Exercises — Power — average and instantaneous, units

Kaunsa tool: "ek time interval mein work kiya" → average power, $P_{\text{avg}}=W/\Delta t$. Average word aur ek poora interval (na ki ek instant) — yahi pehchaan hai. $P_{\text{avg}}=\frac{6000}{4}=1500\text{W}=1.5\text{kW}.$

Kaunsa tool: $P=W/\Delta t$ ko rearrange karo $W=P\Delta t$ mein, kyunki ab hamein power pata hai aur energy chahiye. $1$ minute $=60$ s. $W=P\Delta t=60\times 60=3600\text{J}.$ Ek $60$ W bulb har second $60$ J jalata hai — $60$ s mein yeh $3600$ J ho jaata hai.

Kaunsa tool: hamen power chahiye abhi is waqt force aur speed ke saath, toh $P=Fv\cos \theta$. Same direction matlab $\vec{F}$ aur $\vec{v}$ ke beech angle $\theta =0^{\circ }$ hai, aur $\cos 0^{\circ }=1$. $P=Fv\cos 0^{\circ }=2000\times 15\times 1=30000\text{W}=30\text{kW}.$

Step 1 — Kya aur kyun: constant speed par lifting force weight ko balance karti hai, toh work gravity ke against ki gayi work hai, $W=mgh$. (Yeh Work — definition and W = F·d cosθ ke ideas use karta hai.) $W=mgh=300\times 10\times 20=60000\text{J}.$ Step 2: ise time par rate karo. $P_{\text{avg}}=\frac{W}{\Delta t}=\frac{60000}{50}=1200\text{W}=1.2\text{kW}.$

Kaunsa tool: $P=Fv\cos \theta$, kyunki $\theta =60^{\circ }$ tilted push $\vec{F}$ aur horizontal velocity $\vec{v}$ ke beech ka angle hai. Yeh exactly dot product $\vec{F}\cdot \vec{v}$ hai — yeh sirf $\vec{F}$ ka wo component rakhta hai jo $\vec{v}$ ke along point karta hai. $P=Fv\cos 60^{\circ }=50\times 2\times \frac{1}{2}=50\text{W}.$ Push ka vertical component ($F\sin 60^{\circ }$) motion ke perpendicular hai, isliye yeh zero power deliver karta hai — yehi wajah hai ki formula mein $\sin \theta$ nahi, $\cos \theta$ aata hai.

Step 1 — Kinetic energy kyun use karein: work done kinetic energy ke change ke barabar hai (work–energy theorem), kyunki saari work car ko speed dene mein gayi. $W=\Delta KE=\frac{1}{2}m{v}^2-0=\frac{1}{2}(1000)(20{)}^2=200000\text{J}.$ Step 2: $P_{\text{avg}}=\frac{W}{\Delta t}=\frac{200000}{8}=25000\text{W}=25\text{kW}.$

Force (constant): $F=ma=2\times 4=8\text{N}$. $t=3$ par velocity: $v=at=4\times 3=12\text{m/s}$.

(a) $t=3$ par instantaneous power: $P_{\text{inst}}(3)=Fv=8\times 12=96\text{W}.$ Kyun badhti hai: force fixed hai, lekin $v$ time ke saath badhti hai, isliye $P=Fv$ bhi badhta hai. Actually $P(t)=F(at)=8\times 4t=32t$ — origin se ek straight line (figure mein red line).

(b) Average power: kinetic energy se total work, $W=\frac{1}{2}m{v}^2=\frac{1}{2}(2)(12{)}^2=144\text{J}$. $P_{\text{avg}}=\frac{144}{3}=48\text{W}.$ Note karo $48=\frac{1}{2}\times 96$: kyunki power linearly $0$ se $96$ tak badhi, uska average exactly midpoint hai (figure mein shaded triangle ki area ÷ base).

Integrate kyun karein: power $dW/dt$ hai, work karne ki rate. Varying rate se total work recover karne ke liye hamen $Pdt$ ko interval par add karna hoga — yeh ek integral hai (equivalently, $P$–$t$ graph ke neeche ka area). $W={\int }_0^{5}Pdt={\int }_0^{5}6tdt={\left[3t^2\right]}_0^5=3(25)=75\text{J}.$ Geometric check: $P=6t$ ka graph ek triangle hai jiska base $5$ aur height $6\times 5=30$ hai; area $=\frac{1}{2}(5)(30)=75$ J. ✓

Steady speed par drive force resistance ke barabar hoti hai, isliye $P=Fv$. $P_1=30\times 8=240\text{W},P_2=30\times 16=480\text{W}.$ $\frac{P_2}{P_1}=\frac{480}{240}=2.$ Force fixed rakhne par, power speed ke saath linearly scale karti hai — speed double, power double.

