Work-energy theorem — derivation from Newton's second law: Visual walkthrough

1.3.3 · D2 · Physics › Work, Energy & Power › Work-energy theorem — derivation from Newton's second law

Hum shuru karte hain bilkul zero se — yahan tak ki "work" ka matlab bhi pehle nahi pata. Har symbol pehle kamaya jaata hai, phir use kiya jaata hai.

Step 1 — Hum maap kya rahe hain?

Figure — Work-energy theorem — derivation from Newton's second law

KYA: hum ek horizontal wire par bead set up karte hain taaki sab kuch ek seedhi line mein ho. Is tarah ek akela number ( $+$ ya $-$ sign ke saath) har quantity ko capture kar leta hai — abhi koi angles nahi.

KYU: poora theorem 1-D mein dekhna sabse aasaan hai. Ek baar dekh lene ke baad, 3-D version (Step 8) wohi story hai bas arrows ke saath.

PICTURE: bead position $x$ par baitha hai; orange arrow force $F$ hai; teal arrow velocity $v$ hai. Abhi dono ek hi direction mein point kar rahe hain (speed up ho raha hai).

Step 2 — Newton's law: ek cheez jo humein assume karni hai

Naya symbol padhte hain. $\frac{d v}{d t}$ ka matlab hai "time ke ek tiny sliver $d t$ mein velocity ka change $d v$ " — woh rate jis par speed change ho rahi hai. Yahi rate exactly acceleration kehlata hai.

KYU yeh tool? Humein koi na koi law chahiye jo force ko motion se connect kare, aur Newton's Second Law sabse gehri law hai jo hamare paas hai. Neeche sab kuch is ek equation se nikala gaya hai — hum kuch naya invent nahi karte.

PICTURE: bead ke do snapshots ek tiny time $d t$ apart. Velocity arrow thoda bada hua; woh thodi si growth $d v$ hai. Newton kehta hai force ki size set karti hai ki $d v$ per unit time kitna bada hai.

Step 3 — $d x$ multiply karke distance introduce karo

KYA: humne dono sides ko same tiny distance $d x$ se multiply kiya — ek legal move, jaise $3 = 3$ ke dono sides ko $5$ se multiply karna.

KYU: left side par ab $F d x$ hai, jo force over distance hai — work ka raw material. Hum deliberately $\int F d x$ ki taraf steer kar rahe hain.

PICTURE: bead thoda aage nudge hota hai ek sliver $d x$ se (plum). Us sliver ke dauran, force $F$ "work" ka ek patla rectangle banata hai — height $F$ , width $d x$ . Saare slivers add karo aur total work milta hai; yahi woh shaded strip hai.

Step 4 — Chain-rule swap: time khatam, distance raha

Yahan poori cheez ka clever core hai. Right side dekho: $m \frac{d v}{d t} d x$ . $d t$ aur $d x$ bas tiny numbers hain, toh hum inhe shuffle kar sakte hain:

KYU yeh aur kuch nahi? Hum $d t$ hatana chahte the. $d x$ ko $\frac{1}{d t}$ ke saath pair karke hum $\frac{d x}{d t} = v$ banate hain, aur time disappear ho jaata hai. Jo bacha, $v d v$ , sirf speed ki language mein baat karta hai — koi clocks nahi.

PICTURE: ise ek viewpoint change ki tarah socho. Left par humne poocha "speed har second kitni change hoti hai?" Right par hum poochh rahe hain "speed har metre mein kitni change hoti hai?" Figure usi badhti velocity arrow ko ek distance ruler ke against dikhata hai stopwatch ki jagah — metres exactly wahi hain jo work count karta hai.

Step 5 — Saare slivers add karo: net work define karo

KYA: hum Step 4 ki equation ke dono sides ko poore safar par integrate karte hain.

$\int_{x_{i}}^{x_{f}} F_{n e t} d x = \int_{v_{i}}^{v_{f}} m v d v$

KYU limits change hoti hain — change-of-variables rule. Yeh sloppy relabelling nahi hai; yeh ek genuine substitution hai. Ise dhire padhо:

KYU (piling-up wala hissa): ek akela sliver $F d x$ use karne ke liye bahut chota hai. Real quantity — poore path par work — saare slivers ka stack hai, aur $\int$ woh machine hai jo inhe stack karta hai.

PICTURE: force graph. Har patla plum rectangle ek $F d x$ hai; total orange area $W_{n e t}$ hai. Yeh exactly Work done by a variable force ki picture hai — area, "force times distance" nahi, honest definition hai.

