Worked examples — Conservative forces — path-independent work, potential energy defined

1.3.6 · D3 · Physics › Work, Energy & Power › Conservative forces — path-independent work, potential energ

Shuru karne se pehle, do tools ki ek yaad-dahaani jinpar hum poori tarah rely karte hain:

Scenario matrix

Is topic ka har problem in mein se kisi ek cell mein aata hai. Aage ke examples tagged hain us cell ke saath jo woh cover karte hain.

Example 1 — Gravity upar aur neeche (cells A, B)

Forecast: Gravity ke kaam ka sign upar jaate waqt aur neeche aate waqt guess karo. Kya woh cancel ho jaate hain?

1. Har height par $U$ likho. Ye step kyun? Humne parent mein $U_{\text{grav}}=mgy$ derive ki thi, toh sirf do piggy-bank values padhni hain. $U_{\text{floor}}=mg(0)=0,U_{\text{shelf}}=mg(3)=2\cdot 10\cdot 3=60\text{J}.$
2. Upar ki leg (cell A): Ye step kyun? $W=U_A-U_B$ use karo jahan $A=$floor, $B=$shelf. $W_{\uparrow }=U_{\text{floor}}-U_{\text{shelf}}=0-60=-60\text{J}.$ Gravity negative kaam karti hai — woh lift se ladhti hai, toh 60 J piggy bank mein daal deti hai.
3. Neeche ki leg (cell B): Ye step kyun? Ab $A=$shelf, $B=$floor — endpoints swap ho gaye. $W_{\downarrow }=U_{\text{shelf}}-U_{\text{floor}}=60-0=+60\text{J}.$ Gravity ab positive kaam karti hai — piggy bank apne 60 J wapas deta hai.
4. Round trip: $W_{\uparrow }+W_{\downarrow }=-60+60=0.$

Verify: Round trip ek closed loop hai, aur gravity conservative hai, toh $\oint \vec{F}\cdot d\vec{r}$ zaroor $0$ hona chahiye. Hai. ✔ Units: $\text{kg}\cdot {\text{m/s}}^2\cdot \text{m}=\text{N}\cdot \text{m}=\text{J}$. ✔

Neeche ki figure ye see-saw draw karti hai: magenta arrow padhho (upar, $W=-60$ J, gravity ladhti hui) aur orange arrow (neeche, $W=+60$ J, gravity wapas deti hui). Baayein taraf ke do $U$-labels notice karo — $0$ J at the floor, $60$ J at the shelf — aur dekho ki arrows ek doosre ko exactly undo karte hain, isliye loop zero hota hai.

Example 2 — Same start, same end, alag raaste (cell C)

Forecast: Zyada steps, zyada distance — kya gravity zyada charge karti hai?

1. Horizontal legs khatam karo. Ye step kyun? Gravity $\vec{F}=-mg\hat{y}$ hai; kisi bhi horizontal move par $\vec{F}\cdot d\vec{r}=0$ hota hai, toh teeno "daayein" legs kuch contribute nahi karti.
2. Vertical legs jodo. Ye step kyun? Har "upar $1$ m" leg $W=-mg(1)=-20\text{J}$ karti hai, aur teeno hain. $W_{\text{stairs}}=3\times (-20)=-60\text{J}.$
3. Ex 1 ki seedhi lift se compare karo: $W_{\uparrow }=-60\text{J}$. Bilkul same.

Verify: Path-independence conservative ki definition hi hai — same endpoints ($0\to 3$ m) par $-mgh=-2\cdot 10\cdot 3=-60\text{J}$ milna chahiye, chahe kitne bhi corners turn karein. ✔

Example 3 — Spring stretched vs compressed (cell D)

Forecast: Compression negative $x$ use karta hai. Kya isse $U$ negative ho jaata hai?

