Non-conservative forces — friction, air drag: Question bank

1.3.7 · D5 · Physics › Work, Energy & Power › Non-conservative forces — friction, air drag

Questions se pehle, teen anchors taaki neeche har symbol already samjha hua ho:

Work $W = F \cdot d r$ ka matlab hai "force times the distance moved along the force." Opposing force ⇒ negative work.
Mechanical energy $E_{m ec h} = K E + U$ : motion energy plus stored (height/spring) energy.
Non-conservative = ek aisi force jiska total work path length par depend karta hai, sirf start aur end point par nahi. Dekho Work–Energy theorem aur Conservative forces & potential energy.

Sach ya jhooth — justify karo

Friction hamesha ek moving object par negative work karta hai.

Jhooth. Jis surface par sliding ho rahi hai uske liye negative hota hai, lekin friction ek decelerating truck mein aage carry ho rahe passenger par positive work kar sakta hai, ya tumhare haath ke neeche wale block par — sign decide karta hai relative-motion ka direction, word "friction" nahi.

Ek rough floor par round trip ka net friction work zero hoga kyunki displacement zero hai.

Jhooth. Friction work ==total path length

L

== use karta hai, net displacement nahi. Ek round trip mein

L = 2 d \neq = 0

hota hai, isliye

W_{f r i c} = - μ_{k} N (2 d) < 0

Agar koi force kisi ek particular closed loop par zero work karta hai, to woh conservative hona chahiye.

Jhooth. Conservative hone ke liye zaroori hai ki

\oint F \cdot d r = 0

har loop ke liye. Ek accidental zero (jaise friction ek aisi loop par jahan tum actually move hi nahi karte) kuch prove nahi karta.

Air drag non-conservative hai usi wajah se jis wajah se friction hai.

Sach. Drag hamesha velocity ke opposite point karta hai, isliye kisi bhi closed loop par woh har segment mein subtract karta hai — loop integral strictly negative hota hai, bilkul friction ki tarah.

Terminal velocity par ek skydiver ki mechanical energy constant rehti hai.

Jhooth. Height girti rehti hai isliye

U

girta hai, jabki

K E

constant hai — total

E_{m ec h}

ghatta hai, aur sab kuch air ko heat karne mein chala jaata hai. Dekho Terminal velocity & projectile with drag.

Kyunki energy hamesha conserved hoti hai, isliye bullet-into-block collision mein kinetic energy conserved hoti hai.

Jhooth. Total energy conserved hoti hai, lekin kinetic energy nahi — kaafi heat aur deformation mein jaati hai. Sirf momentum conserved hota hai, isliye pehle momentum use karo. Dekho Inelastic collisions.

Speed double karne se quadratic drag force double ho jaati hai.

Jhooth. Quadratic drag hai

F_{d} = \frac{1}{2} C_{d} ρ A v^{2}

, isliye

v

double karne par force chaar guna ho jaati hai. Sirf linear (Stokes) drag

F_{d} = b v

double hota hai.

Friction se generate hua heat potential energy ki loss ke barabar hota hai.

Jhooth. Heat

= - W_{n c} = Δ K E + Δ U

. Yeh sirf U ka nahi, balki KE aur U dono ke change ka hisaab rakhta hai. Dekho Heat & first law of thermodynamics.

Agar

μ_{k} > tan θ

ho, to ek rough incline par already sliding kar raha block speed up karega.

Jhooth. Bracket

(sin θ - μ_{k} cos θ) < 0

hai, isliye slope ke saath gravity ki pull friction se harti hai — block decelerate karta hai (aur rest se kabhi start hi nahi karta).

Static friction kaam kar sakta hai lekin kabhi mechanical energy ko heat ke roop mein nahi nikalta.

Sach (ideal no-slip case mein). Relative sliding nahi hone par, surfaces rub nahi karte, isliye heat generate nahi hoti; static friction phir bhi energy transfer kar sakta hai (jaise ek rolling wheel) bina dissipate kiye. Dekho Friction — static & kinetic.

Error dhundo

"Block distance

d

slide karta hai aur waapis aata hai; friction ne

- μ m g d

forward aur

+ μ m g d

back kiya, isliye net friction work zero hai."

Error: friction displacement direction ke saath sign nahi badalta — woh velocity ka virodh karne ke liye badalta hai, isliye return trip bhi

- μ m g d

hi hai. Net

= - 2 μ m g d

, non-conservative ka loop-test signature.

"Terminal velocity: maine drag times distance ko

m g

ke barabar set kiya."

Error: ek force ko ek energy statement ke saath mix kar rahe ho. Terminal velocity forces balance karne se aati hai

m g = \frac{1}{2} C_{d} ρ A v_{t}^{2}

, kisi work equation se nahi.

"Incline example mein block

m g L sin θ

friction ko deta hai."

Error: gravity

m g L sin θ

release karti hai; friction sirf

μ_{k} m g cos θ L

leta hai. Baaki KE ban jaata hai. Released PE ko friction loss samajhna galat hai.

"Main ek skydiver ke liye atmosphere mein girte hue

W = - F_{d} L

use karunga."

Error:

F_{d}

fall ke dauran constant nahi hota — jaise

v

badhta hai waise

F_{d}

bhi badhta hai. Sirf terminal velocity par (constant

v

) tum constant-force product use kar sakte ho.

