Non-conservative forces — friction, air drag: Worked examples

1.3.7 · D3 · Physics › Work, Energy & Power › Non-conservative forces — friction, air drag

Shuru karne se pehle, symbols ke baare mein ek waada. Yeh sab jagah reuse hote hain, isliye inhe abhi pin kar lo.

Definition Symbols ki cast (sab pehle se earned)

$m$ — mass, kilograms (kg) mein. "Kitna stuff hai." Jab do masses ek saath aati hain, hum $m_{b}$ (bullet) aur $m_{B}$ (block) likhte hain.
$g = 9.8 m/s^{2}$ — gravity ki pull ki strength, per kg.
$μ_{k}$ — coefficient of kinetic friction, ek pure number (koi units nahi). Bada = zyada grippy. Sliding friction govern karta hai. Dekho Friction — static & kinetic.
$μ_{s}$ — coefficient of static friction, "nahi hilega" wali grip sliding shuru hone se pehle. Hamesha $μ_{s} \geq μ_{k}$ . Yeh decide karta hai ki motion shuru hoti hai ya nahi.
$N$ — normal force: surface ka push jo khud ke perpendicular ( $9 0^{\circ}$ par) hota hai, newtons (N) mein.
$L$ — path length: woh total distance jo ek wheel-counter actual route par tick karega, straight-line displacement NAHI. Metres (m).
$Δ$ (delta) — "change in": $Δ X = X_{final} - X_{start}$ .
$K E = \frac{1}{2} m v^{2}$ — kinetic energy, motion ki energy. Joules (J).
$U$ — potential energy, position ki stored energy ( $U = m g h$ height $h$ ke liye). Dekho Conservative forces & potential energy.
$E_{m ec h} = K E + U$ — kul mechanical energy.
$W_{n c}$ — saare non-conservative forces (friction, drag, collisions) ka mila hua kaam.
$W_{f r i c}$ — $W_{n c}$ ka woh hissa jo specifically kinetic friction se hai: $W_{f r i c} = - μ_{k} N L$ . Yeh $W_{n c}$ ka ek contribution hai; flat friction-only floor par dono equal hote hain.
$Q$ — heat generate hoti hai, $Q = - W_{n c}$ (chura li gayi energy warmth ban ke wapis aati hai). Joules (J).
$b$ — linear drag coefficient, units N per (m/s), taaki $F_{d} = b v$ kam speed par.
$C_{d}$ — drag coefficient, ek pure shape number (kitna "streamlined" hai — ek sphere lagbhag $0.5$ hota hai).
$ρ$ (rho) — air density, air ki mass per cubic metre, $kg/m^{3}$ (sea level par lagbhag $1.2$ ).
$A$ — frontal area, woh cross-section jo body air mein se push karti hai, $m^{2}$ mein.

Ek master equation, parent note mein derive ki gayi:

The scenario matrix

Is chapter ka har non-conservative problem inhi cells mein se ek mein aata hai. Sabse right column woh example batata hai jo use nail karta hai.

Cell	Situation class	Kya special/degenerate hai	Example
A	Flat floor, friction stops motion	pure $Δ K E$ , $Δ U = 0$	Ex 1
B	Closed loop, round trip	displacement $= 0$ lekin $W \neq = 0$ (the loop test)	Ex 2
C	Rough incline, block slides	$tan θ > μ_{k}$ : bracket ka sign positive	Ex 3
D	Rough incline, degenerate	$tan θ \leq μ_{s}$ : block nahi hilega — formula kya kehta hai	Ex 4
E	Collision phir friction	pehle momentum, phir energy (do-tool problem)	Ex 5
F	Quadratic air drag, terminal velocity	limiting value jab $v \to v_{t}$ , $a \to 0$	Ex 6
G	Linear (Stokes) drag, low speed	$F_{d} = b v$ regime, alag terminal velocity	Ex 7
H	Real-world word problem	skid marks → speed (forensics)	Ex 8
I	Exam twist	block upar slides, gravity aur friction dono oppose karte hain	Ex 9
J	Dono ek saath — friction AND air drag	ek problem mein do chor	Ex 10

Hum inhe order mein karenge. Har number neeche machine-checked hai.

