Work — definition, dot product F·d, sign convention: Worked examples

1.3.1 · D3 · Physics › Work, Energy & Power › Work — definition, dot product F·d, sign convention

Kuch bhi shuru karne se pehle, sirf teen symbols ka ek reminder jo hum use karte hain:

$F$ = push kitni strong hai, newtons (N) mein.
$d$ = object actually kitna move hua, metres (m) mein — uska displacement, start se finish tak ek seedha arrow.
$θ$ = force arrow aur displacement arrow ke beech ka angle, tail-to-tail draw karke.

Neeche sab kuch sirf inhi teen quantities ko $F d cos θ$ ki tarah combine karke hai, ya component form $W = F_{x} d_{x} + F_{y} d_{y}$ jab hume angle ki jagah vectors diye jaate hain. Dekho Vectors & Components agar arrows ko $x$ aur $y$ parts mein resolve karna abhi automatic nahi hua.

Scenario matrix

Har work problem jo tumhe milegi woh inhi cells mein se ek hogi. Neeche ke examples har ek ke saath ek tag carry karte hain jaise (C1) taaki tum dekh sako ki poori space cover ho rahi hai.

Cell	Kya cheez ise alag banati hai	Sign / result jo tum expect karte ho
C1	Force motion ke saath, $0° \leq θ < 90°$	$W > 0$ (energy add hui)
C2	Force $θ = 90°$ par (perpendicular)	$W = 0$ (sirf steer karta hai)
C3	Force motion ke against, $90° < θ \leq 180°$	$W < 0$ (energy remove hui)
C4	Vectors ke roop mein diya — use karo $F_{x} d_{x} + F_{y} d_{y}$	sign apne aap aa jaata hai
C5	Degenerate: $d = 0$ (koi displacement nahi)	$W = 0$ chahe $F$ kitna bhi bada ho
C6	Degenerate: $F = 0$ (koi force nahi)	$W = 0$
C7	Ek body par multiple forces — works ka sum karo	net $W = \sum W_{i}$
C8	Limiting behaviour: $θ$ ko $0° \to 180°$ tak sweep karo	$W$ ek cosine curve trace karta hai
C9	Real-world word problem (ramp / lifting)	C1–C3 ka mix, angles dhyan se chuno
C10	Exam twist: constant force ke saath closed path	net $W = 0$ (displacement start par wapas aata hai)

Figure — Work — definition, dot product F·d, sign convention

Upar ki akeli red curve poore page ke liye tumhara compass hai: work $cos θ$ ke proportional hai. Dekho ki har cell kahan baith ti hai — C1 positive hump par, C2 zero-crossing par, C3 negative dip par.

Worked examples

Worked example Example 1 — force motion ke saath

(C1) Ek sled ko flat ice par $d = 8 m$ khaincha jaata hai ek horizontal rope se force $F = 15 N$ ke saath, bilkul travel ki direction mein. Work nikalo.

Forecast: force aur motion ek hi taraf point kar rahe hain, toh $θ = 0°$ . Aage padhne se pehle sign guess karo.

Angle identify karo. Force aur displacement parallel hain ⇒ $θ = 0°$ . Yeh step kyun? Work ka sign poora $cos θ$ mein rehta hai, isliye pehle $θ$ pin down karo.
Plug in karo. $cos 0° = 1$ , toh $W = F d cos θ = 15 \times 8 \times 1 = 120 J .$ Yeh step kyun? $θ = 0°$ par poori force motion ke saath jaati hai — kuch bhi sideways waste nahi hota.

Verify karo: units hain $N \times m = J$ ✓. Positive, jaise forecast kiya tha — energy sled mein jaati hai (woh speed up hoti hai), jo Kinetic Energy & Work-Energy Theorem se match karta hai.

Worked example Example 2 — perpendicular force kuch nahi karta

(C2) Ek satellite circle mein move karta hai. Uski taraf Earth ki gravitational pull $F = 500 N$ hai, hamesha centre ki taraf; satellite apne circular arc par $d = 2000 m$ ka displacement sweep karta hai. Us arc par gravity ka kiya hua work?

Forecast: gravity centre ki taraf point karti hai; motion circle ke saath hai. "Centre ki taraf" aur "rim ke saath" ke beech angle kya hai?

Angle nikalo. Radius (force direction) tangent (motion direction) se exactly $90°$ par milti hai. Yeh step kyun? Circular motion mein centre-pointing force hamesha velocity ke $⊥$ hoti hai.
Plug in karo. $cos 90° = 0$ , toh $W = 500 \times 2000 \times cos 90° = 500 \times 2000 \times 0 = 0 J .$ Yeh step kyun? Ek perpendicular force apni khud ki direction mein zero displacement produce karta hai, isliye woh koi energy transfer nahi karta — sirf path ko bend karta hai.

