Work done by variable force — integration: Worked examples

1.3.2 · D3 · Physics › Work, Energy & Power › Work done by variable force — integration

Shuru karne se pehle, ek reminder un symbols ke baare mein jo hum use karte hain, taaki koi bhi pehli line se hi confused na ho:

Scenario matrix

Har variable-force work problem in cells mein se ek (ya inka blend) hoti hai. Right column us example ka naam batata hai jo use cover karta hai.

#	Case class	Isme tricky kya hai	Covered by
C1	Force formula se, poori tarah positive	routine integrate-and-plug	Ex 1
C2	Force poori tarah negative (motion ko oppose kare)	answer ka sign	Ex 2
C3	Force mid-path mein sign change kare	upar aur neeche ka area cancel hota hai	Ex 3
C4	Force sirf graph ke roop mein di gayi (piecewise)	geometric area compute karo, formula nahi	Ex 4
C5	Spring / Hooke's law — "spring dwara" vs "spring ke against"	opposite signs, $\frac{1}{2} k x^{2}$	Ex 5
C6	Reversed limits (peeche ki taraf move karna)	limits swap karne se sign flip hota hai	Ex 6
C7	Degenerate / limiting inputs ( $F = 0$ , zero displacement, $k \to \infty$ )	edge behaviour, sanity checks	Ex 7
C8	Angle par force — $cos θ$ dot product	sirf along-motion component kaam karta hai	Ex 8
C9	Real-world word problem (chipchipa floor)	words ko → integral mein translate karna	Ex 9
C10	Exam twist: inverse-square force to infinity	improper integral, $\infty$ par limit	Ex 10

Ab hum har cell ko hit karte hain.

C1 — Force formula se, poori tarah positive

Worked example Ex 1 — ek badhti hui force

$F (x) = (4 x + 3) N$ ek block ko $x = 0$ se $x = 5 m$ tak push karti hai, force hamesha motion ke saath. $W$ nikalo.

Forecast: Pehle guess karo — force $3$ N se shuru hoti hai aur $23$ N par khatam hoti hai. Average kahin $13$ N ke aas paas hai $5$ m par, toh roughly $65$ J expect karo. Yeh number yaad rakho.

$W = \int_{0}^{5} (4 x + 3) d x$ likho. Yeh step kyun? $F$ , $x$ par depend karta hai, toh ek " $F d$ " ek jhooth hai; hum slice-push-sum karte hain, jo hai hi integral.
Term-by-term antiderivative: $\int 4 x d x = 2 x^{2}$ , $\int 3 d x = 3 x$ . Toh $W = [2 x^{2} + 3 x]_{0}^{5}$ . Yeh step kyun? Antiderivative running-total-of-area function hai; ends par evaluate karne se net area milta hai — Fundamental Theorem of Calculus.
Plug karo: $(2 \cdot 25 + 3 \cdot 5) - (0) = 50 + 15 = 65 J$ .

Verify karo: Average force $= \frac{3 + 23}{2} = 13$ N (valid kyunki $F$ linear hai), times $5$ m $= 65$ J. Step 3 aur forecast se match karta hai. Units: N·m = J. ✓

C2 — Force poori tarah negative

C3 — Force mid-path mein sign change kare

Yahi woh case hai jo students miss kar dete hain: journey ke ek hisse mein force help karti hai, doosre mein hinder karti hai. Figure dekho — green area (helping) aur red area (hindering) partly cancel ho jaate hain.

Figure — Work done by variable force — integration

Worked example Ex 3 — pehle help, phir hinder

$F (x) = (6 - 3 x) N$ , body $x = 0 \to x = 4 m$ move karti hai. $W$ nikalo. Force ka sign kahan flip hota hai?

Forecast: $F = 0$ jab $6 - 3 x = 0 \Rightarrow x = 2$ . $x = 2$ se pehle, $F > 0$ (helps); baad mein, $F < 0$ (hinders). Dono areas almost cancel ho sakte hain — kuch chhota expect karo.

