Exercises — Conservation of mechanical energy — derivation

Kinetic energy ::: $K=\frac{1}{2}m{v}^2$ Gravitational PE ::: $U=mgh$ (ek chosen zero height se measure karo) Spring PE ::: $U=\frac{1}{2}k{x}^2$ (derived here) Conservation ::: $K_1+U_1=K_2+U_2$ jab sirf conservative forces kaam karein; subscript 1 = start, 2 = end With friction ::: $K_2+U_2=K_1+U_1+W_{nc}$, jahan $W_{nc}<0$ (why)

Jab block neeche jaata hai, $U=mgh$ ghatta hai (height kam hoti hai) aur $K=\frac{1}{2}m{v}^2$ badhta hai (speed badh jaati hai). Akela koi bhi constant nahi hai — yahi woh pocket-swapping hai jo upar figure mein dikhaya gaya hai. Kyunki ramp frictionless hai aur normal force motion ke perpendicular hai (koi kaam nahi karti), sirf gravity kaam karti hai — jo conservative hai. To $E=K+U$ constant hai.

Answer: mechanical energy $E$.

Jhooth. Gravity upar jaate waqt negative work karti hai, lekin woh khoyi hui kinetic energy gravitational PE ke roop mein store ho jaati hai — mechanical total se nahi hati. Paisa "moving" pocket se "stored-up" pocket mein jaata hai. $E=K+U$ frozen rehta hai (koi friction, koi air drag assume nahi).

Sirf gravity kaam karti hai ⇒ conservation. State 1 = upar (rest mein), state 2 = zameen par. Zameen ko $U=0$ lo. $K_1+U_1=K_2+U_2\Rightarrow 0+mgh=\frac{1}{2}m{v}^2+0$ "Mass cancel kyun hoti hai": har surviving term mein $m$ hai, to equation $mgh=\frac{1}{2}m{v}^2$ ban jaati hai. Dono sides ko $m$ se divide karo — ye allowed hai kyunki $m=2\text{kg}\ne 0$, isliye $m$ ek nonzero common factor hai: $\frac{mgh}{m}=\frac{\frac{1}{2}m{v}^2}{m}\Rightarrow gh=\frac{1}{2}{v}^2$ $v=\sqrt{2gh}=\sqrt{2(9.8)(20)}=\sqrt{392}\approx 19.8\text{m/s}$ Dhyaan do mass gayab ho gayi: ek bhaari pathar aur ek halka pathar same speed se zameen par girte hain.

Stored spring PE (state 1) kinetic energy ban jaati hai (state 2). Floor level hai to $U_{grav}$ change nahi hoti aur drop out ho jaati hai. $\frac{1}{2}k{x}^2=\frac{1}{2}m{v}^2\Rightarrow v=x\sqrt{\frac{k}{m}}=0.10\sqrt{\frac{800}{0.50}}=0.10\sqrt{1600}=0.10(40)=4.0\text{m/s}$

Tension har jagah velocity ke perpendicular hai ⇒ zero work karta hai ⇒ sirf gravity kaam karti hai. State 1 = release (string horizontal, rest mein), state 2 = lowest point.

Drop exactly $h=L$ kyun hai: neeche di gayi figure dekho. Pivot fixed hai, release point pivot ke level par hai (string horizontal), aur lowest point pivot se ek poori string-length seedhe neeche hai. To bob ka state 1 se state 2 tak vertical fall, pivot ke level se string ke bottom tak ki vertical distance hai — jo exactly $L$ hai. Horizontal shift energy mein bilkul enter nahi hoti; sirf vertical drop $U=mgh$ ko feed karta hai.

$mgL=\frac{1}{2}m{v}^2$ Dono sides ko $m\ne 0$ se divide karo: $gL=\frac{1}{2}{v}^2$, to $v=\sqrt{2gL}=\sqrt{2(9.8)(1.25)}=\sqrt{24.5}\approx 4.95\text{m/s}$

Friction non-conservative hai, to mechanical energy conserved nahi hai. Corrected law use karo (state 1 = upar, state 2 = neeche): $K_2+U_2=K_1+U_1+W_{friction}$ $U_2=0$, $K_1=0$, $U_1=mgh$ ke saath: $\frac{1}{2}m{v}^2=mgh+W_{friction}=(2.0)(9.8)(5.0)+(-30)=98-30=68\text{J}$ Yahaan hum $m$ cancel nahi karte, kyunki friction term $-30\text{J}$ mein $m$ ka koi factor nahi hai — ab masses har term mein nahi hain. Seedha solve karo: $v=\sqrt{\frac{2(68)}{2.0}}=\sqrt{68}\approx 8.25\text{m/s}$

Neeche di gayi figure mein, loop ka centre height $R$ par hai aur uska top height $2R=6.0\text{m}$ par hai (track ke upar ek puri diameter). Frictionless ⇒ start (state 1, rest, height $h$) aur loop-top (state 2, height $2R$) ke beech conservation: $mgh=\frac{1}{2}m{v}_{top}^2+mg(2R)$ Hum $m$ se divide kyun kar sakte hain: har term mein $m$ hai ($mg(2R)$ term mein bhi), to $m\ne 0$ poori equation ka common factor hai. Dono sides ko $m$ se divide karo: $gh=\frac{1}{2}{v}_{top}^2+g(2R)\Rightarrow \frac{1}{2}{v}_{top}^2=g(h-2R)=9.8(10-6)=39.2$ $v_{top}=\sqrt{78.4}\approx 8.85\text{m/s}$

