Kinetic energy — derivation: Exercises — Hinglish deep dive

1.3.4 · D4 · Physics › Work, Energy & Power › Kinetic energy — derivation

Yeh page teen tools par tikaa hai jo tumhare paas pehle se hain:

$K = \frac{1}{2} m v^{2}$ — mass $m$ ki kinetic energy speed $v$ par.
$W_{n e t} = K_{f} - K_{i}$ — Work–Energy Theorem (net work, kinetic energy mein change ke barabar).
$W = F \cdot d$ — work ki definition (force times displacement ka along-motion part).

Intuition Work ke edge cases — jab

$W = 0$ ho ya $W$ ka sign palat jaaye Work $W = F d cos θ$ sirf "force thi" waali baat nahi hai. Teen degenerate cases dekho:

Perpendicular force ( $θ = 9 0^{\circ}$ ): $cos 9 0^{\circ} = 0$ , isliye $W = 0$ . Motion ke right angles par laga force (jaise horizontally sliding puck par gravity, ya ghoomti ball par string tension) koi work nahi karta — yeh direction badalta hai, speed nahi.
Zero displacement ( $d = 0$ ): ek wall ko push karna jo hilti nahi, $W = 0$ deta hai, chahe kitni bhi force lagao. Motion nahi, toh energy transfer nahi.
Opposing force ( $9 0^{\circ} < θ \leq 18 0^{\circ}$ ): $cos θ < 0$ , isliye $W < 0$ — energy nikal jaati hai (friction, braking). Agar koi force beech mein direction palat le (sign flip), toh uske do parts mein kiya work cancel hokar zero ya chota ho sakta hai.

Yaad rakho: kinetic energy tabhi badlti hai jab net work non-zero ho.

Level 1 — Recognition

Kya tum $K = \frac{1}{2} m v^{2}$ mein numbers daal sakte ho aur formula ko ulta padh sakte ho?

Recall Solution L1·1

KYA: definition seedha apply karo. KYUN: $m$ aur $v$ pata hai, kuch aur chahiye nahi. $K = \frac{1}{2} m v^{2} = \frac{1}{2} (4) (5)^{2} = \frac{1}{2} (4) (25) = 50 J$ Answer: $== 50 J ==$ .

Recall Solution L1·2

KYA: $K = \frac{1}{2} m v^{2}$ ko $v$ ke liye solve karo. KYUN: $v$ unknown hai, toh use isolate karo. $v = \frac{2 K}{m} = \frac{2 ( 64 )}{8} = 16 = 4 m/s$ Answer: $== 4 m/s ==$ .

Recall Solution L1·3

KYA: formula mein se units carry karo. KYUN: confirm hota hai ki formula dimensionally sahi hai. $[K] = kg \times (m/s)^{2} = kg \cdot \frac{m ^{2}}{s ^{2}} = J .$ Ek joule literally "kilogram-metre-squared per second-squared" hota hai.

Level 2 — Application

Ab kinetic energy ko Work–Energy Theorem mein daalo.

Recall Solution L2·1

KYA: $W_{n e t} = K_{f} - K_{i}$ use karo. KYUN: work poochha gaya, speeds di gayi hain — theorem inhe link karta hai. $W_{n e t} = \frac{1}{2} m (v_{f}^{2} - v_{i}^{2}) = \frac{1}{2} (1200) (2 0^{2} - 8^{2})$ $= \frac{1}{2} (1200) (400 - 64) = 600 \times 336 = 201600 J$ Answer: $== 201600 J \approx 2.02 \times 1 0^{5} J ==$ (positive — energy add hui).

Recall Solution L2·2

KYA: friction yahan ek hi horizontal force hai, isliye uska work $W_{friction}$ net work ke barabar hai, yaani $Δ K$ . KYUN: work–energy theorem phir se, lekin final speed chhoti hai. $W_{friction} = W_{n e t} = \frac{1}{2} (0.2) (2^{2} - 6^{2}) = \frac{1}{2} (0.2) (4 - 36) = 0.1 \times (- 32) = - 3.2 J$ Answer: $== - 3.2 J ==$ . Minus sign ka matlab hai energy puck se nikal gayi — friction motion ke opposite hai ( $θ = 18 0^{\circ}$ , isliye $cos θ = - 1$ ), toh uska work negative hai.

