Conservation of mechanical energy — derivation: Worked examples

1.3.8 · D3 · Physics › Work, Energy & Power › Conservation of mechanical energy — derivation

Do symbols jinhe hum poori tarah use karenge:

$K = \frac{1}{2} m v^{2}$ = "moving-money" pocket (kinetic energy).
$U$ = "stored-money" pocket (potential energy): $U_{g r a v} = m g h$ height ke liye, $U_{s p r in g} = \frac{1}{2} k x^{2}$ dabaye/khiche hue spring ke liye.

Master sentence, words mein: moving pocket plus stored pocket, start mein joda, end mein bhi same sum hoga — jab tak koi non-conservative force (friction, ek push) beech mein money add ya remove na kare.

$start total K_{1} + U_{1} + non-conservative forces ka money add/remove W_{n c} = end total K_{2} + U_{2}$

Yahan $W_{n c}$ non-conservative forces (friction, applied pushes) dwara kiya gaya work hai. Jab $W_{n c} = 0$ hota hai, yeh clean law $K_{1} + U_{1} = K_{2} + U_{2}$ mein collapse ho jaata hai.

The scenario matrix

Is topic ke har problem inn cells mein se ek hai. Neeche har example us cell ke saath tagged hai jise woh cover karta hai.

Cell	Kya special hai	Example
A. Pure drop	sirf gravity, seedhi line, rest se start	Ex 1
B. Path-independence	curved/slope path, gravity phir bhi conservative	Ex 2
C. Spring only	conservative spring, koi height change nahi	Ex 3
D. Mixed height + spring	do stored pockets swap kar rahe hain	Ex 4
E. Non-zero start speed	$K_{1} \neq = 0$ — kuch bhi rest se start nahi	Ex 5
F. Friction (the thief)	$W_{n c} < 0$ , mechanical energy conserved NAHI	Ex 6
G. Sign / direction trap	"upar" kaun sa direction hai, $U$ + ya − hai, upar jaana vs neeche jaana	Ex 7
H. Degenerate / limiting	$h \to 0$ , $v_{1} = v_{2}$ , $k \to \infty$ , zero mass check	Ex 8
I. Real-world word problem	roller-coaster / vertical loop, minimum-speed condition	Ex 9
J. Exam twist	unknown height/length nikalna, back-solve karna	Ex 10

Example 1 — Pure drop (Cell A)

Worked example Statement

$m = 2 kg$ mass ka ek pathar height $h = 20 m$ se rest se giraya jaata hai. $g = 10 m/s^{2}$ lo. Ground se theek pehle uski speed nikalo.

Forecast: Padhne se pehle guess karo: kya answer mass par depend karta hai? $10 m/s$ se zyada ya kam speed?

Step 1 — Do states aur $U = 0$ kahaan hai yeh choose karo. State 1 = top (rest mein, $v_{1} = 0$ , $h_{1} = 20$ ). State 2 = ground ( $h_{2} = 0$ ). Reference $U = 0$ ground par rakho. Yeh step kyun? Energy conservation do snapshots compare karta hai, journey nahi. Hume unhe name karna hoga aur height ke liye ek "sea level" fix karna hoga taaki $U = m g h$ ka koi meaning ho.

Step 2 — Check karo ki law apply hota hai. Sirf gravity kaam karta hai (air resistance ignore kiya). Gravity conservative hai $\Rightarrow W_{n c} = 0 \Rightarrow K_{1} + U_{1} = K_{2} + U_{2}$ . Yeh step kyun? Agar hum yeh check skip karein toh hum galti se clean law apply kar sakte hain jab koi chor present ho.

Step 3 — Likho aur solve karo. $0 + m g h = \frac{1}{2} m v^{2} + 0 \Rightarrow v = 2 g h$ Mass cancel ho jaata hai — mass ke baare mein forecast ka jawaab mil gaya: yeh matter nahi karta. $v = 2 \cdot 10 \cdot 20 = 400 = 20 m/s$

Verify: Units: $(m/s^{2}) (m) = m^{2} / s^{2} = m/s$ . ✓ Kinematics se cross-check $v^{2} = 2 g h = 400$ . ✓

Example 2 — Path independence: slide vs fall (Cell B)

Worked example Statement

Do identical balls ( $m = 1 kg$ ) same height $h = 5 m$ se rest se start karte hain. Ball P seedha neeche girta hai; ball Q ek frictionless curved ramp se slide karke same ground level par aata hai. Unki ground speeds compare karo. ( $g = 10$ .)