$W = \int_{0}^{4} (6 - 3 x) d x = [6 x - \frac{3}{2} x^{2}]_{0}^{4}$ . Yeh step kyun? Ek integral automatically dono signs handle karta hai — positive area axis ke upar, negative neeche, apne signs ke saath add hote hain. Split karne ki zaroorat nahi.
$= (24 - \frac{3}{2} \cdot 16) - 0 = 24 - 24 = 0 J$ . Yeh step kyun? Helping triangle aur hindering triangle yahan equal hote hain, toh net work exactly zero hai.

Verify — split karke check karo:

$0 \to 2$ : triangle $\frac{1}{2} (2) (6) = + 6$ J (axis ke upar, green).
$2 \to 4$ : triangle $\frac{1}{2} (2) (- 6) = - 6$ J (axis ke neeche, red).
Sum $= 0$ J. ✓ Step 2 se match karta hai, "areas cancel" reading confirm karta hai.

C4 — Force sirf graph ke roop mein di gayi

Worked example Ex 4 — picture se area padho

Ek force neeche diye graph ko follow karti hai: $x = 0 \to 2$ m se $10$ N par hold karti hai, phir $x = 5$ m par linearly $0$ tak ramp down hoti hai. $W$ nikalo.

Forecast: Koi formula nahi diya — lekin work area hai, aur area hum rulers se compute kar sakte hain. Ek rectangle plus ek triangle expect karo.

Rectangle ( $0 \to 2$ ): $10 N \times 2 m = 20$ J. Yeh step kyun? Is stretch par $F$ constant hai, toh iska area simply length × height hai ( $F d$ wala special case).
Triangle ( $2 \to 5$ ): $\frac{1}{2} \times base \times height = \frac{1}{2} (3) (10) = 15$ J. Yeh step kyun? Linear ramp ek triangle enclose karta hai; ek straight line ka integral hai hi triangle ka area.
Total $W = 20 + 15 = 35$ J.

Verify karo: Formula ke roop mein, $2 \to 5$ par line hai $F = 10 - \frac{10}{3} (x - 2)$ , aur $\int_{2}^{5} F d x = 15$ J (average height $5$ N × base $3$ m). Rectangle add karo: $35$ J. ✓

C5 — Spring: "spring dwara" vs "spring ke against"

Worked example Ex 5 — do-sign wala spring

Ek spring ka $k = 200 N/m$ hai. Ise $x = 0$ se $x_{0} = 0.1 m$ tak stretch kiya jaata hai. (a) Tumhare dwara kiya gaya work jab tum ise kheenchte ho. (b) Spring ke dwara kiya gaya work.

Forecast: Yeh dono equal aur opposite hone chahiye — tum energy daalo, spring resist kare, use store kare. Dekho Hooke's law and spring potential energy.

Spring force hai $F_{spring} = - k x$ ( $x = 0$ ki taraf wapas kheenchti hai). Tumhe $F_{you} = + k x$ apply karna hoga use balance karne ke liye. Yeh step kyun? Kaun si force integrate karni hai yeh define karna — yahi poori ladaai hai yahan.
(a) $W_{you} = \int_{0}^{0.1} k x d x = \frac{1}{2} k x_{0}^{2} = \frac{1}{2} (200) (0.1)^{2} = 1 J$ . Yeh step kyun? Positive displacement ke saath positive force → positive work, spring energy ke roop mein store hota hai.
(b) $W_{spring} = \int_{0}^{0.1} (- k x) d x = - \frac{1}{2} k x_{0}^{2} = - 1 J$ . Yeh step kyun? Spring force stretch ko oppose karti hai, toh woh negative work karti hai.

Verify karo: $W_{you} = + 1$ J, $W_{spring} = - 1$ J — equal aur opposite jaise forecast tha, sum zero (agar dheere kheencho toh koi net kinetic gain nahi). Geometrically dono triangle $\frac{1}{2} x_{0} \cdot (k x_{0}) = \frac{1}{2} (0.1) (20) = 1$ J hain. ✓

C6 — Reversed limits (peeche ki taraf move karna)

Worked example Ex 6 — path par ulti disha mein chalna

Ex 1 se $F (x) = 4 x + 3$ use karo, ab block $x = 5$ se wapas $x = 0$ tak move karta hai. $W$ nikalo.

Forecast: Same force, opposite direction of travel — work Ex 1 ka negative hona chahiye, toh $- 65$ J.

