Work-energy theorem — derivation from Newton's second law: Worked examples

1.3.3 · D3 · Physics › Work, Energy & Power › Work-energy theorem — derivation from Newton's second law

Pehle: un sab integral symbols ka matlab kya hai?

Kisi bhi example se pehle, hume ek notation clearly samajhni hogi jo neeche har scenario mein reuse hoti hai. Parent note ne work teen tariyon se likha — $\int F \cdot d r$ , $\int F d x$ , aur $F d cos θ$ . Ek beginner ko dekhna chahiye kyun yeh sab same cheez hain.

Definition Vector line integral se

$\int F d x$ tak $d r$ kya hai? Jab particle apne path par chalta hai, $d r$ ek tiny displacement step hai — ek chhoti si arrow jo us direction mein point karti hai jisme particle abhi ja raha hai, jis ki length us step mein cover ki gayi tiny distance ke barabar hai. Poore path ko lakho aisi tiny arrows mein tod lo.

Ek step ka work wahan ki force aur us tiny step ka dot product hai: $d W = F \cdot d r .$ Total work path par har step ko add karta hai — path-ke-saath-summation hi woh hai jo $\int F \cdot d r$ symbol matlab rakhta hai (yeh ek line integral hai: ek integral jo actual route par travel karte hue liya jaata hai).

Dot product ko components mein likhna. Agar $F = (F_{x}, F_{y}, F_{z})$ aur tiny step $d r = (d x, d y, d z)$ hai, to dot products ke component rule se: $F \cdot d r = F_{x} d x + F_{y} d y + F_{z} d z .$ Yahi bridge hai: abstract $F \cdot d r$ literally " $x$ mein force times $x$ mein step, plus $y$ aur $z$ ke liye waisa hi" hai.

1-D collapse. Agar motion sirf $x$ -axis ke along hai, to $d y = d z = 0$ aur sirf pehla term bachta hai: $F \cdot d r = F_{x} d x ⟶ W = \int F d x,$ jahan ab hum motion ki direction ke along (single) force component ke liye $F$ likhte hain. Isliye parent ka $\int F d x$ wahi cheez hai jaise $\int F \cdot d r$ — yeh bas 3-D form hai jisme teen mein se do pieces band ho gayi hain.

Angle form. $F \cdot d r = ∣ F ∣ ∣ d r ∣ cos θ$ use karte hue, aur tiny distance (ek positive scalar) ke liye $d s = ∣ d r ∣$ likhte hue aur force aur motion ke beech angle ke liye $θ$ : $d W = F cos θ d s ⟶ W = \int F cos θ d s .$ Agar additionally force constant hai aur uska angle $θ$ nahi badalta, to integrand $F cos θ$ ek fixed number hai aur sum se bahar aa jaata hai, jo chhodta hai: $W = F cos θ \int d s = F d cos θ,$ jahan $d = \int d s$ total distance travelled hai (ek single scalar magnitude, yahan is tarah define ki gayi hai taaki hum kabhi bare $d$ na likhen bina matlab ke). Yeh aakhri form familiar $W = F d cos θ$ hai — lekin ab aap dekh sakte hain yeh sabse special case hai, definition nahi.

Recall "Work" ke chaar chehere

Same quantity, chaar notations — situation ke hisaab se chuno. $\int F \cdot d r$ ::: general definition, koi bhi path, koi bhi geometry. $\int (F_{x} d x + F_{y} d y)$ ::: component form, useful jab aap $F$ aur $d r$ coordinate-wise jaante ho. $\int F d x$ ::: $x$ ke along 1-D motion; $F$ $x$ ke along force component hai. $F d cos θ$ ::: force constant AUR angle fixed ho — integral se bahar aa jaata hai.

"Net work" ka actually kya matlab hai?

Neeche har example mein $W_{n e t}$ hai. Aao exactly pin down karein yeh kya hai — kyunki is page par aadhe traps is ek symbol ko galat padhne se aate hain.

