Spring potential energy — derivation: Worked examples

1.3.12 · D3 · Physics › Work, Energy & Power › Spring potential energy — derivation

Scenario matrix

Har spring-energy problem in cells mein se ek hoti hai (ya kuch ka blend). Hum sab cover karenge.

#	Cell	Kya alag hai ismein	Example
A	Pure stretch, $x > 0$	Baseline plug-in	Ex 1
B	Pure compression, $x < 0$	Sign khatam ho jaata hai (squared)	Ex 2
C	Zero displacement, $x = 0$	Degenerate — energy exactly $0$ hoti hai	Ex 2 (aside)
D	Change of state $Δ U$	Energy difference $U_{2} - U_{1}$ , $U$ nahi	Ex 3
E	Spring → kinetic (launch)	Energy conservation, Conservation of mechanical energy	Ex 4
F	Spring → height (vertical)	Elastic PE ↔ Gravitational potential energy	Ex 5
G	Solve backwards for $x$ ya $k$	Square undo karo → , do roots	Ex 6
H	Limiting / scaling behaviour	"double $x$ ", "double $k$ " — $x^{2}$ law	Ex 7
I	Exam twist: partial release	Energy sirf $x_{1}$ se $x_{2}$ tak, friction	Ex 8

Neeche sab kuch force–displacement triangle se linked hai — energy hamesha $F = k x$ line ke neeche ka area hoti hai.

Figure — Spring potential energy — derivation

Worked examples

Worked example Ex 1 — Cell A · pure stretch

Ek spring jisme $k = 300 N/m$ hai, use $x = 0.05 m$ stretch kiya gaya. Stored energy find karo.

Forecast: Compute karne se pehle order of magnitude guess karo. Kya yeh $0.4 J$ ke kareeb hoga ya $4 J$ ke? Likh lo.

Formula likho: $U = \frac{1}{2} k x^{2}$ . Yeh step kyun? Stored energy woh triangle area hai jo $F = k x$ ke neeche hoti hai; woh area hai $\frac{1}{2} \cdot base \cdot height = \frac{1}{2} x (k x) = \frac{1}{2} k x^{2}$ .
Substitute karo: $U = \frac{1}{2} (300) (0.05)^{2}$ . Yeh step kyun? $x$ already relaxed length se displacement hai, toh koi conversion nahi chahiye.
Pehle square compute karo: $(0.05)^{2} = 0.0025$ . Yeh step kyun? Multiply karne se pehle square karo taaki chhoti number control mein rahe.
Finish karo: $U = \frac{1}{2} (300) (0.0025) = 0.375 J$ .

Verify: Units: $N/m \times m^{2} = N \cdot m = J$ ✓. Magnitude $0.375 J$ zyada kareeb hai $0.4 J$ ke — modest hai, kyunki stretch bahut chhoti hai.

Worked example Ex 2 — Cells B & C · compression aur zero

Wahi spring ( $k = 300 N/m$ ) ab compressed hai $0.05 m$ se, yaani $x = - 0.05 m$ . Phir, alag se, poochho: $x = 0$ par $U$ kya hai?

Forecast: Kya compression se negative energy milegi? Padhne se pehle jawab do.

$U = \frac{1}{2} k x^{2} = \frac{1}{2} (300) (- 0.05)^{2}$ . Yeh step kyun? Formula bilkul same hai — compression ke liye kuch special nahi.
Square sign khatam kar deta hai: $(- 0.05)^{2} = + 0.0025$ . Yeh step kyun? Kisi bhi real number ko square karo toh $\geq 0$ milta hai; yahi wajah hai ki compressed spring positive energy store karti hai.
$U = \frac{1}{2} (300) (0.0025) = 0.375 J$ — Ex 1 se identical. Yeh step kyun? Figure dekho: $x = 0$ ke left wale triangle ka area right wale ke barabar hai. Same $∣ x ∣$ , same energy.
Zero case (Cell C): $x = 0$ par, $U = \frac{1}{2} k (0)^{2} = 0 J$ . Yeh step kyun? Relaxed spring kuch bhi store nahi karti — triangle ka base zero hai.

Verify: Equal amount se stretch aur compression dono equal $U$ dete hain ( $0.375 = 0.375$ ) ✓. Lowest possible $U$ hai $0$ , jo sirf equilibrium par milta hai ✓.

Worked example Ex 3 — Cell D · stored energy mein change

Ek spring ( $k = 400 N/m$ ) ko $x_{1} = 0.10 m$ se $x_{2} = 0.20 m$ tak stretch kiya gaya. Tumne kitni extra energy add ki?

