Exercises — Conservative forces — path-independent work, potential energy defined

KYA: Closed loop ke around har leg ka kaam add karo. KYUN: Conservative $\iff$ $\oint \vec{F}\cdot d\vec{r}=0$. $W_{\text{loop}}=W_{A\to B}+W_{B\to A}=(+6)+(-6)=0\text{J}.$ Dono paths $A\to B$ ne same $+6\text{J}$ diya (path-independent) aur loop zero hai. Haan, yeh ek conservative force ke saath consistent hai.

Friction (c). Iska kaam path-dependent hai — yeh hamesha motion ka oppose karta hai, isliye lamba path zyada energy lose karta hai aur loop work $\ne 0$ hota hai. Gravity aur springs path-independent hain, isliye dono ke paas $U$ hai. Answer: friction. Dekho Friction and dissipation.

Floor reference: $U=mgy=(2.0)(9.8)(1.5)=29.4\text{J}$. Ceiling reference: height ceiling se measure karo, isliye shelf $y^{\prime }=1.5-3.0=-1.5\text{m}$ par hai: $U^{\prime }=mg{y}^{\prime }=(2.0)(9.8)(-1.5)=-29.4\text{J}.$ Floor par drop:

• Floor ref: $\Delta U=U_{\text{floor}}-U_{\text{shelf}}=0-29.4=-29.4\text{J}$.
• Ceiling ref: floor $y^{\prime }=-3.0$ par hai, isliye $U_{\text{floor}}^{\prime }=(2.0)(9.8)(-3.0)=-58.8\text{J}$; $\Delta U=-58.8-(-29.4)=-29.4\text{J}$.

Same $\Delta U=-29.4\text{J}$ chahe absolute $U$ values alag hon. Sirf differences hi physical hain. Dekho Gravitational potential energy.

(a) $U=\frac{1}{2}k{x}^2=\frac{1}{2}(200)(0.10{)}^2=1.0\text{J}$. (b) $F=-\frac{dU}{dx}=-kx=-(200)(0.10)=-20\text{N}$. Minus ka matlab hai force $x=0$ ki taraf wapas point karta hai (restoring). Magnitude $20\text{N}$. Dekho Spring / elastic potential energy.

Constant force ke liye, $W=\vec{F}\cdot \Delta \vec{r}$ path se independent hota hai (yahi toh baat hai). Path 1: $\Delta \vec{r}=(2,1)$, isliye $W_1=(3)(2)+(4)(1)=6+4=10\text{J}$. Path 2, leg 1 $(0,0)\to (2,0)$: $\Delta \vec{r}=(2,0)$, $W=(3)(2)+(4)(0)=6\text{J}$. Path 2, leg 2 $(2,0)\to (2,1)$: $\Delta \vec{r}=(0,1)$, $W=(3)(0)+(4)(1)=4\text{J}$. $W_2=6+4=10\text{J}=W_1.$ ✔ Path-independent, isliye yeh constant force conservative hai.

Friction: dono legs par motion ko oppose karta hai, isliye har baar negative hai: $W_f=-fd-fd=-(5.0)(3.0)-(5.0)(3.0)=-30\text{J}\ne 0.$ Loop work non-zero $\Rightarrow$ non-conservative, koi $U$ nahi. Gravity horizontal loop par: force vertical hai, motion horizontal, $\vec{F}\cdot d\vec{r}=0$ har jagah, isliye $W_{\text{grav}}=0$. Gravity zero loop work deta hai (jaisa ek conservative force ke liye hona chahiye).

(a) KYA: $F=-dU/dx$ ko undo karo, yaani $dU=-Fdx$. KYUN: $U$, $-F$ ka anti-derivative hai (dekho Work done by a variable force). $U(x)=-\int Fdx=-\int (-kx+a{x}^3)dx=\frac{1}{2}k{x}^2-\frac{1}{4}a{x}^4+C.$ $U(0)=0$ ke saath, $C=0$: $U(x)=\frac{1}{2}(8)x^2-\frac{1}{4}(2)x^4=4x^2-\frac{1}{2}{x}^4.$ (b) Force zero: $F=0\Rightarrow -8x+2x^3=0\Rightarrow 2x(x^2-4)=0\Rightarrow x=0,+2,-2$ (metres mein).