"Kyunki friction non-conservative hai, main yahan work–energy theorem use nahi kar sakta."

Error: work–energy theorem

W_{t o t a l} = Δ K E

hamesha sach hai. Non-conservative forces ka bas yehi matlab hai ki tum unka work potential energy

U

mein nahi fold kar sakte.

"Ek conservative force ko loop par hamesha zero hona chahiye, isliye gravity koi kaam nahi karti jab main ek book uthata hoon."

Error: loop par net aur ek leg par ko confuse kar rahe ho. Gravity upar uthate waqt negative work karti hai aur neeche aate waqt positive; round-trip sum zero hota hai, lekin ek akela leg nahi.

Why questions

Friction ka kaam path par kyun depend karta hai lekin gravity ka nahi?

Gravity sirf height difference (endpoints) par depend karti hai, isliye ek potential energy

U

exist karta hai. Friction har step par motion ka virodh karta hai, isliye lambe/tedhe-medhe paths zyada ghisate hain — loss endpoints nahi, total distance set karta hai.

Friction ke liye potential energy function impossible kyun hai?

Ek potential energy tumhe har position par ek number

U

assign karne deti, history se independent. Lekin friction ka kaam depend karta hai tum wahan kaise pahunche (path length) par, isliye koi bhi sirf-position-wala function use nahi ho sakta.

Inelastic collision mein energy nahi, momentum kyun use karna chahiye?

Kinetic energy heat/deformation mein ek unknown amount se gayab ho jaati hai, isliye energy bookkeeping mein ek missing term hota hai. Momentum mein aisa koi loss nahi hota (koi external impulse nahi), jo ek clean, solvable equation deta hai.

Speed badhne par drag linear se quadratic mein kyun change hota hai?

Kam speed par viscous shear dominate karta hai (momentum loss

\propto v

). Zyada speed par tum per second air ka ek mass

\propto v

side mein karte ho, har parcel

\propto v

momentum carry karta hai, jo

\propto v^{2}

deta hai.

Q = - W_{n c}

kyun hai na ki

Q = + W_{n c}

Non-conservative forces system par negative work karte hain (

W_{n c} < 0

). Produce hua heat ek positive quantity hai jo removed energy ke barabar hai, isliye

Q = - W_{n c} > 0

Ek block ko incline par

tan θ > μ_{k}

ki zaroorat kyun padti hai sirf move karne ke liye?

Slope ke saath gravity component

m g sin θ

hai aur maximum friction

μ_{s} m g cos θ

hai. Motion tab shuru hoti hai jab

sin θ > μ_{s} cos θ

, yaani

tan θ > μ_{s}

Edge cases

Jab ek block rough floor par still baitta ho to friction ka kaam kya hoga?

Zero — koi sliding nahi matlab koi path length traverse nahi hoti, isliye

W_{f r i c} = - μ_{k} N L

mein

L = 0

. Motion ke bina static friction kuch dissipate nahi karta.

v = 0

par drag work kya hoga (rest se drop kiya gaya)?

Uss instant drag exactly zero hota hai (

F_{d} \propto v

v^{2}

), isliye initially fall pure free-fall hai; drag tab hi kaam aata hai jab speed badhti hai. Dekho Terminal velocity & projectile with drag.

Ek incline par jahan

μ_{k} = tan θ

exactly ho, block kaunsi final speed reach karta hai?

Bracket

(sin θ - μ_{k} cos θ) = 0

ho jaata hai, isliye

v = 0

— agar already move kar raha hai to constant velocity par coast karta hai (net force zero); rest se kabhi start nahi karta. Ek perfect knife-edge.

Ek perfectly horizontal round trip (

N = m g

constant) ke liye, friction loss usi displacement ki ek hilly round trip se kaise compare karta hai?

Hilly trip mein usually zyada path length aur varying

N

hota hai, isliye generally zyada loss — baat ye hai ki sirf displacement se friction work nahi pata kiya ja sakta; tumhe actual traversed path chahiye.

Frictionless, drag-free world ki limit mein, kya modified equation

W_{n c} = Δ E_{m ec h}

phir bhi hold karta hai?

Haan, trivially:

W_{n c} = 0

, isliye

Δ E_{m ec h} = 0

— mechanical energy conserved ho jaati hai, Mechanical energy conservation ko special case ke roop mein recover karte hue.

Agar drag velocity ka virodh kare lekin speed ke saath chhota hota jaye, to kya woh phir bhi non-conservative ho sakta hai?

Haan — non-conservativeness aati hai force ke hamesha opposing motion se (loop integral negative), is baat se nahi ki magnitude

v

ke saath kaise scale karta hai. Direction rule akele potential energy ko forbid karta hai.

Recall One-line reflex checks

Friction work path length use karta hai displacement nahi ::: kyunki woh har leg par velocity ka phir se virodh karta hai. Collision mein pehla move ::: momentum conserve karo, kabhi kinetic energy nahi. Heat produced ::: $Q = - W_{n c} = Δ K E + Δ U$ . Block sliding tab shuru karta hai jab ::: $tan θ > μ_{s}$ .

Connections

Parent topic — full derivations jin traps ko ye test karte hain.
Work–Energy theorem · Mechanical energy conservation · Conservative forces & potential energy — honest counterparts.
Friction — static & kinetic · Terminal velocity & projectile with drag · Inelastic collisions · Heat & first law of thermodynamics — jahan ye traps phir milte hain.