Definition Net work = sab forces ke works ka sum

Maan lo kuch forces $F_{1}, F_{2}, \dots, F_{n}$ ek particle par ek saath act karti hain (tumhara push, friction, gravity, normal force...). Zaroori baat yeh hai ki yeh sab same particle par same single path par act karti hain, isliye har ek tiny steps $d r$ ke same sequence par integrate hoti hai. Har ek apna work karti hai $W_{i} = \int F_{i} \cdot d r$ . Net work unka sum hai: $W_{n e t} \equiv \sum_{i} W_{i} = \sum_{i} \int F_{i} \cdot d r = \int (F_{n e t} \sum_{i} F_{i}) \cdot d r = \int F_{n e t} \cdot d r .$ Left se right padho: har force ka work add karna waisa hi hai jaise pehle forces ko ek net force $F_{n e t}$ mein add karo aur phir uska work nikalo. Yeh sirf isliye kaam karta hai kyunki path sab forces ke liye common hai — same route par travel karti cheezein ke sum ka integral un integrals ke sum ke barabar hota hai.

(Baad ke liye caveat: kuch forces ke liye — friction — work actual path par depend karta hai, sirf start aur end points par nahi; doosron ke liye — gravity, springs — sirf endpoints par depend karta hai. Woh path-dependence upar ke sum ko kabhi nahi todti, lekin yeh matter karta hai jab hum Conservation of Mechanical Energy se milte hain.)

To do equivalent recipes hain, aur hum woh use karte hain jo easier ho:

Recipe 1 (works add karo): $W_{1}, W_{2}, \dots$ alag alag compute karo, phir add karo — best jab aap har force jaante ho. (Ex 6, 7, 9 mein use hota hai.)
Recipe 2 (pehle forces add karo): $F_{n e t}$ nikalo, phir ek baar integrate karo — best jab forces simply combine hoti hain. (Kinematics checks mein use hota hai.)

Theorem phir puri tarah padhti hai: $W_{n e t} = \int F_{n e t} \cdot d r = Δ K = \frac{1}{2} m v_{f}^{2} - \frac{1}{2} m v_{i}^{2} .$

Scenario matrix

Har cheez jo ek work–energy problem aap par throw kar sakti hai in cells mein se kisi ek mein aati hai. Hum neeche har cell hit karenge.

#	Case class	Isme kya tricky hai	Example
A	Constant force, parallel, speeding up	"textbook" baseline	Ex 1
B	Constant force, opposing motion (negative work)	$F \cdot d r$ ka sign	Ex 2
C	Variable force (integrate karna zaroori)	$F d$ use nahi ho sakta	Ex 3
D	Force jo path ke beech mein sign badlti hai	work partly cancel ho sakta hai	Ex 4
E	Force jo motion se angle par hai	sirf parallel part kaam karta hai	Ex 5
F	Multiple forces — net work chahiye	ek force doosri ko cancel karti hai	Ex 6
G	Degenerate: zero net work lekin motion hoti hai	$Δ K = 0$ trap	Ex 7
H	Limiting / real-world word problem (stopping distance)	theorem se symbolically reason karo	Ex 8
I	Exam twist: curved path, neeche speed	direction + variable geometry combined	Ex 9

Prerequisites jinpar hum lean karte hain: Newton's Second Law, Kinetic Energy, Work done by a variable force, aur bilkul end mein Power relation.

Examples se pehle, ek reminder jo neeche har sign question ko unlock karta hai.

Figure padho (s01): same rightward displacement arrow (grey, labelled "step $d r$ ") ke teen copies. Left par, ek coral force arrow motion ke along flat lie karta hai ( $θ = 0^{\circ}$ ) — label padhta hai "positive W". Middle mein, ek butter-yellow force seedha upar khada hai, motion ke perpendicular ( $θ = 9 0^{\circ}$ ) — "zero W". Right par, ek mint force seedha peeche motion ke against point karta hai ( $θ = 18 0^{\circ}$ ) — "negative W". Har base par mint arc badhta hua angle dikhata hai. Jaise force flat se, upright par, reversed par swing karti hai, $cos θ$ $+ 1$ se $0$ se $- 1$ par slide karta hai, aur work ka sign uske saath chalta hai.