$W = \int_{5}^{0} (4 x + 3) d x$ . Yeh step kyun? Motion ki direction limits ka order set karti hai; yahan hum $5$ se shuru karte hain aur $0$ par khatam.
Limits swap karne se sign flip hota hai: $\int_{5}^{0} = - \int_{0}^{5} = - 65 J$ . Yeh step kyun? $\int_{b}^{a} f = - \int_{a}^{b} f$ — definite integral ki ek definition hai, aur physically displacement $d r$ ab doosri taraf point karta hai, har $F d x$ ko flip karta hai.

Verify karo: Ex 1 ne $+ 65$ J forward diya; reverse karne par $- 65$ J milta hai. Dono ka sum $0$ hai — ek poore round trip mein position-only force ka zero net work hota hai. ✓

C7 — Degenerate & limiting inputs

Worked example Ex 7 — edge cases (inpar kabhi mat tripna)

Teen degenerate scenarios check karo.

Forecast: Har ek ko ek "obvious" answer par collapse karna chahiye; agar machine sahi hai, toh agree karti hai.

Zero force, $F (x) = 0$ kisi bhi path par: $W = \int_{a}^{b} 0 d x = 0$ J. Yeh step kyun? Koi push nahi ⇒ koi work nahi, distance chahe kuch bhi ho. $x$ -axis line ke neeche area zero hai.
Zero displacement, koi bhi $F$ , $a = b$ : $W = \int_{a}^{a} F d x = 0$ J. Yeh step kyun? Har slice ki width zero hai ⇒ koi area nahi. Tum ek wall pakad ke chaahe jitna zyada strain karo: zero mechanical work.
Stiff-spring limit: $W = \frac{1}{2} k x_{0}^{2}$ ke liye, energy $E$ fix rakho aur $k \to \infty$ hone do. Toh $x_{0} = 2 E / k \to 0$ . Yeh step kyun? Infinitely stiff spring muskil se stretch hoti hai; same energy store karne ke liye use almost koi displacement nahi chahiye. Consistent hai, paradoxical nahi.

Verify karo: (1) $0$ J, (2) $0$ J, (3) $x_{0} = 2 E / k \to 0$ as $k \to \infty$ . Teeno physical intuition ki demand ke hisaab se behave karte hain. ✓

C8 — Angle par force ( $cos θ$ )

Force ka sirf woh hissa jo motion ke saath ho woh work karta hai. Figure $F$ ko ek useful (along-path) piece aur ek waste (perpendicular) piece mein resolve karta hai — dekho Dot product and components of vectors.

Worked example Ex 8 — ek sled ko angle par kheenchna

Tum ek sled ko $d = 10$ m door kheenchte ho ek rope se jo horizontal se $θ = 6 0^{\circ}$ upar hai. Tension constant nahi hai: $T (x) = (2 x + 5) N$ jahan $x$ travel ki gayi distance hai. Motion horizontal hai. $W$ nikalo.

Forecast: Sirf $cos 6 0^{\circ} = \frac{1}{2}$ tension forward kheenchti hai; vertical part zero work karta hai. Toh straight pull se aadha expect karo.

General work: $d W = F \cdot d r = T (x) cos θ d x$ . Yeh step kyun? Dot product sirf force ka woh component rakhta hai jo displacement ke saath hai; $cos 6 0^{\circ} = \frac{1}{2}$ .
$W = \int_{0}^{10} (2 x + 5) (\frac{1}{2}) d x = \frac{1}{2} [x^{2} + 5 x]_{0}^{10}$ . Yeh step kyun? $cos θ$ yahan constant hai, toh woh integral ke baahir aa jaata hai.
$= \frac{1}{2} (100 + 50) = \frac{1}{2} (150) = 75 J$ .

Verify karo: Seedha ( $θ = 0$ ) pull $\int_{0}^{10} (2 x + 5) d x = 150$ J deta; $6 0^{\circ}$ par hume aadha milta hai, $75$ J, jaise forecast tha. Vertical component $0$ J karta hai (horizontal motion ke perpendicular hai). ✓

C9 — Real-world word problem

Worked example Ex 9 — zyada zyada chipchipa hota floor

Tum ek cart ko ek aisi floor par push karte ho jisme friction distance ke saath badhti hai: tumhe jo force apply karni hoti hai woh hai $F (x) = (8 + 2 x) N$ pehle $6$ m mein ( $x$ metres mein). Tum kitna work karte ho?