Forecast: Guess karo: kya Q slower hai kyunki usne "zyada lamba raasta liya"?

Step 1 — Yaad karo gravity conservative kyun hai. Gravity dwara work sirf vertical drop par depend karta hai, path par nahi (yeh Conservative and non-conservative forces se definition hai). $W_{g r a v} = m g Δ h$ horizontal wiggles se independent hai. Yeh step kyun? Yahi woh fact hai jo dono balls ko ek equation share karne deta hai.

Step 2 — Same start, same end, same law. Dono ke liye: $m g h = \frac{1}{2} m v^{2}$ . Ramp ki shape kabhi aai hi nahi. Yeh step kyun? Yeh dikhata hai ki path-length ek red herring hai; sirf height difference $U$ ko feed karta hai.

Step 3 — Solve karo. $v = 2 g h = 2 \cdot 10 \cdot 5 = 100 = 10 m/s (dono ke liye)$

Verify: Dono $10 m/s$ hain. Forecast ("Q slower hai") galat tha — path independence hi poora point hai. Units Ex 1 ki tarah check hote hain. ✓

Example 3 — Spring only, koi height change nahi (Cell C)

Worked example Statement

Ek block $m = 0.5 kg$ frictionless horizontal floor par ek spring ( $k = 200 N/m$ ) ke against push kiya jaata hai, jisse woh $x = 0.10 m$ compress hoti hai, phir release hota hai. Launch speed nikalo.

Forecast: Yahan koi height change nahi — kinetic energy kahaan se aati hai?

Step 1 — Do stored pockets identify karo. Koi height change nahi ⇒ $U_{g r a v}$ dono states mein same hai, cancel ho jaata hai. Sirf changing $U$ spring ka hai: $U_{s p r in g} = \frac{1}{2} k x^{2}$ (parent note mein build kiya gaya). Yeh step kyun? Tumhe pata hona chahiye kaun sa potential energy change ho raha hai; constant terms drop out ho jaate hain.

Step 2 — Conservation apply karo. State 1 = compressed, rest mein. State 2 = spring natural length par, block moving. $\frac{1}{2} k x^{2} = \frac{1}{2} m v^{2} \Rightarrow v = x \frac{k}{m}$ Yeh step kyun? Spring ka stored money block ka moving money ban jaata hai.

Step 3 — Numbers. $v = 0.10 \frac{200}{0.5} = 0.10 400 = 0.10 \cdot 20 = 2 m/s$

Verify: Stored energy $= \frac{1}{2} (200) (0.10)^{2} = 1 J$ ; kinetic $= \frac{1}{2} (0.5) (2)^{2} = 1 J$ . Equal. ✓ Units: $m (N/m) / kg = m s^{- 2} = m/s$ . ✓ Dekho Spring potential energy (Hooke's law).

Example 4 — Mixed: ramp phir spring (Cell D)

Worked example Statement

Ek ball $m = 1 kg$ height $h = 0.8 m$ se frictionless ramp neeche roll karti hai aur ek horizontal spring ko maximum $x = 0.20 m$ compress karti hai. $k$ nikalo. ( $g = 10$ .)

Forecast: Maximum compression par, kya ball move kar rahi hai ya momentarily still hai?

Step 1 — Key snapshot: maximum compression. Maximum compression par ball momentarily at rest hoti hai ( $K = 0$ ): woh bounce back karne se pehle ruk gayi hai. Yeh step kyun? Is special instant ko choose karne se $K_{2} = 0$ ho jaata hai, isliye sara gravitational money spring money ban jaata hai.

Step 2 — Do stored pockets swap karte hain. $m g h = \frac{1}{2} k x^{2} \Rightarrow k = \frac{2 m g h}{x ^{2}}$ Yeh step kyun? $U_{g r a v}$ top par poora $U_{s p r in g}$ bottom par ban jaata hai — dono ends par kinetic zero hai.

