Exercises — Non-conservative forces — friction, air drag

Test hamesha yahi hai: kya ek closed loop mein work zero aata hai? ($\oint$ par chhota circle ka matlab hai "ek poore loop mein $\vec{F}\cdot d\vec{r}$ add karo wapas start tak").

• (a) gravity — C. Upar uthao, neeche wapas aao: PE exactly return ho jaati hai. Loop work $=0$.
• (b) kinetic friction — NC. Yeh hamesha velocity ke opposite direction mein hoti hai, isliye round trip mein jaate waqt bhi subtract karti hai aur aate waqt bhi. Loop work $<0$.
• (c) spring $-kx$ — C. Iska potential energy $U=\frac{1}{2}k{x}^2$ hota hai; khiincho aur chodo, energy wapas aati hai. Loop work $=0$.
• (d) air drag — NC. Friction jaisi hi wajah: hamesha motion ke opposite, hamesha energy churaata hai.
• (e) normal force — C-ish / neither. Flat floor par normal force motion ke perpendicular hoti hai, isliye $\vec{N}\cdot d\vec{r}=0$: yeh har segment par zero work karti hai. Yeh kuch churaati nahi, isliye "chor" nahi hai — bas bahar baith jaati hai.

Pehle sign: friction motion ke opposite hoti hai, isliye $W_{fric}$ negative hoga — mnemonic "$W_{fric}$ hamesha $-$ hota hai". Flat floor par normal force: $N=mg=2.0\times 9.8=19.6\text{N}$. $W_{fric}=-{\mu }_{k}NL=-(0.30)(19.6)(3.0)=-17.64\text{J}.$ Toh floor ne $17.64\text{J}$ churaa liya, use heat mein badal diya.

Incline par weight $mg$ do perpendicular pieces mein split hoti hai (figure dekho): ek slope ke along, $mg\sin \theta$, jo slide drive karta hai; ek slope ke andar, $mg\cos \theta$, jise surface wapas push karta hai — isliye $N=mg\cos \theta$. Master equation $W_{nc}=\Delta KE+\Delta U$ use karo:

• Height drop $=L\sin \theta$, isliye $\Delta U=-mgL\sin \theta$.
• Friction: $W_{nc}=-{\mu }_{k}NL=-{\mu }_{k}mg\cos \theta L$. $-{\mu }_{k}mg\cos \theta L=\frac{1}{2}m{v}^2-0+(-mgL\sin \theta )$ $m$ cancel karo, $v$ ke liye solve karo — yeh exactly parent ka result hai: $v=\sqrt{2gL(\sin \theta -{\mu }_k\cos \theta )}.$ Numbers: $\sin 30^{\circ }=0.5$, $\cos 30^{\circ }=0.866$. $v=\sqrt{2(9.8)(5.0)(0.5-0.20\times 0.866)}=\sqrt{98(0.3268)}=\sqrt{32.02}=5.66\text{m/s}.$

Bracket $(\sin \theta -{\mu }_k\cos \theta )$ sab decide karta hai. Yeh positive tabhi hoga jab $\tan \theta >{\mu }_k$. $\tan 10^{\circ }=0.176,{\mu }_k=\mathrm{0.20.}$ Kyunki $0.176<0.20$ hai, bracket negative hai → formula mein $v^2<0$ aayega, jo impossible hai → box accelerate nahi karta; woh wahi ruk jaata hai (slope ke along gravity ka pull friction se zyada weak hai). Yeh wohi condition hai jaise parent ka ${\mu }_k\ge \tan \theta$ stopping rule — energy aur statics dono agree karte hain.

Terminal velocity woh hoti hai jahan drag exactly gravity ko cancel kare, isliye acceleration ruk jaata hai: $mg=\frac{1}{2}{C}_d\rho A{v}_t^2\Rightarrow v_t=\sqrt{\frac{2mg}{C_d\rho A}}.$ $v_t=\sqrt{\frac{2(80)(9.8)}{0.25}}=\sqrt{6272}=79.2\text{m/s}.$

(a) Start aur end same point hain ⇒ net displacement $=0$. (b) Path length $L=2.0+2.0=4.0\text{m}$ — yeh distance travelled hai, jo friction ko matter karta hai. (c) $N=mg=1.5\times 9.8=14.7\text{N}$. $W_{fric}=-{\mu }_{k}NL=-(0.25)(14.7)(4.0)=-14.7\text{J}.$ Kyunki loop work $\ne 0$ hai (strictly negative), friction conservative nahi ho sakti aur iska koi potential energy nahi hota. Ek conservative force zero displacement ke saath zero net work deti; friction ne $-14.7\text{J}$ diya. Woh $14.7\text{J}$ heat ban gayi.

