Work — definition, dot product F·d, sign convention: Visual walkthrough

1.3.1 · D2 · Physics › Work, Energy & Power › Work — definition, dot product F·d, sign convention

Step 1 — "Push" kya hai aur "move" kya hai?

WHAT. Is kahani mein do alag cheezein hain. Pehli, ek force — ek push ya pull, jiske saath ek strength (kitna hard) aur ek direction (kis taraf) hota hai. Hum ise ek arrow ki tarah draw karte hain; arrow jitna lamba, push utna harder. Doosri, ek displacement — woh seedhi-line wali trip jo object ne actually start se finish tak ki, yeh bhi ek arrow hai: length = kitna door, direction = kis taraf khatam hua.

WHY. Do cheezein combine karne se pehle hume yaqeen karna hai ki woh do alag cheezein hain. Ek common trap yeh hai ki log sochte hain work sirf "force" hai, ya sirf "tum kitna chale" — yeh dono galat hai. Yeh dono ka ek milan hai, aur yeh milan unki relative direction pe depend karta hai. Isliye hume dono arrows ek saath table pe rakhne padte hain.

PICTURE. Blue arrow force $F$ hai. Green arrow displacement $d$ hai. Dhyan do ki dono ko tail-to-tail draw kiya gaya hai (ek hi dot se shuru ho rahe hain) — dono ke beech ka angle padhne ka yahi sahi tarika hai.

Figure — Work — definition, dot product F·d, sign convention

Step 2 — Hum simply $F$ times $d$ multiply kyun nahi kar sakte?

WHAT. Aao ek lazy guess test karte hain: "work $= F \times d$ ." Ek box ko seedha aage slide karo jabki use seedha upar push kar rahe ho. Woh 5 m aage move karta hai; tum 20 N upar push karte ho. Kya tumhare upar push ne use aage jane mein kuch bhi help ki?

WHY. Upar push aur aage ki trip bilkul alag directions mein point karti hain. Tumhara push motion mein "involve" nahi tha — box kabhi upar nahi gaya, isliye tumhari upar wali mehnat kabhi aage ki progress mein "cash in" nahi hui. Yeh lazy guess ko khatam karta hai: raw $F \times d$ direction ignore karta hai, aur direction hi sab kuch hai.

PICTURE. Blue force seedha upar point karta hai; green displacement seedha dahine point karta hai. Yeh right angles par hain ( $θ = 90°$ ). Force aage ki trip ke liye kuch nahi karta — object iske neeche se nikal jaata hai.

Step 3 — Force ko "along" aur "across" mein split karna

WHAT. Tilted force $F$ lo aur use do arrows mein kato jo milke wahi bante hain: ek displacement ke along (ise $F_{∥}$ , "F-parallel" kaho), ek iske across, right angle par ( $F_{⊥}$ , "F-perp"). Dono, tip-to-tail milke, $F$ ko exactly reproduce karte hain.

WHY. Humne abhi seekha ki sirf along-part trip par kharch hota hai. Toh poore tilted arrow se ladne ki jagah hum useful piece ko isolate karte hain. Do perpendicular pieces mein split karna sabse saaf split hai kyunki dono pieces kabhi interfere nahi karte — across-part aage ki progress mein literally zero contribute karta hai, jaisa Step 2 ne dikhaya.

PICTURE. Blue $F$ tilted draw kiya gaya hai. Motion ki green line par seedha uska shadow $F_{∥}$ (yellow) hai. Bacha hua, line se seedha upar khada, $F_{⊥}$ (red) hai. Teeno arrows ek right triangle banate hain: $F$ slanted side hai, $F_{∥}$ aur $F_{⊥}$ do legs hain.

Step 4 — Cosine kyun? Right triangle padhna

WHAT. Hume yellow arrow $F_{∥}$ ki length chahiye. Step 3 ke right triangle ko dekho. Poora force $F$ hypotenuse hai (sabse lamba, slanted side). Tail par angle $θ$ hai. Yellow leg $F_{∥}$ us angle ke bilkul saath baithti hai — yeh adjacent side hai.

WHY cosine aur sine nahi? Yeh tool choice hai, clearly batata hoon. Kisi bhi right triangle mein, ratio $\frac{adjacent}{hypotenuse}$ ko $cos θ$ kaha jaata hai — yahi cosine ki definition hai. Yeh exactly woh sawaal answer karta hai jo hum pooch rahe hain: "poore slanted arrow mein se kitna fraction base ke along hai?" Hume along leg chahiye, aur along leg adjacent leg hai, isliye tool cosine hai. (Sine across leg $F_{⊥}$ deta, jise hum discard kar rahe hain.)

PICTURE. Triangle labelled hai: hypotenuse $= F$ , adjacent leg (along motion) $= F_{∥}$ , opposite leg (across) $= F_{⊥}$ , angle $= θ$ . Cosine ratio adjacent leg par likhi gayi hai.

Step 5 — Worker ko trip se multiply karna

WHAT. Work hai "along-motion force, poori trip mein kharch." Toh worker piece $F_{∥}$ ko trip length $d$ se multiply karo: $W = F_{∥} \cdot d = (F cos θ) d = F d cos θ .$

WHY. Sirf force energy nahi hai — ek force jo wall se chipki ho aur kabhi move na ho, kuch nahi deliver karta. Energy sirf tab deliver hoti hai jab object us force ke neeche travel karta hai. Toh hum useful force lete hain aur use travelled distance par stretch karte hain. Term by term: $F$ (kitna hard) $\times$ $d$ (kitna door) $\times$ $cos θ$ (push ka kitna fraction sahi taraf point karta tha).