Step 3 — Numbers. $k = \frac{2 \cdot 1 \cdot 10 \cdot 0.8}{( 0.20 ) ^{2}} = \frac{16}{0.04} = 400 N/m$

Verify: $U_{g r a v} = m g h = 8 J$ ; $U_{s p r in g} = \frac{1}{2} (400) (0.20)^{2} = 8 J$ . Equal. ✓ $k$ ke units: $J/m^{2} = N/m$ . ✓

Example 5 — Non-zero starting speed (Cell E)

Worked example Statement

Ek ball $m = 1 kg$ height $h = 5 m$ se neeche ki taraf $v_{1} = 3 m/s$ se throw ki jaati hai. Uski ground speed nikalo. ( $g = 10$ .)

Forecast: Kya answer giraye gaye ball ke $10 m/s$ (Ex 2) se zyada hona chahiye?

Step 1 — Poora $K_{1}$ term rakho. Yahan kuch bhi rest se start nahi hota, isliye $K_{1} = \frac{1}{2} m v_{1}^{2}$ carry karna hoga, zero drop nahi karna. Yeh step kyun? Sabse common galti $K_{1} = 0$ auto-set karna hai. Yeh tabhi karo jab "from rest" bola gaya ho.

Step 2 — Dono kinetic terms ke saath conservation. $\frac{1}{2} m v_{1}^{2} + m g h = \frac{1}{2} m v_{2}^{2} \Rightarrow v_{2} = v_{1}^{2} + 2 g h$ Yeh step kyun? Dhyaan do ki throw ki direction ( $3 m/s$ neeche) kabhi matter nahi karti — energy ek scalar hai; sirf magnitude $v_{1}^{2}$ ke through enter karta hai.

Step 3 — Numbers. $v_{2} = 3^{2} + 2 \cdot 10 \cdot 5 = 9 + 100 = 109 \approx 10.44 m/s$

Verify: $10 m/s$ (dropped case) se zyada hai jaise forecast kiya tha — extra $9 J$ initial KE dikhta hai. ✓ $v_{1}^{2} + 2 g h$ ke units: $m/s$ . ✓

Example 6 — The thief: friction (Cell F)

Worked example Statement

Ek block $m = 2 kg$ ek $3 0^{\circ}$ ramp par length $L = 4 m$ slide karta hai, rest se start karke. Kinetic friction coefficient $μ = 0.25$ . Bottom par speed nikalo. ( $g = 10$ .)

Forecast: Kya speed frictionless answer $2 g h$ se zyada hogi ya kam?

Step 1 — Height dropped. Vertical drop $h = L sin 3 0^{\circ} = 4 \cdot 0.5 = 2 m$ . Yeh step kyun? Sirf displacement ka vertical component $U_{g r a v}$ change karta hai.

Step 2 — Friction work (woh money jo chor remove karta hai). Normal force $N = m g cos 3 0^{\circ}$ . Friction $f = μ N$ , motion ke opposite, distance $L$ par: $W_{n c} = - μ m g cos 3 0^{\circ} \cdot L = - 0.25 \cdot 2 \cdot 10 \cdot \frac{3}{2} \cdot 4 \approx - 17.32 J$ Yeh step kyun? Friction non-conservative hai — uska work $W_{n c}$ ke roop mein aana chahiye, potential energy ke roop mein NAHI. Dekho Energy lost to friction.

Step 3 — Modified conservation. $K_{1} + U_{1} + W_{n c} = K_{2} + U_{2} \Rightarrow 0 + m g h + W_{n c} = \frac{1}{2} m v^{2} + 0$ $m g h + W_{n c} = 40 + (- 17.32) = 22.68 J = \frac{1}{2} m v^{2}$ $v = \frac{2 \cdot 22.68}{2} = 22.68 \approx 4.76 m/s$

Verify: Frictionless mein $v = 2 g h = 40 \approx 6.32 m/s$ hota. Hamara answer chota hai jaise forecast kiya tha — chor ne $17.32 J$ le liye. ✓ $W_{n c}$ ke units: $N \cdot m = J$ . ✓

Example 7 — Sign / direction trap: upar throw karna (Cell G)

Worked example Statement

Ek ball $m = 0.5 kg$ seedha upar $v_{1} = 8 m/s$ se throw ki jaati hai. Maximum height nikalo. ( $g = 10$ .)

Forecast: Sabse uunche point par, kya ball move kar rahi hai? Wahan $K$ kya hai?

Step 1 — Top ko sahi se padho. Maximum height par ball momentarily rest mein hoti hai ( $v_{2} = 0$ , $K_{2} = 0$ ). Yahi trap hai: bahut log top par kinetic term rakh lete hain. Yeh step kyun? Special instant "highest point" zero velocity se define hota hai.