1. Stretch ke liye $U_{\text{spring}}=\frac{1}{2}k{x}^2$ apply karo. Ye step kyun? Ye woh formula hai jo humne $F=-kx$ integrate karke derive ki thi. $U_{+}=\frac{1}{2}(200)(0.10{)}^2=\frac{1}{2}(200)(0.01)=1\text{J}.$
2. Ab compress karo ($x=-0.10$). Ye step kyun? Negative $x$ plug in karo — lekin woh squared hai. $U_{-}=\frac{1}{2}(200)(-0.10{)}^2=\frac{1}{2}(200)(0.01)=1\text{J}.$ Same energy! $x^2$ ki wajah se $U$ ek symmetric bowl hai $x=0$ ke baare mein — chahe kheeencho ya dhaako, energy store hoti hai.
3. Forces direction mein alag hain. Ye step kyun? $F=-kx$ use karo, jo $x$ ka sign rakhta hai. $F_{+}=-200(0.10)=-20\text{N}(\text{wapas kheenchta hai, }0\text{ ki taraf}),F_{-}=-200(-0.10)=+20\text{N}(\text{bahar dhakelta hai, }0\text{ ki taraf}).$

Verify: Dono cases mein force $x=0$ ki taraf point karti hai — $U$-bowl ka bottom, yaani potential slope ke neeche — bilkul waise jaisa $F=-dU/dx$ demand karta hai. Energies symmetry se match karti hain $(-x{)}^2=x^2$. ✔ Units: $\text{N/m}\cdot {\text{m}}^2=\text{N}\cdot \text{m}=\text{J}$. ✔ Dekho Spring / elastic potential energy.

Figure parabolic $U$-bowl dikhata hai. Magenta dot (stretch $+0.10$) aur orange dot (compress $-0.10$) same height $U=1$ J par hain — ye $(-x{)}^2=x^2$ symmetry hai. Neeche ke paas chhote arrows andar ki taraf point karte hain, dikhate hain ki force mass ko hamesha $x=0$ ki taraf — bowl ke base ki taraf — dhakelta hai.

Example 4 — Zero badlo, number badlo (cell E)

Forecast: $U$ ke do alag sets of numbers. Kya physics (drop) badlegi?

1. Floor reference. Ye step kyun? $U=mgy$ jahan $y$ floor se measure kiya gaya hai. $U_{\text{table}}=2\cdot 10\cdot 1=20\text{J},U_{\text{shelf}}=2\cdot 10\cdot 3=60\text{J}.$
2. Table reference. Ye step kyun? Ab $y^{\prime }=y-1$; table $y^{\prime }=0$ par hai, shelf $y^{\prime }=2$ par. $U_{\text{table}}^{\prime }=2\cdot 10\cdot 0=0\text{J},U_{\text{shelf}}^{\prime }=2\cdot 10\cdot 2=40\text{J}.$ Har number $20\text{J}$ kam ho gaya — reference switch karne se $U$ mein har jagah ek constant add ho gaya.
3. Table→shelf ka drop compare karo. Ye step kyun? Sirf $\Delta U$ physical hai. $\Delta U=U_{\text{shelf}}-U_{\text{table}}=60-20=40\text{J}=40-0=U_{\text{shelf}}^{\prime }-U_{\text{table}}^{\prime }.$

Verify: Reference switch karne par jo constant add hua use $U_0$ kaho (yahan $U_0=-20$ J, kyunki har value $20$ J drop hui). Kyunki $U_0$ ek constant hai, uski derivative zero hai, toh $F=-dU/dx$ unchanged rehta hai; aur kisi bhi difference mein woh cancel ho jaata hai, toh $\Delta U$ bhi unchanged rehta hai — dono mein $40\text{J}$ aaya. ✔ Lesson: hamesha apna reference batao.

Example 5 — Ek force jo position ke saath change hoti hai (cell F)

Forecast: Force $x$ ke saath tezi se badhti hai. Kya hum sirf force ko distance se multiply kar sakte hain? (Nahi — kaunsa tool chahiye?)

