Potential energy — definition, gravitational (mgh and −GMm - r), elastic (½kx²): Worked examples

1.3.5 · D3 · Physics › Work, Energy & Power › Potential energy — definition, gravitational (mgh and −GMm -

Yeh page ek complete tour of cases hai. Parent note ne teen formulas build kiye the. Yahan hum ensure karte hain ki tum koi bhi aisi situation kabhi nahi dekho jo tumne pehle solve hoti nahi dekhi: har sign, har reference choice, degenerate inputs (zero height, zero stretch, infinite distance), limiting behaviours, ek real-world word problem, aur ek exam-style twist.

Pehle, woh toolbox jo hum use kar sakte hain (teeno parent mein earn kiye gaye hain):

Scenario matrix

Is topic ke har problem ka answer in cells mein se kisi ek mein hai. Example column batata hai ki neeche konsa worked case isko cover karta hai.

Cell	Kya vary karta hai	Trap / edge behaviour	Example
G1	Flat gravity, upar jaana	$Δ U > 0$	(a)
G2	Flat gravity, neeche jaana / origin ke neeche	$Δ U < 0$ , $h$ negative	(a)
G3	Full gravity, inward vs outward move karna	$U < 0$ hamesha, lekin $Δ U$ ka sign flip hota hai	(b)
G4	Reference choice ( $h = 0$ vs $r = \infty$ )	same physics, alag $U$ values	(c)
G5	Limiting: $- GM m / r \to m g h$ jab $h ≪ R$	approximation, equality nahi	(d)
S1	Spring stretched ( $x > 0$ )	$U > 0$	(e)
S2	Spring compressed ( $x < 0$ )	$U > 0$ phir bhi, symmetry	(e)
S3	Spring degenerate ( $x = 0$ )	$U = 0$ , force $= 0$	(e)
F1	Ek general $U (x)$ se force nikalna	$- d U / d x$ ka sign, equilibria	(f)
W1	Real-world word problem (energy conservation)	multiple stores add up hote hain	(g)
X1	Exam twist (mixed spring + gravity, ya graph)	kaun sa reference, double-count mat karo	(h)

Worked examples

Worked example (a) Flat gravity — upar

aur floor ke neeche · cells G1, G2 Ek $2 kg$ ka box $h = 0$ par ek table pe rakha hai. (i) Use $h = + 1.5 m$ tak uthaya jaata hai. (ii) Table se use ek pit mein $h = - 0.8 m$ tak neecha kiya jaata hai. Dono ke liye $Δ U$ find karo ( $g = 9.8$ ).

Forecast: Pehle signs guess karo. Upar jaane se energy store honi chahiye ( $+$ ); reference ke neeche girne se negative PE aani chahiye. Kya numbers agree karte hain?

Up: $Δ U_{i} = m g h = 2 \times 9.8 \times (+ 1.5) = + 29.4 J$ . Yeh step kyun? Displacement upar hai, gravity neeche hai → gravity negative work karti hai → hum positive energy bank karte hain.
Down: $Δ U_{ii} = m g h = 2 \times 9.8 \times (- 0.8) = - 15.68 J$ . Yeh step kyun? Reference ke neeche $h$ negative hota hai. Formula mein kuch special nahi hota — $h$ ka sign physics carry karta hai. Apne chosen zero ke neeche PE genuinely negative hoti hai.

Verify: Units: $kg \cdot (m/s^{2}) \cdot m = J$ . ✓ Case (i) positive, (ii) negative — forecast se match karta hai. Pit mein rakha box table ki taraf upar release hone par speed up karega, jo neeche lower PE ke consistent hai.

Worked example (b) Full gravity — inward vs outward · cell G3

Earth: $GM = 3.986 \times 1 0^{14} m^{3} / s^{2}$ , radius $R = 6.371 \times 1 0^{6} m$ . Ek $1000 kg$ satellite $r_{1} = 7.0 \times 1 0^{6} m$ se inward move karke $r_{2} = 6.6 \times 1 0^{6} m$ tak jaata hai. $Δ U$ find karo.

