Conservation of mechanical energy — derivation: Visual walkthrough

1.3.8 · D2 · Physics › Work, Energy & Power › Conservation of mechanical energy — derivation

Step 1 — "Energy" actually hai kya (do pockets)

KYA. Koi bhi algebra se pehle, un do quantities se milo jinhe hum track karte hain.

Kinetic energy $K$ — "moving money." Jo cheez move karti hai uske paas hoti hai; jo cheez rest mein hai uske paas bilkul nahi. Iska size hai $K = \frac{1}{2} m v^{2}$ , jahan $m$ hai kitna stuff hai (mass) aur $v$ hai kitni tez move kar raha hai (speed).
Potential energy $U$ — "stored money." Upar uthaya hua ball, ya dabaya hua spring, loaded hota hai — release karo aur woh move karega. Iska size position par depend karta hai, speed par nahi.

YEH DO KYUN. Yeh do hi tarike hain jisme ek mechanical system energy hold kar sakta hai: apni motion mein ya apni configuration mein. Niche sab kuch in do pockets ke beech bookkeeping hai.

PICTURE. Do jars. Abhi ball upar aur still hai: $U$ jar full hai, $K$ jar khali hai. Total liquid woh hai jo hum prove karenge ki kabhi nahi badlta.

Step 2 — "Work" kya measure karta hai (force times kitni door yeh aapke saath push karti hai)

KYA. Ek force $F$ jo tab act karti hai jab object ek tiny distance $d x$ move karta hai, woh ek tiny amount ka work karta hai $d W = F d x$ . Work woh currency hai jo do jars ke beech money move karti hai.

YEH QUANTITY KYUN. Humein ek aisa number chahiye jo kahe "is force ne motion ko kitna change kiya?" Jo force motion ke along push karti hai woh speed badhati hai; jo against push karti hai woh slow karti hai. $F d x$ exactly yahi capture karta hai: positive jab force aur motion agree karein, negative jab ladhein.

PICTURE. Ek blue arrow (force) jo ek block ko right side mein ek chhote gap $d x$ ke upar push kar raha hai. Shaded strip kaam ka ek sliver hai. Poore path mein saare slivers add karo → total work $W$ .

Step 3 — Work–energy theorem (net work moving jar ko bharta hai)

KYA. Hum dikhate hain: net work (saari forces add karke) kinetic energy mein change ke barabar hota hai.

Newton's second law se shuru karo — force acceleration cause karti hai: $F_{n e t} = ma = m \frac{d v}{d t} .$ Yahan $\frac{d v}{d t}$ hai kitni tez speed khud change ho rahi hai (acceleration). Ise work sliver mein dalo: $d W_{n e t} = F_{n e t} d x = m \frac{d v}{d t} d x .$

CHAIN-RULE SWAP KYUN. Right side abhi bhi time $t$ mention karta hai, lekin $K$ sirf speed $v$ ki parwah karta hai. Hum time hatana chahte hain. Kyunki ek tiny step speed times ek tiny time hota hai, $d x = v d t$ : $m \frac{d v}{d t} d x = m \frac{d v}{d t} (v d t) = m v d v .$ Do $d t$ cancel ho jaate hain — time gayab ho gaya, exactly jo hum chahte the, sirf $v$ bach gaya.

PICTURE. Speed ki ek staircase: jaise object $v_{1}$ se $v_{2}$ tak speed karta hai, har patli vertical strip $m v d v$ hai; line $y = m v$ ke niche poora triangle-ish area total hai, $\frac{1}{2} m v_{2}^{2} - \frac{1}{2} m v_{1}^{2}$ .

$W_{n e t} = \int_{v_{1}}^{v_{2}} m v d v = K_{2} \frac{1}{2} m v_{2}^{2} - K_{1} \frac{1}{2} m v_{1}^{2} = Δ K .$

Step 4 — Ek force conservative kab hoti hai (path matter nahi karta)

KYA. Ek conservative force woh work karta hai jo sirf shuru aur khatam hone ki jagah par depend karta hai, kabhi route par nahi. Equivalent roop se, shuru ki jagah wapas ek round trip mein zero net work hota hai.

YEH KYUN MATTER KARTA HAI. Agar work sirf do endpoints par depend karta hai, toh hum ise position se attached ek number ke roop mein bottle kar sakte hain — ek stored quantity. Woh bottled number potential energy hai. Path-dependent forces (friction) ko is tarah bottle nahi kiya ja sakta: jitni lambi road, utna zyada woh churaate hain.

PICTURE. Gravity ke under A se B tak do paths — ek seedha girna aur ek lazy curve. Gravity ka work dono ke liye identical hai (pink labels equal hain). Saath mein, ek friction path: lamba wiggly route (blue) zyada khota hai, toh friction conservative nahi hai.

Step 5 — Potential energy banana (conservative work ko bottle karna)

KYA. Ek conservative force ke liye hum uska potential energy $U$ define karte hain taaki woh work jo woh karta hai $U$ ki drop ke barabar ho: $W_{co n s} = - Δ U = - (U_{2} - U_{1}) .$

MINUS SIGN KYUN. Ball ko upar uthao: gravity niche khichti hai jabki ball upar jaati hai, toh gravity negative work karti hai — phir bhi ball ab zyada loaded hai, toh $U$ badha. Dono ke opposite signs hone chahiye, toh hum definition mein minus bake karte hain. Sanity check: ball girta hai ⇒ gravity $+$ work karti hai ⇒ $U$ girti hai. ✓ (Dekho Potential energy.)