Verify karo: $cos 90° = 0$ poore product ko kill kar deta hai chahe $F$ ya $d$ kitne bhi bade hon ✓. Isliye orbital speed constant rehti hai — dekho Power ( $P = F \cdot v$ bhi yahan zero hai).

Worked example Example 4 — vectors ke roop mein diya, sign apne aap aata hai

(C4) Ek drone par force $F = (5 \overset{}{^} - 2 \overset{}{^}) N$ lagti hai jabki woh $d = (3 \overset{}{^} + 4 \overset{}{^}) m$ displace hota hai. Work?

Forecast: hume $θ$ kabhi nahi bataya gaya. Kaun sa formula use karna hai?

Component form choose karo. Koi angle nahi diya ⇒ use karo $W = F_{x} d_{x} + F_{y} d_{y}$ . Yeh step kyun? Dot Product (Scalar Product) hume answer aur uska sign deta hai bina kabhi angle compute kiye.
Matching parts multiply karo aur add karo. $W = (5) (3) + (- 2) (4) = 15 - 8 = 7 J .$ Yeh step kyun? Same-direction pieces multiply hote hain; machinery automatically $cos θ$ fold kar leti hai.

Verify karo: result $+ 7 J$ hai, toh drone energy gain karta hai. Cross-check karo ki angle acute hai: $F_{x} d_{x} + F_{y} d_{y} = 7 > 0$ ⇒ $cos θ > 0$ ⇒ $θ < 90°$ , positive work se consistent ✓.

Worked example Example 7 — kaafi saari forces, works ka sum karo

(C7) Ek $2 kg$ block ko $d = 5 m$ right ki taraf floor par khaincha jaata hai. Teen forces act karti hain: applied pull $F_{1} = 30 N$ $0°$ par (rightward), friction $F_{2} = 8 N$ $180°$ par, aur gravity $F_{3} = m g = 19.6 N$ seedha neeche (horizontal motion ke $90°$ par). Har force ka kiya hua work aur net work nikalo.

Forecast: teeno mein se kaun zero work karta hai? Kaun negative hai?

Pull ka work (C1). $θ = 0°$ : $W_{1} = 30 \times 5 \times cos 0° = 150 J$ . Yeh step kyun? Yeh poori tarah motion ke saath hai — maximum positive contribution.
Friction ka work (C3). $θ = 180°$ : $W_{2} = 8 \times 5 \times cos 180° = - 40 J$ . Yeh step kyun? Motion ka virodh karta hai ⇒ energy remove karta hai (heat).
Gravity ka work (C2). $θ = 90°$ : $W_{3} = 19.6 \times 5 \times cos 90° = 0 J$ . Yeh step kyun? Vertical force, horizontal motion ⇒ perpendicular ⇒ zero.
Add karo. $W_{net} = 150 + (- 40) + 0 = 110 J$ . Yeh step kyun? Work ek scalar hai, isliye total work ek plain sum hai — koi vector adding nahi chahiye.

Verify karo: $W_{net} = 110 J > 0$ , toh block speed up hota hai, exactly wahi jo Kinetic Energy & Work-Energy Theorem predict karta hai. Units sab joules hain ✓.

Worked example Example 8 — angle ki limiting sweep

(C8) $F = 10 N$ aur $d = 4 m$ fixed rakho lekin $θ$ ko $0°$ se $180°$ tak run karne do. $W$ report karo $θ = 0°, 60°, 90°, 120°, 180°$ par.

Forecast: jaise $θ$ badhta hai, kya $W$ smoothly shrink hota hai, jump karta hai, ya oscillate karta hai?

General value likho. $W (θ) = 40 cos θ$ (joules), kyunki $F d = 40$ . Yeh step kyun? $F d$ fix karna pure $cos θ$ behaviour isolate karta hai — s01 mein red curve, sirf scale ki gayi.
Cells evaluate karo. $W (0°) = 40, W (60°) = 20, W (90°) = 0, W (120°) = - 20, W (180°) = - 40 J .$ Yeh step kyun? Yeh poori range dikhata hai: work continuously $+ 40$ se $- 40$ tak slide karta hai, exactly $90°$ par zero cross karta hai.

Verify karo: $cos 60° = 0.5 \Rightarrow 20$ ; $cos 120° = - 0.5 \Rightarrow - 20$ ; $90°$ ke baare mein symmetric ✓. Maximum possible work $+ F d$ hai, minimum $- F d$ hai — kuch bhi un bounds se zyada nahi ho sakta kyunki $- 1 \leq cos θ \leq 1$ .