Figure — Work-energy theorem — derivation from Newton's second law

cos θ

sign

positive for

θ < 9 0^{\circ}

, zero at

9 0^{\circ}

, negative for

θ > 9 0^{\circ}

Example 1 — Cell A: constant force, speeding up

Worked example Baseline: frictionless ice par sled

Ek $4 kg$ ka sled rest se start karta hai. Ek constant force $12 N$ use frictionless ice ke $3 m$ par (motion ki direction mein) push karti hai. Final speed nikalo.

Forecast: Compute karne se pehle guess karo — kya answer $2$ , $3$ , ya $6 m/s$ ke aaspaas hoga? Likh lo.

Step 1. Motion $x$ ke along 1-D hai, isliye general $\int F \cdot d r$ collapse ho jaata hai $\int_{0}^{3} F d x$ par. Force constant hai ( $F = 12 N$ ) aur parallel hai ( $θ = 0$ , $cos θ = + 1$ ), isliye yeh bahar aa jaata hai: $W_{n e t} = \int_{0}^{3} F cos θ d x = F cos 0^{\circ} \int_{0}^{3} d x = 12 (1) (3) = 36 J .$ Yeh step kyun? Sirf ek force motion ke along act karti hai, isliye $W_{n e t} = W_{p u s h}$ . Kyunki woh force constant aur parallel hai, integrand ek fixed number hai aur line integral plain product $F d$ ban jaata hai distance $d = 3 m$ par — yeh ek hi situation hai jahan $W = F d$ puri $\int F \cdot d r$ ki jagah leta hai.

Step 2. $Δ K = W_{n e t} = 36 J$ . Kyunki $K_{i} = \frac{1}{2} (4) (0)^{2} = 0$ , hume $K_{f} = 36 J$ milta hai. Yeh step kyun? Theorem: net work hai kinetic energy mein change. Rest se start karna matlab sab kuch final $K$ ban jaata hai.

Step 3. $\frac{1}{2} (4) v_{f}^{2} = 36 \Rightarrow v_{f}^{2} = 18 \Rightarrow v_{f} = 18 = 32 \approx 4.24 m/s$ . Yeh step kyun? Speed recover karne ke liye $K = \frac{1}{2} m v^{2}$ invert karo.

Figure padho (s02, single panel): ek mint sled grey ice line par rest par ( $v_{i} = 0$ ); ek coral force arrow labelled $F = 12 N$ motion ke along point karta hai, aur ek grey displacement bracket $3 m$ path mark karta hai. Axis ke along small grey tick-arrows tiny steps $d r$ hain — force har ek ke parallel lie karta hai, isliye har step ka $cos θ = 1$ hai, jo geometrically $W = F d$ ka reason hai.

Verify: Kinematics se cross-check karo. $a = F / m = 12/4 = 3 m/s^{2}$ , aur $v_{f}^{2} = v_{i}^{2} + 2 a d = 0 + 2 (3) (3) = 18$ ✓. Units: $[J] = [N\cdotm] = [kg\cdotm^{2} / s^{2}]$ , isliye $2 K / m$ ki units $m/s$ hain ✓.

Example 2 — Cell B: constant force opposing motion

Worked example Sirf friction

Ek $1000 kg$ ki car $20 m/s$ par chal rahi hai aur ek rough patch se takraati hai jahan friction constant $5000 N$ hai jo backward direct hai. Yeh $30 m$ skid karti hai. Patch ke end par speed nikalo.

Forecast: Kya yeh $30 m$ se pehle ruk jaayegi, ya phir bhi roll kar rahi hogi? Guess karo.

Step 1. $W_{f r i c} = \int_{0}^{30} F cos 18 0^{\circ} d x = 5000 \times 30 \times (- 1) = - 150, 000 J$ . Yeh step kyun? Friction motion ke opposite point karta hai, isliye har tiny step par $θ = 18 0^{\circ}$ aur $cos θ = - 1$ . Force constant hai isliye integral se bahar aa jaata hai. Yeh ek hi horizontal force hai, isliye $W_{n e t} = W_{f r i c}$ , aur yeh negative hai — yeh kinetic energy remove karta hai.

