Work — definition, dot product F·d, sign convention: Question bank

1.3.1 · D5 · Physics › Work, Energy & Power › Work — definition, dot product F·d, sign convention

In machine ka reminder jo hum test kar rahe hain, taaki koi symbol unexplained na rahe:

$W = F \cdot d = F d cos θ$ — work equals (force magnitude) $\times$ (displacement magnitude) $\times$ ( $cos$ of the angle $θ$ between them, drawn tail-to-tail).
$θ$ hamesha $F$ aur $d$ ke beech ka angle hai; $W$ ka sign sirf $cos θ$ decide karta hai.
$cos θ$ : $0° \leq θ < 90°$ ke liye positive, $90°$ par zero, $90° < θ \leq 180°$ ke liye negative.

Har trap ke peeche ek hi picture

Questions se pehle, geometry apni aankhon mein fix karo. $F$ aur $d$ ko ek hi starting point se draw karo (tail-to-tail) — jo angle woh kholta hai woh $θ$ hai. Ab $F$ ko seedha $d$ ki line par drop karo: jo shadow woh $d$ ke along dalta hai uski length $F cos θ$ hai (yeh parallel part $F_{∥}$ hai), aur jo sideways nikalta hai woh $F sin θ$ hai (perpendicular part $F_{⊥}$ ).

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

Teen regimes ek nazar mein notice karo: jab $F$ motion ki taraf jhukta hai toh shadow $d$ ke same direction mein point karta hai (positive work); jab $F$ bilkul across hota hai, shadow ek point mein shrink ho jaata hai (zero work); jab $F$ motion ke khilaf jhukta hai toh shadow $d$ ke opposite point karta hai (negative work). Yahi ek shadow poori sign convention hai.

Aur yahan mathematical reason hai kyun perpendicular part vanish hota hai, sirf "useless lagta hai" nahi. Work aslaan tiny force-dots-displacement steps ka running sum (integral) hai, $W = \int F \cdot d r$ . Har tiny step ki force ko $F_{∥}$ (path ke along) aur $F_{⊥}$ (path ke across) mein split karo. Kyunki $d r$ path ke along point karta hai, $F_{⊥} \cdot d r = F_{⊥} d r cos 90° = 0$ har step mein — perpendicular piece $cos 90° = 0$ se multiply hoti hai aur sum ke har term se drop out ho jaati hai. Sirf $\int F_{∥} d r$ bachta hai.

True ya false — justify karo

Ye har ek ek-sentence ka claim hai. True ya false decide karo, phir reason do.

Ek large magnitude waali force hamesha large work karti hai.

False. Work

F d cos θ

hai; agar

θ = 90°

toh

cos θ = 0

aur

W = 0

chahe

F

kitna bhi bada ho, aur agar

θ > 90°

toh work negative hoga.

Agar ek object apne starting point par wapas aata hai, toh har force dwara us par kiya gaya total work zero hona chahiye.

False. Sirf woh forces jinka work sirf start/end points par depend karta hai (jaise gravity) closed loop mein zero dete hain; friction hamesha motion ko oppose karta hai, isliye woh har leg par negative work karta hai aur total nonzero (negative) value deta hai.

Object par zero net work ka matlab hai ki object hila nahi.

False. Woh bahut move kar sakta hai lekin speed gain nahi karta: e.g. circular orbit mein ek satellite lagaataar move karta hai lekin gravity hamesha uski velocity ke perpendicular hai, isliye net work zero hai aur uski kinetic energy unchanged rehti hai.

Negative work ka matlab hai force choti ya kamzor hai.

False. Negative work ka sirf yeh matlab hai ki

cos θ < 0

, yaani force ka ek component motion ke khilaf hai (

90° < θ \leq 180°

); ek bahut bada braking force large-magnitude negative work karta hai.

Flat floor par slide karte box par normal force dwara kiya gaya work zero hai.

True. Normal force floor ke perpendicular point karta hai (horizontal displacement ke perpendicular), isliye

θ = 90°

cos 90° = 0

, aur

W = 0

Velocity ke perpendicular force kisi object ki kinetic energy kabhi change nahi kar sakti.