Teen energy states hain, lekin hume sirf pehli (upar, rest mein — state 1) aur aakhri (max compression, momentarily rest mein — state 2) chahiye. Max compression par $K=0$ kyunki ball ruk gayi hai. Saari gravitational PE spring PE ban gayi hai: $mgh=\frac{1}{2}k{x}^2\Rightarrow k=\frac{2mgh}{x^2}=\frac{2(0.40)(9.8)(1.8)}{(0.12{)}^2}=\frac{14.112}{0.0144}=980\text{N/m}$

Is problem mein do alag reference points hain aur hume unhe theek rakha chahiye — neeche di figure dekho.

• Spring apni stored energy apni natural (uncompressed) length se measure karta hai: $U_{spring}=\frac{1}{2}k{x}^2$ jahan $x$ compression hai. Release par spring $x=0.20\text{m}$ compress hai, to $\frac{1}{2}k{x}^2$ store hai. Jab ball spring chhod deti hai, spring natural length par wapas aati hai aur kuch store nahi karti.
• Gravity apni stored energy ek height-zero se measure karti hai jo hum choose karte hain. Hum gravity ka zero release point par rakhte hain (compressed position). To release par $U_{grav}=0$; highest point par $U_{grav}=mgH$.

Ye do references independent hain — ek spring ka, ek gravity ka — aur ye theek hai, kyunki har energy store alag bookkeep hoti hai. "Spring natural length" aur "release point" ke beech offset exactly compression $x$ hai, lekin woh kabhi explicitly appear nahi hota: hume sirf spring ki stored energy release par (state 1) aur gravitational PE top par (state 2) chahiye. Dono states mein $K=0$ hai (ball momentarily rest mein).

State 1 (release): $K=0$, spring stores $\frac{1}{2}k{x}^2$, $U_{grav}=0$. State 2 (top): $K=0$, spring stores $0$, $U_{grav}=mgH$. $\frac{1}{2}k{x}^2=mgH$ $H=\frac{k{x}^2}{2mg}=\frac{200(0.20{)}^2}{2(0.50)(9.8)}=\frac{200(0.04)}{9.8}=\frac{8}{9.8}\approx 0.816\text{m}$ ($H$ release point se measure kiya gaya hai, exactly jaisa pucha gaya था।)

"Just completes" ka matlab hai top par minimum speed jahan track still zero push kar sake — gravity saari centripetal force provide kare: $mg=\frac{m{v}_{top}^2}{R}$ Dono sides ko $m\ne 0$ se divide karo: $g=\frac{v_{top}^2}{R}$, to $v_{top}^2=gR$. Ab start (rest at $h$, state 1) se loop-top (height $2R$, state 2) tak conservation: $mg{h}_{min}=\frac{1}{2}m{v}_{top}^2+mg(2R)=\frac{1}{2}m(gR)+2mgR=\frac{5}{2}mgR$ $m\ne 0$ se divide karo: $\boxed{h_{min}=\frac{5}{2}R=2.5R}$ Mass aur $g$ se independent — ek clean, famous result. $R=3.0\text{m}$ ke liye: $h_{min}=7.5\text{m}$.

Total energy turning point ($x=A$, saari spring PE, $v=0$) par energy ke barabar hai: $E=\frac{1}{2}k{A}^2$. General $x$ par: $E=\frac{1}{2}m{v}^2+\frac{1}{2}k{x}^2$. Barabar rakho: $\frac{1}{2}k{A}^2=\frac{1}{2}m{v}^2+\frac{1}{2}k{x}^2\Rightarrow m{v}^2=k(A^2-x^2)$ $v=\sqrt{\frac{k}{m}}\sqrt{A^2-x^2}$ Check $x=0$ (equilibrium): $v=\sqrt{k/m}A$ — ye maximum speed hai, sahi hai kyunki wahan saari energy kinetic hai. Check $x=A$ (extreme): $v=0$ — sahi hai, mass turning point par momentarily rest mein hai. Ye SHM ka energy view hai.

Energy mein koi direction nahi hoti — $K=\frac{1}{2}m{v}^2$ mein sirf speed ka magnitude jaata hai. To $u$ upar point kare ya neeche, same $K_1$ milta hai. $\frac{1}{2}m{u}^2+mgh=\frac{1}{2}m{v}^2\Rightarrow v=\sqrt{u^2+2gh}=\sqrt{25+2(9.8)(15)}=\sqrt{25+294}=\sqrt{319}\approx 17.9\text{m/s}$ Same speed $5.0\text{m/s}$ se upar phenka jaaye: wapas neeche aate waqt $h=15\text{m}$ par exactly $5.0\text{m/s}$ se guzarega (energy restore ho jaati hai), phir aage badhta hai — same landing speed. Energy direction ki parwah nahi karti; ye iska superpower hai.