Question bank — Conservative forces — path-independent work, potential energy defined

False — gravity upar jaane par negative work karti hai lekin neeche aane par positive work; ek trip mein work ka sign conservativeness decide nahi karta, path-dependence karta hai.

False — loop work ko har possible closed loop ke liye zero hona chahiye; ek lucky loop kuch prove nahi karta.

True — $U=mgy$ apne chosen $y=0$ reference ke neeche negative hoti hai; sirf differences $\Delta U$ physical hain, isliye $U$ ka sign khud ek bookkeeping choice hai.

False — ye decrease ke barabar hota hai: $W_{\text{cons}}=-\Delta U=U_A-U_B$, toh jab ball girta hai, $U$ ghatta hai aur gravity positive work karti hai.

False — ye hamesha zero work karti hai (jaise magnetic force ya normal force), toh koi energy nahi jaati, lekin "zero work" ek alag property hai "path-independent" se; aisi forces ka simply koi associated potential energy nahi hota aur ye definition ke point nahi hain.

False — force $-dU/dx$ hai aur constant ki derivative zero hoti hai, isliye physics unchanged rehti hai; yahi freedom hai jiske karan hum koi bhi $U=0$ reference choose kar sakte hain.

False — ek out-and-back loop mein friction $-2fd\ne 0$ karta hai kyunki ye dono legs mein motion ka oppose karta hai; starting point par wapas aana loss ko undo nahi karta.

False — ye sirf ek additive constant tak determined hoti hai jo tumhare reference choice se fix hoti hai; ek hi physical point different references ke under alag $U$ values carry kar sakta hai.

False — 2D test apply karo $\partial F_x/\partial y$ vs $\partial F_y/\partial x$: yahan $\partial (-y)/\partial y=-1$ lekin $\partial (x)/\partial x=+1$, ye agree nahi karte, isliye field swirl karti hai aur iska loop work nonzero hai.

True — ek conservative force $\vec{F}=-\nabla U$ hoti hai, aur $U$ ke mixed second partials equal hote hain (${\partial }^{2}U/\partial x\partial y={\partial }^{2}U/\partial y\partial x$), jo is curl-free condition ko enforce karta hai.

Galat — gravitational work sirf height change $h$ par depend karta hai, path par nahi; dono routes same $mgh$ store karte hain. Geometrically, sirf har step ka vertical projection count hota hai, isliye same endpoints wale saare paths ek hi answer share karte hain — yahi conservative field ka visual signature hai.

Sign error — sahi relation hai $F=-dU/dx$ isliye force $U$ graph par downhill lower $U$ ki taraf point karta hai; spring par test karo: $U=\frac{1}{2}k{x}^2$ se restoring $F=-kx$ milni chahiye, jo sirf minus sign deliver karta hai.

Galat test — conservativeness path-dependence se decide hoti hai, force ke naam se nahi; inextensible string par ideal tension zero work karta hai aur simply iska koi potential energy nahi hota.

Bakwaas hai — friction ka koi potential-energy function nahi hota kyunki iska loop work nonzero hai; woh $50\text{ J}$ heat ban gaya, stored recoverable energy nahi.

Sign slip — return leg opposite sense mein traverse hota hai, isliye ye $-W_2$ hai; loop work $W_1-W_2$ hai, jo gravity ke liye zero hai kyunki $W_1=W_2$.

Do ideas confuse ho rahe hain — zero work karna conservative hone ke barabar nahi hai; normal force ko bilkul bhi potential energy assign nahi hoti, ye simply kabhi energy budget mein appear hi nahi hoti.

Galat loop choose kiya — seedhe out-and-back par ye cancel ho sakta hai, lekin ek circle par force hamesha motion ke along point karti hai, har step par positive work deti hai aur nonzero loop deta hai, jo ise non-conservative expose karta hai.

Taaki force potential slope ke neeche smaller $U$ ki taraf point kare, jo hamari intuition se match karta hai ki ball downhill roll karti hai; plus sign predict karta ki objects upar hill ki taraf bhaagein.

Ye fundamental theorem for gradients hai: agar $\vec{F}=-\nabla U$, toh ${\int }_A^B\vec{F}\cdot d\vec{r}=U_A-U_B$ chahe path kuch bhi ho, bilkul waisi tarah jaise ${\int }_a^b{g}^{\prime }(x)dx=g(b)-g(a)$ endpoints ke beech jo hota hai use ignore karta hai.

Friction hamesha instantaneous motion ka oppose karta hai, isliye iska work $-fds$ hai jahan $ds$ arc length (actually travelled distance) hai, jo har meter mein accumulate hota hai; gravity ka work sirf net vertical drop track karta hai, isliye wiggles cancel ho jaate hain.

Experiments work aur force measure karte hain, aur dono sirf $U$ ke change par depend karte hain ($-dU/dx$ aur $-\Delta U$); absolute level arbitrary reference se set hoti hai, isliye ye kabhi physically appear nahi hoti.

Gravity vertically point karti hai, isliye $\vec{F}\cdot d\vec{r}=-mgdy$; horizontal displacement mein $dy=0$ hota hai, jo zero dot product deta hai chahe tum sideways kitna bhi move karo.

Ek non-conservative force accidentally ek symmetric loop par zero de sakti hai jabki doosre par nonzero deti hai; sirf sabhi loops ka vanishing hona guarantee karta hai ki ek consistent potential exist karta hai.

Green's theorem se loop work equal hota hai $(\partial F_y/\partial x-\partial F_x/\partial y)$ ke area integral ke; agar ye "swirl" har jagah zero hai, toh har loop integral vanish hota hai aur ek potential exist karta hai.

Reference shift karne se $U$ har jagah same constant add hota hai, jo har $\Delta U$ mein cancel hota hai aur $F=-dU/dx$ mein derivative ke under vanish hota hai.

Force zero hai — ye ek equilibrium point hai ($U$ curve par ek flat spot); ye stable hai ya unstable ye is par depend karta hai ki $U$ uske around upar curve karta hai ya neeche.

Definition se exactly zero, kyunki $U=mgy$ mein $y=0$ wahan hai; ye ek choice hai, us location ke baare mein koi physical fact nahi.

Nahi — conservative part zero contribute karta hai, lekin friction ek negative amount contribute karta hai, isliye total loop work negative hai, jo non-conservative piece ko reveal karta hai.

Nahi — work force times net displacement nahi hai balki path integral $\int \vec{f}\cdot d\vec{r}$ hai; friction poore path par loss accumulate karta hai chahe net displacement vanish ho.

Saari, $\frac{1}{2}k{x}^2$, kyunki spring force conservative hai aur iska loop work zero hai — stored energy poori tarah kinetic mein convert ho jaati hai.

Haan, trivially conservative hai (sabhi loop work zero hai), aur iska potential energy koi bhi constant hai — ek perfectly flat $U$, $F=-dU/dx=0$ se consistent.

Ye har path par zero work karta hai (hamesha velocity ke perpendicular), isliye trivially iska zero loop work hai, lekin ye velocity-dependent hai aur position-only $U$ se derivable nahi, isliye ye usual conservative-force framework ke bahar hai.

Ye $2\pi$ deta hai (Green's theorem se enclosed area ka double), ek nonzero value — concrete proof ki ye swirling field non-conservative hai chahe ye smooth dikhti ho.