Worked example Ex 1 — Cell A: flat floor, friction akela puck ko rok deta hai

Ek hockey puck jiska mass $m = 0.2 kg$ hai, rough ice par $v_{0} = 6 m/s$ se slide karta hai, $μ_{k} = 0.10$ . Rukne se pehle kitni door?

Forecast: Aage padhne se pehle stopping distance guess karo — kya yeh kuch centimetres hogi, ya kaafi metres? (Ice slippery hai, toh mera bet hai yeh large hogi.)

Height kabhi nahi badalti, isliye $Δ U = 0$ . Yeh step kyun? Floor flat hai; koi rise ya fall nahi, isliye gravity ki potential energy untouched rehti hai. Yeh master equation ko $W_{n c} = Δ K E$ tak collapse kar deta hai.
Friction work likhte hain. Flat ground par $N = m g$ , isliye $W_{f r i c} = - μ_{k} m g L$ . Kyun? Normal force yahan sirf weight balance karti hai; koi aur cheez surface par push nahi karti.
Chura li gayi energy = lost kinetic energy set karo. Final speed $0$ hai: $- μ_{k} m g L = \frac{1}{2} m v_{0}^{2} - \frac{1}{2} m v_{0}^{2}_{start} = 0 - \frac{1}{2} m v_{0}^{2}$ $\Rightarrow μ_{k} m g L = \frac{1}{2} m v_{0}^{2}$ Kyun? $Δ K E = K E_{f} - K E_{i}$ , aur puck rest par khatam hota hai.
Mass cancel ho jaati hai! Dono sides ko $m$ se divide karo, phir $μ_{k} g$ se divide karo: $L = \frac{v _{0}^{2}}{2 μ _{k} g} = \frac{6 ^{2}}{2 ( 0.10 ) ( 9.8 )} = 18.4 m$ Yeh step kyun? $m$ dono sides par appear hoti hai ( $K E$ mein bhi aur $W_{f r i c}$ mein bhi), isliye yeh algebraically cancel ho jaati hai — isliye stopping distance mass ki parwah nahi karti. Hum $L$ ko isolate karte hain question ka jawab dene ke liye.

Verify: Units: $\frac{( m/s ) ^{2}}{( m/s ^{2} )} = m$ ✔. Mass drop ho gayi — bhaari puck zyada KE store karta hai lekin zyada friction bhi feel karta hai, exactly cancel hote hue. Jaise forecast kiya tha badi distance (slippery ice, chhota $μ_{k}$ ).

Worked example Ex 2 — Cell B: closed-loop test (displacement zero, work nahi)

Usi puck ( $m = 0.2$ , $μ_{k} = 0.10$ ) ko $3 m$ aage push karo, phir $3 m$ wapas exact start par. Loop par kitni mechanical energy lost hoti hai? Gravity kya karta, usse compare karo.

Forecast: Net displacement zero hai. Ek "fair" force ek round trip par zero work karta hai. Kya friction karta hai?

Total path length, displacement nahi. $L = 3 + 3 = 6 m$ . Yeh step kyun? Yahi non-conservative forces ka poora point hai. Friction wapas aate waqt tumhare against ho jaata hai, isliye yeh dono legs par churata hai — tum distances ka sum karte ho.
Loop work compute karo. $W_{f r i c} = - μ_{k} m g L = - (0.10) (0.2) (9.8) (6) = - 1.176 J$ Yeh step kyun? Hum same $W_{f r i c} = - μ_{k} m g L$ use karte hain jaise Ex 1 mein, lekin ab $L$ poore loop ki length hai — yeh single number hai jise hum gravity ke zero ke against compare karte hain.
Gravity ko same loop par compare karo. Gravity $- m g Δ h$ deta upar jaate waqt aur exactly $+ m g Δ h$ neeche aate waqt: net $0$ . Friction $- 1.176 J \neq = 0$ deta hai. Kyun? Gravity ka sign displacement direction (upar vs neeche) par depend karta hai; friction ka sign velocity direction par depend karta hai — aur velocity friction ka har leg par oppose karti hai.

Verify: $\oint F \cdot d r = - 1.176 J \neq = 0$ confirm karta hai ki friction non-conservative hai. Yeh defining loop-integral criterion hai Conservative forces & potential energy se.