Recall Solution L2·3

KYA: box kinetic energy gain karta hai; woh work se aayi. KYUN: force unknown hai, distance aur speeds pata hain — work ko $W_{n e t} = Δ K$ se combine karo. Kyunki force yahan displacement ke saath point kar rahi hai, unke beech angle $θ = 0^{\circ}$ hai, isliye $cos θ = 1$ aur general work formula ek special case mein collapse ho jaata hai: $W_{n e t} = F s cos θ = F s cos 0^{\circ} = F s .$ (Yeh koi alag formula nahi hai — yeh $W = F d cos θ$ hai jahan $θ = 0^{\circ}$ hai. Agar force tilted hoti, toh $cos θ$ rakhte, bilkul L3·2 ki tarah.) Ab: $W_{n e t} = \frac{1}{2} (5) (1 0^{2} - 0) = 250 J, F = \frac{W _{n e t}}{s} = \frac{250}{4} = 62.5 N$ Answer: $== 62.5 N ==$ .

Level 3 — Analysis

Proportions, ratios, aur geometry ke baare mein sochna.

Recall Solution L3·1

KYA: $K \propto v^{2}$ use karke energies compare karo. KYUN: masses cancel ho jaate hain, toh sirf speed ratio bachta hai. $\frac{K _{B}}{K _{A}} = \frac{\frac{1}{2} m v _{B}^{2}}{\frac{1}{2} m v _{A}^{2}} = (\frac{v _{B}}{v _{A}})^{2} = 3^{2} = 9$ Answer: $== 9 ==$ . Kyunki braking friction stopping distance $d$ par fixed force karta hai (isliye $K = F d$ ), nau guna energy ke liye nau guna stopping distance chahiye. Figure s01 dekho: energy bars speed ke square ki tarah badhte hain.

Recall Solution L3·2

KYA: force ka sirf along-motion part work karta hai, $W = F d cos θ$ se. KYUN: dot product sirf horizontal component rakhta hai; vertical part motion ke perpendicular hai, isliye zero work karta hai. Figure s02 dekho. $W_{rope} = F d cos θ = 20 \times 10 \times cos 6 0^{\circ} = 200 \times 0.5 = 100 J$ Frictionless ice par rope hi ek force hai jo work kar rahi hai, isliye $W_{n e t} = 100 J$ . Yeh kinetic energy ban jaata hai: $100 = \frac{1}{2} (3) v^{2} \Rightarrow v^{2} = \frac{200}{3} = 66.67 \Rightarrow v = 8.165 m/s$ Answer: $== v \approx 8.16 m/s ==$ .

Recall Solution L3·3

KYA: energies equal set karo aur speed ratio nikalo. KYUN: equal $K$ dono speeds ko unki masses ke through link karta hai. $\frac{1}{2} (2) v_{2}^{2} = \frac{1}{2} (8) v_{8}^{2} \Rightarrow 2 v_{2}^{2} = 8 v_{8}^{2} \Rightarrow \frac{v _{2}^{2}}{v _{8}^{2}} = 4 \Rightarrow \frac{v _{2}}{v _{8}} = 2$ Answer: $== 2 ==$ . Lighter mass ko same energy carry karne ke liye do guna fast chalna padta hai.

Level 4 — Synthesis

Kinetic energy ko doosre chapter tools ke saath chain karo.

Recall Solution L4·1

KYA: stored potential energy $U = m g h$ ball ke girne par kinetic energy mein convert hoti hai. KYUN: yeh energy conservation hai — lost height speed khareedti hai. Gravity ek hi acting force hai, isliye saara $U$ , $K$ ban jaata hai. $U = m g h = \frac{1}{2} m v^{2} \Rightarrow v = 2 g h = 2 (10) (5) = 100 = 10 m/s$ Answer: $== 10 m/s ==$ . Dhyan do mass cancel ho gaya — free fall mein saare objects same speed gain karte hain.