Forecast: Dono $U$ values negative hain (well). Inward move karna matlab aur gehre girna → $Δ U$ negative hona chahiye. Magnitude guess karo: order $1 0^{9} J$ ?

$U (r) = - \frac{GM m}{r}$ , to $U_{1} = - \frac{3.986 \times 1 0 ^{14} \times 1000}{7.0 \times 1 0 ^{6}} = - 5.694 \times 1 0^{10} J$ . Yeh step kyun? Reference infinity hai, isliye bound satellites ka $U$ negative hota hai.
$U_{2} = - \frac{3.986 \times 1 0 ^{14} \times 1000}{6.6 \times 1 0 ^{6}} = - 6.039 \times 1 0^{10} J$ .
$Δ U = U_{2} - U_{1} = - 6.039 \times 1 0^{10} - (- 5.694 \times 1 0^{10}) = - 3.45 \times 1 0^{9} J$ . Yeh step kyun? Well mein aur gehre = aur zyada negative = energy release hoti hai ( $Δ U < 0$ ), jo girne ki kinetic energy ban jaati hai.

Verify: $Δ U$ negative hai ✓ (inward move kiya). Magnitude $\sim 3 \times 1 0^{9} J$ , forecast se match karta hai. Dono $U$ values negative ✓.

Worked example (c) Same physics, do references · cell G4

Dikhao ki surface se $h = 100 km$ tak PE ka change bilkul same hai chahe $m g h$ use karo (reference surface par) ya $- GM m / r$ (reference infinity par). Use karo $m = 1 kg$ , $g = 9.8$ , $GM = 3.986 \times 1 0^{14}$ , $R = 6.371 \times 1 0^{6} m$ , $h = 1.0 \times 1 0^{5} m$ .

Forecast: $U$ ki values bahut alag hongi (ek zero se shuru hoti hai, doosri ek bade negative number se). Lekin difference $Δ U$ almost match karna chahiye — sirf differences hi physical hote hain.

Flat: $Δ U_{flat} = m g h = 1 \times 9.8 \times 1.0 \times 1 0^{5} = 9.80 \times 1 0^{5} J$ .
Full: $Δ U_{full} = - GM m (\frac{1}{R + h} - \frac{1}{R})$ $= - 3.986 \times 1 0^{14} (\frac{1}{6.471 \times 1 0 ^{6}} - \frac{1}{6.371 \times 1 0 ^{6}}) = 9.66 \times 1 0^{5} J$ . Yeh step kyun? Absolute $U$ reference par depend karta hai, lekin subtract karne se constant offset cancel ho jaata hai — physical quantity survive karti hai.

Verify: $9.80 \times 1 0^{5}$ vs $9.66 \times 1 0^{5}$ — $\sim 1.5%$ tak agree karte hain. ✓ Chhota gap isliye hai kyunki $g$ altitude ke saath thoda weak hota jaata hai, jo flat formula ignore karta hai. Yeh wahi reconciliation hai jo parent note ne promise ki thi, numeric form mein.

Worked example (d) Limiting case, honestly samjhaya · cell G5

Parent note ne likha tha $- GM m / r \approx m g h$ chhote $h$ ke liye. "Small" ko quantify karo: kis height par flat formula $1%$ se galat hota hai? Use karo $R = 6.371 \times 1 0^{6} m$ .

Forecast: Error $h / R$ ke saath badhti hai. $1%$ error ke liye $h$ ko $R$ ke kuch percent ke aas-paas hona chahiye — tens se ek sau km.

Exact factor: full $Δ U = m g h \cdot \frac{R}{R + h}$ ( $\frac{1}{R} - \frac{1}{R + h} = \frac{h}{R ( R + h )}$ ka algebra, times $GM m = g R^{2} m$ ). Yeh step kyun? Exact result ko (flat) $\times$ ek correction factor $\frac{R}{R + h}$ ke roop mein likhne se exactly pata chalta hai ki approximation kitni galat hai.
Fractional error $= 1 - \frac{R}{R + h} = \frac{h}{R + h}$ . $0.01$ ke barabar set karo: $h = \frac{0.01}{0.99} R = 6.44 \times 1 0^{4} m$ . Yeh step kyun? Threshold ke liye solve karna "small" ko ek real number mein badal deta hai.