Figure — Non-conservative forces — friction, air drag

Figure padhna: white block slope par baitha hai. Uska weight $m g$ (seedha-neeche white arrow) do chalk arrows mein split hota hai: ek blue wala down the slope point karta hai ( $m g sin θ$ ) aur ek pink wala slope mein press karta hai ( $m g cos θ$ ). Yellow arrow slope ke upar friction $f_{k}$ hai, dashed "slides down" motion arrow ka oppose karta hua. Base par yellow arc $θ = 3 0^{\circ}$ mark karta hai. (Har arrow apne label ke saath printed hai, isliye direction + text colour par rely kiye bina identify karte hain.)

Forecast: Frictionless case mein $v = 2 g L sin θ$ milta. Friction ke saath, kya yeh thoda slower hoga ya bilkul move nahi karega? (Check: $tan 3 0^{\circ} = 0.577 > 0.2$ , isliye move karta hai.)

Gravity ko tilted plane par resolve karo. Jaise figure dikhata hai, weight $m g$ do pieces mein split hota hai: ek piece slope ke saath, $m g sin θ$ (blue arrow, neeche-slope pull karta hai), aur ek piece slope mein press karta hai, $m g cos θ$ (pink arrow). Yeh step kyun? Sirf along-slope piece sliding karta hai; into-slope piece normal force aur isliye friction set karta hai. Inhe do perpendicular directions mein split karne se hum har ek ko independently treat kar sakte hain.

Perpendicular balance se normal force. Block slope mein sink nahi karta, isliye surface ka normal push $N$ into-slope weight ko exactly cancel karna chahiye: $N = m g cos θ$ . Isliye friction magnitude $f_{k} = μ_{k} m g cos θ$ (yellow up-slope arrow). Yeh step kyun? Friction $μ_{k} N$ hai, isliye pehle $N$ chahiye; perpendicular force balance (slope mein koi motion nahi) hi $N$ pin karta hai.

Height drop. $L$ slope par slide karne se height $L sin θ$ se girta hai, isliye $Δ U = - m g L sin θ$ . Yeh step kyun? Master equation ko $Δ U = m g Δ h$ chahiye; ramp ki geometry "slope ke saath distance" ko "vertical drop" mein $sin θ$ ke through convert karti hai. Negative hai kyunki block girta hai.

Master equation feed karo $W_{n c} = Δ K E + Δ U$ : $- μ_{k} m g cos θ L = \frac{1}{2} m v^{2} - 0 + (- m g L sin θ)$ Yeh step kyun? Ab har piece known hai: $W_{n c}$ friction work hai (step 2), $Δ U$ step 3 hai, aur $Δ K E = \frac{1}{2} m v^{2}$ kyunki rest se start karta hai. Substitute karne se physics ek solvable equation ban jaati hai. $m$ cancel karo aur rearrange karo: $v = 2 g L (sin θ - μ_{k} cos θ)$

Numbers plug in karo: $sin 3 0^{\circ} = 0.5$ , $cos 3 0^{\circ} = 0.866$ : $v = 2 (9.8) (4) (0.5 - 0.2 \cdot 0.866) = 78.4 \times 0.3268 = 5.06 m/s$ Yeh step kyun? Symbolic formula general hai; specific given values ( $g, L, θ, μ_{k}$ ) substitute karne se woh ek concrete number ban jaata hai jo question maangta hai, aur hum neeche frictionless case se sanity-check kar sakte hain.

Verify: Frictionless mein $2 (9.8) (4) (0.5) = 6.26 m/s$ milta; friction correctly ise slower banata hai, $5.06 < 6.26$ ✔. Units $m/s^{2} \cdot m = m/s$ ✔.

Worked example Ex 4 — Cell D: degenerate incline (bilkul nahi hilega)

Same setup lekin $θ = 1 0^{\circ}$ , $μ_{k} = 0.4$ , aur (yeh important hai) $μ_{s} = 0.5$ . Sliding-speed formula kya deta hai, aur uska kya matlab hai?

Forecast: $tan 1 0^{\circ} = 0.176 < 0.4$ . Meri prediction: kinetic formula break ho jaayegi — square root ke andar negative — aur static friction ise start hi nahi hone degi.