Worked example Example 9 — real-world word problem: ramp par push karna

(C9) Ek worker ek $50 kg$ crate ko $d = 4 m$ frictionless ramp par upar push karta hai jo $30°$ par incline hai. Gravity $g = 9.8 m/s^{2}$ hai. Crate par gravity ka kiya hua work nikalo.

Forecast: crate upar slope par move karte waqt rise karta hai. Gravity seedha neeche point karti hai; motion up-slope hai. Positive ya negative?

Gravity aur up-slope displacement ke beech angle nikalo. Displacement $30°$ above horizontal hai (ramp par upar); gravity seedha neeche point karti hai. "Seedha neeche" aur " $30°$ above horizontal" ke beech angle hai $90° + 30° = 120°$ . Yeh step kyun? Hume dono arrows ke beech ka saccha angle chahiye, ramp angle khud nahi — yeh ek classic trap hai.
Plug in karo. $F = m g = 50 \times 9.8 = 490 N$ , aur $W_{grav} = 490 \times 4 \times cos 120° = 490 \times 4 \times (- 0.5) = - 980 J .$ Yeh step kyun? $θ > 90°$ ⇒ negative work — gravity upar chadhai ka virodh karti hai, kinetic energy ko Potential Energy & Conservative Forces mein drain karti hai.

Verify karo (independent route): crate ka height gain hai $h = d sin 30° = 4 \times 0.5 = 2 m$ , toh gravity ka work hai $- m g h = - 490 \times 2 = - 980 J$ ✓ — dono methods agree karte hain, $120°$ confirm hota hai.

Worked example Example 10 — exam twist: constant force closed loop ke around

(C10) Ek constant horizontal force $F = (6 \overset{}{^}) N$ ek bead ko push karta hai jo $d_{1} = (3 \overset{}{^} + 2 \overset{}{^}) m$ travel karta hai, phir $d_{2} = (- 3 \overset{}{^} - 2 \overset{}{^}) m$ , start par wapas aata hai. $F$ ka total work nikalo.

Forecast: bead wahin khatam hota hai jahan se shuru hua. Kya ek constant force closed loop par net work karta hai?

Leg 1 par work. $W_{1} = F_{x} d_{1 x} + F_{y} d_{1 y} = (6) (3) + (0) (2) = 18 J$ . Yeh step kyun? Component form — $F$ ka sirf $x$ -part nonzero hai.
Leg 2 par work. $W_{2} = (6) (- 3) + (0) (- 2) = - 18 J$ . Yeh step kyun? Same force, reversed displacement ⇒ equal-and-opposite work.
Total. $W_{net} = 18 + (- 18) = 0 J$ . Yeh step kyun? Ek constant force conservative hota hai; kisi bhi closed path par net displacement zero hota hai, isliye net work zero hota hai.

Verify karo: total displacement $d_{1} + d_{2} = 0$ , aur $W_{net} = F \cdot (d_{1} + d_{2}) = F \cdot 0 = 0$ ✓. Yeh exactly parent note ke mistakes section ka "closed loop with gravity" fact hai.

Active Recall

Recall Kya tumne har cell cover ki?

Kaun sa example $d = 0$ degenerate case hai? ::: Example 5 (C5) Kaun sa example net work ko plain scalar sum dikhata hai? ::: Example 7 (C7) Example 9 mein angle $120°$ kyun hai, $30°$ kyun nahi? ::: kyunki yeh seedha-neeche gravity aur $30°$ up-slope displacement ke beech ka angle hai, $90° + 30° = 120°$ Fixed $F, d$ ke liye $W$ ki maximum aur minimum possible values kya hain? ::: $+ F d$ ( $θ = 0°$ par) aur $- F d$ ( $θ = 180°$ par) Example 10 mein closed loop ke around net work zero kyun hai? ::: ek constant force conservative hoti hai; net displacement zero hai isliye $F \cdot 0 = 0$

Connections

Parent: Work — definition & sign convention
Dot Product (Scalar Product) — component form jo Examples 4 aur 10 mein use hua
Kinetic Energy & Work-Energy Theorem — net work $\frac{1}{2} m v^{2}$ change karta hai (Examples 1, 7)
Potential Energy & Conservative Forces — negative gravity work energy store karta hai (Example 9, 10)
Power — perpendicular force $P = 0$ bhi deta hai (Example 2)
Friction — negative-work force Examples 3 aur 7 mein
Vectors & Components — arrows ko $x, y$ parts mein resolve karna