1. Sahi tool chuno. Ye step kyun? $F$ $x$ ke saath vary karti hai, toh $W=F\Delta x$ illegal hai; sahi machinery hai integral $U(x)=-\int Fdx$ (ye $F=-dU/dx$ ko ulta padhna hai). Ye exactly Work done by a variable force waali situation hai.
2. Integrate karo. Ye step kyun? Derivative undo karo $U$ recover karne ke liye. $U(x)=-\int (-b{x}^3)dx=b\int x^{3}dx=\frac{b{x}^4}{4}+C.$ $U(0)=0$ choose karo $\Rightarrow C=0$, toh $U(x)=\frac{b{x}^4}{4}=x^4$ (kyunki $b/4=1$).
3. Dono endpoints evaluate karo. Ye step kyun? Energy released = $U$ mein drop = $U_A-U_B$. $U(1)=\frac{4\cdot 1^4}{4}=1\text{J},U(0)=0.$ $W_{1\to 0}=U(1)-U(0)=1-0=+1\text{J}.$

Verify: Force differentiate karke check karo: $-\frac{dU}{dx}=-\frac{d}{dx}\left(\frac{b{x}^4}{4}\right)=-b{x}^3=F(x)$. ✔ Direct work integral bhi agree karta hai: $W={\int }_1^0(-b{x}^3)dx={\left[-\frac{b{x}^4}{4}\right]}_1^0=0-(-1)=+1\text{J}$. ✔ Units: ${\text{N/m}}^3\cdot {\text{m}}^4=\text{N}\cdot \text{m}=\text{J}$. ✔

Example 6 — Friction ek loop mein (cell G)

Forecast: Ye ek closed loop hai. Gravity ke liye woh $0$ tha (Ex 1). Kya friction mein bhi hoga?

1. Bahar ki leg. Ye step kyun? Friction hamesha motion ko oppose karti hai, toh woh backward point karti hai jabki block aage move karta hai: $W_{\text{out}}=-fd=-3\cdot 5=-15\text{J}.$
2. Wapas ki leg. Ye step kyun? Block ab doosri taraf move karta hai, toh friction bhi flip ho jaati hai aur phir bhi oppose karti hai: $W_{\text{back}}=-fd=-15\text{J}.$ Ye cancel nahi ho sakta — ye har leg par negative hai.
3. Total loop work. $W_{\text{loop}}=-15+(-15)=-30\text{J}\ne 0.$

Verify: Conservative force ke liye loop $0$ dena chahiye (Ex 1 ne diya tha). Yahan $W_{\text{loop}}=-30\text{J}\ne 0$, toh friction non-conservative hai aur koi $U(x)$ exist nahi kar sakta. Churaaye gaye $30$ J heat ban gaye. ✔ Dekho Friction and dissipation.

Figure "hamesha negative" wala point visually dikhata hai. Upar waale track par block bahar jaata hai (motion →) lekin magenta friction arrow ← point karta hai; neeche waale track par woh wapas aata hai (motion ←) phir bhi orange friction arrow ab → point karta hai — friction ne motion oppose karne ke liye phir flip kar liya. Dono legs $-15$ J cost karti hain, toh loop total ($-30$ J, violet mein) kabhi zero nahi pahunch sakta.

Example 7 — Woh force jo kuch nahi karti (cell H)

Forecast: Poora chakkar laga ke ek bada $6$ N force — zaroor bahut zyada kaam hoga?

1. Geometry dekho. Ye step kyun? Kaam hai $\vec{F}\cdot d\vec{r}=|F||dr|\cos \theta$; tension aur motion ke beech ka angle chahiye. Ball circle ke saath move karta hai (tangent); tension centre ki taraf point karti hai (radius). Tangent ⟂ radius, toh $\theta =90^{\circ }$.
2. Dot product evaluate karo. Ye step kyun? $\cos 90^{\circ }=0$. $dW=T|dr|\cos 90^{\circ }=0\text{har instant par}.$
3. Loop ke around integrate karo. $W=\oint dW=0.$

Verify: Ye parent ka "normal force / tension" waala point hai: ek force jo hamesha motion ke perpendicular hoti hai zero kaam karti hai — uske naam ki wajah se nahi balki isliye kyunki $\vec{F}\cdot d\vec{r}=0$. Toh ye kabhi dissipate nahi karti aur kabhi energy store nahi karti. ✔ (Compare karo Gradient and vector fields se: displacement ke perpendicular ek force line integral mein kuch contribute nahi karti.)