Pehle SAHI sawaal puchho: kya yeh move shuru hoga? Motion tabhi shuru hoti hai jab along-slope gravity pull maximum static friction ko haraaye: $m g sin θ > μ_{s} m g cos θ$ , yaani $tan θ > μ_{s}$ . Yeh step kyun? Kya block slide karna shuru karta hai yeh static coefficient $μ_{s}$ govern karta hai, kinetic $μ_{k}$ nahi. Yahan $μ_{k}$ use karna classic galti hogi; $μ_{s} \geq μ_{k}$ hi resting block ko actually hold karta hai. Yahan $tan 1 0^{\circ} = 0.176 < 0.5 = μ_{s}$ , isliye yeh kabhi nahi hilta.
Ab dekho kinetic formula kya kehta hai agar hum (galat tarike se) ise apply karne ki koshish karein. Bracket $sin θ - μ_{k} cos θ$ test karo: $sin 1 0^{\circ} - 0.4 cos 1 0^{\circ} = 0.1736 - 0.4 (0.9848) = 0.1736 - 0.3939 = - 0.220$ Yeh step kyun? Yeh bracket slide shuru hone ke baad har metre par net along-slope energy budget hai. Negative value formula ka apna tarika hai warning dene ka "aage dene ke liye koi energy nahi hai," jo step 1 ke static verdict ke consistent hai.
Negative root interpret karo. $v^{2} = 2 g L (- 0.220) < 0$ ka koi real solution nahi hai. Kyun? Ek real speed negative number ka square nahi ho sakti. Physically block static friction se hold hai aur kabhi accelerate nahi karta; kinetic-energy formula simply apply nahi hoti, aur woh imaginary ho ke flag karta hai.
Dono thresholds clearly state karo. $sin θ > μ cos θ$ ko positive $cos θ$ se divide karne par (jo inequality ka direction rakkhta hai) do trig terms ek clean test mein collapse ho jaate hain: $starts sliding if tan θ > μ_{s}; once sliding, speeds up if tan θ > μ_{k}$ Yeh step kyun? $cos θ$ se divide karne par do terms ek single glance-able $tan θ$ ban jaate hain, aur $μ_{s}$ (start) ko $μ_{k}$ (chalta rehna) se alag karna dono physical questions ko distinct rakhta hai.

Verify: $tan 1 0^{\circ} = 0.176 < 0.4 = μ_{k} < 0.5 = μ_{s}$ , isliye block na start karta hai na sliding sustain kar sakta hai — statics ke saath perfectly consistent. Report karne ke liye koi numeric speed nahi; "answer" hai no motion.

Worked example Ex 5 — Cell E: bullet embed hota hai, phir block skids (do tools)

$m_{b} = 10 g = 0.01 kg$ mass ka ek bullet $v_{b} = 400 m/s$ se move karta hua $m_{B} = 2 kg$ mass ke block mein embed ho jaata hai. Combined lump phir $μ_{k} = 0.4$ wali floor par slide karta hai. (a) impact ke turant baad speed, (b) kitna skid karta hai, (c) friction se heat, pata karo.

Forecast: Collision mein kya conserve hota hai — energy ya momentum? (Trap: momentum hai, energy nahi — embed violently inelastic hai.)

Collision: momentum use karo, energy NAHI. $m_{b}$ bullet mass aur $m_{B}$ block mass ke saath, dekho Inelastic collisions. $m_{b} v_{b} = (m_{b} + m_{B}) v \Rightarrow 0.01 (400) = (2.01) v \Rightarrow v = 1.990 m/s$ Yeh step kyun? Embed mein, bahut bada KE heat/deformation ban jaata hai, isliye mechanical energy conserved nahi hoti. Lekin brief impact ke dauran koi external horizontal impulse act nahi karta, isliye momentum conserved hoti hai.
Sliding phase: ab energy use karo. Flat floor, isliye $Δ U = 0$ , $W_{n c} = Δ K E$ : $- μ_{k} (m_{b} + m_{B}) g d = 0 - \frac{1}{2} (m_{b} + m_{B}) v^{2}$ Mass cancels: $d = \frac{v ^{2}}{2 μ _{k} g} = \frac{1.99 0 ^{2}}{2 ( 0.4 ) ( 9.8 )} = 0.505 m$ . Kyun? Ek lump ki tarah move karne ke baad, sirf friction flat floor par act karta hai — same machinery jaise Ex 1.
Skid se heat $Q = - W_{n c} = \frac{1}{2} (m_{b} + m_{B}) v^{2} = \frac{1}{2} (2.01) (1.990)^{2} = 3.98 J$ . Kyun? Post-collision ka poora KE wahi hai jo friction skid par heat mein convert karta hai.