KYUN second derivative: $F=-dU/dx$, isliye equilibrium ke paas force ek spring jaisi hai $F\approx -(U^{\prime \prime })\delta$. Agar $U^{\prime \prime }>0$ (valley) toh force tumhe wapas push karta hai → stable; agar $U^{\prime \prime }<0$ (hill) toh yeh tumhe door push karta hai → unstable. $U^{\prime }(x)=8x-2x^3,U^{\prime \prime }(x)=8-6x^2.$

• $x=0$: $U^{\prime \prime }=8>0$ → stable (valley ka bottom).
• $x=\pm 2$: $U^{\prime \prime }=8-6(4)=-16<0$ → unstable (do hilltops).

Figure dekho: curve $x=0$ par dip karta hai aur $x=\pm 2$ par crest karta hai — exactly crest par rakha ball roll off kar jaata hai. Yeh Gradient and vector fields se connect hota hai (force = $-\nabla U$).

KYA: Chaar edges par chalo, har ek par $\vec{F}\cdot d\vec{r}$ add karte hue.

• $O\to A$: $x$ ke along, $y=0$, $d\vec{r}=(dx,0)$, $\vec{F}\cdot d\vec{r}=ydx=0$. → $0$.
• $A\to B$: $y$ ke along, $x=1$, $\vec{F}\cdot d\vec{r}=xdy=1dy$, ${\int }_0^1=1$.
• $B\to C$: $x$ ke along 1→0, $y=1$, $\vec{F}\cdot d\vec{r}=ydx=1dx$, ${\int }_1^0=-1$.
• $C\to O$: $y$ ke along 1→0, $x=0$, $\vec{F}\cdot d\vec{r}=xdy=0$. → $0$. $\oint =0+1-1+0=0.$ Zero loop work → conservative. $U$ nikalo: chahiye $-\partial U/\partial x=F_x=y$ aur $-\partial U/\partial y=F_y=x$. Pehle ko integrate karo: $U=-xy+h(y)$. Phir $-\partial U/\partial y=x-h^{\prime }(y)=x\Rightarrow h^{\prime }(y)=0$. Isliye $U(x,y)=-xy+C.$ Check: $\vec{F}=-\nabla U=-(-y,-x)=(y,x)$ ✔.

KYUN energy conservation: incline frictionless hai aur spring aur gravity dono conservative hain, isliye mechanical energy conserved hai (dekho Conservation of mechanical energy). Spring energy poori gravitational PE mein convert hoti hai highest point par (speed wahan $=0$): $\frac{1}{2}k{x}^2=mg\Delta h,\Delta h=s\sin 30^{\circ }.$ Left side: $\frac{1}{2}(400)(0.20{)}^2=\frac{1}{2}(400)(0.04)=8.0\text{J}$. $s$ solve karo: $s=\frac{\frac{1}{2}k{x}^2}{mg\sin 30^{\circ }}=\frac{8.0}{(0.50)(9.8)(0.5)}=\frac{8.0}{2.45}\approx 3.27\text{m}.$ Block $\approx 3.27\text{m}$ slope par upar slide karta hai. (Agar tum total climb compressed position se measure karna chahte ho, toh $0.20\text{m}$ add karo jo spring ne incline ke through push kiya.)

Q: Kaunsa ek test decide karta hai ki kisi force ke paas potential energy hai? ::: Kya uska kaam path-independent hai — equivalently, kya $\oint \vec{F}\cdot d\vec{r}=0$ hai? Agar haan, toh $U$ exist karta hai $\vec{F}=-\nabla U$ ke saath.