Step 1 — useful output power: useful kaam gravity ke against constant speed par uthana hai, isliye lifting force weight $mg$ hai aur useful power $P_{\text{out}}=(mg)v$ hai. (Energy conservation and efficiency dekho.) $P_{\text{out}}=mgv=20\times 10\times 2=400\text{W}.$ Step 2 — efficiency = useful out ÷ total in. $\eta =\frac{P_{\text{out}}}{P_{\text{in}}}=\frac{400}{500}=0.8=80\%.$ Baaki $100$ W motor ke andar heat aur sound ban jaata hai.

Do forces kyun add karein: constant speed par koi kinetic-energy change nahi, isliye engine ki drive force ko gravity ke slope-down component aur resistance dono ko balance karna hoga.

• Slope ke along gravity ka pull: $mg\sin 30^{\circ }=1200\times 10\times 0.5=6000\text{N}$.
• Resistance: $500\text{N}$.

Total drive force $F=6000+500=6500\text{N}$, aur yeh force motion ke along hai, isliye $\theta =0$: $P=Fv=6500\times 10=65000\text{W}=65\text{kW}.$

Step 1 — girte paani se aane wali power. Har second, mass $\dot{m}=2000$ kg gravitational energy $mgh$ kho deta hai. Toh energy aane ki rate hai $P_{\text{in}}=\dot{m}gh=2000\times 10\times 50=1000000\text{W}=1\text{MW}.$ "Per second" yahan kyun kaam karta hai: power energy per time hai, aur $2000$ kg wo mass hai jo har second deliver hota hai, isliye $\dot{m}gh$ already ek energy-per-second hai, yaani power. Step 2 — efficiency apply karo: $P_{\text{out}}=0.9\times 1000000=900000\text{W}=0.9\text{MW}.$

Kyun: top speed par acceleration zero hai, isliye drive force resistance ke barabar hai, $F=600$ N. Maximum power par engine $P=75000$ W deta hai, aur $P=Fv$, isliye $v_{\max }=\frac{P}{F}=\frac{75000}{600}=125\text{m/s}.$ Insight: top speed power ÷ resistance se set hoti hai, sirf force ke size se nahi. (Haqeeqat mein drag speed ke saath badhti hai, jo ise bahut kam rakti hai — lekin di gayi constant resistance ke saath, $125$ m/s answer hai. Mass $1500$ kg ek distractor hai: yeh steady-speed power balance mein kabhi nahi aata.)

(a) Energy in kWh — power kW mein aur time hours mein rakho taaki units kWh hi bane: $E=P\times t=3\text{kW}\times (5\times 30)\text{h}=3\times 150=450\text{kWh}.$ Joules mein, $1\text{kWh}=3.6\times 10^6$ J use karte hue: $E=450\times 3.6\times 10^6=1.62\times 10^9\text{J}.$ (b) Cost: $\text{cost}=450\times 8=₹3600.$

Calculus kyun: hamen $P=Fv$ chahiye, aur $F$ aur $v$ dono time ke saath change hote hain, isliye differentiate karna hoga. Velocity position ke change ki rate hai (Velocity and instantaneous rate (Kinematics) dekho); acceleration velocity ke change ki rate hai.

• Velocity: $v=\frac{dx}{dt}=3t^2$. $t=2$ par: $v=3(4)=12\text{m/s}$.
• Acceleration: $a=\frac{dv}{dt}=6t$. $t=2$ par: $a=12{\text{m/s}}^2$.
• Force: $F=ma=2\times 12=24\text{N}$.

Force aur velocity same direction mein hain (dono $+x$ ke along), isliye $\theta =0$: $P=Fv=24\times 12=288\text{W}.$

Gravity ki force $\vec{F}=mg=0.5\times 10=5\text{N}$ hai, neeche ki taraf. Velocity (upar positive): $v(t)=20-10t$.

(a) $t=1$ s par: $v=20-10=10\text{m/s}$ (upar). Force neeche hai, velocity upar, isliye $\theta =180^{\circ }$, $\cos 180^{\circ }=-1$: $P=Fv\cos 180^{\circ }=5\times 10\times (-1)=-50\text{W}.$ Negative power: gravity ball se energy nikal rahi hai jab yeh chadh rahi hai.

(b) $t=3$ s par: $v=20-30=-10\text{m/s}$ (yaani $10$ m/s neeche). Ab force aur velocity dono neeche hain, $\theta =0$: $P=Fv\cos 0^{\circ }=5\times 10\times 1=+50\text{W}.$ Positive power: gravity ab ball ko girtey waqt energy wapas de rahi hai. Power ka sign tumhe energy flow ki direction batata hai.

• $60$ W bulb se $60$ s mein kitni energy? ::: $3600$ J.
• $50$ N push $60^{\circ }$ par motion se, speed $2$ m/s, power? ::: $50$ W.
• Constant force ke liye rest se, $P_{\text{avg}}$ vs $P_{\text{peak}}$? ::: $P_{\text{avg}}=\frac{1}{2}{P}_{\text{peak}}$.
• Upar jaati ball par gravity ki power ka sign? ::: Negative (energy nikal rahi hai).
• Power aur resistance se top speed? ::: $v_{\max }=P/F_{\text{resist}}$.