Step 2. $K_{i} = \frac{1}{2} (1000) (20)^{2} = 200, 000 J$ . Yeh step kyun? Hume starting kinetic energy chahiye taaki dekh sakein kitni bachi hai.

Step 3. $K_{f} = K_{i} + W_{n e t} = 200, 000 - 150, 000 = 50, 000 J$ . Yeh step kyun? Rearranged theorem: $K_{f} = K_{i} + W_{n e t}$ . Kyunki $K_{f} > 0$ , car abhi bhi chal rahi hai.

Step 4. $\frac{1}{2} (1000) v_{f}^{2} = 50, 000 \Rightarrow v_{f}^{2} = 100 \Rightarrow v_{f} = 10 m/s$ .

Figure padho (s03): road line, car right ja rahi hai (grey velocity arrow, $v_{i} = 20 m/s$ ) jabki ek coral friction arrow left point karta hai, motion ke against. Yeh head-to-tail opposition $θ = 18 0^{\circ}$ visible hai — arrow har tiny step $d r$ se "fight" karta hai, jo bilkul wahi reason hai kyun uske work mein minus sign aata hai.

Verify: Kinetic energy quarter ho gayi, isliye speed half honi chahiye ( $1/4 = 1/2$ ): $20 \to 10$ ✓. Sign sanity: work negative ⇒ slower ⇒ braking intuition se match ✓.

Example 3 — Cell C: variable force (integrate karna zaroori)

Worked example Ek force jo position ke saath badhti hai

Ek force $F (x) = 8 x N$ (jahan $x$ metres mein hai), hamesha motion ke along point karti hai, ek $2 kg$ mass par act karti hai jo rest se start hai, $x = 0$ se $x = 3 m$ tak. Final speed nikalo.

Forecast: Kya $F (3) \times 3 = 24 \times 3 = 72 J$ use karna sahi hoga ya bahut bada hoga? Predict karo.

Step 1. Motion 1-D hai aur force uske along point karti hai, isliye $θ = 0$ aur $cos θ = 1$ har point par — general $\int F \cdot d r$ ban jaata hai $\int F (x) d x$ : $W = \int_{0}^{3} 8 x d x = [4 x^{2}]_{0}^{3} = 4 (9) - 0 = 36 J .$ Yeh step kyun? Yahan do cheezein hoti hain. Pehli, kyunki force hamesha motion ke parallel hai hum $cos θ = 1$ set kar sakte hain aur vector notation drop kar sakte hain — yahi hai jo $F \cdot d r$ ko $F d x$ mein turn karta hai. Doosri, force constant nahi hai, isliye $F cos θ$ ko bahar nahi nikala ja sakta; shortcut $F d$ illegal hai aur hume zaroori hai ise integral ke andar rakhna, variable-force rule ke mutabiq. ( $72 J$ guess karna overcount karta hai kyunki yeh final, sabse badi force har jagah use karta hai.)

Step 2. $Δ K = 36 J$ , aur $K_{i} = 0$ , isliye $K_{f} = 36 J$ . Yeh step kyun? Work–energy theorem variable forces par bhi unchanged apply hoti hai — integral hi net work hai.

Step 3. $\frac{1}{2} (2) v_{f}^{2} = 36 \Rightarrow v_{f}^{2} = 36 \Rightarrow v_{f} = 6 m/s$ .

Verify: Work $36 J$ line $F = 8 x$ ke neeche $0$ se $3$ tak triangle ka area hai: base $3$ , height $F (3) = 24$ , area $\frac{1}{2} (3) (24) = 36$ ✓ — bina calculus ke ek picture-check.

Example 4 — Cell D: force jo path ke beech sign badlti hai

Worked example Push phir pull

Ek $1 kg$ block, initially rest par $x = 0$ par, $F (x) = (4 - 2 x) N$ feel karta hai (hamesha $x$ -axis ke along directed) $x = 0$ se $x = 4 m$ tak. Notice karo $F > 0$ for $x < 2$ aur $F < 0$ for $x > 2$ . $x = 4$ par speed nikalo.

Forecast: Force pehle aadhe ke liye forward push karti hai aur doosre aadhe ke liye backward. Kya block chalte rehega, ya rest par wapas aa jaayega?

