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

1.3.5 · Physics › Work, Energy & Power

WHAT is potential energy?

WHY the minus sign? Agar koi force kisi object par positive kaam karti hai (use move karne mein help karti hai), toh woh stored energy kharach karti hai, isliye $U$ neeche jaata hai. Iske ulta, agar tum force ke against push karo, toh tum energy store karte ho aur $U$ upar jaata hai. Minus sign yeh encode karta hai ki "stored energy = force ke against kiya gaya kaam."

Inverse relation (energy se force): $F_{x} = - \frac{d U}{d x} F = - \nabla U$ WHY: force energy landscape mein "downhill" point karta hai, lower PE ki taraf — jaise ek ball valley mein roll karti hai.

HOW to derive each case (from scratch)

1. Gravitational PE near Earth's surface — $m g h$

Derivation. Surface ke paas gravitational force constant hoti hai: $F = - m g \overset{y}{^}$ (neeche ki taraf). Ek object ko $y = 0$ se $y = h$ tak uthao. Force yeh kaam karti hai: $W_{grav} = \int_{0}^{h} (- m g) d y = - m g h$ Phir $Δ U = - W_{grav} = - (- m g h) = m g h . ✓$ Why this step? Displacement upar hai ( $+ d y$ ) jabki gravity neeche point karti hai, isliye $F \cdot d r < 0$ — gravity negative kaam karti hai, aur hum $+ m g h$ store karte hain.

2. General gravitational PE — $- \frac{GM m}{r}$

Derivation. Ab force vary karti hai: $F = - \frac{GM m}{r ^{2}} \overset{r}{^}$ (attractive, andar ki taraf). Ek mass ko infinity se distance $r$ tak move karo: $W = \int_{\infty}^{r} (- \frac{GM m}{r ^{'2}}) d r^{'} = - GM m [- \frac{1}{r ^{'}}]_{\infty}^{r} = - GM m (- \frac{1}{r} + 0) = \frac{GM m}{r}$ $U (r) = - W = - \frac{GM m}{r} . ✓$ Why negative? Infinity par $U = 0$ hai. Jaise mass andar girta hai, gravity positive kaam karti hai, isliye $U$ zero se neeche gir jaata hai. Object ek "well" mein hai. Infinity tak wapas chadne ke liye tumhe energy add karni padegi.

3. Elastic (spring) PE — $\frac{1}{2} k x^{2}$

Derivation. Hooke's law: spring force $F = - k x$ hai (restoring, stretch ka virodh karti hai). $0$ se $x$ tak dheere dheere stretch karo: $W_{spring} = \int_{0}^{x} (- k x^{'}) d x^{'} = - \frac{1}{2} k x^{2}$ $U = - W_{spring} = \frac{1}{2} k x^{2} . ✓$ Why $x^{2}$ (not $x$ )? Kyunki force stretch ke saath badhti hai — aakhri bit kheeenchna pehli bit se zyada cost karta hai. Ek linear force ko integrate karne par quadratic energy milti hai. Dhyan do ki $U (+ x) = U (- x)$ : compress karna utna hi store karta hai jitna stretch karna.

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

Worked Examples

Common Mistakes (Steel-manned)

Flashcards

Define potential energy in one sentence.

Conservative force dwara system ko reference configuration se current configuration tak move karne mein kiya gaya kaam ka negative;

Δ U = - W_{co n s}

Why does

Δ U = - W

carry a minus sign?

Force dwara positive kaam stored energy kharach karta hai, isliye

U

kam hoti hai; force ke against kaam karne se energy store hoti hai.

Why do only conservative forces have a PE?

Unka kaam path-independent hota hai, isliye

U

position ka single-valued function ho sakta hai; friction ka kaam path par depend karta hai.

Derive

m g h

Δ U = - \int_{0}^{h} (- m g) d y = m g h

(constant downward force, upward displacement).

Derive

- GM m / r

and state the reference.

U = - \int_{\infty}^{r} (- GM m / r^{'2}) d r^{'} = - GM m / r

; reference

U = 0

r = \infty

Why is bound gravitational PE negative?

Reference infinity par hai; andar girne par gravity positive kaam karti hai,

U

ko zero se neeche "well" mein le jaati hai.

Show

- GM m / r

reduces to

m g h

Δ U \approx GM m h / R^{2} = m g h

using

g = GM / R^{2}

for

h ≪ R

Derive elastic PE.

U = - \int_{0}^{x} (- k x^{'}) d x^{'} = \frac{1}{2} k x^{2}

Why is spring PE quadratic, not linear?

Force

k x

stretch ke saath badhti hai; linear force ko integrate karne par quadratic energy milti hai.

How do you get force from PE?

F = - d U / d x

(force lower PE ki taraf point karta hai).

Energy to escape Earth from surface?

GM m / R

(well ki poori depth, kyunki

U_{\infty} - U_{R} = GM m / R

Whom does PE belong to?

Poore system ko (do interacting objects), kisi ek akele object ko nahi.

Recall Feynman: explain to a 12-year-old

Playground mein ek slide imagine karo. Jab tum upar chadhte ho, tum khud ko uthane ke liye kaam kar rahe ho — tum woh effort store karte ho. Upar pahunchke tumhare paas bahut saara "stored go" hota hai. Neeche slide karo aur stored energy speed mein badal jaati hai. Woh stored climbing effort hi potential energy hai. Ek stretched rubber band bhi same hai: tum apna pull store karte ho, aur woh snap back karta hai. Gravity ise store karti hai jab tum upar jaate ho ( $m g h$ ); ek spring ise store karta hai jab tum use daba te ya kheenchte ho ( $\frac{1}{2} k x^{2}$ ). Deep-space version ek planet ke aas-paas "space mein ek hole" hai — jitna zyada tum andar girte ho, utna gehra (zyada negative) hote ho, aur bahar nikaln ke liye tumhe energy chahiye.

Connections

Work done by a force — PE bas banked kaam hai.
Conservative vs non-conservative forces — sirf conservative forces PE define karti hain.
Conservation of mechanical energy — $K + U = const$ .
Hooke's Law — $F = - k x$ deta hai, $\frac{1}{2} k x^{2}$ mein integrate hota hai.
Gravitation — Newton's law — $- GM m / r$ ka source.
Escape velocity — well depth $GM m / R$ use karta hai.
Force from potential — F = -dU/dx — inverse operation.