Recall Solution L4·2

KYA: average power = deliver ki gayi energy ÷ time. KYUN: engine ka saara work kinetic energy ban gaya. $K = \frac{1}{2} (1000) (3 0^{2}) = \frac{1}{2} (1000) (900) = 450000 J$ $P_{a v g} = \frac{K}{t} = \frac{450000}{12} = 37500 W = 37.5 kW$ Answer: $== 37500 W = 37.5 kW ==$ .

Recall Solution L4·3

KYA: $K$ ko momentum ke terms mein likhna. KYUN: momentum $p = m v$ diya hai, speed nahi, isliye $v = p / m$ use karke $K$ express karo. $K = \frac{1}{2} m (\frac{p}{m})^{2} = \frac{p ^{2}}{2 m}$ $K_{X} = \frac{1 2 ^{2}}{2 ( 2 )} = \frac{144}{4} = 36 J, K_{Y} = \frac{144}{2 ( 6 )} = \frac{144}{12} = 12 J$ $\frac{K _{X}}{K _{Y}} = \frac{36}{12} = 3$ Answer: lighter cart X ki energy $== 3 ==$ guna zyada hai. Equal momentum ke liye, $K \propto 1/ m$ : lighter matlab faster matlab zyada energy.

Level 5 — Mastery

Multi-step problems jahan decide karna padta hai ki kaun sa tool use karein.

Recall Solution L5·1

KYA: ramp par potential energy $U = m g h$ kinetic energy ban jaati hai; floor par friction sab nikal deta hai. KYUN: poore trip mein energy accounting — final $K = 0$ . Bottom par kinetic energy (frictionless ramp par di gayi potential energy ke barabar): $K_{bottom} = U = m g h = 2 \times 10 \times 3 = 60 J$ Floor par friction negative work $W_{friction} = - f d$ karta hai jab tak block ruk na jaaye. Theorem deta hai $W_{friction} = K_{f} - K_{i} = 0 - 60$ : $0 - 60 = - f d \Rightarrow d = \frac{60}{f} = \frac{60}{4} = 15 m$ Answer: $== 15 m ==$ .

Recall Solution L5·2

KYA: force position ke saath vary karta hai, isliye work ek integral hai. KYUN: $W_{n e t} = \int F d x$ kisi bhi force history ko handle karta hai — exactly parent derivation ki variable-force branch. $W_{n e t} = \int_{0}^{4} 6 x d x = [3 x^{2}]_{0}^{4} = 3 (16) = 48 J$ Phir $W_{n e t} = \frac{1}{2} m v^{2}$ : $48 = \frac{1}{2} (3) v^{2} \Rightarrow v^{2} = 32 \Rightarrow v = 32 = 42 \approx 5.657 m/s$ Answer: $== v = 42 \approx 5.66 m/s ==$ . Figure s03 dikhata hai work ko force–position line ke neeche area ke roop mein.

Recall Solution L5·3

KYA: total mechanical energy $K + U$ (kinetic plus potential) constant rehti hai. KYUN: sirf gravity act karta hai, isliye Conservation of Mechanical Energy hold karta hai. Initial energy (sab kinetic, kyunki ground level se shuru hoti hai jahan $U = 0$ ): $E = \frac{1}{2} (1) (8^{2}) = 32 J$ (a) Top par, saari energy potential hai ( $K = 0$ ): $U = m g h_{ma x} = 32 \Rightarrow h_{ma x} = \frac{32}{1 \times 10} = 3.2 m$ (b) $h = 2 m$ par: $U = m g h = 1 \times 10 \times 2 = 20 J$ , isliye $K = 32 - 20 = 12 J$ : $\frac{1}{2} (1) v^{2} = 12 \Rightarrow v^{2} = 24 \Rightarrow v = 24 = 26 \approx 4.899 m/s$ Answers: $== h_{ma x} = 3.2 m ==$ , $== v = 26 \approx 4.90 m/s ==$ .

Exercises — Kinetic energy — derivation