Forecast: Stretch $0.10$ se $0.20$ ho gayi. Kya stored energy double hui? Guess karo.

Har state mein energy: $U_{1} = \frac{1}{2} k x_{1}^{2}$ , $U_{2} = \frac{1}{2} k x_{2}^{2}$ . Yeh step kyun? Jo work tum supply karte ho woh difference $Δ U = U_{2} - U_{1}$ ke barabar hai, $U_{2}$ akele ke nahi — tum pehle se stretched shuru kar rahe the.
$U_{1} = \frac{1}{2} (400) (0.10)^{2} = \frac{1}{2} (400) (0.01) = 2 J$ .
$U_{2} = \frac{1}{2} (400) (0.20)^{2} = \frac{1}{2} (400) (0.04) = 8 J$ . Yeh step kyun? $x$ double hua, aur kyunki $U \propto x^{2}$ hai, $U$ 4× badh gaya ( $2 \to 8$ ).
$Δ U = 8 - 2 = 6 J$ . Yeh step kyun? Yeh extra area hai — triangle ki $x_{1}$ aur $x_{2}$ ke beech ki trapezoid strip.

Verify: $Δ U = 6 J$ . Sanity check: yeh $U_{1}$ se zyada hai, kyunki us outer slice mein force badi hoti hai ✓.

Worked example Ex 4 — Cell E · spring ek block launch karti hai

Ek spring ( $k = 800 N/m$ ) ko $0.10 m$ compress karke frictionless floor par $0.5 kg$ ke block ke against release kiya gaya. Launch speed find karo.

Forecast: $5 m/s$ se fast ya slow? Guess karo.

Stored energy: $U = \frac{1}{2} (800) (0.10)^{2} = \frac{1}{2} (800) (0.01) = 4 J$ . Yeh step kyun? Yahi energy spring hand over karegi.
Saari $U$ kinetic energy ban jaati hai: $\frac{1}{2} k x^{2} = \frac{1}{2} m v^{2}$ . Yeh step kyun? Conservation of mechanical energy — koi friction nahi, koi height change nahi, toh elastic PE → kinetic energy bina kisi loss ke.
$4 = \frac{1}{2} (0.5) v^{2} = 0.25 v^{2}$ .
$v^{2} = \frac{4}{0.25} = 16 \Rightarrow v = 4 m/s$ . Yeh step kyun? Positive root lo — speed non-negative hoti hai.

Verify: $\frac{1}{2} (0.5) (4)^{2} = 0.25 \times 16 = 4 J$ — stored $4 J$ se match karta hai ✓. $5 m/s$ guess se slow hai kyunki block kaafi heavy hai.

Worked example Ex 5 — Cell F · vertical launch (spring → height)

Ek spring ( $k = 1000 N/m$ ) ko $0.1 kg$ ki ball ke neeche $0.06 m$ compress karke seedha upar fire kiya gaya. Ball release point ke upar kitna ooncha jaati hai? $g = 10 m/s^{2}$ use karo.

Forecast: $2 m$ se upar ya neeche? Guess karo.

Elastic energy: $U = \frac{1}{2} (1000) (0.06)^{2} = \frac{1}{2} (1000) (0.0036) = 1.8 J$ . Yeh step kyun? Yahi poora energy budget hai.
Top par ball momentarily ruk jaati hai, toh saari $U$ gravitational PE ban jaati hai: $U = m g h$ . Yeh step kyun? Elastic PE $\to$ Gravitational potential energy; peak par kinetic energy zero hoti hai.
$1.8 = (0.1) (10) h = 1 \cdot h$ .
$h = 1.8 m$ . Yeh step kyun? Height ke liye single linear equation solve karo.

Verify: $m g h = 0.1 \times 10 \times 1.8 = 1.8 J$ ✓. $2 m$ guess se neeche hai. (Strictly ball spring travel ke $0.06 m$ bhi energy lete hue rise karti hai, lekin "release point ke upar" ka matlab hai hum wahin se measure karte hain.)

Worked example Ex 7 — Cell H · scaling laws (limiting behaviour)

Ek spring kisi stretch $x_{0}$ par $U_{0} = 2 J$ store karti hai. (a) Agar tum stretch triple kar do, toh nayi energy kya hogi? (b) Agar instead $x_{0}$ same rakho lekin double $k$ wali spring use karo, toh nayi energy kya hogi?