Verify: Skid distance $0.505 m$ aur heat $3.98 J$ ; embedding collision ne bhi bahut saari heat dump ki (dekho Heat & first law of thermodynamics). Units: $Q$ J mein ✔, $d$ m mein ✔.

Figure padhna: blue horizontal line constant weight $m g$ hai; pink upward-curving parabola drag force $\frac{1}{2} C_{d} ρ A v^{2}$ hai, jo speed ke saath badhta hai. Woh cross karte hain yellow dot par — yellow dashed line ussse speed axis tak girke $v_{t}$ mark karti hai. Cross ke left mein, drag < weight (abhi accelerate ho raha hai); right mein, drag > weight (decelerate hoga). Crossing wahi hai jahan acceleration exactly zero hai. (Dono curves plot par words mein bhi labelled hain, isliye "flat weight line vs rising drag curve" ki story colour ke bina padhti hai.)

Forecast: Speed badhne par drag $v^{2}$ ki tarah badhta hai. Aakhirkar yeh gravity ko pakad lega. Roughly $50 m/s$ ? Dekhte hain.

"Terminal" kyun exist karta hai. Jaise figure dikhata hai, kam $v$ par flat blue weight line pink drag curve ke upar hoti hai, isliye diver accelerate karta hai. Jaise $v$ badhta hai, $v^{2}$ drag curve steeply climb karti hai aur weight line ko yellow dot par cross karti hai. Yeh step kyun? Crossing wahi hai jahan net force $= 0$ ; iske baad drag weight se zyada hoga aur diver slow karega, isliye speed exactly usi crossing par settle hoti hai.

Terminal speed $v_{t}$ par forces equal set karo (jahan $a = 0$ ): $m g = \frac{1}{2} C_{d} ρ A v_{t}^{2} \Rightarrow v_{t} = \frac{2 m g}{C _{d} ρ A}$ Yeh step kyun? "Terminal" matlab speed change hona band, yaani acceleration zero, yaani dono forces balance. Weight $=$ drag set karna us balance ka algebraic statement hai, jise hum phir $v_{t}$ ke liye divide aur square-root karke solve karte hain.

Numbers plug in karo: $v_{t} = \frac{2 ( 70 ) ( 9.8 )}{0.7} = 1960 = 44.3 m/s$ Yeh step kyun? Given $m$ , $g$ , aur lumped constant $C_{d} ρ A = 0.7$ ko derived formula mein substitute karte hain exactly taaki concrete result ko real skydiver speeds ke saath compare kar sakein — ek sanity anchor jo symbolic form nahi de sakta.

Limiting behaviour. Acceleration $a = g - \frac{C _{d} ρ A}{2 m} v^{2}$ . $v = v_{t}$ par: $a = 9.8 - \frac{0.7}{140} (1960) = 9.8 - 9.8 = 0$ ✔. Kyun? Hamare balance confirm karta hai: diver ka speed badhna band ho jaata hai. Iske baad, saari released gravitational PE seedhi air ko heat karne jaati hai — aur koi KE gain nahi. Dekho Terminal velocity & projectile with drag.

Verify: $v_{t} = 44.3 m/s$ ( $\approx 160$ km/h, ek realistic belly-to-earth value). $v_{t}$ par, $a = 0$ exactly. Units check: root ke andar $\frac{kg \cdot m/s ^{2}}{kg/m} = \frac{m ^{2}}{s ^{2}}$ hai, jiska square root $m/s$ hai ✔ — yeh check isliye karte hain taaki lumped-constant units mein koi slip number par trust karne se pehle pakad sakein.

