Visual walkthrough — Work done by variable force — integration

1.3.2 · D2 · Physics › Work, Energy & Power › Work done by variable force — integration

Step 1 — "Work" ka matlab jab push kabhi change nahi hoti

KYA. Hum sabse simple situation se shuru karte hain: ek constant push. Aap ek box ko ek steady force $F$ (newtons mein measure hoti hai, N — "kitna zyada push karte ho" ki unit) se ek distance $d$ (metres mein, m) ke across dhakelte ho. Work done defined hai: $W=F\times d.$

Main har piece ka naam theek wahin likhta hoon jahan woh hai:

• $F$ — steady push, poore raaste ek hi number (N).
• $d$ — box kitna door gaya jab aapne push kiya (m).
• $W$ — work, "deliver kiya gaya total effort" (unit: newton·metre = joule, J).

YEH PEHLE KYUN. Aage ki har cheez isi se bani hai. $W=Fd$ ka ek solid mental image chahiye hamare dimag mein, isse pehle ki hum $F$ ko wobble karne dein.

PICTURE. Push $F$ ko vertical axis par aur position $x$ ko horizontal axis par rakhein. Ek constant push ek flat horizontal line hoti hai. Work $W=F\times d$ exactly rectangle ka area hai neeche: height $F$, width $d$. Red rectangle ko dekho.

Step 2 — Masla: real forces move karte waqt change karti hain

KYA. Ab push ko position par depend karne do. Hum ise $F(x)$ likhte hain, padho "F of x" — matlab "jab box position $x$ par ho tab ki force." $x=1$ par force $5$ N ho sakti hai; $x=2$ par $9$ N ho sakti hai. Wohi box, alag jagah alag push.

Term-by-term:

• $x$ — box ki position uske path ke saath (m).
• $F(x)$ — push us position par evaluate ki gayi. Chota "$(x)$" multiplication nahi hai; matlab hai "is $x$ ke liye force dekho."

YEH KYUN MATTER KARTA HAI. Step 1 ke rectangle ko ek height chahiye. Lekin ab top edge slope karti hai aur curve karti hai — koi ek height nahi hai. Formula $W=Fd$ mein $F$ ke liye koi honest number nahi hai. Yeh toot gaya.

PICTURE. Wahi axes jaise pehle, lekin ab top edge ek rising curve hai. Apne aap se poocho: "ek curved roof ke neeche ka area kya hai?" Ek single rectangle ise cover nahi kar sakta — ya toh bahar nikalta hai ya gaps chhod deta hai (red mismatch dikhata hai).

Step 3 — Ek honest move: itna zoom in karo ki curve flat lage

KYA. Path ka ek tiny sliver chuno, $x$ se $x+\Delta x$ tak. Yahan $\Delta x$ (padho "delta x", the Greek capital delta) ka matlab bas $x$ mein ek choti si width hai — ek little step. Itne chote step par, curve muskil se hilti hai, isliye force practically ek number hai, $F(x)$.

Us single sliver par kiya gaya work ek thin rectangle hai: $\Delta W\approx F(x)\Delta x.$

Har symbol:

• $\Delta x$ — is ek step ki tiny width (m).
• $F(x)$ — step ke start par force, (almost) constant height ki tarah use hoti hai (N).
• $\Delta W$ — us step ko cross karte waqt kiya gaya thoda sa work (J).
• $\approx$ — "approximately equal," kyunki curve $\Delta x$ par perfectly flat nahi hai — lekin lagbhag hai.

YEH ALLOWED KYUN HAI. Yahi is poore chapter ka key idea hai: ek curve, kaafi zyada zoom karo, seedhi lagti hai. Itne narrow slice par, $F(x)$ muskil se change karta hai, isliye Step 1 ka rectangle us slice par legal hai. Humne ek impossible problem (curved roof) ko kai easy problems (tiny flat-topped rectangles) se trade kar liya.

PICTURE. Curve ke neeche ek single thin strip mein zoom karo. Uski top almost flat hai; yeh genuinely ek chota rectangle hai height $F(x)$ aur width $\Delta x$ ke saath (red mein dikhaya gaya hai). Uska area us step ke liye work hai.

