Power — average and instantaneous, units: Worked examples

1.3.9 · D3 · Physics › Work, Energy & Power › Power — average and instantaneous, units

Ye Power — average and instantaneous, units ka companion hai. Agar koi symbol yahan unfamiliar lage, toh wo parent note mein build kiya gaya tha; hum yahan sirf tools use kar rahe hain.

The scenario matrix

$P = F v cos θ$ ke through power mein exactly kuch "knobs" hain: $cos θ$ ka sign (force aur velocity ke beech ka angle), kya $v = 0$ hai (degenerate case), kya power constant hai ya time mein changing, aur kya problem units/billing ke roop mein dress ki gayi hai ya exam twist ke roop mein. Yahan har wo cell hai jo hume cover karni hai.

Cell	Kya special hai	$P$ ka sign / behaviour	Covered by
A. $θ = 0$	force motion ke parallel	$P = + F v$ , maximum positive	Ex 1
B. $0 < θ < 90°$	force partly along motion	$P > 0$ lekin reduced	Ex 2
C. $θ = 90°$	force perpendicular	$P = 0$ (circular motion!)	Ex 3
D. $90° < θ \leq 180°$	force motion ko oppose karta hai	$P < 0$ (braking)	Ex 4
E. $v = 0$ degenerate	kuch bhi move nahi ho raha	$P = 0$ regardless of $F$	Ex 5
F. Time-varying $P$	$v$ badhta hai, toh $P$ badhta hai	$P_{inst} \neq = P_{avg}$	Ex 6
G. Limiting behaviour	engine ki max speed	$P$ fixed $\Rightarrow v_{m a x} = P / F$	Ex 7
H. Units / real-world	billing in kWh	energy = power × time	Ex 8
I. Exam twist	efficiency + do powers	useful/input ratio	Ex 9

Sabse important geometric idea angle $θ$ hai, isliye hum pehle use ek picture ke roop mein dekhte hain.

Figure — Power — average and instantaneous, units

$cos$ kyun, $sin$ nahi? Kyunki hume $v$ ke along component chahiye. Right triangle par, $θ$ ke adjacent side ( $v$ ke flat along bicha shadow) ko hypotenuse ( $F$ ) se divide karo toh exactly $cos θ$ milta hai. Perpendicular part ( $F sin θ$ ) koi work per second nahi karta — wo sirf path ko bend karta hai.

Cell A — force fully along motion ( $θ = 0$ )

Worked example Ex 1 · Sprinter pushing off the blocks

Ek sprinter apni motion ki direction ke perfectly along constant $600$ N force se aage drive karti hai. Jis instant uski speed $5$ m/s hai, wo kitna power deliver kar rahi hai? (Cell A)

Forecast: $θ = 0 \Rightarrow cos θ = 1$ , toh power sirf $F \times v$ hona chahiye — is force aur speed ke liye jo ho sakta hai sabse bada. Value padhne se pehle guess karo.

Angle identify karo: force aur velocity ek hi direction mein point kar rahe hain, toh $θ = 0°$ . Ye step kyun? $cos θ$ sab kuch control karta hai; pehle use pin down karo.
$cos 0° = 1$ , toh $P = F v cos θ = F v$ . Ye step kyun? Ye maximum-efficiency case hai — saari force motion ke along push karti hai.
$P = 600 \times 5 \times 1 = 3000 W = 3 kW$ .

Verify: Units: $N \times m/s = kg m s^{- 2} \times m s^{- 1} = kg m^{2} s^{- 3} = W$ . ✓ Expected ke anusaar positive (energy runner ki motion mein flow ho rahi hai).

Cell B — force at an oblique angle ( $0 < θ < 90°$ )

Worked example Ex 2 · Child pulling a sled by a rope

Ek child ek sled ko $50$ N force se ek rope ke along kheenchta hai jo horizontal ground se $60°$ upar tilt hai. Sled horizontally $2$ m/s par slide karta hai. Sled ko deliver kiya gaya power? (Cell B)

Forecast: Sirf force ka horizontal shadow hi work karta hai, toh power $50 \times 2 = 100$ W se kam hona chahiye. Kitna kam, guess karo.

Velocity horizontal hai; force uske $60°$ upar hai, toh $θ = 60°$ . Ye step kyun? $θ$ $F$ aur $v$ ke beech measure hota hai, ground se arbitrarily nahi — lekin yahan ground hi velocity direction hai.
Motion ke along force ka component: $F cos θ = 50 cos 60° = 50 \times 0.5 = 25$ N. Ye step kyun? Vertical part ( $F sin 60°$ ) sirf ground normal ke against lift karta hai — ye $v$ ke along kuch bhi move nahi karta, toh koi work per second nahi karta.
$P = (F cos θ) v = 25 \times 2 = 50 W$ .

Verify: $50$ W "naive" $100$ W ka exactly aadha hai, jo $cos 60° = 0.5$ se match karta hai. ✓ Positive aur reduced, jaise forecast kiya tha.

