Spring potential energy — derivation: Question bank

1.3.12 · D5 · Physics › Work, Energy & Power › Spring potential energy — derivation

True ya false — justify karo

Har line ek claim hai. Reveal karne se pehle true/false aur reason decide karo.

x

se compressed spring, same

x

se stretched spring se kam energy store karti hai.

False —

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

mein

x

squared hai, toh

x

ka sign khatam ho jaata hai; equal-magnitude stretch aur compression identical energy store karte hain.

Stretch double karne se stored energy double ho jaati hai.

False —

U \propto x^{2}

, toh

x

double karne se

U

quadruple ho jaata hai (

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

\frac{1}{2} k (2 x)^{2} = 4 \cdot \frac{1}{2} k x^{2}

Spring force tumpe positive work karti hai jab tum use stretch karte ho.

False — restoring force

- k x

tumhari outward motion ke opposite point karti hai, toh ye moving end par negative work karti hai; tum woh positive work karte ho jo

U

ke roop mein store hoti hai.

Equilibrium position par stored spring energy zero hoti hai.

True —

x = 0

par,

U = \frac{1}{2} k (0)^{2} = 0

; spring relaxed hai aur koi elastic energy nahi rakhti, chahe ye oscillating mass ki maximum speed ka point kyun na ho.

Stiffer spring (bada

k

) hamesha given displacement ke liye zyada energy store karti hai.

True —

x

fixed hone par,

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

linearly

k

ke saath badhta hai, toh stiffer spring same stretch par proportionally zyada energy store karti hai.

Formula

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

chahe kitna bhi spring stretch karo, valid rehta hai.

False — ye Hooke's law (

F = - k x

) ke valid rehne par depend karta hai; real springs apni elastic limit se aage deviate ya permanently deform ho jaate hain, aur wahan linear force law (aur hence formula) fail ho jaata hai. Dekho Hooke's Law.

Spring potential energy spring se attached block mein store hoti hai.

False — energy deformed spring mein hi store hoti hai (uske stretched/compressed bonds mein); block bas release hone par use kinetic energy ke roop mein le jaata hai.

Agar compressed spring release karo, toh uski energy sirf kinetic energy banti hai.

Generally False, sirf ideal frictionless case mein True — tab poora

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

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

ban jaata hai, lekin friction hone par kuch heat ban jaata hai, toh "sirf kinetic" tab fail hota hai jab losses hoon. Dekho Conservation of mechanical energy.

Force–displacement graph ke neeche ka area stored energy deta hai.

True — work hai

\int F d x

, jo

F

–

x

line ke neeche ka area hai; spring ke liye woh area triangle hai

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

. Dekho Work done by a variable force.

Error dhundho

Har line mein ek flawed argument hai. Flaw ko ek sentence mein name karo.

"Full stretch par force hai

k x

aur distance hai

x

, toh work

= k x \cdot x = k x^{2}

Flaw ye hai ki final (maximum) force poore path par use ki gayi hai; force zero se shuru hui thi, toh tumhe average

\frac{k x}{2}

use karni chahiye, jo deta hai

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

"Compress hone par

x < 0

, toh

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

negative hai."

Flaw ye hai ki

x

squared hai —

x^{2} \geq 0

hamesha, toh

U

kabhi negative nahi hoti chahe stretch ya compression ka sign kuch bhi ho.

"Spring pull back karti hai, toh uska force storing karta hai — force hai

+ k x

Flaw galat sign hai: spring ki restoring force

- k x

hai aur stretch karte waqt negative work karti hai, toh tumhari applied force hi energy store karti hai.

"Maine spring ki total length

0.30 m

x

mein plug in ki."

Flaw absolute length use karna hai;

x

natural relaxed length se displacement hai, spring ki poori length nahi.

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

aur

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

jaisi hai, toh ½ usi reason se aata hai."

Flaw shared origin assume karna hai; spring ka ½ triangle area / average-force factor hai, jabki kinetic energy ka ½

m v d v

integrate karne se aata hai — form ki coincidence hai, cause ki nahi.

"Ek tiny slice mein force constant hai, toh poore stretch par work hai

F x

Flaw ye hai ki force sirf ek infinitesimal slice ke andar constant hai; poore stretch mein

F

change karta hai, toh tumhe slices ko sum (integrate) karna hoga, ek baar multiply nahi.

"Spring force conservative hai, toh ye kisi bhi path par zero net work karti hai."

Flaw conservative ko zero-work se confuse karna hai; conservative ka matlab hai work sirf endpoints par depend karta hai (closed loop around zero), na ki har path par zero.