Step 2 — Conservation, $U = 0$ launch point par choose karo. $\frac{1}{2} m v_{1}^{2} + 0 = 0 + m g h_{ma x} \Rightarrow h_{ma x} = \frac{v _{1}^{2}}{2 g}$ Yeh step kyun? Upar jaate waqt, kinetic money stored (height) money mein convert hoti hai — fall se opposite direction, lekin same equation. Sign automatically handle hota hai kyunki $U = m g h$ $h$ ke badhne ke saath badhta hai.

Step 3 — Numbers. $h_{ma x} = \frac{8 ^{2}}{2 \cdot 10} = \frac{64}{20} = 3.2 m$

Verify: Kinematics $v^{2} = v_{1}^{2} - 2 g h_{ma x} \Rightarrow 0 = 64 - 20 h_{ma x} \Rightarrow h_{ma x} = 3.2 m$ . ✓ Mass cancel hua — sahi hai, gravity sab masses ko equally accelerate karti hai. Units: $\frac{( m/s ) ^{2}}{m/s ^{2}} = m$ . ✓

Example 8 — Degenerate aur limiting cases (Cell H)

Worked example Statement

Master equation ko teen "edge" situations mein check karo taaki pata chale yeh sahi behave karta hai: (a) $h \to 0$ : zero height se drop. (b) $v_{1} = v_{2}$ : dono states par same speed. (c) $k \to \infty$ : ek rigid wall spring.

Forecast: Kya inn mein se kisi se finite, sensible answer aani chahiye ya blow up honi chahiye?

Step 1 — (a) Zero height. $v = 2 g h$ mein $h = 0$ rakhne par $v = 0$ aata hai. Sensible: koi drop nahi, koi speed gain nahi. Yeh step kyun? Ek sahi law ko trivial input par trivial result dena chahiye.

Step 2 — (b) Equal speeds. Agar $v_{1} = v_{2}$ toh $Δ K = 0$ , isliye $Δ U = 0$ : koi net height change nahi (jaise ball launch height par same speed magnitude ke saath wapas aaye). Ek poori upar-phir-neeche flight ke saath consistent hai jisme friction ignore kiya gaya ho. Yeh step kyun? Confirm karta hai ki energy conservation symmetric flight predict karta hai.

Step 3 — (c) Rigid spring. Spring launch ke liye $v = x k / m$ . Target $v$ ke liye zaruri compression: $x = v m / k$ . Jaise $k \to \infty$ , $x \to 0$ : infinitely stiff spring ko same energy deliver karne ke liye (almost) koi compression nahi chahiye — lekin stored energy $\frac{1}{2} k x^{2}$ finite tabhi rehti hai jab $x$ shrink ho. Concretely $v = 2, m = 0.5$ ke saath: $x = 2 0.5/ k$ . $k = 800$ par: $x = 2 0.000625 = 0.05 m$ . Yeh step kyun? Limiting behaviour reveal karta hai ki koi division-by-zero ya blow-up nahi — physics finite rehti hai.

Verify: (a) $0$ . (b) $Δ K = 0$ . (c) $k = 800$ par $x = 0.05 m$ ; energy check $\frac{1}{2} (800) (0.05)^{2} = 1 J = \frac{1}{2} (0.5) (2)^{2}$ . ✓

Example 9 — Real-world: vertical loop minimum speed (Cell I)

Worked example Statement

Ek frictionless track par cart ko radius $R = 2 m$ ka vertical circular loop complete karna hai. Loop ke bottom se minimum kitni height $h$ par use rest se start karna hoga taaki woh top se just nikal sake? ( $g = 10$ .)

Forecast: Loop ke bilkul top par, centripetal force kaun provide karta hai — aur minimum speed kya hai jo cart ko track par rakhti hai?

Step 1 — Top par minimum-speed condition. "Just makes it" ka matlab hai gravity akele centripetal pull provide karti hai, track force $= 0$ : $m g = \frac{m v _{t o p}^{2}}{R} \Rightarrow v_{t o p}^{2} = g R$ . Yeh step kyun? Yeh extra physics (circular motion) hai jo woh speed set karti hai jise hum phir energy conservation mein feed karte hain.