Step 1. Force $x$ -axis ke along hai, isliye phir $F \cdot d r = F (x) d x$ aur hum component ko directly integrate karte hain: $W = \int_{0}^{4} (4 - 2 x) d x = [4 x - x^{2}]_{0}^{4} = (16 - 16) - 0 = 0 J .$ Yeh step kyun? Kyunki force $x$ -axis ke parallel (ya anti-parallel) rehti hai, uska component signed value $4 - 2 x$ hi hai, isliye koi alag $cos θ$ factor nahi chahiye — $F (x)$ ka sign khud direction carry karta hai. Sign change ke through integrate karna automatic hai: $0 \to 2$ se positive work bilkul $2 \to 4$ se negative work ko cancel kar deta hai.

Step 2. $Δ K = 0 \Rightarrow K_{f} = K_{i} = 0 \Rightarrow v_{f} = 0$ . Yeh step kyun? Zero net work matlab zero change in kinetic energy. Yeh speed up hua phir slow down hoke rest par aa gaya.

Figure padho (s04): lavender line force $F (x) = 4 - 2 x$ hai; yeh $+ 4 N$ se start hoti hai aur slope down karti hai, $x = 2 m$ par zero cross karti hai (arrow "force flips sign at x = 2 m" se marked). Mint shaded triangle ( $0$ se $2$ tak) positive work hai, $+ 4 J$ ; coral triangle ( $2$ se $4$ tak) axis ke neeche hang karta hai aur negative work hai, $- 4 J$ . Dono triangles mirror images hain — equal size, opposite sign — isliye unke areas cancel hote hain aur net work exactly zero hai. Woh cancellation hi is example ki puri kahani hai.

Verify: Integral split karo. Forward part: $\int_{0}^{2} (4 - 2 x) d x = [4 x - x^{2}]_{0}^{2} = 8 - 4 = 4 J$ . Backward part: $\int_{2}^{4} (4 - 2 x) d x = [4 x - x^{2}]_{2}^{4} = (16 - 16) - (8 - 4) = - 4 J$ . Sum $= 0$ ✓.

Example 5 — Cell E: force angle par

Worked example Suitcase ko handle se kheenchna

Aap ek $10 kg$ ka suitcase smooth floor par $50 N$ force se kheenchte ho jo horizontal se $6 0^{\circ}$ upar directed hai, $4 m$ tak. Yeh rest se start karta hai. (Floor frictionless hai; tumhari pull ka vertical part reduced normal force se le liya jaata hai, jo koi work nahi karta kyunki koi vertical motion nahi hai.) Final speed nikalo.

Forecast: Tumhare $50 N$ ka sirf kuch hissa ise forward move karne mein madad karta hai. $50$ ke aadhe se zyaada ya kam?

Step 1. Displacement horizontal hai, force tilted hai. Component form $F \cdot d r = F_{x} d x + F_{y} d y$ use karte hue $d y = 0$ (koi vertical motion nahi) ke saath, sirf horizontal component $F_{x} = F cos θ$ bachta hai: $W = \int_{0}^{4} F cos θ d x = F d cos θ = 50 \times 4 \times cos 6 0^{\circ} = 200 \times 0.5 = 100 J .$ Yeh step kyun? Component form ise explicit banata hai: force ka vertical part ( $F_{y}$ ) $d y = 0$ ko multiply karta hai aur kuch contribute nahi karta, isliye sirf $F cos θ$ work karta hai. Force aur angle dono constant hain, isliye $F cos θ$ bahar aa jaata hai aur $F d cos θ$ deta hai. Yahi dot product apna kaam kar raha hai — projection ke liye figure dekho.

Step 2. $Δ K = 100 J$ , $K_{i} = 0 \Rightarrow K_{f} = 100 J$ .

Step 3. $\frac{1}{2} (10) v_{f}^{2} = 100 \Rightarrow v_{f}^{2} = 20 \Rightarrow v_{f} = 20 = 25 \approx 4.47 m/s$ .

