1.3.2
Intuition The big pictureJab force constant hoti hai, toh work sirf W = F d change hoti hain jaise tum move karte ho: ek spring jitna zyada stretch karo utna harder pull karti hai, gravity altitude ke saath weak hoti hai, friction surface ke saath vary karti hai.
Trick yeh hai: path ko itne chote pieces mein chop karo ki har piece pe force practically constant ho. Har piece pe work ek chota rectangle F d x saare slivers add karo — yahi adding-up exactly woh hai jo ek integral karta hai.
Definition Work (general definition)Ek force F d r d W = F ⋅ d r A B total work in saari choti contributions ka sum hota hai:
W = ∫ A B F ⋅ d r
WHAT hai problem? W = F d F d F F ( x ) ek jhooth hai — sivaaye ek vanishingly small step pe.
WHY integration isko fix karta hai? Kyunki F ( x ) d x F ( x ) d x ∫ F ( x ) d x define hota hai un rectangles ko summing karne ki limit ke roop mein — yeh literally "F x
HOW hum ise scratch se derive karte hain? (1-D case, force along motion.)
Worked example 1 — Stretching a spring (Hooke's law)Ek spring F applied = k x 0 x 0 W = ∫ 0 x 0 k x d x Yeh step kyun? Force stretch ke saath linearly grow karta hai, toh hum F d W = k [ 2 x 2 ] 0 x 0 = 2 1 k x 0 2 Yeh answer sense kyun deta hai: stretch pe average force hai 2 0 + k x 0 = 2 1 k x 0 = 2 1 k x 0 ⋅ x 0 = 2 1 k x 0 2 F = k x triangle hai.
Worked example 2 — A position-dependent forceF ( x ) = ( 3 x 2 + 2 x ) N x = 1 x = 3 W W = ∫ 1 3 ( 3 x 2 + 2 x ) d x Yeh step kyun? F x = [ x 3 + x 2 ] 1 3 = ( 27 + 9 ) − ( 1 + 1 ) = 36 − 2 = 34 J Yeh limits kyun? Lower limit = start position, upper = end. Order matter karta hai: inhe swap karna sign flip kar deta hai (force negative work karta hai agar tum reverse karo).
Worked example 3 — Force given as a graphEk force dikhaye anusaar vary karta hai: yeh linearly 0 20 0 → 4 4 → 6 20 Graph kyun formula ki jagah? Work = area under the curve , toh bas geometric area compute karo.
Triangle: 2 1 ( 4 ) ( 20 ) = 40
Rectangle: ( 6 − 4 ) ( 20 ) = 40
Total W = 80 Yeh kyun kaam karta hai: area hi limit mein Riemann sum hai — same cheez hai jaise integrate karna.
Worked example 4 — When force opposes motion (negative work)Ek retarding force F = − b v F ( x ) = − 5 x N 0 → 2 W = ∫ 0 2 ( − 5 x ) d x = − 5 [ 2 x 2 ] 0 2 = − 5 ( 2 ) = − 10 J Negative kyun? Force poore time displacement ke opposite point karta hai, toh har F d x bahar jaati hai.
W = F d Yeh sahi kyun lagta hai: W = F d Yeh galat kyun hai: F initial ya final force se multiply karna saari beech ki values ko ignore karta hai.
Fix: Jaise hi F x v t W = ∫ F d x F d F
Common mistake Forgetting thecos θ Yeh sahi kyun lagta hai: 1-D problems mein force aur motion aligned hote hain, toh log angle drop kar dete hain.
Yeh galat kyun hai: d W = F ⋅ d r = F d r cos θ Fix: cos θ zero work karte hain.
Common mistake Mixing up "work by spring" vs "work to stretch spring"Yeh sahi kyun lagta hai: dono mein 2 1 k x 2 Yeh galat kyun hai: spring force − k x spring dwara kiya work − 2 1 k x 2 tumhara uske against kiya work + 2 1 k x 2 Fix: clearly batao ki tum kaunsa force integrate kar rahe ho.
Recall Feynman: explain it to a 12-year-oldSocho tum ek shopping cart push kar rahe ho, lekin floor jitna aage jaao utna stickier hoti jaati hai. Start mein easy hai, baad mein bahut hard push karna padta hai. Total effort nikalne ke liye tum ek push ko distance se multiply nahi kar sakte — push badhti rehti hai! Toh tum pretend karte ho ki tum ek tiny step at a time chalt ho, note karte ho ki us step pe kitna push kiya, tiny step se multiply karte ho, aur likh lete ho. Yeh har tiny step ke liye karo aur saari choti numbers add karo. "Push × tiny step" ka yeh bada addition exactly wahi hai jise mathematicians integral kehte hain, aur yeh total work deta hai. Yeh us picture ka area bhi hai jo tum draw karte agar "push" upar jaata aur "distance" across jaata.
"Slice it, push it, sum it."
Slice the path → push (force) over each slice → sum the slivers = ∫ F d x "Variable force? Reach for the integral sign."
What is the general definition of work for a variable force? W = ∫ A B F ⋅ d r , path pe
F ⋅ d r ka sum.
Why can't we use W = F d Kyunki F F d
How does the integral ∫ F d x Yeh Force vs position curve ke neeche ka area hai.
Derive the work to stretch a spring from 0 to x 0 W = ∫ 0 x 0 k x d x = 2 1 k x 0 2
What does the area under an F x Force dwara kiya gaya work.
For F ( x ) = 3 x 2 + 2 x x = 1 x = 3 W [ x 3 + x 2 ] 1 3 = 36 − 2 = 34
When is work done by a force negative? Jab force ka ek component displacement ke opposite ho (cos θ < 0
Why does a perpendicular force (e.g. normal/centripetal) do zero work? What is the Riemann-sum origin of W = ∫ F d x W = lim Δ x → 0 ∑ F ( x i ) Δ x i
Force varies with position F of x
Sliver work F of x times dx
Integral W = integral F dx