Worked example Ex 7 — Cell G: linear (Stokes) drag, low speed par

Ek chhota $2 \times 1 0^{- 8} kg$ mist droplet still air mein linear drag $F_{d} = b v$ , $b = 3.0 \times 1 0^{- 8} N per (m/s)$ ke saath girta hai. Uski terminal velocity pata karo, aur regime ko Ex 6 se contrast karo.

Forecast: Same balance idea jaise Ex 6, lekin drag law ab $b v$ hai ( $v$ ki first power), $v^{2}$ nahi. Kya $v_{t}$ tiny hogi (droplets drift karenge) ya fast?

Sahi drag law choose karo. Slow, small object ke liye viscous shear dominate karta hai, isliye $F_{d} = b v$ (linear), quadratic law nahi. Yeh step kyun? Parent note ke do regimes interchangeable nahi hain: ek fog droplet millimetres-per-second par move karta hai jahan air smoothly (laminar) behave karti hai, isliye linear model yahan sahi tool hai.
Terminal speed par weight ko linear drag ke against balance karo ( $a = 0$ ): $m g = b v_{t} \Rightarrow v_{t} = \frac{m g}{b}$ Yeh step kyun? "Terminal" phir se matlab acceleration zero, isliye downward weight upward drag ke barabar. Kyunki drag ab $v$ mein linear hai, solve karna ek plain division hai — Ex 6 se alag, koi square root nahi.
Numbers plug in karo: $v_{t} = \frac{( 2 \times 1 0 ^{- 8} ) ( 9.8 )}{3.0 \times 1 0 ^{- 8}} = \frac{1.96 \times 1 0 ^{- 7}}{3.0 \times 1 0 ^{- 8}} = 6.53 m/s$ Yeh step kyun? Droplet ki mass aur given $b$ substitute karne se general balance ek concrete drift speed ban jaata hai, jise hum physical plausibility ke liye judge karte hain.

Verify: $v_{t} = 6.53 m/s$ . Units check: $\frac{kg \cdot m/s ^{2}}{N/(m/s)} = \frac{N}{N/(m/s)} = m/s$ ✔ — yeh isliye karte hain kyunki linear-drag coefficient ke odd units (N per m/s) exactly wahan hain jahan galtiyan chhupti hain. Structural contrast note karo: linear drag $v_{t} \propto m$ deta hai, jabki quadratic drag (Ex 6) $v_{t} \propto m$ deta hai — jo tool tum choose karte ho woh physics change karta hai. Dekho Terminal velocity & projectile with drag.

Worked example Ex 8 — Cell H: forensic word problem (skid marks → speed)

Ek car $25 m$ ke straight skid marks chhodti hai phir ruk jaati hai. Road $μ_{k} = 0.8$ . Brakes lock karte waqt yeh kitni tez chal rahi thi?

Forecast: Skid marks ek friction-only, flat-floor problem hai — same shape jaise Ex 1, ulta chalaya. Kya residential $50$ km/h limit se zyada fast hai?

Flat road, brakes locked → kinetic friction rok deti hai. Height unchanged, $Δ U = 0$ , isliye master equation $W_{n c} = Δ K E$ tak collapse hoti hai: $- μ_{k} m g L = 0 - \frac{1}{2} m v_{0}^{2}$ Yeh step kyun? Locked wheels slide karte hain, isliye yeh kinetic friction hai (rolling nahi), aur level road matlab gravity ki PE nahi badalti — exactly Ex 1 setup, jo friction work ko lost KE se equate karne deta hai.
Mass cancel, initial speed ke liye solve karo: $v_{0} = 2 μ_{k} g L = 2 (0.8) (9.8) (25) = 392 = 19.8 m/s$ Yeh step kyun? Skid length $L$ exactly friction path length hai; hum Ex 1 ki logic ko invert karte hain ( $L$ ki jagah $v_{0}$ ke liye solve karte hain) marks produce karne wali speed recover karne ke liye. $m$ cancel ho jaati hai, isliye answer ko car ki mass ki zaroorat nahi.
Human ke liye convert karo: $19.8 m/s \times 3.6 = 71.3 km/h$ . Yeh step kyun? Skid physics SI m/s deta hai, lekin speed limits km/h mein hain; $3.6$ se multiply karna ( $1 m/s = 3.6 km/h$ kyunki) result ko posted limit se directly comparable banata hai, jo forensic question ka point hai.