Cell C — force perpendicular to motion ( $θ = 90°$ )

Worked example Ex 3 · Ball on a string (circular motion)

Ek ball horizontal circle mein constant speed $3$ m/s par whirl karta hai. String tension $12$ N hai, hamesha centre ki taraf point karta hai. Tension kitna power deliver karta hai? (Cell C)

Forecast: String har instant motion ke sideways kheenchti hai. Kya ye ball ko speed up karti hai? Power guess karo.

Velocity circle ka tangent hai; tension centre ki taraf point karta hai — ye perpendicular hain, toh $θ = 90°$ . Ye step kyun? Uniform circular motion mein centripetal force hamesha velocity ke right angles par hota hai.
$cos 90° = 0$ , toh $P = F v cos 90° = 0$ . Ye step kyun? Ek perpendicular force sirf direction change karta hai, kabhi speed nahi — koi energy per second add nahi hoti.
$P = 12 \times 3 \times 0 = 0 W$ .

Verify: Constant speed ke saath consistent: constant speed ⇒ constant kinetic energy ⇒ zero net work ⇒ zero power. ✓ Pehle figure mein imagine karo blue force seedha upar point kar rahi hai ( $v$ ke perpendicular): $v$ par uski dashed shadow zero length tak collapse ho jaati hai — wo zero-length shadow exactly ye zero power hai.

Cell D — force opposes motion ( $90° < θ \leq 180°$ )

Worked example Ex 4 · Car braking (negative power)

Ek car $15$ m/s par chalti hai. Brakes aur friction total $2000$ N force directly backwards apply karte hain, motion ke opposite. Ye forces kitna power deliver karte hain? (Cell D)

Forecast: Force velocity ko oppose kar raha hai — kya power positive, zero, ya negative hogi? Guess karo.

Force velocity ke anti-parallel hai, toh $θ = 180°$ . Ye step kyun? Opposite arrows sabse bada possible angle dete hain.
$cos 180° = - 1$ , toh $P = F v cos 180° = - F v$ . Ye step kyun? Negative sign ka matlab hai energy car ki motion se remove ho rahi hai (brakes mein heat ban rahi hai).
$P = 2000 \times 15 \times (- 1) = - 30000 W = - 30 kW$ .

Verify: Negative sign physically correct hai — car kinetic energy kho rahi hai. Magnitude $30$ kW ka matlab hai har second $30$ kJ KE dissipate ho rahi hai. ✓

Cell E — velocity is zero (degenerate)

Cell F — power changing in time

Worked example Ex 6 · Constant force, growing speed

Ek $2$ kg trolley constant $6$ N force ke under rest se start hoti hai. (a) $t = 4$ s par instantaneous power aur (b) $0$ – $4$ s par average power find karo. (Cell F)

Forecast: Force constant hai — kya power bhi constant hai? Guess karo kya $P_{avg}$ , $P_{inst} (4)$ ke barabar hai.

Acceleration: $a = F / m = 6/2 = 3 m/s^{2}$ . Ye step kyun? Hume $v (t)$ chahiye, jo acceleration se aata hai.
$t = 4$ par velocity: $v = a t = 3 \times 4 = 12$ m/s (rest se start kiya). Ye step kyun? Instantaneous power current velocity use karta hai.
$P_{inst} (4) = F v = 6 \times 12 = 72 W$ . Ye step kyun? Force aur motion yahan parallel hain, toh $P = F v$ .
KE ke through total work: $W = \frac{1}{2} m v^{2} = \frac{1}{2} (2) (1 2^{2}) = 144$ J. Ye step kyun? Average power ko interval par total work chahiye; work–energy theorem use cleanly deta hai.
$P_{avg} = W /Δ t = 144/4 = 36 W$ .

Verify: $P$ linearly $0$ se $72$ W tak bada, toh uska time-average midpoint $\frac{1}{2} (72) = 36$ W hai. ✓ Constant force ka matlab constant power nahi — ye is cell ka poora point hai.

Agla figure is concretely banata hai: ye instantaneous power $P = F v$ ko time ke against plot karta hai. Kyunki $v = a t$ steadily rest se upar jaata hai, blue power line ek straight ramp hai jo $0$ se $t = 4$ s par peak $72$ W tak rise karti hai. Green dashed line average $36$ W mark karti hai — exactly ramp ke aadhe mein — isliye $P_{avg} = \frac{1}{2} P_{peak}$ yahan.

Cell G — limiting behaviour (top speed)

Worked example Ex 7 · Car's maximum speed at rated power

Ek car engine maximum constant power $60$ kW deliver karta hai. Level road par total resistance (drag + friction) high speed par $1500$ N hai. Car ki top speed kya hai? (Cell G)

Forecast: Top speed par car aur accelerate nahi kar sakti. Kya balance kya karta hai? $v_{m a x}$ guess karo.

Top speed par, acceleration $= 0$ , toh engine ka drive force $F_{drive}$ resistance ke exactly equal hai: $F_{drive} = 1500$ N. Ye step kyun? Koi acceleration nahi ⇒ net force zero ⇒ forward drive force magnitude mein backward resistive force ke equal hai.
Engine ka power car ko forward move karta hai, toh drive force (motion ke parallel) hum $P = F_{drive} v$ mein daalte hain, giving $v = P / F_{drive}$ . Ye step kyun? Hume woh force use karna hai jo motion ke along positive work karta hai — drive force, resistance nahi. Ye size mein equal hote hain yahan, lekin ye opposite directions mein point karte alag-alag forces hain.
$v_{m a x} = \frac{60000}{1500} = 40 m/s$ .