Step 6 — Right side evaluate karo: kinetic energy appear hoti hai

Humein $\int_{v_{i}}^{v_{f}} m v d v$ ki value chahiye. Kyunki $m$ ek constant hai (hamari Step 1 ki assumption), yeh summing ka hissa nahi hai — hum ise integral sign ke aage slide kar sakte hain:

$\int_{v_{i}}^{v_{f}} m v d v = m \int_{v_{i}}^{v_{f}} v d v .$

Yeh allowed kyu hai? Ek fixed number ko sum se bahar nikalna sirf distributive law hai: $ma + mb + \dots = m (a + b + \dots)$ . Agar $m$ bead ke saath move karte change hota, toh yeh andar trapped hota aur yeh step illegal hota — exactly isliye constant-mass assumption ne apni jagah banaayi. Ab Step 4 ka pocket note use karo ( $v d v$ se $\frac{1}{2} v^{2}$ banta hai):

Step 5 aur Step 6 ek saath rakhne par:

$W_{n e t} = \frac{1}{2} m v_{f}^{2} - \frac{1}{2} m v_{i}^{2} = Δ K ■$

PICTURE: parabola $K = \frac{1}{2} m v^{2}$ . Horizontal axis par $v_{i}$ aur $v_{f}$ mark karo; curve par do heights ke beech ka vertical gap $Δ K$ hai — aur woh gap Step 5 ke orange work-area ke barabar hai. Theorem yeh hai ki yeh do pictures same number hain. Kinetic Energy dekho yeh samajhne ke liye ki $K$ hamesha $\geq 0$ kyun hota hai.

Step 7 — Har case: signs, zero, aur ulta chalna

Boxed result ko saare scenarios mein survive karna chahiye. Ek figure par har ek check karte hain.

Case A — Force motion ke saath ( $F > 0$ , $+ x$ direction mein move): slivers $F d x$ positive hain ( $F > 0$ , $d x > 0$ ), isliye $W_{n e t} > 0$ , isliye $Δ K > 0$ : bead speed up hoti hai. (Left panel.)

Case B — Force motion ke against ( $F < 0$ , $+ x$ direction mein move): har sliver $F d x$ negative hai ( $F < 0$ , $d x > 0$ ), isliye $W_{n e t} < 0$ , isliye $Δ K < 0$ : bead slow down hoti hai. Yeh braking/friction hai. Work ka negative hona koi paradox nahi — force sirf kinetic energy cheen leta hai. (Middle panel.)

Case C — Constant velocity ( $Δ K = 0$ ): agar speed kabhi nahi badlti, toh positive aur negative slivers exactly cancel hone chahiye, isliye $W_{n e t} = 0$ chahe individual forces bahut kaam karte rahe. (Right panel: ek dragged block — tumhara push aur friction cancel ho jaate hain.)

Case D — Bead $- x$ direction mein move kar raha hai ( $v < 0$ ). Yeh woh case hai jo upar ke panels ne soch-samajhkar careful thought ke liye chhoda. Jab bead left slide karta hai, har step $d x < 0$ hota hai. Sliver $F d x$ ka sign ab do signs ka product hai:

Force bhi left point kar rahi hai ( $F < 0$ ) aur leftward motion ( $d x < 0$ ): $F d x = (-) (-) = +$ ⇒ positive work, bead leftward direction mein speed up hoti hai. Sahi — leftward-moving bead par leftward push use energise karna chahiye.

Force right point kar rahi hai ( $F > 0$ ) leftward motion ( $d x < 0$ ) ke against: $F d x = (+) (-) = -$ ⇒ negative work, bead slow down hoti hai. Sahi — yeh motion ka oppose kar raha hai.

Dhyan do ki $K = \frac{1}{2} m v^{2}$ mein $v^{2}$ use hota hai, isliye $5$ m/s speed se leftward-moving bead ka $K$ ek rightward bead ke same hi hai. Theorem travel ki direction ki parwah nahi karta — sirf $F d x$ ka sahi sign milna chahiye, jo hamesha hota hai kyunki $F$ aur $d x$ dono apne signs carry karte hain.

Degenerate case — zero displacement ( $d x = 0$ ): agar bead hilti nahi, toh har sliver ki width $0$ hai, isliye $W_{n e t} = 0$ chahe tum kitna bhi push karo. Wall ko push karna koi kinetic energy change nahi karta.

Degenerate case — rest par start aur end ( $v_{i} = v_{f}$ ): toh $Δ K = 0$ , isliye trip par total net work zero hai, chahe bead beech mein speed up aur slow down hoti rahi ho (pehle positive phir negative work).

Step 8 — 3-D version: arrows aur dot product

Real life mein motion space mein curve karti hai. Position ek vector $r$ ban jaata hai, velocity $v$ , force $F$ . Tiny step $d r$ hai. Lekin work ek single number rehna chahiye, isliye humein do arrows ko ek number mein multiply karne ka tarika chahiye — woh tool hai dot product $F \cdot d r$ .