Verify: $h \approx 64 km$ — $R$ ka kuch percent, forecast se match karta hai. Everest ( $\sim 9$ km) ke neeche error $0.15%$ se kam hai, isliye $m g h$ everyday problems ke liye excellent hai. ✓

Worked example (e) Spring — stretch, compress, aur natural length · cells S1, S2, S3

Ek spring $k = 200 N/m$ . (i) stretched $x = + 0.10 m$ ke liye, (ii) compressed $x = - 0.10 m$ ke liye, (iii) natural length par $x = 0$ ke liye $U$ find karo. Figure dekho.

Forecast: Kyunki $U = \frac{1}{2} k x^{2}$ mein $x$ square hota hai, (i) aur (ii) ka same positive number aana chahiye, aur (iii) exactly zero.

Stretched: $U = \frac{1}{2} (200) (0.10)^{2} = 1.0 J$ .
Compressed: $U = \frac{1}{2} (200) (- 0.10)^{2} = 1.0 J$ . Yeh step kyun? $(- 0.10)^{2} = (+ 0.10)^{2}$ . Parabola $U (x)$ symmetric hai — andar push karna utni hi energy store karta hai jitni bahar pull karna (blue curve dekho, $x = 0$ ke dono taraf equal heights).
Natural length: $U = \frac{1}{2} (200) (0)^{2} = 0 J$ ; force $F = - k x = 0$ . Yeh step kyun? Yeh degenerate input hai — well ka bottom. Zero stored energy aur zero force (flat tangent, orange arrow gayab ho jaata hai).

Verify: Symmetry (i)=(ii)=1.0 J ✓, degenerate (iii)=0 ✓. Units: $(N/m) \cdot m^{2} = N\cdotm = J$ ✓.

Worked example (f) General potential se force nikalna — equilibria dhundhna · cell F1

Ek particle ka $U (x) = x^{3} - 3 x$ hai (joules, $x$ metres mein). Force $F (x)$ find karo, aur equilibrium points locate + classify karo. Figure dekho.

Forecast: Equilibria wahan hain jahan landscape flat hoti hai ( $F = 0$ ). Ek cubic $U$ mein ek hill aur ek valley hoti hai, isliye do equilibria expect karo — ek stable (valley), ek unstable (hill).

$F = - \frac{d U}{d x} = - (3 x^{2} - 3) = 3 - 3 x^{2}$ . Yeh step kyun? Force $U$ ki slope ka minus hoti hai — yeh downhill point karti hai.
Equilibria: $F = 0 \Rightarrow 3 - 3 x^{2} = 0 \Rightarrow x = \pm 1$ . Yeh step kyun? No force matlab no slope — ek resting configuration.
$U$ ki shape se classify karo: $x = - 1$ par, $U = (- 1) - 3 (- 1) = 2$ (ek peak, unstable); $x = + 1$ par, $U = 1 - 3 = - 2$ (ek valley, stable). Yeh step kyun? Peak se nudge kiya hua ball roll away karega (unstable); valley se nudge kiya hua ball roll back karega (stable). Figure mein green valley vs red peak dekho.

Verify: $F (\pm 1) = 3 - 3 (1) = 0$ ✓. $U (- 1) = 2 > U (+ 1) = - 2$ , isliye $x = + 1$ lower (stable) point hai ✓, forecast ke ek hill + ek valley se match karta hai.