Forecast: (a) ke liye kya yeh $6 J$ hogi ya $18 J$ ? Padhne se pehle guess karo.

(a) $U \propto x^{2}$ hai, toh $x$ triple karne se $U$ $3^{2} = 9$ se multiply ho jaata hai. Yeh step kyun? Square ke andar $x$ poora scaling carry karta hai.
Nayi energy $= 9 \times 2 = 18 J$ . Yeh step kyun? $6 J$ nahi — trap yeh sochna hai ki energy $x$ mein linear hai.
(b) $U \propto k$ (linearly) hai, toh $k$ double karne se $U$ double hoti hai: $2 \times 2 = 4 J$ . Yeh step kyun? $k$ first power mein appear karta hai — same stretch par stiffer spring proportionally zyada store karti hai.

Verify: $k = 100, x_{0} = 0.2$ ke saath: $U_{0} = \frac{1}{2} (100) (0.2)^{2} = 2 J$ ✓; triple $x$ : $\frac{1}{2} (100) (0.6)^{2} = 18 J$ ✓; double $k$ : $\frac{1}{2} (200) (0.2)^{2} = 4 J$ ✓.

Worked example Ex 8 — Cell I · exam twist · partial release with friction

Ek spring ( $k = 600 N/m$ ) $x_{1} = 0.20 m$ tak compressed hai. Yeh ek $1 kg$ block ko push karti hai aur tab chhodti hai jab compression sirf $x_{2} = 0.10 m$ reh jaati hai (spring ka natural length block ko aage nahi le jaata). Friction block ke against $1.5 J$ work karta hai. Release par block ki speed find karo.

Forecast: $3 m/s$ se upar ya neeche? Guess karo.

Spring dwara di gayi energy stored energies ka difference hai, $Δ U = U_{1} - U_{2}$ . Yeh step kyun? Release par spring abhi bhi $U_{2}$ hold karti hai; sirf $x_{1}$ aur $x_{2}$ ke beech ka drop deliver hua tha (Work done by a variable force).
$U_{1} = \frac{1}{2} (600) (0.20)^{2} = \frac{1}{2} (600) (0.04) = 12 J$ .
$U_{2} = \frac{1}{2} (600) (0.10)^{2} = \frac{1}{2} (600) (0.01) = 3 J$ .
Released energy $= 12 - 3 = 9 J$ . Yeh step kyun? $x_{2}$ aur $x_{1}$ ke beech ki trapezoid strip — wahi " $Δ U$ " idea jo Ex 3 mein tha.
Friction minus karo, baaki kinetic ban jaata hai: $\frac{1}{2} m v^{2} = 9 - 1.5 = 7.5 J$ . Yeh step kyun? Energy accounting — friction ne $1.5 J$ heat ke roop mein remove kiya, pure mechanical conservation tod ke.
$\frac{1}{2} (1) v^{2} = 7.5 \Rightarrow v^{2} = 15 \Rightarrow v = 15 \approx 3.87 m/s$ .

Verify: $\frac{1}{2} (1) (3.873)^{2} \approx 7.5 J$ , plus friction $1.5 J$ , gives $9 J = Δ U$ ✓. $3 m/s$ se upar hai.

Recall Cell-by-cell self-check

Kaunsi cell square ke sign khatam karne ko test karti hai? ::: Cell B (compression, Ex 2). Kaunsi cell mein $Δ U = U_{2} - U_{1}$ chahiye? ::: Cells D aur I (Ex 3, Ex 8). Kaunsi cell mein $\pm$ square root force hoti hai? ::: Cell G (Ex 6). $x$ triple karne par $U$ kitne se multiply hoti hai? ::: $9$ (Cell H, Ex 7). Ex 8 mein mechanical energy conserve kyun nahi hoti? ::: Friction $1.5 J$ heat ke roop mein remove karta hai.

Connections

1.3.12 Spring potential energy — derivation — parent derivation jise yeh examples exercise karte hain.
Hooke's Law — $F = - k x$ line jiska triangle hum baar baar measure karte hain.
Work done by a variable force — Ex 3 aur 8 mein $Δ U$ (trapezoid strip) reasoning.
Conservative forces and potential energy — spring ke against work kyun stored $U$ ke barabar hai.
Conservation of mechanical energy — Ex 4, 5, 8.
Gravitational potential energy — Ex 5, elastic-to-height swap.
Simple Harmonic Motion — wahi $\frac{1}{2} k x^{2}$ jo yeh launches release karte hain.