Ab hum same chaar moves run karte hain pehle ki tarah, lekin har ek ka apna KYU hai kyunki arrows plain numbers se thoda alag behave karte hain:

Integrate karna — aur limits kahan jaati hain. Path par start point se end point tak har tiny piece stack karna: $\int_{start}^{end} F \cdot d r = \int_{v_{i}}^{v_{f}} m v \cdot d v = \frac{1}{2} m v_{f}^{2} - \frac{1}{2} m v_{i}^{2} = Δ K .$ KYU limits transfer hoti hain: bilkul Step 5 jaisi change-of-variables logic — left integral path along position par li gayi hai, isliye uske endpoints start aur end points hain; right integral velocity par hai, isliye uske endpoints velocities $v_{i}, v_{f}$ hain unhi do instants par. Final answer mein sirf speeds $v_{i}, v_{f}$ bachte hain kyunki $v \cdot v = v^{2}$ direction throw away kar deta hai.

PICTURE: ek curved path. Ek point par force arrow do pieces mein split hai — ek piece motion ke along (yeh work karta hai) aur ek piece cross mein (kuch nahi karta). Sirf along-piece dot product mein bachta hai — woh geometry jo 1-D mein sign quietly handle kar raha tha.

Ek-picture summary

Sab kuch ek canvas par: $F = ma$ (time-language) se shuru → $d x$ se multiply karo → chain-rule swap karke $v d v$ (distance-language) → integrate karo → left par force-area right par $K$ -gap ke barabar hai.

Recall Feynman retelling — ek 12-saal ke bachche ko batao

Socho tum ek bead ko ek wire par push kar rahe ho. Newton ka rule batata hai tumhara push bead ki speed ko har second kitna change karta hai — lekin tum jaanna chahte ho effect har metre ka, kyunki "work" metres ke baare mein hai. Toh tum ek tiny swap khelte ho: speed ko stopwatch ke against track karne ki jagah, tum ise ek ruler ke against track karte ho. Jab tum yeh swap honestly karte ho, toh messy "speed-change-per-second" ek tidy " $v$ times a nudge in $v$ " mein ban jaata hai. Saare tiny nudges stack karo (aur kyunki bead puri time same mass rakhti hai, tum mass ko cleanly bahar factor kar sakte ho) aur tumhe exactly one-half-mass-times-speed-squared milta hai — go-fast energy. Toh total pushing jo tumne poori distance par ki (tumhare force ke neeche shaded area) exactly bead ki go-fast energy ka change hai. Aage push karo, gain hota hai; peeche push karo, lose hota hai; side mein push karo, kuch nahi hota; chahe bead left slide kar rahi ho, plus-and-minus signs khud sort ho jaate hain. Yahi poora theorem hai — motion ke liye bookkeeping.

Recall checkpoints

Recall Self-test

Newton's law kaunsi language bolti hai, aur humein kaunsi chahiye? ::: Newton time bolti hai ( $d v / d t$ ); work ko distance chahiye ( $\int F d x$ ). $d x$ se multiply karna kya set up karta hai? ::: Left side $\int F d x$ ban jaata hai, net work ki definition. Kaunsi hidden assumption $m$ ko integral ke bahar nikalne deti hai? ::: Mass puri motion mein constant hai. Chain-rule swap kya eliminate karta hai? ::: Time — $\frac{d v}{d t} d x = v d v$ mein koi $d t$ nahi bachta. $\int F d x$ ko $\int m v d v$ mein change karte waqt limits ka kya hota hai? ::: Woh unhi do instants par speeds ban jaati hain: $x_{i} \to v_{i}$ , $x_{f} \to v_{f}$ . Ek bead $- x$ direction mein slide kar rahi hai ek leftward force ke under — kya work positive hai? ::: Haan: $F < 0$ aur $d x < 0$ se $F d x > 0$ milta hai, isliye kinetic energy gain hoti hai. Sideways force work kyun nahi karta? ::: $F \cdot d r = ∣ F ∣∣ d r ∣ cos 9 0^{\circ} = 0$ .

Connections

Parent topic — wo symbolic derivation jo yeh page illustrate karta hai.
Newton's Second Law — Step 2, humara ek assumed input.
Kinetic Energy — woh $\frac{1}{2} m v^{2}$ jo Step 6 mein emerge hota hai.
Work done by a variable force — Step 5, kyun work ek area hai, product nahi.
Power — woh rate $\frac{d W}{d t} = F \cdot v$ , agar tum time-clock rakhte ho drop karne ki jagah.
Impulse–Momentum Theorem — sibling derivation jo $d x$ ki jagah $d t$ se multiply karta hai.
Conservation of Mechanical Energy — theorem kya banta hai jab forces conservative hoti hain.