Worked example (g) Real-world word problem — teen energy stores · cell W1

Ek $0.5 kg$ ball ek spring ( $k = 200 N/m$ ) par rakhi hai jo $x = 0.10 m$ compressed hai. Release hone par, yeh seedha upar fly karti hai. Air ignore karte hue, release point se kitna upar jaati hai? ( $g = 9.8$ )

Forecast: Spring ki saari stored energy top par gravitational PE ban jaati hai (wahan speed = 0 hoti hai). Kuch tens of centimetres ki height expect karo.

Spring PE released: $U_{s} = \frac{1}{2} k x^{2} = \frac{1}{2} (200) (0.10)^{2} = 1.0 J$ . Yeh step kyun? Yeh flight ke liye total energy budget hai.
Energy conservation: highest point par sab gravitational hai, $m g h = 1.0$ . Yeh step kyun? No friction → recoverable spring PE fully convert hoti hai. Peak par KE $= 0$ .
$h = \frac{1.0}{m g} = \frac{1.0}{0.5 \times 9.8} = 0.204 m$ .

Verify: $h \approx 0.20 m$ ✓ (tens of cm, forecast ke anusaar). Units: $J / (kg \cdot m/s^{2}) = m$ ✓. Subtlety: technically ball spring par rahte hue bhi $0.10$ m rise karti hai; yahan $h$ release point se measure kiya gaya hai, isliye hum sirf spring→gravity conversion count karte hain.

Worked example (h) Exam twist — spring + gravity, no double-counting · cell X1

Ek $0.5 kg$ bead ek vertical spring ( $k = 200 N/m$ ) par hai, use natural length se $x = 0.10 m$ neeche compress karke release kiya jaata hai. Natural-length position se pass karte waqt uski speed find karo.

Forecast: Do PE stores change hote hain: spring ( $\frac{1}{2} k x^{2}$ release karta hai) aur gravity (bead $x$ upar rise karti hai, jo $m g x$ cost karti hai). Net energy → KE. Speed parent ke example (c) se pure-spring $2 m/s$ se thodi kam honi chahiye, kyunki gravity kuch le jaati hai.

Spring energy released: $\frac{1}{2} k x^{2} = \frac{1}{2} (200) (0.10)^{2} = 1.0 J$ . Yeh step kyun? Spring compressed se natural length tak jaati hai → uski saari PE free ho jaati hai.
Gravity PE gained (bead $x$ height upar gayi): $m g x = 0.5 \times 9.8 \times 0.10 = 0.49 J$ . Yeh step kyun? Upar jaane mein energy lagti hai; hume ise subtract karna hai, add nahi — yeh classic exam trap hai double-counting ya wrong sign ka.
KE at natural length: $\frac{1}{2} m v^{2} = 1.0 - 0.49 = 0.51 J$ , isliye $v = \frac{2 ( 0.51 )}{0.5} = 1.428 m/s$ .

Verify: $v \approx 1.43 m/s$ , horizontal case ke $2 m/s$ se kam ✓ (gravity ne $0.49$ J liya). Units check: $J / kg = m/s$ ✓.

Recall Matrix par quick self-test

Kaun sa cell aisa hai jahan $U < 0$ hai phir bhi $Δ U$ kisi bhi sign ka ho sakta hai? ::: G3 — full gravity: $U$ hamesha negative hota hai, lekin inward move karne par $Δ U < 0$ , outward move karne par $Δ U > 0$ . Spring ko equally compress karne par utni hi energy kyun store hoti hai jitni stretch karne par? ::: $U = \frac{1}{2} k x^{2}$ $x^{2}$ par depend karta hai, isliye $x$ ka sign matter nahi karta (cell S2 = S1). Example (h) mein $m g x$ subtract kyun kiya? ::: Bead upar gayi, isliye gravitational PE badhi — woh energy KE ke liye available nahi hai (cell X1, double-counting avoid karo).

Connections

1.3.05 Potential energy — definition, gravitational (mgh and −GMm - r), elastic (½kx²) (Hinglish)
Force from potential — F = -dU/dx
Conservation of mechanical energy
Hooke's Law
Gravitation — Newton's law
Escape velocity
Work done by a force
Conservative vs non-conservative forces