Example 8 — Real-world energy budget (cell I)

Forecast: Friction ke bina woh saara $mgh$ $KE$ mein convert kar leti. Chor use kitna nuksaan pahunchaata hai?

1. Piggy bank jo energy release karta hai. Ye step kyun? Gravity conservative hai; released energy drop $U_{\text{top}}-U_{\text{bottom}}=mgh$ hai. $mgh=30\cdot 10\cdot 2=600\text{J}.$
2. Chor ko minus karo. Ye step kyun? Friction non-conservative hai aur $60\text{J}$ heat ke roop mein le jaati hai, toh actual kinetic energy jo milti hai woh baaki bachi hai (ye Work–Energy theorem + Conservation of mechanical energy hai loss term ke saath). $K{E}_{\text{bottom}}=600-60=540\text{J}.$
3. Speed nikalo. Ye step kyun? $KE=\frac{1}{2}m{v}^2\Rightarrow v=\sqrt{2KE/m}$. $v=\sqrt{\frac{2\cdot 540}{30}}=\sqrt{36}=6\text{m/s}.$

Verify: Friction ke bina: $v_{\max }=\sqrt{2gh}=\sqrt{40}\approx 6.32\text{m/s}$, toh friction thoda kam dena chahiye — aur $6<6.32$. ✔ Units: $\sqrt{\text{J/kg}}=\sqrt{{\text{m}}^2/{\text{s}}^2}=\text{m/s}$. ✔

Example 9 — Exam twist: kya ye force conservative hai? (cell J)

Forecast: Ye sirf ek direction mein push hai — harmless aur conservative lagta hai. Vibe par nahi, loop par trust karo.

1. Leg $A\to B$ ($x$ ke saath, $y=0$). Ye step kyun? $\vec{F}=(y,0)=(0,0)$ yahan, toh $\int \vec{F}\cdot d\vec{r}=0$.
2. Leg $B\to C$ ($y$ ke saath, motion $\hat{y}$ hai). Ye step kyun? $\vec{F}=(y,0)$ mein koi $y$-component nahi, toh $\vec{F}\cdot d\vec{r}=0$. Contribution $0$.
3. Leg $C\to D$ ($x$ ke saath, peeche, $y=1$). Ye step kyun? Ab $\vec{F}=(1,0)$, motion $-\hat{x}$ hai length $1$ par: $\int =(1)(-1)=-1\text{J}$.
4. Leg $D\to A$ ($y$ ke saath, neeche). Ye step kyun? Phir force mein koi $y$-component nahi, contribution $0$.
5. Loop jodo. $\oint \vec{F}\cdot d\vec{r}=0+0+(-1)+0=-1\text{J}\ne 0.$

Conclusion: Loop work $-1\text{J}\ne 0$ hai, toh ye innocent-dikhne waali single-direction push non-conservative hai — iske liye koi potential energy $U$ exist nahi karti.

Verify: Gradient and vector fields curl test se cross-check karo: conservative 2D field $\vec{F}=(F_x,F_y)$ ke liye hume $\partial F_x/\partial y=\partial F_y/\partial x$ chahiye. Yahan $\partial (y)/\partial y=1$ lekin $\partial (0)/\partial x=0$, aur $1\ne 0$, toh field test fail kar deti hai — confirm karta hai ki ye non-conservative hai. ✔ Dono methods (loop integral aur curl) agree karte hain.

Active recall

Connections

• Parent topic (Hinglish)
• Work done by a variable force
• Work–Energy theorem
• Conservation of mechanical energy
• Gravitational potential energy
• Spring / elastic potential energy
• Friction and dissipation
• Gradient and vector fields

Case Map