Verify: $P = F_{drive} v = 1500 \times 40 = 60000$ W = $60$ kW ✓. Limiting insight: kyunki $P$ capped hai, car jitni fast jaati hai utna kam drive force muster kar sakti hai — $F_{drive} = P / v$ shrink hota hai jaise $v$ badhta hai jab tak ye sirf drag ko match kare. Isliye top speed exist karta hai.

Cell H — units aur billing (real-world word problem)

Worked example Ex 8 · Monthly electricity bill

Ek $2$ kW heater $30$ days ke liye din mein $3$ ghante chalta hai. kWh aur joules mein use ki gayi energy find karo. (Cell H)

Forecast: kWh energy hai ya power? Time se multiply ya divide karo? Start karne se pehle guess karo.

Total running time: $3 h/day \times 30 days = 90$ h. Ye step kyun? Energy = power × time, toh hume total time chahiye.
kWh mein energy: $E = P \times t = 2 kW \times 90 h = 180$ kWh. Ye step kyun? kW aur hours ko saath rakhne se directly bill-friendly kWh milta hai.
$1 kWh = 3.6 \times 1 0^{6}$ J use karke joules mein convert karo: $E = 180 \times 3.6 \times 1 0^{6} = 6.48 \times 1 0^{8}$ J. Ye step kyun? SI answers joules mein hone chahiye; $1 kWh = 1000 W \times 3600 s$ .

Verify: SI se cross-check: $P = 2000$ W, $t = 90 \times 3600 = 324000$ s, $E = 2000 \times 324000 = 6.48 \times 1 0^{8}$ J ✓. kWh energy hai, confirm hua, power nahi.

Cell I — exam twist (do powers ki efficiency)

Worked example Ex 9 · Pump with efficiency

Ek water pump $5$ kW input power rated hai lekin sirf $80%$ efficient hai. Ye density $1000 kg/m^{3}$ wala paani $10$ m height tak lift karta hai. Ye kitne litres per second raise kar sakta hai? $g = 10 m/s^{2}$ use karo (acceleration due to gravity, upar symbol list mein defined). (Cell I)

Forecast: Useful power $5$ kW se kam hai. Raise kiye gaye paani ko compute karne ke liye hum kaun sa power use karte hain? Guess karo.

Useful (output) power: $P_{out} = η P_{in} = 0.80 \times 5000 = 4000$ W. Ye step kyun? Efficiency $η = P_{out} / P_{in}$ — sirf useful part paani uthata hai.
Useful power har second mass $m$ ko gravity ke against raise karta hai: $P_{out} = \frac{m g h}{t}$ , toh mass rate hai $\frac{m}{t} = \frac{P _{out}}{g h}$ . Ye step kyun? Lift karne ka work = $m g h$ (weight $m g$ times height $h$ ); power = wo work per second.
$\frac{m}{t} = \frac{4000}{10 \times 10} = 40 kg/s$ .
Kyunki $1000$ kg paani $= 1000$ L, $40$ kg/s $= 40$ L/s. Ye step kyun? Density $1000 kg/m^{3}$ ka matlab hai paani ke liye $1$ L $= 1$ kg.

Verify: Power wapas check karo: $40 kg/s$ ko $10$ m tak lift karne mein $40 \times 10 \times 10 = 4000$ W useful power chahiye ✓, aur $4000/0.8 = 5000$ W input ✓, rating se match karta hai. Dekho Energy conservation and efficiency.

Active Recall

Recall Cover the answers

Jab force motion ko oppose kare toh power ka sign? ::: Negative — energy remove ho rahi hai. Uniform circular motion mein centripetal force ki taraf se deliver kiya gaya power? ::: Zero, kyunki $θ = 90°$ . Agar force constant hai lekin object accelerate kar raha hai, kya power constant hai? ::: Nahi — $P = F v$ badhta hai jaise $v$ badhta hai. Fixed engine power $P$ ke saath resistance $F$ ke against car ki top speed ka formula? ::: $v_{m a x} = P / F$ . kWh energy hai ya power, aur kyun? ::: Energy — ye power ko time se multiply kiya hota hai. Useful work rate find karne ke liye kaun sa power (input ya output) use karte ho? ::: Output (useful) power = $η \times$ input.

Connections

Power — average and instantaneous, units — core formulas wala parent note.
Work — definition and W = F·d cosθ — $cos θ$ projection yahan reuse kiya gaya.
Kinetic Energy and Work-Energy Theorem — average power ke liye total work (Ex 6).
Velocity and instantaneous rate (Kinematics) — velocity jo time mein badhti hai (Cell F).
Dot product of vectors — $F \cdot v$ ke peeche ki geometry (Cell A–D).
Energy conservation and efficiency — input vs output power (Ex 9).

Worked examples — Power — average and instantaneous, units