Forecast: Shuru mein aasaan ( $8$ N), ant mein mushkil ( $20$ N). Average karib $14$ N over $6$ m ⇒ roughly $84$ J.

Words ko → integral mein translate karo: $W = \int_{0}^{6} (8 + 2 x) d x$ . Yeh step kyun? "Force position ke saath change hoti rehti hai" yeh literal signal hai $\int F d x$ reach karne ka (parent-note mnemonic: variable force? integral sign pakdo).
$= [8 x + x^{2}]_{0}^{6} = 48 + 36 = 84 J$ . Yeh step kyun? $8$ ka antiderivative $8 x$ hai; $2 x$ ka $x^{2}$ hai; ends par evaluate karo.

Verify karo: Linear force ⇒ average $= \frac{8 + 20}{2} = 14$ N, times $6$ m $= 84$ J. Forecast se match karta hai. Units N·m = J. ✓

C10 — Exam twist: force to infinity (improper integral)

Worked example Ex 10 —

$1/ x^{2}$ ke against infinity tak escape karna Ek repulsive force $F (x) = \frac{c}{x ^{2}}$ jisme $c = 8 N\cdotm^{2}$ hai, ek particle ko $x$ ke saath outward push karta hai. Is force dwara kiya gaya work nikalo jab particle $x = 2$ m se $x = \infty$ tak move karta hai.

Forecast: Force tezi se weak hoti hai ( $\propto 1/ x^{2}$ ), toh chahe distance infinite ho, curve ke neeche ka area finite ho sakta hai. Ek modest number expect karo.

$W = \int_{2}^{\infty} \frac{c}{x ^{2}} d x$ . Yeh step kyun? Ab bhi slice-push-sum hai, lekin upper limit infinite hai — ek improper integral, ise limit lekar handle karte hain.
$x^{- 2}$ ka antiderivative $- x^{- 1}$ hai: $W = c [- \frac{1}{x}]_{2}^{\infty} = c (lim_{b \to \infty} (- \frac{1}{b}) - (- \frac{1}{2}))$ . Yeh step kyun? $- 1/ x$ running-area function hai; top end ko limit ke roop mein evaluate karte hain.
Jaise $b \to \infty$ , $- 1/ b \to 0$ . Toh $W = c (0 + \frac{1}{2}) = \frac{c}{2} = \frac{8}{2} = 4 J$ . Yeh step kyun? Limit mein tail kuch contribute nahi karta; poora infinite journey sirf $4$ J cost karta hai.

Verify karo: $\int_{2}^{\infty} 8 x^{- 2} d x = 8 \cdot \frac{1}{2} = 4$ J — infinite range ke bawajood finite, exactly jaise forecast tha. Yahi mechanism hai Conservative forces and potential energy mein escape energy ke peeche. ✓

Recall Quick self-test (answers dhako)

Ex 3 net work, aur kyun? ::: $0$ J — equal helping aur hindering triangles cancel ho jaate hain. Ex 6 aur Ex 1 mein kya relation hai? ::: Reversed limits sign flip karte hain: $- 65$ J vs $+ 65$ J. Ex 8 mein $\frac{1}{2}$ ka factor kyun? ::: $cos 6 0^{\circ} = \frac{1}{2}$ ; sirf along-motion component kaam karta hai. Ex 10 ka answer finite kyun hai? ::: Force $\propto 1/ x^{2}$ itni tezi se decay karti hai ki infinite-range area converge kar jaata hai. Zero displacement par work? ::: Hamesha $0$ J — har slice ki width zero hoti hai.

Connections

Parent: Work done by a variable force — woh machine jin examples ko exercise karti hai.
Work done by a constant force — $F =$ const cell (Ex 4 ka rectangle piece).
Hooke's law and spring potential energy — Ex 5.
Work–Energy Theorem — upar ke kisi bhi $W$ ko kinetic energy mein change mein convert karo.
Area under curves and the definite integral — graph/area reading (Ex 3, 4).
Conservative forces and potential energy — Ex 10 ka improper integral ek escape-energy calculation hai.
Dot product and components of vectors — Ex 8 mein $cos θ$ .

Worked examples — Work done by variable force — integration