or a body of weight mg) through height h every cycle bills work mgh per cycle; P = mgh/cycle-time. Template two — cruise: a vehicle at constant speed has zero acceleration, so the engine force equals total resistance f, and P = F v = f v; capped power therefore caps speed at vmax = P/f — the template behind every "engine power 50 kW, resistance 2500 N" numerical. Both templates are the work ledger wearing a stopwatch.
3. Visualising: One Second of a 100 W Motor
Figure 5.7 — Power is work per second, visible as a production line
Top: the motor's force and speed — P = F v = 200 × 0.5 = 100 W. Bottom: the same 100 W as a per-second delivery: each second adds one 100 J block.
Exam read-out: two views, one quantity: the force-speed product upstairs, the joules-per-second production line downstairs. Every pump, engine and heater numerical in NTA papers is one of the two views with the numbers rescaled.
4. Solved Examples
Example 1 — The water pump (per-minute input!)
A pump lifts 60 kg of water per minute through a height of 10 m. Find its average power (g = 10 m s⁻²).
Solution (step by step): Work per minute: W = mgh = 60 × 10 × 10 = 6000 J. The minute is the trap: convert to per-second — 60 kg/min means 1 kg/s. P = W/t = 6000/60 = 100 W (equivalently 1 kg lifted 10 m each second: 1 × 10 × 10 = 100 J/s). Check the rating line: a real pump would need extra power for pipe friction and inefficiency, but NCERT asks for the ideal ledger — 100 W.
Example 2 — Cruise power and the horsepower conversion
A truck cruises at a steady 72 km h⁻¹ on a level road with its engine exerting a 2000 N driving force. Find the engine's power output in kW and in hp.
Solution: Steady speed ⇒ no acceleration ⇒ the ledger needs force × speed only. 72 km h⁻¹ = 20 m s⁻¹. P = F v = 2000 × 20 = 40,000 W = 40 kW. Horsepower: P = 40000/746 ≈ 53.6 hp. The two conversions — 72 km/h → 20 m/s and W → hp ÷ 746 — are exactly the Speed-Squared Bank and unit bank moves from the chapter hub; NTA expects them cold.
5. Practice Questions
Q1. Convert 2 kWh into joules. If a household consumes 2 kWh daily, how many joules per week is that?
ANSWER: 1 kWh = 3.6 × 10⁶ J, so 2 kWh = 7.2 × 10⁶ J per day; per week = 7 × 7.2 × 10⁶ = 5.04 × 10⁷ J. The conversion is the whole question — and it is also the sanity check that a household's daily energy is a few million joules, not a few watts.
Q2 (MCQ). Which of the following is a unit of energy? (a) watt (b) kilowatt (c) kilowatt-hour (d) horsepower
ANSWER: (c) — the kilowatt-hour is power × time = energy (3.6 × 10⁶ J). Watt, kilowatt and horsepower are all units of power (rates); the "hour" hiding in kWh is what converts it to the energy side of the ledger — the single most reliable MCQ gift in this chapter.
Q3. A motor of power 10 kW drags a load at constant speed against a total resistive force of 500 N. Find the maximum steady speed.
ANSWER: Constant speed ⇒ engine force = resistance: P = f v → vmax = P/f = 10000/500 = 20 m/s = 72 km/h. Note the structure: capped power against fixed resistance is a division, and any option quoting 50 m/s or 2 m/s has f and P swapped or the kW un-converted.
6. Key Formulas & Takeaways
| Relation | Condition / remark |
|---|---|
| Pavg = W/t | Average rate over an interval; unit watt |
| P = dW/dt | Instantaneous rate; 1 W = 1 J s⁻¹ |
| P = F v cos θ | θ between force and velocity; ⊥ force bills zero power |
| vmax = P/f | Engine at power P against resistance f, terminal (constant) speed |
| Ppump = mgh/cycle ÷ time | Lifting template; convert per-minute inputs to per-second |
| 1 hp = 746 W | Nameplate conversions; ÷746 to reach SI |
| 1 kWh = 3.6 × 10⁶ J | Unit of ENERGY (1 kW for 1 h); the bill's "unit" |
Power questions are conversion questions wearing a costume: km/h to m/s, minutes to seconds, hp to W, kWh to J. Fix the units first and the physics — one multiplication or division — finishes itself.
7. Frequently Asked Questions
Define average power and instantaneous power.
Average power is the total work done divided by the time taken: P_avg = W/t — it prices the whole job at one rate. Instantaneous power is the work rate at a moment: P = dW/dt, the limit of ΔW/Δt as the interval shrinks. When the rate varies — a cyclist sprinting, a car accelerating — the two differ, and NTA's questions usually give one and ask for the other. Both carry the watt as their unit, 1 W = 1 J s⁻¹.
Why is the instantaneous power of a force equal to F v cosθ?
In a tiny time dt a force F does work dW = F·ds = F cosθ · ds, where ds = v dt is the displacement in that instant and θ is the angle between force and velocity. Dividing by dt gives P = dW/dt = F v cosθ. The cosθ matters: only the force component along the motion bills power, so a normal force (θ = 90°) delivers zero power at any speed. This single formula answers every 'engine force at cruise speed' question in one multiplication.
Is the kilowatt-hour a unit of power or of energy?
Of energy — and the distinction is NTA's favourite conversion trap. A kilowatt-hour is the energy delivered by a 1 kW device running for one hour: 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J. Electricity bills sell kilowatt-hours (commonly called 'units'), never kilowatts; the kilowatt on the bill's tariff line is a power, the 40 units consumed last month were energy. Convert before comparing: a 2 kWh daily consumption is 7.2 × 10⁶ J per day.
How do you find the maximum speed of an engine-driven vehicle?
At top speed the vehicle stops accelerating, so the engine's driving force exactly balances the total resistance f (air drag plus friction). The engine at power P driving at speed v delivers force F = P/v, and setting P/v = f gives v_max = P/f. Example: a 50 kW engine against 2500 N of resistance tops out at 50000/2500 = 20 m/s = 72 km/h. The physics: with power capped, pushing harder means moving slower — force and speed trade along the hyperbola P = F v.
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