Step 2 — Start se energy (rest, height $h$ ) loop top tak (height $2 R$ ). $m g h = \frac{1}{2} m v_{t o p}^{2} + m g (2 R)$ Yeh step kyun? Loop top bottom se $2 R$ height par hai; conservation dono heights ko link karta hai.

Step 3 — $h$ ke liye solve karo. $g h = \frac{1}{2} g R + 2 g R = \frac{5}{2} g R \Rightarrow h = \frac{5}{2} R = 2.5 \cdot 2 = 5 m$

Verify: $v_{t o p}^{2} = g R = 20$ , isliye $v_{t o p} = 20 \approx 4.47 m/s$ . Energy: start $m g h = 100 m$ ; top $\frac{1}{2} m (20) + m g (4) = 10 m + 40 m = 50 m$ ... ruko recompute karein: $m g (2 R) = m \cdot 10 \cdot 4 = 40 m$ , $\frac{1}{2} m (20) = 10 m$ , total $50 m$ . Hmm start $50 m$ ke equal hona chahiye, jisse $h = 5$ milta hai. ✓ Consistent ( $100 m \neq =$ ... restate karte hain: $m g h = 100 \cdot 0.5$ ? mass cancel hota hai; $g h = 10 \cdot 5 = 50 = \frac{5}{2} (10) (2) = 50$ ). ✓

Example 10 — Exam twist: unknown back-solve karna (Cell J)

Worked example Statement

Ek block frictionless incline par upar launch hota hai aur rukne se pehle slope ke along $d$ distance travel karta hai. Woh $v_{1} = 6 m/s$ speed se start hua tha aur incline $3 7^{\circ}$ par hai ( $sin 3 7^{\circ} = 0.6$ ). $d$ nikalo. ( $g = 10$ .)

Forecast: Unknown slope ke along distance hai, height nahi. Dono kaise related hain?

Step 1 — Slope distance ko height se relate karo. Height gained $h = d sin 3 7^{\circ} = 0.6 d$ . Yeh step kyun? $U_{g r a v}$ vertical rise par depend karta hai, isliye hume along-slope distance ko height mein convert karna hoga.

Step 2 — Conservation (block rukta hai ⇒ $K_{2} = 0$ ). $\frac{1}{2} m v_{1}^{2} = m g h = m g (0.6 d) \Rightarrow d = \frac{v _{1}^{2}}{2 g ( 0.6 )}$ Yeh step kyun? Saara initial kinetic money climbing mein kharach hota hai; final $K = 0$ set karne se equation close ho jaata hai.

Step 3 — Numbers. $d = \frac{6 ^{2}}{2 \cdot 10 \cdot 0.6} = \frac{36}{12} = 3 m$

Verify: Height gained $h = 0.6 \cdot 3 = 1.8 m$ ; $m g h = m \cdot 10 \cdot 1.8 = 18 m$ ; initial $\frac{1}{2} m (36) = 18 m$ . Equal. ✓ Units: $\frac{( m/s ) ^{2}}{m/s ^{2}} = m$ . ✓

Recall Matrix par quick self-test

Kaun sa cell $K_{1} + U_{1} = K_{2} + U_{2}$ directly use karne se rokta hai? ::: Cell F — friction, kyunki $W_{n c} \neq = 0$ . Spring-only problem mein $U_{g r a v}$ equation se kyun gayab ho jaata hai? ::: Height change nahi hoti, isliye $U_{g r a v}$ dono states mein equal hota hai aur cancel ho jaata hai. Maximum spring compression (ya throw ke peak) par kinetic energy kya hoti hai? ::: Zero — object momentarily rest mein hota hai. Kya giraye gaye ball ki speed ke liye mass matter karta hai? ::: Nahi — yeh cancel ho jaata hai, $v = 2 g h$ deta hai.

Worked examples — Conservation of mechanical energy — derivation

The scenario matrix

Example 1 — Pure drop (Cell A)

Example 2 — Path independence: slide vs fall (Cell B)

Example 3 — Spring only, koi height change nahi (Cell C)

Example 4 — Mixed: ramp phir spring (Cell D)

Example 5 — Non-zero starting speed (Cell E)

Example 6 — The thief: friction (Cell F)

Example 7 — Sign / direction trap: upar throw karna (Cell G)

Example 8 — Degenerate aur limiting cases (Cell H)

Example 9 — Real-world: vertical loop minimum speed (Cell I)

Example 10 — Exam twist: unknown back-solve karna (Cell J)

Connections