Why questions

Sirf fact nahi, reason explain karo. (Notation pe ek note: jab derivation stretch ke upar sum karta hai, toh ye ek dummy variable use karta hai jise $x^{'}$ likhte hain — padhte hain "x-prime" — moving position ke liye pull ke dauraan, taaki fixed final displacement ka naam $x$ reh sake; $d x^{'}$ bas us moving position ka ek tiny step hai.)

Hum spring ke liye ek single

F

ke saath

W = F \times d

kyun nahi use kar sakte?

Kyunki force

- k x

position ke saath change hoti hai (linearly badhti hai), toh koi ek value of

F

poore stretch ko describe nahi karti — humein integral ke zariye tiny constant-force slices add karni hoti hain.

Derivation mein

\frac{1}{2}

ka factor kyun aata hai?

Force

0

k x

tak linearly badhti hai, toh uska average $\frac{k x}{2}$ hai; work = average force × distance =

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

, equivalently triangle ka half.

Har slice ko infinitesimally small (

d x^{'}

) banane se hum us par

W = F d

kyun use kar sakte hain?

Infinitely thin slice mein moving position

x^{'}

barely change hoti hai, toh force

k x^{'}

effectively constant hai, aur simple constant-force work rule exactly apply hoti hai.

Spring ke against kiya hua work potential energy gain ke barabar kyun hota hai?

Kyunki spring force conservative hai — koi energy heat mein nahi jaati, toh har joule jo tum us par push in karte ho store ho jaati hai aur fully recoverable hai. Dekho Conservative forces and potential energy.

Simple Harmonic Motion mein wahi formula

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

kyun dobara aata hai?

Kyunki SHM hi mass-on-a-spring hai; stored

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

aur

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

ke saath uska trade oscillation drive karta hai, continuously dono ke beech swap karta hua. Dekho Simple Harmonic Motion.

Spring PE,

x

mein quadratic kyun hai jabki gravitational PE

m g h

, height mein sirf linear hai?

Gravity ek constant force deta hai (

m g

), toh uska work force × distance (linear) hai; spring ki force $x$ ke saath badhti hai, toh badhti hui line integrate karne se quadratic milta hai. Dekho Gravitational potential energy.

Edge cases

Boundary aur degenerate situations jahan sloppy thinking break ho jaati hai.

x = 0

par (spring natural length par)

U

kya hai?

Exactly zero —

\frac{1}{2} k (0)^{2} = 0

; relaxed spring koi elastic energy nahi rakhti, jo

U

measure karne ka reference point hai.

Spring par oscillate karte mass ke liye

U

maximum kahan hai aur kinetic energy maximum kahan hai?

U

turning points par maximum hai (sabse bada

∣ x ∣

, momentarily at rest) aur kinetic energy equilibrium par maximum hai (

x = 0

, sabse fast); ye trade off karte hain constant total ke saath. Dekho Conservation of mechanical energy.

Jab

k \to 0

(infinitely floppy spring) toh

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

ka kya hota hai?

Kisi bhi finite

x

ke liye

U \to 0

— zero stiffness wali spring koi force exert nahi karti, toh use move karne ke liye koi work nahi chahiye aur kuch store nahi hota.

Agar spring elastic limit se zyada stretch ho, toh kya

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

tab bhi valid hai?

Nahi — Hooke's law break down ho jaata hai (force ab

- k x

nahi rehti), kuch energy permanent deformation/heat mein jaati hai, toh neat triangle formula ab apply nahi hota.

0

x

tak stretch karne mein stored energy aur

x

2 x

tak stored energy compare karo.

Ye equal nahi hain:

0 \to x

stores

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

, lekin

x \to 2 x

stores

\frac{1}{2} k (2 x)^{2} - \frac{1}{2} k x^{2} = \frac{3}{2} k x^{2}

— teen guna zyada, kyunki

U

quadratically badhta hai.

Ek poore oscillation cycle (wapas start par) mein spring ka net work kya hai?

Zero — spring force conservative hai, toh closed path par same

x

par wapas aane par net work exactly zero hai.

Connections

Hooke's Law — linear force law jo ye traps baar baar test karti hain.
Work done by a variable force — kyun integration, multiplication nahi.
Conservative forces and potential energy — "work against = stored" logic.
Conservation of mechanical energy — PE↔KE trade-off edge cases.
Simple Harmonic Motion — jahan $\frac{1}{2} k x^{2}$ dobara aata hai.
Gravitational potential energy — linear-vs-quadratic contrast.
Spring potential energy — derivation — parent note jismein full build hai.