1.3.1
Work ek concept ke roop mein kyun exist karta hai? Poore din wall ko dhakelnaa tumhe thaka deta hai, phir bhi kuch nahi hilta —
physics kehti hai tumne wall par zero work kiya . Work effort ya paseene ke baare mein nahi hai; yeh
ek force ke through displacement ke zariye energy ke transfer ke baare mein hai. Gehra idea yeh hai: force ka sirf woh hissa jo motion ke along point karta hai, object ki energy badalta hai. Jo bhi perpendicular hai, woh sirf steer karta hai — woh kabhi speed up ya slow down nahi karta.
Constant force F d scalar product (dot product) hota hai:
W = F ⋅ d = F d cos θ θ force aur displacement vectors ke beech ka angle hai.
Work ek scalar hai jiska SI unit joule hai ( 1 J = 1 N⋅m )
Symbols ka MATLAB kya hai?
F d displacement , na ki distance/path lengthθ
Humein kaise pata ki sirf along-motion wala hissa count hota hai? Force ko do hisson mein resolve karo:
ek parallel displacement ke, ek perpendicular usके. Perpendicular piece kabhi bhi object ko apni khud ki direction mein koi distance travel nahi kara sakta (object us taraf nahi jaata), isliye woh kuch contribute nahi karta. Sirf parallel piece displacement ke saath "saath chalta" hai.
Step 1 — Force ko Resolve Karo.
F d F ∥ = F cos θ F ⊥ = F sin θ Yeh step kyun? Kyunki energy transfer path ke along ek-dimensional hoti hai; decompose karna us hisse ko isolate karta hai jo actually object ke saath move karta hai.
Step 2 — Work ko (along-motion force) × (displacement) ke roop mein define karo.
W = F ∥ d = ( F cos θ ) d = F d cos θ Yeh step kyun? Perpendicular component F ⊥ apni direction mein zero displacement produce karta hai, isliye woh zero work karta hai. Sirf F ∥
Step 3 — Recognize karo ki yeh dot product hai.
Scalar product ki definition se hi, F ⋅ d = F d cos θ cos θ W = F ⋅ d = F x d x + F y d y + F z d z Yeh step kyun? Component form tumhe θ
W poori tarah cos θ
Worked example Pehle Forecast karo, phir Verify: chaar situationsForecast karo sign, answer padhne se pehle.
Girte hue seb par gravity → force neeche, motion neeche, θ = 0° cos 0 = + 1 + work ✓
Seedha upar phenke gaye ball par gravity → force neeche, motion upar, θ = 180° − work ✓
Circular motion mein ball par string ka tension → hamesha ⊥ θ = 90° zero work ✓
Sliding box par friction → motion ka virodh karta hai, θ = 180° − work ✓ (energy heat mein gayi)
Worked example Example 1 — angled pullEk box ko rope se d = 5 m 30° F = 20 N
W = F d cos θ = 20 × 5 × cos 30° = 100 × 0.866 = 86.6 J sin 30° cos 30° horizontal hai aur θ F F
Worked example Example 2 — vector componentsF = ( 3 ^ + 4 ^ ) N d = ( 2 ^ − ^ ) m W = F x d x + F y d y = ( 3 ) ( 2 ) + ( 4 ) ( − 1 ) = 6 − 4 = 2 J Component form kyun use karein? Humein angle kabhi nahi bataya gaya; dot product seedha answer de deta hai aur sign bhi automatically de deta hai.
Worked example Example 3 — bag ko horizontally carry karnaTum 10 m 2 kg Tumhare dwara bag par kiya gaya work?
Upar ki taraf holding force ≈ m g = 19.6 N vertical hai; displacement
horizontal hai, θ = 90° W = F d cos 90° = 0 J Thakane ke bawajood zero kyun? Thakaan biological hai (muscle fibres fire ho rahi hain), physics work nahi. Koi vertical displacement nahi ⇒ vertical force dwara koi work nahi.
Common mistake "Work mein distance use hoti hai, toh lamba path matlab zyada work."Kyun sahi lagta hai: zyada chalna zyada effort lagta hai. Fix: W force par project ki gayi displacement use karta hai, total path length nahi (jab tak force constant aur path ke along na ho). Gravity ke saath closed loop mein, displacement start par wapas aa jaati hai → net gravity work = 0
Common mistake "Agar main zyada push karun, toh main hamesha positive work karta hun."Kyun sahi lagta hai: badi force = bada effort. Fix: sign angle par depend karta hai, magnitude par nahi. Aage jaate cart par peeche push karo (θ = 180° negative work karte ho, chahe kitna bhi zyada push karo.
Common mistake "Jab force ek angle par ho tohsin θ Kyun sahi lagta hai: hum sin Fix: work ko displacement ke along component chahiye = F cos θ θ F d beech ka angle hai, isliye yeh hamesha cos
Recall Feynman: ek 12-saal ke bacche ko explain karoSocho tum ek toy car dhakeel rahe ho. Agar tum use usi direction mein push karo jis direction mein woh pehle se ja rahi hai , tum use tez jaane mein help karte ho — yeh positive work hai, tumne use energy di. Agar tum use rokne ke liye against push karo, yeh negative work hai, tumne energy le li. Agar tum use sideways push karo — sirf moda, speed up nahi ki — tumne koi work nahi kiya . Work bas "tumhari push ne moving cheez ko kitni energy di" hai. Kisi cheez ko still pakde rehna, chahe tumhare arms dard karein, use kuch nahi deta — isliye yeh zero work hai!
"Cos for the boss (force), along the road (displacement)."
Work sirf force ke us hisse ki parwah karta hai jo uss raaste ke along ho jis par tum chalte ho → hamesha cos θ PNZ : P arallel→Positive, ninety→Z ero, opposite→N egative.
Bag ko horizontally carry karna zero work kyun karta hai?
Kaunsa angle maximum positive work deta hai, aur tab W
Kya 1000 N
Constant force dwara kiya gaya work define hota hai Work ka SI unit aur uska base-unit equivalent joule, 1 J = 1 N⋅m
Force ka perpendicular component work mein kyun ignore kiya jaata hai woh apni direction mein koi displacement produce nahi karta, isliye koi energy transfer nahi hoti
Work positive hota hai jab force aur displacement ke beech ka angle 0° 90° cos θ > 0
Work zero hota hai jab force, displacement ke perpendicular ho (θ = 90°
Upar phenke gaye object par gravity dwara kiye gaye work ka sign negative (force neeche, motion upar, θ = 180°
Sliding body par kinetic friction dwara kiye gaye work ka sign negative (θ = 180°
F d W = F x d x + F y d y + F z d z
Work ek scalar kyun hai jabki do vectors involve hain do vectors ka dot product ek scalar produce karta hai
Uniform circular motion mein centripetal tension dwara kiya gaya work zero, kyunki tension hamesha velocity ke perpendicular hoti hai
positive speeds up, zero steers
Resolve force into components
Component form Fx dx + Fy dy + Fz dz