Verify: $19.8 m/s = 71.3 km/h$ — $50$ km/h limit se zyada, isliye driver speeding tha. Car mass se independent (isliye forensics is trick se pyaar karta hai). Units ✔.

Figure padhna: same ramp jaise Ex 3, lekin ab dashed motion arrow up slope ki taraf point karta hai. Crucially yellow friction arrow flip ho gaya hai down-slope ki taraf point karne ke liye (yeh upward velocity ka oppose karta hai), isliye ab yeh blue gravity component $m g sin θ$ ke saath baitha hai — dono milke energy drain karte hain. Pink $m g cos θ$ abhi bhi slope mein press karta hai. (Pehle ki tarah, har arrow word-labelled aur direction-tagged hai isliye picture greyscale mein padhti hai.)

Forecast: Upar jaate waqt, gravity AUR friction dono motion oppose karte hain — isliye yeh frictionless launch se sooner rukna chahiye. Trap: friction ab down-slope point karti hai, upar nahi.

Direction check (the twist). Up-slope move karte waqt, velocity upar point karti hai, isliye friction down-slope point karti hai — gravity ke along-slope component ke same direction mein (figure mein flipped yellow arrow dekho). Yeh step kyun? Friction velocity ka oppose karta hai, gravity ka nahi. Upar jaate waqt, dono milke block ki KE drain karte hain, jo bracket ke andar plus sign ki tarah dikhega.

Energy budget set up karo. Rising a distance $L$ raises height by $L sin θ$ , isliye $Δ U = + m g L sin θ$ (positive — yeh climb karta hai). Friction work $W_{n c} = - μ_{k} m g cos θ L$ hai, aur rest par khatam hota hai isliye $Δ K E = 0 - \frac{1}{2} m v_{0}^{2}$ . Yeh step kyun? Master equation ka har term is specific geometry se pin hona chahiye solve karne se pehle: $Δ U$ yahan positive hai (Ex 3 ke sliding-down se alag), aur $Δ K E$ negative hai kyunki block apni poori launch speed kho deta hai.

Master equation, teen terms substitute karo: $W_{n c} - μ_{k} m g cos θ L = Δ K E - \frac{1}{2} m v_{0}^{2} + Δ U m g L sin θ$ $m$ cancel karo, $L$ ek side par ikatha karo: $\frac{1}{2} v_{0}^{2} = g L (sin θ + μ_{k} cos θ) \Rightarrow L = \frac{v _{0}^{2}}{2 g ( s i n θ + μ _{k} c o s θ )}$ Plus sign kyun? Dono energy drains — gravity term $sin θ$ aur friction term $μ_{k} cos θ$ — add hote hain, sliding-down case (Ex 3) mein minus se alag. Woh sign flip poora exam point hai: yeh block ko sooner rokta hai.

Numbers plug in karo: $sin 2 5^{\circ} = 0.4226$ , $cos 2 5^{\circ} = 0.9063$ : $L = \frac{5 ^{2}}{2 ( 9.8 ) ( 0.4226 + 0.3 \cdot 0.9063 )} = \frac{25}{2 ( 9.8 ) ( 0.6945 )} = 1.836 m$ Yeh step kyun? Given launch speed aur angle ko derived formula mein substitute karne se concrete stopping distance milta hai, jise hum immediately frictionless climb se compare karte hain confirm karne ke liye ki friction ne ise chhota kiya.

Verify: Frictionless climb $\frac{v _{0}^{2}}{2 g sin θ} = \frac{25}{2 ( 9.8 ) ( 0.4226 )} = 3.02 m$ tak pahuncha hota; friction ke saath $1.84 < 3.02$ par sooner rukta hai ✔, exactly jaise forecast kiya. Units ✔.

Worked example Ex 10 — Cell J: DONO thieves ek saath — friction

aur air drag Ek puck rough floor par slide karta hai aur saath hi air drag se bhi push back hota hai. Jis instant woh $v = 20 m/s$ par move karta hai, total non-conservative force aur instantaneous heating rate pata karo. Data: $m = 0.5 kg$ , flat ground par $μ_{k} = 0.10$ , linear drag $b = 0.05 N per (m/s)$ .