PICTURE. Ek shaded rectangle: iska height yellow $F_{∥}$ hai, iska width green trip $d$ hai. Iska area work $W$ hai. Bada force, lambi trip, ya seedha alignment (bada $cos θ$ ) → bada rectangle → zyada work.

Step 6 — Sign $cos θ$ ke andar rehta hai: teeno cases

WHAT. Jaise hum $F$ ko ghoomayenge, $cos θ$ sign badlega, aur work bhi. Hume har direction check karni chahiye, sirf friendly forward wali nahi.

WHY. Ek formula jise tum sirf ek case mein test karo woh ek trap hai. Cosine chhote angles par positive hai, right angle par zero, right angle ke baad negative — aur har ek ek real physical story se correspond karta hai (speed up / sirf steering / slow down). Ek case miss karo aur test mein sign galat ho jaayega.

PICTURE. Teeno panels ek hi green displacement share karte hain.

Left ( $0° \leq θ < 90°$ ): force aage lean karta hai, $F_{∥}$ trip ke saath point karta hai, $cos θ > 0$ → $W > 0$ , energy add hui.
Middle ( $θ = 90°$ ): force bilkul across hai, $F_{∥} = 0$ , $cos 90° = 0$ → $W = 0$ , sirf steering.
Right ( $90° < θ \leq 180°$ ): force peechhe lean karta hai, $F_{∥}$ trip ke against point karta hai, $cos θ < 0$ → $W < 0$ , energy remove hui.

Formula Sign summary

Angle $θ$	$cos θ$	$W$	Story
$0° \leq θ < 90°$	$+$	positive	force isse speed up karta hai (KE gain)
$θ = 90°$	$0$	zero	force sirf isko turn karta hai
$90° < θ \leq 180°$	$-$	negative	force isse slow down karta hai (friction, heat lost)

Step 7 — Degenerate cases (inhe kabhi skip mat karo)

WHAT. Formula ko uski edges par push karo, jahan ek arrow zero ho jaata hai.

WHY. Ek achha formula tab bhi sensibly behave karna chahiye jab koi input zero ho. Agar wahan nonsense deta, toh hum use kahin bhi trust nahi karte.

PICTURE. Do mini-panels:

No trip ( $d = 0$ ): object kabhi move nahi karta (wall ko push karna). $W = F \cdot 0 \cdot cos θ = 0$ . Step 5 ke rectangle ki width zero hai → zero area. Tum paseena baha sakte ho, par tumne koi physics work nahi kiya.
No force ( $F = 0$ ): ek coasting puck freely drift karta hai. $W = 0 \cdot d \cdot cos θ = 0$ . Zero height rectangle → zero area.

Recall

Zero-trip ya zero-force ::: dono mein $W = 0$ milta hai — work rectangle dono taraf se zero area tak collapse ho jaata hai. "No trip" case kaun si real situation hai? ::: ek immovable wall ko hard push karna — force bada hai, displacement zero hai.

Ek-picture summary

Sab ek saath: tilted blue force $F$ , uska yellow along-piece $F_{∥} = F cos θ$ green road $d$ par baitha hua, discarded red across-piece $F_{⊥}$ , aur shaded work rectangle jiska area = $F d cos θ$ hai. Blue arrow ko ghoomao aur dekho shaded area badhta hai, $90°$ par vanish hota hai, aur $90°$ ke baad negative ho jaata hai (road ke neeche flip karta hai).

Recall Feynman: poora walkthrough plain words mein

Do arrows draw karo: kitna hard push karte ho (blue) aur cheez kahan gayi (green). Sirf tumhari push ka woh slice jo trip ke saath same direction mein lean karta hai actually help karta hai — ise yellow slice kaho. Yellow slice ki length jaanney ke liye, apne push ka right triangle dekho: yellow slice woh side hai jo angle ke saath snug hai, aur "snug side over slanted side" woh cheez hai jo maths cosine kehta hai. Toh yellow slice $F cos θ$ hai. Ab woh helpful slice poori travelled distance par spread karo — $d$ se multiply karo — aur tumhe deliver ki gayi energy milti hai: $W = F d cos θ$ . Agar tumhari push trip ke saath lean karti hai toh energy add hoti hai (plus); seedha sideways kuch nahi add karta (zero); trip ke against lean karna energy churaata hai (minus). Aur agar kuch move nahi hota, ya tum push nahi karte, toh rectangle ka koi area nahi — bilkul bhi koi work nahi.

Connections

Dot Product (Scalar Product) — $F \cdot d$ exactly yahi along-fraction machinery hai.
Vectors & Components — Step 3 ka parallel/perpendicular split.
Kinetic Energy & Work-Energy Theorem — positive work wahan jaata hai jahan energy jaati hai.
Friction — classic negative-work force.
Potential Energy & Conservative Forces — jab work sirf start aur end par depend karta hai.
Power — same product, par per second: $P = F \cdot v$ .