Ab gravity ka $U$ concretely banao. Earth ke paas force hai $F = - m g$ (minus = "niche point karta hai"). Height $h_{1}$ se $h_{2}$ tak move karte hue: $W_{g r a v} = \int_{h_{1}}^{h_{2}} (- m g) d h = - m g (h_{2} - h_{1}) = - (U_{2} m g h_{2} - U_{1} m g h_{1}) .$ $W_{co n s} = - Δ U$ se match karo aur $U = m g h$ padho — derived, assumed nahi (Gravitational potential energy). Wahi recipe spring $F = - k x$ par $U = \frac{1}{2} k x^{2}$ deta hai (Spring potential energy (Hooke's law)).

PICTURE. Ek ball chadh rahi hai: down-arrow (gravity, negative work) ek barhte hue green $U$ column ke saath jo $m g h$ label hai. Opposite signs, ek see-saw ke roop mein draw ki gayi.

Step 6 — Do halvon ko ek saath snap karna

KYA. Suppose karo ki sirf conservative force work kar rahi hai. Tab Step 3 ka "net work" Step 5 ka "conservative work" hi hai: $Step 3 Δ K = W_{n e t} = W_{co n s} = Step 5 - Δ U .$

Sab kuch ek side move karo: $Δ K + Δ U = 0 ⟹ Δ (K + U) = 0.$

YEH POORA GAME KYUN HAI. $Δ (kuch bhi) = 0$ matlab woh kuch bhi change nahi hota. Toh $K + U$ frozen hai — yahi conservation of mechanical energy hai.

PICTURE. Wahi do jars, girne ke beech: liquid $U$ jar se seedha $K$ jar mein ghus raha hai, combined total ka level ek flat dashed line se mark kiya gaya jo kabhi move nahi karti.

Step 7 — Degenerate case: friction aata hai (leak)

KYA. Ab ek non-conservative force bhi act karne do — friction. Net work ab do parts mein hai: $W_{n e t} = W_{co n s} + W_{n c} .$ Ise work–energy theorem mein daalo: $Δ K = W_{co n s} + W_{n c} = - Δ U + W_{n c} \Rightarrow Δ (K + U) = W_{n c} .$

ISKO IGNORE KYUN NAHI KAR SAKTE. Friction hamesha motion ke opposite hota hai, toh $W_{n c} < 0$ : combined jar girti hai. Gaya hua liquid destroy nahi hota — woh heat ban gaya, jo $K + U$ ledger ke bahar rehta hai (dekho Energy lost to friction). Yahi exactly "no friction, no problem" caveat hai.

PICTURE. Wahi do jars, lekin ab niche se "heat" label ka ek chhota drip leak ho raha hai — flat total line dheere dheere niche jhukti hai.

Step 8 — Degenerate case: zero-work forces (kyun tension invisible hai)

KYA. Ek force bahut badi ho sakti hai phir bhi koi work nahi karta, agar woh motion ke perpendicular point kare. Work hai $d W = F d x$ motion ke along measure kiya gaya; ek sideways force kuch contribute nahi karti.

YEH KYUN MATTER KARTA HAI. Ek pendulum par string tension badi hoti hai, lekin woh hamesha string ke along point karti hai — bob ke swing ke perpendicular. Toh $W_{t e n s i o n} = 0$ , sirf gravity work karti hai, aur energy conserved rehti hai. Yahi hai jo ek pendulum par conservation use karne ki permission deta hai.

PICTURE. Ek swinging bob: velocity arrow (yellow) arc ke along, tension arrow (blue) right angle par pivot ki taraf point karta hua — chhota right-angle square unka $9 0^{\circ}$ dikhata hai, hence zero work.

Recall Quick checks

Ball height $h$ se rest se drop ki gayi: ground par speed? ::: $m g h = \frac{1}{2} m v^{2} \Rightarrow v = 2 g h$ Spring $x$ compress hoke block ( $m$ ) launch karta hai: speed? ::: $\frac{1}{2} k x^{2} = \frac{1}{2} m v^{2} \Rightarrow v = x k / m$ Ball $h$ se roll karke max $x$ tak spring compress karta hai: $k$ nikalo? ::: max compression par $K = 0$ , toh $m g h = \frac{1}{2} k x^{2} \Rightarrow k = 2 m g h / x^{2}$

Ek-picture summary

Upar sab kuch ek diagram hai: Newton's law → work–energy theorem $Δ K$ deta hai; $U$ ki definition $- Δ U$ deti hai; jab sirf conservative forces work karein toh woh same number hain, toh $K + U$ ek flat line hai. Friction add karo aur line jhuk jaati hai.

Recall Feynman: plain words mein poora walkthrough

Aap do pockets mein paisa carry karte ho — ek "moving" pocket aur ek "stored" pocket. Work hi ek tarika hai paisa apni moving pocket mein laane ka, aur Newton's law promise karta hai ki har bit of net work wahan exactly land karta hai (Step 3). Nice forces jaise gravity aur springs ke liye, humne notice kiya ki work sirf is par depend karta hai ki aap kahan jaate ho, kaise nahi — toh humne use stored money ke roop mein bottle kiya, ise minus sign ke saath define kiya taaki arithmetic line up ho (Step 5). Ab agar sirf woh nice forces aapko push karein, stored pocket se nikla hua paisa precisely wahi paisa hai jo moving pocket mein aa raha hai — aapka grand total kabhi nahi badlta (Step 6). Paisa khone ka ek hi tarika hai ek chor jiska naam friction hai, jo quietly kuch heat ke roop mein siphon kar leta hai jo aapki do pockets ke bahar rehti hai (Step 7). Aur dhyan raho: kuch bade-lagnewaale forces, jaise string ka pull, sideways push karte hain aur koi bhi paisa move nahi karte (Step 8). Koi chor nahi, koi sideways surprises nahi — total hamesha ke liye frozen.

Visual walkthrough — Conservation of mechanical energy — derivation