True. Perpendicular force zero work karta hai (

cos 90° = 0

), aur work–energy idea se (dekho Kinetic Energy & Work-Energy Theorem) zero work ka matlab

\frac{1}{2} m v^{2}

mein koi change nahi; yeh sirf motion ko redirect karta hai.

Agar do forces act karein aur ek

+ 8 J

kare jabki doosri

- 8 J

kare, toh object ki speed unchanged rehti hai.

True. Net work

8 + (- 8) = 0 J

hai, aur net work kinetic energy mein change ke barabar hai, isliye end mein speed start ki speed ke barabar hai.

Kisi object par positive work karna hamesha use speed up karta hai.

Careful — False as stated. Ek force dwara positive work us force ke through energy add karta hai, lekin doosri force zyada remove kar sakti hai; sirf net work decide karta hai ki speed badhti hai ya nahi. Ek single positive contribution outweigh ho sakta hai.

Work ki unit, joule, force times distance ki unit ke barabar hai.

True.

1 J = 1 N\cdotm

by definition, kyunki

W = F d cos θ

ek force (N) ko ek length (m) aur ek dimensionless cosine se multiply karta hai.

Error dhundho

Har line mein ek reasoning hai jisme ek galti chupi hai. Error ka naam lo aur use correct karo.

"Rope

30°

par pull karta hai, isliye useful force

F sin 30°

hai."

Error: galat trig function.

θ

F

aur horizontal displacement ke beech ka angle hai, aur work ko displacement ke along component chahiye, jo

F cos 30°

hai.

sin

vertical part dega, jo koi horizontal work nahi karta.

"Main usi box ko usi height tak le jaate hue lambe, curvy path par chala, isliye maine gravity ke khilaf zyada work kiya."

Error: path length displacement nahi hai. Gravity ka work sirf vertical displacement (start-to-end height change) par depend karta hai, liye gaye path par nahi; same height change ka matlab same gravitational work hai.

"Friction ne sliding block par

+ 15 J

kiya kyunki friction ek strong real force hai."

Error: sign ignore kiya. Kinetic friction motion ke opposite point karta hai (

θ = 180°

cos 180° = - 1

), isliye uska work negative hai; strength magnitude affect karta hai, kabhi sign nahi. Dekho Friction.

F d cos θ

mein angle

30°

hai kyunki ramp

30°

par inclined hai."

Error: galat angle.

θ

specific force aur specific displacement ke beech ka angle hai jo tail-to-tail draw kiye gaye hoon, automatically ramp ka tilt nahi; tumhe pehle har vector ki direction identify karni hogi (dekho Vectors & Components).

"Kyunki maine dumbbell ek minute tak bilkul still pakde rakha, maine bahut work kiya — mera arm thak gaya tha."

Error: effort, work nahi hai. Displacement zero hai, isliye

W = F d cos θ = 0

; muscular fatigue biological hai (fibres baar baar fire karte hain), physics work nahi.

"Gravity horizontally thrown ball par zero work karta hai kyunki woh horizontally move karta hai."

Error: displacement purely horizontal nahi hai. Ek projectile bhi girta hai, isliye gravity ke saath aligned downward displacement hai; us vertical part ke liye

cos θ

positive hai, jo positive gravity work deta hai.

"Work ek vector hai kyunki yeh do vectors

F

aur

d

se bana hai."

Error: dot product output. Do vectors ka scalar (dot) product ek scalar deta hai; work isliye magnitude aur sign rakhta hai lekin koi direction nahi. Dekho Dot Product (Scalar Product).

Why questions

Underlying mechanism ke saath jawab do, sirf rule nahi.

Sirf force ka displacement ke along component work mein kyun contribute karta hai?

Kyunki work integral

W = \int F \cdot d r

hai;

F

F_{∥}

aur

F_{⊥}

mein split karo, har perpendicular term

F_{⊥} d r cos 90° = 0

ban jaati hai aur term by term drop out ho jaati hai, sirf

\int F_{∥} d r

bachta hai. Physically,

F_{⊥}

apni khud ki direction mein koi displacement produce nahi karta, isliye woh koi energy transfer nahi karta.

W = F d cos θ

mein

sin θ

ki jagah

cos θ

kyun sahi factor hai?

cos θ

F

ka displacement direction par projection extract karta hai (angle ki adjacent side), jo exactly along-motion share hai jo work karta hai;

sin θ

useless perpendicular share dega.