Figure padho (s05): floor ke along grey arrow $4 m$ displacement hai. Coral arrow $6 0^{\circ}$ upar tilted (mint arc angle mark karta hai) poora $50 N$ pull hai. Uski tip se floor tak ek dashed vertical line daalo: lavender arrow ground ke along force ka shadow (projection $F_{x} = F cos 60$ ) hai motion ki direction par, aur uski length $25 N$ hai. Sirf woh shadow work karta hai — jitna zyaada tilt, utna chhota shadow, utna kam work.

Verify: Horizontal force $= 50 cos 60 = 25 N$ , isliye $a = 25/10 = 2.5 m/s^{2}$ ; kinematics $v_{f}^{2} = 2 a d = 2 (2.5) (4) = 20$ ✓. Sanity: $100 J$ exactly aadha hai jo straight $50 N$ pull ( $200 J$ ) deta, $cos 60 = 0.5$ se match ✓.

Example 6 — Cell F: multiple forces, NET work use karo

Worked example Friction ke against kheeencha gaya

Ek $6 kg$ ka crate rest par hai aur $40 N$ horizontal rope se $5 m$ par kheeencha jaata hai. Kinetic friction ise $16 N$ se oppose karti hai. Final speed nikalo.

Forecast: Applied work $40 \times 5 = 200 J$ hai. Kya final $K$ $200 J$ hogi ya kam?

Step 1. $W_{p u l l} = \int_{0}^{5} 40 d x = + 40 \times 5 = + 200 J$ (motion ke parallel, $cos 0 = 1$ ). Yeh step kyun? Rope motion ke along kheenchti hai, constant aur parallel, isliye uska work positive $F d$ hai.

Step 2. $W_{f r i c} = \int_{0}^{5} 16 cos 18 0^{\circ} d x = - 16 \times 5 = - 80 J$ (motion ko oppose karta hai). Yeh step kyun? Friction backward point karta hai, $cos 180 = - 1$ , negative work.

Step 3. $W_{n e t} = W_{p u l l} + W_{f r i c} = 200 + (- 80) = 120 J$ . Yeh step kyun? Yahan hum net-work definition se Recipe 1 use karte hain: har force ka work add karo. Theorem net work use karta hai — sab forces par sum, sirf jo aap apply karo us par nahi. (Gravity aur normal force vertical hain, motion horizontal hai, isliye unka $F \cdot d r = F_{y} d y = 0$ aur woh sum se bahar nikal jaate hain.)

Step 4. $Δ K = W_{n e t} = 120 J$ , aur $K_{i} = 0$ , isliye $K_{f} = 120 J$ ; phir $\frac{1}{2} (6) v_{f}^{2} = 120 \Rightarrow v_{f}^{2} = 40 \Rightarrow v_{f} = 40 = 210 \approx 6.32 m/s$ . Yeh step kyun? Theorem apply karo: net work kinetic energy mein change hai; rest se, $K_{f} = W_{n e t}$ , phir speed ke liye $K = \frac{1}{2} m v^{2}$ invert karo.

Verify: Recipe 2 se cross-check karo (pehle forces add karo): net force $= 40 - 16 = 24 N$ , $a = 24/6 = 4 m/s^{2}$ , $v_{f}^{2} = 2 a d = 2 (4) (5) = 40$ ✓ — dono recipes agree karti hain, jaise definition ne promise kiya tha.

Example 7 — Cell G: zero net work, phir bhi motion hai (degenerate)

Worked example Classic trap: constant velocity

Ek $20 kg$ ka box floor ke along $8 m$ constant velocity par horizontal force se kheeencha jaata hai. Friction $30 N$ hai. Nikalo (a) applied force, (b) aap jo work karte ho, (c) net work.

Forecast: Agar yeh chal raha hai, to surely net work positive hai... kya hai? Dhyan rakhna.

Step 1. Constant velocity ⇒ zero acceleration ⇒ net force $= 0$ , isliye applied force $=$ friction $= 30 N$ . Yeh step kyun? Newton's Second Law: $Δ v = 0 \Rightarrow a = 0 \Rightarrow F_{n e t} = 0$ .

Step 2. Tumhara work: $W_{y o u} = + 30 \times 8 = + 240 J$ . Friction ka work: $W_{f r i c} = - 30 \times 8 = - 240 J$ . Yeh step kyun? Aap forward push karte ho (positive), friction same distance par peeche push karta hai (negative), equal magnitudes.