Step 4 — Poore path ko slivers se tile karo aur add karo

KYA. Step 3 har jagah karo. Path ko start $x_A$ se end $x_B$ tak $N$ slivers mein kaato. Unhe $i=1,2,\dots ,N$ number karo. Sliver $i$ ki width $\Delta x_i$ aur force $F(x_i)$ hai, toh uska work $F(x_i)\Delta x_i$ hai. Har sliver add karo: $W\approx {\sum }_{i=1}^{N}F(x_i)\Delta x_i.$

Naye symbols padhna:

• $x_A,x_B$ — box ki start aur end positions (m).
• $N$ — hum path ko kitne slices mein kaatein (ek plain counting number).
• $i$ — ek slice ka label, $1$ se $N$ tak.
• $\sum _{i=1}^N$ — summation sign; "daayein wali cheez ko $i=1$ ke liye add karo, phir $2$, $N$ tak."
• $F(x_i)\Delta x_i$ — $i$-th rectangle ka area: uski height times uski width.

KYUN. Total work total effort hai, aur effort additive hai: poora path karna matlab sliver 1 karo, phir sliver 2, phir sliver 3… Toh hum bas sliver-rectangles add karte hain. Is total ko Riemann sum kehte hain — curve ko hug karne wale rectangles ka ek staircase.

PICTURE. Curve ke neeche region fill karne wale rectangles ka ek staircase. Har red step ek $F(x_i)\Delta x_i$ hai. Saath mein woh true curved area ko approximate karte hain — lekin aap phir bhi staircase aur curve ke beech chote triangular gaps dekh sakte ho.

Step 5 — Slivers ko zero tak shrink karo: staircase curve ban jaata hai

KYA. Step 4 ke woh chote gaps error hain. Unhe zyada, patley slivers use karke khatam karo: $N\to \infty$ hone do (infinitely many) taaki har $\Delta x_i\to 0$ ho jaaye (infinitely thin). Us limit mein staircase curve par perfectly press karta hai aur sum definite integral ban jaata hai: $W={\lim }_{\Delta x_i\to 0}{\sum }_{i=1}^{N}F(x_i)\Delta x_i={\int }_{x_A}^{x_B}F(x)dx.$

Integral ko piece by piece decode karo — yeh summation hai, transform hua:

• $\int$ — Sum ke liye ek stretched "S"; $\sum$ ki limit.
• $x_A$ (bottom) aur $x_B$ (top) — path ka start aur end; yeh $i=1$ aur $i=N$ ki jagah lete hain.
• $F(x)$ — height, exactly waise hi jaise pehle.
• $dx$ — infinitely thin width, jo $\Delta x_i$ limit mein bana.
• Put together: "continuously $x_A$ se $x_B$ tak (height $F(x)$) × (infinitesimal width $dx$) add karo."

YEH TOOL KYUN AUR KOI NAHI. Humein ek machine chahiye thi jo infinitely many infinitely small products sum kare. Ordinary multiplication ($Fd$) nahi kar sakta — woh ek height use karta hai. Ordinary addition ($\sum$) gaps chhod deta hai. Definite integral defined hai exactly is limit-of-a-sum ki tarah, isliye yeh exactly woh tool hai jo problem maangti hai. (Iska poora machinery Area under curves and the definite integral mein hai.)

PICTURE. Left half: visible gaps wala ek coarse staircase. Right half: bahut saare razor-thin red rectangles — gaps khatam ho gaye hain aur shape curve ke neeche smooth area hai. Staircase curve ka area ban gaya hai.

Step 6 — Jab force aur motion ek line mein nahi: $\cos \theta$ case

KYA. Ab tak push seedha motion ke saath point karti thi. Generally force $\vec{F}$ (ek arrow jisme size aur direction dono hain) aur tiny displacement $d\vec{r}$ (ek tiny move-arrow) ek doosre ke saath angle $\theta$ par ho sakte hain. Tab ek single sliver ka work hai: $dW=\vec{F}\cdot d\vec{r}=Fdr\cos \theta .$

Term-by-term:

• $\vec{F}$ — force arrow (upar chota arrow matlab "isme ek direction hai").
• $d\vec{r}$ — tiny displacement arrow, length $dr$.
• $\theta$ — do arrows ke beech ka angle.
• $\cos \theta$ — force ka woh fraction jo motion ke saath point karta hai (dekho Dot product and components of vectors).
• $\vec{F}\cdot d\vec{r}$ — dot product, jo built hai motion-ke-saath-waala part nikalne ke liye.