Work ka sign poori tarah angle se kyun aata hai, magnitudes

F

d

se nahi?

F

aur

d

magnitudes hain aur kabhi negative nahi hote, isliye sirf

cos θ

sign flip kar sakta hai, jo exactly tab negative hota hai jab force motion ko oppose kare (

θ > 90°

Component form

W = F_{x} d_{x} + F_{y} d_{y} + F_{z} d_{z}

aksar

F d cos θ

se zyada convenient kyun hai?

Yeh angle

θ

dhundhe bina directly vectors ke components se work compute karta hai, aur automatically sahi sign deta hai. Dekho Vectors & Components.

Uniform circular motion mein centripetal (string) tension zero work kyun karta hai?

Tension hamesha centre ki taraf point karta hai, velocity (aur isliye instantaneous displacement) ke perpendicular, isliye

θ = 90°

cos 90° = 0

, aur

W = 0

— yahi wajah hai ki speed constant rehti hai.

Do log dono "wall ko zor se push kar sakte hain" lekin wall par physics work zero kyun ho sakta hai?

Wall ka displacement zero hai, isliye

W = F d cos θ = 0

force ke bawajood; application ke point ki motion ke bina koi energy transfer nahi hoti.

Negative work ka matlab object energy kho raha hai kyun hota hai?

Negative work ka matlab hai force ka ek component motion ke opposite hai, isliye woh object se kinetic energy remove karta hai; work–energy connection se yeh

\frac{1}{2} m v^{2}

mein drop ke roop mein dikhta hai.

Power, work se related lekin alag kyun hai?

Work total transferred energy hai, jabki power

P = F \cdot v

us transfer ki rate hai per unit time; same work zyada fast karna zyada power ka matlab hai. Dekho Power.

Edge cases

Boundary aur degenerate situations jo definition ko phir bhi handle karni chahiye.

Agar displacement

d

exactly zero ho (object move nahi karta) toh work kya hai?

W = F d cos θ = F \cdot 0 \cdot cos θ = 0

; koi displacement nahi toh koi work nahi, chahe force kitna bhi bada ho.

Exactly

θ = 90°

par work kya hai, aur yeh knife-edge case kyun hai?

W = F d cos 90° = 0

; yeh energy-adding forces (

θ < 90°

cos > 0

) aur energy-removing forces (

θ > 90°

cos < 0

) ko alag karne wali boundary hai.

Diye gaye

F

aur

d

ke liye maximum possible work kya hai, aur kis angle par?

Maximum

W = F d

hai,

θ = 0°

par hota hai jahan

cos 0° = 1

aur force poori tarah displacement ke along point karta hai.

Diye gaye

F

aur

d

ke liye most negative possible work kya hai?

W = - F d

θ = 180°

par hota hai jahan

cos 180° = - 1

, yaani force exactly displacement ke opposite point karta hai (jaise friction directly motion ko oppose kare).

Kya ek

1000 N

ki force exactly zero work kar sakti hai?

Haan, jab bhi woh displacement ke perpendicular ho (

θ = 90°

) ya displacement zero ho; large magnitude nonzero work force nahi karta.

Agar ek box right move kare lekin force motion se

135°

par up-and-left point kare, toh work ka sign kya hai?

Negative, kyunki

θ = 135°

(90°, 180°]

mein hai isliye

cos 135° < 0

; force ka ek leftward component rightward displacement ko oppose karta hai.

Seedha upar throw ki gayi aur same height par pakdi gayi ball ke liye, poore trip mein gravity dwara kiya gaya total work kya hai?

Zero: gravity upar jaate time negative work karta hai (

θ = 180°

) aur neeche aate time equal positive work (

θ = 0°

), aur woh cancel ho jaate hain kyunki net vertical displacement zero hai (gravity ki path-independence — dekho Potential Energy & Conservative Forces).

Connections

Parent: Work definition & sign convention
Kinetic Energy & Work-Energy Theorem — net work speed mein change set karta hai
Dot Product (Scalar Product) — kyun work ek scalar hai
Friction — classic negative-work force
Potential Energy & Conservative Forces — closed-loop aur path-independence traps
Power — work per unit time
Vectors & Components — forces resolve karna, sahi angle choose karna