Step 3. $W_{n e t} = W_{y o u} + W_{f r i c} = 240 - 240 = 0 J$ . Yeh step kyun? Works ka sum (Recipe 1) zero hona chahiye, kyunki $Δ K = \frac{1}{2} m (v_{f}^{2} - v_{i}^{2}) = 0$ jab speed nahi badlti. Theorem ne ise predict kar liya tha kuch bhi add karne se pehle.

Verify: $Δ K = 0$ (koi speed change nahi) aur $W_{n e t} = 0$ agree karte hain ✓. Lesson: "object chal gaya" tumhe net work ke baare mein kuch nahi batata; sirf speed mein change batata hai. Yahi degenerate cell hai.

Example 8 — Cell H: limiting / real-world (stopping distance)

Worked example Kyun braking distance speed ke saath quadruple hoti hai

Mass $m$ ki ek car constant friction force $f$ se brake karti hai. Algebraically dikhao ki stopping distance $d$ $v^{2}$ ke proportional hai, phir compute karo: agar $30 m/s$ ko rukne ke liye $45 m$ chahiye, to $60 m/s$ se kitni distance chahiye?

Forecast: Speed double karo — kya stopping distance double, triple, ya quadruple hogi?

Step 1. Rukna matlab $K_{f} = 0$ . Net work sirf friction hai: $W_{n e t} = - f d$ . Yeh step kyun? Sirf friction motion ke along act karti hai; yeh car ko rest par laati hai.

Step 2. Theorem: $- f d = 0 - \frac{1}{2} m v^{2} \Rightarrow d = \frac{m v ^{2}}{2 f}$ . Yeh step kyun? $d$ ke liye solve karo. $m$ aur $f$ fixed ke saath, $d \propto v^{2}$ — yahi key structural result hai.

Step 3. Kyunki $d \propto v^{2}$ , $v$ double karne se $d$ $2^{2} = 4$ se multiply hoti hai: $d_{60} = 4 \times 45 = 180 m$ . Yeh step kyun? Ratio method — unknown constants $m$ aur $f$ cancel ho jaate hain kyunki woh dono cases mein same hain: $\frac{d _{60}}{d _{30}} = \frac{m ( 60 ) ^{2} /2 f}{m ( 30 ) ^{2} /2 f} = (\frac{60}{30})^{2} = 4$ . Hume $m$ ya $f$ ki numerical values ki bilkul zaroorat nahi.

Verify: $m$ aur $f$ guess karne ki bajaye, woh single combination extract karo jo data fix karta hai. $d = \frac{m v ^{2}}{2 f}$ se, deceleration hai $a = \frac{f}{m} = \frac{v ^{2}}{2 d} = \frac{( 30 ) ^{2}}{2 ( 45 )} = \frac{900}{90} = 10 m/s^{2}$ — yeh ek number sirf pehle case se determine hota hai (koi alag $m$ ya $f$ nahi chahiye). $60 m/s$ par reuse karte hue: $d = \frac{v ^{2}}{2 a} = \frac{( 60 ) ^{2}}{2 ( 10 )} = \frac{3600}{20} = 180 m$ ✓. Yahi physics hai "double speed = char guna crash distance" ke peeche.

Example 9 — Cell I: exam twist (genuinely curved path)

Worked example Curved frictionless track se neeche (full line-integral payoff)

Ek $0.5 kg$ ka bead rest se ek frictionless curved wire se height $h = 1.8 m$ se neeche slide karta hai. Wire neeche aate hue left aur right twist karta hai. Gravity $g = 10 m/s^{2}$ hai. Neeche speed nikalo — wire ki shape jaane bina.

Forecast: Wire twist aur turn karta hai; kya hum phir bhi answer nikaal sakte hain? Normal force kya contribute karta hai?