$\cos \theta$ KYUN. Push ka sirf woh hissa jis direction mein aap actually move karte ho work karta hai. Cart ko push karo aur thoda upar bhi? Upar wala hissa kahi useful nahi jaata. $\cos \theta$ exactly "force kitna forward point karta hai" ka dial hai: seedha aage push $\Rightarrow \cos 0^{\circ }=1$; sideways push $\Rightarrow \cos 90^{\circ }=0$; peeche push $\Rightarrow \cos 180^{\circ }=-1$.

PICTURE. Ek force arrow horizontal motion arrow ke saath angle $\theta$ par; uska shadow (red) motion line par daalo. $F\cos \theta$ length ka woh red shadow hi akela count karta hai.

Step 7 — Saare signs: forward, sideways, backward slivers

KYA. Sliver work $F(x)dx$ (ya $Fdr\cos \theta$) positive, zero, ya negative ho sakta hai, aur integral bas yeh signed slivers add karta hai. Teen cases:

Graph par yeh hai: $x$-axis ke upar area positive count hota hai, $x$-axis ke neeche negative.

TEEN KYUN COVER KARO. Reader ko koi aisa scenario nahi milna chahiye jise humne skip kiya. Ek retarding force jaise $F(x)=-5x$ (parent ke Example 4 se) har sliver ko negative sign deti hai, isliye uska integral negative hai — force energy nikalti hai. Travel direction ulti karna integration limits ko swap karta hai aur poora sign flip karta hai.

PICTURE. Ek graph jahan force curve axis ke neeche jaati hai. Axis ke upar area red-positive hai; axis ke neeche area hatched-negative hai. Net work = (positive area) − (negative area).

Step 8 — Spring: triangle ko appear hote dekhna

KYA. Ab machine ko real force par chalao. Ek spring jo aap stretch karte ho wapas kheenchti hai, aur ise hold karne ke liye aapko $F(x)=kx$ apply karna hota hai (from Hooke's law and spring potential energy). Yahan $k$ spring constant hai (stiffness, N/m) aur $x$ stretch hai. $0$ se $x_0$ tak stretch karne ka work: $W={\int }_0^{x_0}kxdx=k[\frac{x^2}{2}{]}_0^{x_0}=\frac{1}{2}k{x}_0^2.$

Term-by-term:

• $k$ — spring kitna stiff hai; bada $k$ = stretch karna zyada mushkil.
• $kx$ — stretch $x$ par force; slope $k$ ke saath origin se ek straight line.
• $\frac{1}{2}k{x}_0^2$ — answer; note karo yeh base $x_0$ aur height $k{x}_0$ ke triangle ka area hai: $\frac{1}{2}(\text{base})(\text{height})=\frac{1}{2}{x}_0(k{x}_0)$.

YEH EK CLEAN TRIANGLE KYUN HAI. Kyunki $F=kx$ ek straight slanted line hai, $0$ se $x_0$ tak uske neeche ka area ek right triangle hai — answer dekhne ke liye koi fancy integral nahi chahiye, aur integral use confirm karta hai.

PICTURE. Line $F=kx$ origin se upar jaati hai; neeche red triangle, base $x_0$, height $k{x}_0$. Uska area $=\frac{1}{2}k{x}_0^2$ work hai.

Ek-picture summary

Sab kuch ek canvas par: ek flat rectangle ($W=Fd$) curved roof par fail karta hai; hum curve ko thin red rectangles se tile karte hain; unhe zero tak shrink karo; sum-sign $\sum$ integral-sign $\int$ ban jaata hai; aur $F$–$x$ curve ke neeche final smooth red area hi work hai $W={\int }_{x_A}^{x_B}F(x)dx$.

Connections

• Parent topic — full note
• Work done by a constant force — flat-roof special case, Step 1.
• Area under curves and the definite integral — Step 5 ki limit-of-sums machinery.
• Hooke's law and spring potential energy — Step 8 ka spring triangle.
• Dot product and components of vectors — Step 6 mein $\cos \theta$ shadow justify karta hai.
• Work–Energy Theorem — yahi integral kinetic energy ke change ke barabar hai.
• Conservative forces and potential energy — jab yeh integral path-independent ho.

Concept Map

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