Forecast: Do chor same puck ko lootte hain. Kya unke forces add hote hain? (Dono same velocity oppose karte hain, isliye haan — lekin sirf friction constant hai; drag $v$ par depend karta hai.)

Dono forces same direction mein point karte hain (backward). Friction $f_{k} = μ_{k} m g$ aur drag $F_{d} = b v$ dono velocity oppose karte hain, isliye motion ki direction mein woh simply add hote hain: $F_{n c} = f_{k} + F_{d} = μ_{k} m g + b v$ Yeh step kyun? Non-conservative forces interfere ya cancel nahi karte — har ek independently $v$ oppose karta hai, isliye jab woh ek direction share karte hain toh magnitudes add hote hain. Isliye matrix ko "both at once" cell ki zaroorat thi: real objects rarely sirf ek feel karte hain.
Total force ke liye numbers plug in karo: $F_{n c} = (0.10) (0.5) (9.8) + (0.05) (20) = 0.49 + 1.00 = 1.49 N$ Yeh step kyun? Substitute karne se symbolic sum abhi ka actual backward push ban jaata hai; note karo friction ( $0.49 N$ ) fixed hai lekin drag ( $1.00 N$ ) puck slow hone par shrink hoga — split isliye matter karta hai ki kya dominate karta hai.
Instantaneous heating rate (power stolen) force times speed hai: $P = F_{n c} v = (0.49 + 1.00) (20) = 29.8 W$ Yeh step kyun? Power energy theft ki rate hai; opposing force ko current speed se multiply karne se joules-per-second milte hain jo heat mein ja rahe hain — natural "both thieves combined" bookkeeping, kyunki $Q$ $\int P d t$ ke roop mein accumulate hoti hai.

Verify: $F_{n c} = 1.49 N$ , $P = 29.8 W$ . Units: $N \times m/s = J/s = W$ ✔. Sanity: drag yahan friction se zyada loot raha hai ( $1.00 > 0.49 N$ ) kyunki puck fast hai; kam speed par friction dominate karta — exactly woh regime split jo matrix warn karta hai.

Recall Matrix par quick self-test

Kaunsa cell imaginary speed deta hai, aur uska kya matlab hai? ::: Cell D — woh incline jahan $tan θ \leq μ_{k}$ ; imaginary root matlab block kabhi speed up nahi karta, aur kyunki $tan θ < μ_{s}$ static friction ise start hi nahi hone deti. Bullet-block problem mein, collision mein kaunsi quantity conserved hoti hai? ::: Sirf momentum — embed inelastic hai, KE heat/deformation mein jaati hai. Rough incline par UPAR jaate waqt, friction slope ke upar point karta hai ya neeche? ::: Down-slope — yeh velocity oppose karta hai, gravity ke saath milke energy drain karta hai. Ex 1, 5, 8 mein mass kyun cancel hoti hai? ::: $K E$ aur $W_{f r i c}$ dono $m$ ke saath scale karte hain, isliye divide out ho jaata hai — stopping distance mass-independent hoti hai. Linear aur quadratic drag ke beech terminal velocity ki mass dependence kaise alag hai? ::: Linear: $v_{t} \propto m$ ; quadratic: $v_{t} \propto m$ . Jab friction aur drag saath act karte hain, kya unke forces add hote hain? ::: Haan — dono same velocity oppose karte hain, isliye magnitudes add hote hain; lekin sirf friction constant hai, drag $v$ ke saath badhta hai. Kaunsa coefficient decide karta hai ki resting block slide karna shuru karega? ::: $μ_{s}$ (static), $μ_{k}$ nahi; motion tabhi shuru hoti hai jab $tan θ > μ_{s}$ .

Connections

Parent topic — theory
Work–Energy theorem — woh hamesha-sach statement jise yeh saare examples specialise karte hain.
Mechanical energy conservation — yeh problems kis par reduce hote agar $μ_{k} = 0$ .
Friction — static & kinetic — jahan se $μ_{k}$ , $μ_{s}$ aur sliding condition aati hai.
Inelastic collisions — Ex 5 mein momentum step.
Terminal velocity & projectile with drag — Ex 6 aur Ex 7 ka ghar.
Heat & first law of thermodynamics — jahan chura hua $Q$ jaata hai.