Step 1. Yahan general $\int F \cdot d r$ apna kaam karta hai: path curved hai, isliye $d r$ har point par direction badlta hai. Do forces act karti hain — gravity $F_{g}$ (seedha neeche) aur wire ki normal force $N$ . Normal force hamesha motion ke perpendicular hai, isliye har tiny step par $N \cdot d r = 0$ : yeh zero work karta hai, chahe wire kaise bhi curve kare. Yeh step kyun? $N$ aur $d r$ ke beech har jagah $θ = 9 0^{\circ}$ ⇒ $cos 90 = 0$ . Isliye shape matter nahi karta — sirf gravity line integral se bachti hai.

Step 2. Gravity ke liye, component form use karo $F_{g} \cdot d r = F_{g, x} d x + F_{g, y} d y$ . Gravity ka koi horizontal component nahi hai ( $F_{g, x} = 0$ ) aur $F_{g, y} = - m g$ , isliye $F_{g} \cdot d r = - m g d y$ . Poore descent par sum karte hue (total drop $h$ , $d y$ negative going down ke saath): $W_{g r a v} = \int (- m g) d y = m g h = 0.5 \times 10 \times 1.8 = 9 J .$ Yeh step kyun? Component form isolate karta hai jo matter karta hai: sirf vertical travel $d y$ gravity ke saath pair karta hai. Sab left–right wandering ( $d x$ ) $F_{g, x} = 0$ ko multiply karta hai aur kuch contribute nahi karta — isliye twisty path wahi work deta hai jo $h$ ka straight vertical drop deta.

Step 3. $W_{n e t} = W_{g r a v} + W_{N} = 9 + 0 = 9 J$ (Recipe 1). Theorem se $Δ K = W_{n e t} = 9 J$ ; $K_{i} = 0$ ke saath, hume $K_{f} = 9 J$ milta hai. Yeh step kyun? Net work = sab forces ke work ka sum; normal force ne zero contribute kiya, isliye gravity ka work poora net work hai.

Step 4. $\frac{1}{2} (0.5) v_{f}^{2} = 9 \Rightarrow v_{f}^{2} = 36 \Rightarrow v_{f} = 6 m/s$ . Yeh step kyun? Neeche final speed ke liye $K = \frac{1}{2} m v^{2}$ invert karo.

Figure padho (s06): ek lavender curved wire height $h = 1.8 m$ se ground tak snake karta hai. Uske along teen points par, grey tiny-step arrows $d r$ wire ke along point karte hain (har ek alag direction mein), ek coral gravity arrow har jagah seedha neeche point karta hai, aur ek mint normal-force arrow wire ke perpendicular point karta hai. Notice karo: mint arrows hamesha grey steps se $9 0^{\circ}$ par hain (zero work), jabki gravity ka overlap har step se sirf usi par depend karta hai ki woh step kitna drop karta hai. Sum karte hue, sirf total vertical drop $h$ count hota hai.

Verify: Mass $\frac{1}{2} m v^{2} = m g h \Rightarrow v = 2 g h = 2 (10) (1.8) = 36 = 6 m/s$ ✓ mein cancel ho jaata hai — wahi $2 g h$ jaise straight vertical drop mein, confirming shape irrelevant hai. Units: $(m/s^{2}) (m) = m^{2} / s^{2} = m/s$ ✓. Yeh Conservation of Mechanical Energy se bridge karta hai.

Recall checkpoints

Recall Yaad rakhne wala ek sentence

Upar ke har example mein, ek hi equation thi $\int F_{n e t} \cdot d r = Δ K$ , jahan net work ka matlab hai har force ka work add karo. Dot product (via $F_{x} d x + F_{y} d y$ se) ne direction handle kiya, integral ne variation handle kiya, aur "net" ka matlab tha sab forces par sum. In teeno par master ho jao aur koi bhi scenario naya nahi hoga.

Connections

Newton's Second Law — har "verify" step $F = ma$ par wapas jaata hai.
Kinetic Energy — har baar $\frac{1}{2} m v^{2}$ solve karte hain.
Work done by a variable force — Examples 3, 4, 9.
Conservation of Mechanical Energy — Example 9 uska gateway hai.
Power — average power ke liye kisi bhi in works ko liya gaya time se divide karo.
Impulse–Momentum Theorem — time-integrated sibling; in distance problems se contrast karo.