Stage 1 — collision (momentum use karo, energy NAHI). Bullet embed ho jaata hai: perfectly inelastic, isliye KE wood ke andar kharch hoti hai, lekin momentum bachta hai (brief hit ke dauran koi external horizontal impulse nahi). $m_{b}u=(m_b+M)v\Rightarrow (0.012)(350)=(1.812)v\Rightarrow v=\frac{4.2}{1.812}=2.318\text{m/s}.$ Stage 2 — sliding (ab energy use karo). Friction combined mass ko distance $d$ par rok deti hai. Total mass $m=1.812\text{kg}$, $N=mg$. $W_{nc}=\Delta KE:-{\mu }_{k}mgd=0-\frac{1}{2}m{v}^2.$ $m$ cancel ho jaata hai: $d=\frac{v^2}{2{\mu }_{k}g}=\frac{(2.318{)}^2}{2(0.35)(9.8)}=\frac{5.373}{6.86}=0.783\text{m}.$

Energy ko start (spring loaded, cart rest mein) se ramp ke top (cart momentarily rest mein) tak track karo. Poore safar par master equation: $W_{nc}=\Delta KE+\Delta U_{grav}+\Delta U_{spring}.$

• Start KE $=0$, end KE $=0$ ⇒ $\Delta KE=0$.
• Spring apni stored energy release karta hai: $\Delta U_{spring}=0-\frac{1}{2}k{x}^2=-\frac{1}{2}(800)(0.15{)}^2=-9.0\text{J}$.
• Gravity: $\Delta U_{grav}=+mgh$ ($h$ tak upar jaata hai).
• Non-conservative: sirf rough patch, $W_{nc}=-{\mu }_{k}mgL=-(0.30)(0.50)(9.8)(1.2)=-1.764\text{J}$.

Plug in: $-1.764=0+mgh+(-9.0)\Rightarrow mgh=9.0-1.764=7.236\text{J}.$ $h=\frac{7.236}{(0.50)(9.8)}=\frac{7.236}{4.9}=1.477\text{m}.$ Spring ne $9.0\text{J}$ diya; friction ne $1.764\text{J}$ heat mein churaa liya; baaki ($7.236\text{J}$) ramp par chadha.

Heat $=$ non-conservative forces ne jo energy li $=-W_{nc}=+1.764\text{J}$ (parent ke $Q=-W_{nc}$ se). Spring energy ka jo fraction heat mein gaya: $\frac{1.764}{9.0}=0.196=19.6\%.$ Toh launch energy ka roughly ek-paanchva hissa rough patch ko garam kiya; char-paanchva gravitational PE mein pahuncha.

(a) Constant speed ⇒ $F_d=bv$ constant hai $=2.0\times 3.0=6.0\text{N}$, aur yeh motion ke opposite hai, isliye $W_d=-F_{d}L=-(6.0)(5.0)=-30.0\text{J}.$ Yahan shortcut legal hai kyunki speed constant hai. (b) Decelerating: $F_d=bv$ har instant badal raha hai, isliye $-F_{d}L$ meaningless hai. Iske bajaye energy use karo — drag hi ek maatra horizontal force hai, isliye $W_d=\Delta KE$. Yeh $6.0\text{m/s}$ se shuru hota hai; kahan rukta hai? Linear drag ke saath object asymptotically slow hota hai lekin, stated final rest (effectively $4$ m baad ruk gaya) lekar, $\Delta KE=0-\frac{1}{2}(1.0)(6.0{)}^2=-18.0\text{J}$. $W_d=\Delta KE=-18.0\text{J}.$ Dhyan do ki humne $4.0\text{m}$ ko constant force ke saath kabhi use nahi kiya — woh galat number deta. Energy differences varying force ko completely bypass kar deti hain.

Along-slope driving force $mg\sin \theta$ hai; sliding ke dauran resisting friction ${\mu }_{k}mg\cos \theta =\tan \theta mg\cos \theta =mg\sin \theta$ hai. Yeh equal hain — net along-slope force slide karte waqt zero hai! Isliye block constant velocity se move karta hai jab tak chal raha hai... lekin isko chalte rehne ke liye koi force nahi aur rokne ke liye bhi nahi, isliye ideally yeh hamesha $1.0\text{m/s}$ par coast karta rahega (net force zero ⇒ zero acceleration). Yeh exactly "tezi se slide hoga" ($\tan \theta >{\mu }_k$) aur "hilega hi nahi" ($\tan \theta <{\mu }_k$) ke beech ki knife-edge hai. Energetically: gravity $mg\sin \theta \cdot L$ release karta hai aur friction ${\mu }_{k}mg\cos \theta \cdot L$ leta hai — exactly equal, isliye $\Delta KE=0$: speed kabhi nahi badlegi. Ek khoobsurat degenerate case jahan block jis bhi speed par chalo use doge, usi par glide karta rahega.

Friction ko path length ki parwah hoti hai, endpoints ki nahi — yahi non-conservative forces ka poora point hai.

• Route 1 (diagonal): length $=\sqrt{2}s=1.414\text{m}$.
• Route 2 (do edges): length $=2s=2.0\text{m}$. $N=mg=0.20\times 9.8=1.96\text{N}$; har metre friction ${\mu }_{k}N=0.196\text{J/m}$ leta hai.
• Route 1 loss: $0.196\times 1.414=0.277\text{J}$.
• Route 2 loss: $0.196\times 2.0=0.392\text{J}$. Extra loss on Route 2: $0.392-0.277=0.115\text{J}$. Lamba path zyada costly hai — yahi path-dependence ki pehchaan hai. Conservative force dono routes ke liye same charge karti.

Kaun sa tool kis kaam ke liye? ::: Collision/sticking → momentum; sliding/climbing → energy $W_{nc}=\Delta KE+\Delta U$; constant-force drag ya friction → $-FL$; varying drag → sirf energy differences. Sign of friction/drag work? ::: Hamesha negative — yeh chor hain. Path length vs displacement for friction? ::: Hamesha total path length $L$; displacement irrelevant hai. Heat produced? ::: $Q=-W_{nc}$.