Work done by variable force — integration: Question bank

1.3.2 · D5 · Physics › Work, Energy & Power › Work done by variable force — integration

Neeche har answer mein reasoning di gayi hai, kyunki physics mein sirf "haan/na" kehna bekar hai — kyun hi asli physics hai.

True or false — justify

Force–position graph ke neeche ka area kiya gaya work ke barabar hota hai.

True — integral

\int F d x

hi rectangles

F (x_{i}) Δ x_{i}

ke sum ka limit hai, aur woh sum exactly

F

–

x

curve ke neeche ka geometric area hai.

Agar force constant ho, toh tum

W = \int F d x

use nahi kar sakte; tumhe

W = F d

use karna padega.

False — ek constant force sirf ek special case hai,

\int_{a}^{b} F d x = F (b - a) = F d

; integral hamesha kaam karta hai, bas simplify hokar ek rectangle ban jaata hai.

F

–

x

graph ke neeche ka area jo axis se neeche jaata hai, woh phir bhi positive work count hota hai.

False — axis ke neeche ka area matlab

F < 0

(wahan force motion ko oppose karta hai), isliye woh slivers

F d x

negative hote hain aur total work se subtract hote hain.

Spring ko

0

x_{0}

tak stretch karne mein kiya gaya work, usi stretch mein spring ke dwara kiye gaye work ke barabar hota hai.

False — magnitude mein equal hain lekin sign opposite hai: tum spring ke against

+ \frac{1}{2} k x_{0}^{2}

karte ho, spring object par

- \frac{1}{2} k x_{0}^{2}

karta hai.

Ek force jo hamesha motion ke perpendicular ho, woh zero work karta hai, chahe kitna bhi bada ho.

True — kyunki

d W = F \cdot d r = F d r cos 9 0^{\circ} = 0

har instant par, isliye total integral zero hai, force ki magnitude chahe kuch bhi ho.

\int_{x_{A}}^{x_{B}} F d x

ke limits swap karne se work unchanged rehta hai.

False — limits swap karne se sign flip hoti hai, jo physically travel ki direction reverse karne ka matlab hai; work tab opposite sign ka ho jaata hai.

Agar ek force graph distance

d

par

F = 0

F = F_{0}

tak ek straight line hai, toh work

F_{0} d

hai.

False — yeh triangle area

\frac{1}{2} F_{0} d

hai; force trip par

F_{0}

nahi balki

\frac{F _{0}}{2}

average karta hai.

d W = F \cdot d r

mein dot product ko one-dimensional problems mein drop kiya ja sakta hai.

Half-true — motion ke along 1-D mein

cos θ = \pm 1

, isliye yeh

\pm F d x

mein reduce hota hai, lekin sign rakhna zaroori hai: ek backward force phir bhi

cos 18 0^{\circ} = - 1

deta hai.

Integration sirf tab zaroori hai jab force position

x

par depend kare.

False — jab bhi force constant nahi hoti tab integrate karte hain; agar

F

time ya velocity par depend kare tab bhi

F \cdot d r

sum karte hain, bas integral ko matching variable mein recast karte hain.

Why questions

Varying force ko constant kyu treat kar sakte hain agar path ko tiny slices mein chop kar lein?

Vanishingly small step

d x

par force ke paas appreciate karne ki jagah nahi hoti, isliye

F (x) d x

ek honest rectangle hai; constant maanne ki error zero ho jaati hai jab slice shrink hoti hai.

Simple multiplication nahi, integral kyun sahi tool hai variable force ke liye?

Multiplication

F d

ke liye ek single force value chahiye, jo tab exist nahi karti jab

F

change ho; integral defined hai infinitely many tiny

F d x

pieces add karne ke limit ke roop mein, exactly "har step par push ka sum" ko match karta hai.

Normal force ya centripetal force zero work kyun karta hai jabki continuously act karta rehta hai?

Dono instantaneous displacement ke perpendicular point karte hain, isliye

cos θ = 0

har point par; har sliver kuch contribute nahi karta, aur running total zero rehta hai.

Work ki sign humein energy flow ke baare mein kyun batati hai?

Positive work ka matlab force motion ke along push karta hai, energy body mein feed karta hai; negative work ka matlab force motion oppose karta hai, energy bahar drain karta hai — sign literally energy transfer ki direction hai.

\frac{1}{2} k x^{2}

quote karte waqt konsi force integrate kar rahe hain yeh kyun batana zaroori hai?

Kyunki "spring dwara work" aur "tumhara spring ke against work" opposite signs rakhte hain; same magnitude

\frac{1}{2} k x^{2}

do alag energy transfers describe karta hai depending on ki kis force ka

F \cdot d r

sum kiya.

Average-force shortcut

W = \overset{ˉ}{F} d

spring ke liye integration replace kyun kar sakta hai lekin har force ke liye nahi?

Kyunki spring force

x

mein linear hai, isliye uska average simply midpoint

\frac{1}{2} k x_{0}

hai;

3 x^{2}

jaisi nonlinear force ke liye average midpoint value nahi hoti, aur sirf integral sahi answer deta hai.

Edge cases

Agar ek body apne start par wapas aati hai (

x_{A} = x_{B}

) position-only force ke under, toh total work kya hai?

Conservative force ke liye round trip zero work deta hai — closed path par integral cancel ho jaata hai; dekho Conservative forces and potential energy.

Agar displacement zero ho lekin ek bada force apply ho, toh kitna work hota hai?

Exactly zero —

d r = 0

hone se har sliver

F \cdot d r

vanish ho jaata hai; work ke liye motion chahiye, sirf force nahi (weight ko still hold karna koi physics-work nahi karta).

W = \int_{0}^{x_{0}} k x d x

x_{0} \to 0

hone par kya hota hai?

Yeh

\frac{1}{2} k x_{0}^{2}

ki tarah zero ho jaata hai (quadratically), isliye tiny stretches negligible work lete hain — consistent hai ki triangle area ek point tak shrink ho jaata hai.

Ek force beech mein sign flip kare: path ke pehle half mein

F > 0

, doosre mein

F < 0

. Total work kya represent karta hai?

Net signed area — positive aur negative contributions partially cancel hote hain, isliye total work difference hai, possibly zero agar areas match ho jayein.

Agar poore path par force zero ho lekin body phir bhi move kare, toh work kya hai?

Zero work — har sliver

0 \cdot d x = 0

hai; bina force ke move karne wali body (free glide) par koi work nahi hota, isliye uski kinetic energy Work–Energy Theorem ke according constant rehti hai.

Agar

F (x)

discontinuous ho (suddenly jump kare, jaise Example 3 ka

4

m par kink), kya tab bhi integrate kar sakte hain?

Haan — curve ke neeche ka area phir bhi well-defined hai; hum sirf jump se pehle wale piece ko jump ke baad wale piece mein add karte hain (triangle plus rectangle), kyunki discontinuity ka ek single point koi area contribute nahi karta.

Connections

Work done by a constant force — woh degenerate case jahan upar ke saare traps $F d$ mein collapse ho jaate hain.
Hooke's law and spring potential energy — $\frac{1}{2} k x^{2}$ sign traps ka source.
Work–Energy Theorem — "net work" ko "change in kinetic energy" mein convert karta hai, sign questions clarify karta hai.
Area under curves and the definite integral (Maths) — kyun signed area hi poori kahani hai.
Dot product and components of vectors — $cos θ$ factor jo ye traps baar baar test karte hain.
Conservative forces and potential energy — closed-path aur round-trip edge cases ke peeche.

Question bank — Work done by variable force — integration