Newton's first law — inertia, operational definition of force: Worked examples

1.2.1 · D3 · Physics › Newton's Laws & Dynamics › Newton's first law — inertia, operational definition of forc

Kuch bhi shuru karne se pehle, woh ek tool yaad karo jo hum har jagah use karte hain:

Scenario matrix

Is topic mein jo bhi problem aa sakti hai woh inhi cells mein se ek hai. Table mein class, kya badalta hai, aur kaun sa example use nail karta hai listed hai.

Cell	$v$ ke saath kya ho raha hai	Net force?	Example
A. True rest	$v = 0$ , $0$ hi rehta hai	$0$	Ex 1
B. Constant velocity, forces cancel	$v$ constant, $\neq = 0$	$0$ (lekin bahut saari forces)	Ex 2
C. Speeding up (same direction)	$v$ ki magnitude badhti hai	$\neq = 0$ , motion ke saath	Ex 3
D. Slowing down (same direction)	$v$ ki magnitude ghatti hai	$\neq = 0$ , motion ke against	Ex 3
E. Direction change, constant speed	$v$ turn karta hai	$\neq = 0$ , sideways (centripetal)	Ex 4
F. Non-inertial frame illusion	lagta hai accel ho raha hai, real force nahi	$0$ actually	Ex 5
G. Degenerate: zero time / zero interval	change ke liye time nahi	$0$ net effect	Ex 6
H. Real-world word problem	mixed	depends	Ex 7
I. Exam twist: partial cancellation	ek axis balanced, doosra nahi	$\neq = 0$ ek axis pe	Ex 8

Cells C aur D ek hi "along-the-line" case ke do signs hain, isliye Ex 3 dono karta hai.

Worked examples

Worked example 1 — Cell A: woh block jo bas wahan baitha hai

Ek steel block level floor pe resting hai. Tum kuch nahi karte. Net force describe karo aur agli second mein uski motion predict karo.

Forecast: Padhne se pehle guess karo — kya net force zero hai kyunki kuch touch nahi kar raha, ya kuch ho raha hai?

Real forces list karo. Gravity neeche pull karti hai weight $W$ se; floor upar push karta hai normal force $N$ se.
- Yeh step kyun? "Kuch nahi ho raha" ek trap hai — do forces present hain. First law sirf unke sum ki parwah karti hai, isliye summing se pehle hum unhe list karte hain.
Vertically sum karo. Block floor mein dhansता nahi aur upar nahi udta, isliye $N = W$ aur vertical net $N - W = 0$ hai.
- Yeh step kyun? Observed motion (wahan rehe rehna) batata hai net zero hona chahiye; first law ulta padhi ja rahi hai — " $v$ constant" se " $F_{net} = 0$ " tak.
Conclude karo. $F_{net} = 0 \Rightarrow v$ $0$ hi rehti hai. Yeh baitha rehta hai.

Verify: Rest " $v = constant$ " hai jahan constant $0$ hai — poori tarah consistent. Units: $N$ aur $W$ dono forces (newtons) hain, isliye unka difference ek force hai; isse $0 N$ set karna dimensionally theek hai. Mass vs weight dekho: weight ek force hai, block ki inertia nahi.

Worked example 2 — Cell B: woh plane jo cruise karta hai (bahut saari forces, zero net)

Ek jet seedha aur level $250 m/s$ pe fly karta hai. Thrust $= 200000 N$ forward, aur woh equilibrium mein hai. Drag kya hai? Net force kya hai?

Forecast: Engines ghar rahe hain — zaroor net force bahut badi aur forward hogi?

Velocity padho. Seedha aur level, steady speed ⇒ $v$ constant hai.
- Yeh step kyun? First law pehle velocity padhti hai. Constant $v$ neeche ka conclusion force karta hai chahe engines kitna bhi shor karen.
Law apply karo. $v$ constant $\Rightarrow F_{net} = 0$ .
- Yeh step kyun? Yahi operational content hai: constant velocity zero net force ka symptom hai.
Drag nikalo. Horizontally, thrust aur drag hi ek maatra forces hain, isliye $drag = thrust = 200000 N$ (backward).
- Yeh step kyun? Har axis pe Net $= 0$ . Horizontal net $= 200000 - drag = 0$ .
Vertically lift weight ko usi tarah balance karta hai — pucha nahi gaya, lekin isliye woh level rehta hai.

Verify: Net force $= (200000 - 200000) N = 0 N$ . $250 m/s$ ki speed force balance mein kabhi nahi aayi — sahi hai, kyunki first law ko sirf yahi chahiye ki $v$ change na kare, uski value nahi.

Worked example 3 — Cells C & D: puck ko push karna aur drag karna (dono signs)

Frictionless ice pe ek puck pehle $+ 3 m/s$ pe move karta hai, phir $2 s$ baad $+ 9 m/s$ pe (Case C). Ek doosre run mein woh $+ 9 m/s$ jaata hai phir $2 s$ baad $+ 3 m/s$ (Case D). Kin cases mein net force act ki, aur woh kis taraf point karti thi?

Forecast: Do runs — har ek mein net force ka sign (direction) guess karo.

Case C, velocity change compute karo. $Δ v = 9 - 3 = + 6 m/s$ (positive).
- Yeh step kyun? Law ka trigger $v$ mein koi bhi change hai. $Δ v$ ka sign humein net force ki direction batata hai.
Case C verdict. $Δ v \neq = 0 \Rightarrow$ net force act ki, $+$ direction mein point karti hui (motion ke saath — isliye speed badhi). Figure mein blue arrow dekho jo puck ke saath point kar raha hai.
Case D, compute karo. $Δ v = 3 - 9 = - 6 m/s$ (negative).
- Yeh step kyun? Wahi trigger, opposite sign. Negative $Δ v$ ⇒ net force $-$ direction mein point karti hai, motion ke against (pink arrow).
Case D verdict. Net force act ki, backward point karti hui (ek braking force). Speed giri.

Verify: $∣Δ v ∣ = 6 m/s \neq = 0$ dono runs mein — first law kehti hai dono mein force act ki, jo sahi hai (frictionless ice khud puck ko slow ya speed nahi karti). Sign flip hota hai ( $+ 6$ vs $- 6$ ): tool "speeding up" ko "slowing down" se sirf sign se distinguish karta hai, bilkul jaisa promise kiya tha. Yahan zero friction ka matlab hai change sirf external push se aa sakta hai.

Worked example 4 — Cell E: constant speed, turning (direction counts)

Ek string pe ek ball radius $R = 0.5 m$ ke horizontal circle mein constant speed $4 m/s$ pe swing karti hai. "East" point pe uski velocity $(0, + 4) m/s$ hai (north point karti hai); quarter-circle baad "north" pe woh $(- 4, 0) m/s$ hai (west point karti hai). Kya net force act ki?

Forecast: Speed kabhi nahi badli — toh zaroor net force zero hai?

Do velocity vectors compare karo, speeds nahi. Start $(0, + 4)$ , baad mein $(- 4, 0)$ .
- Yeh step kyun? First law maangti hai ki poora vector $v$ constant ho, sirf uski length nahi. Law mein "straight line" clause ka poora point yahi hai.
Unhe subtract karo. $Δ v = (- 4, 0) - (0, + 4) = (- 4, - 4) m/s$ .
- Yeh step kyun? Ex 3 jaisa hi trigger, lekin ab 2-D mein. Nonzero $Δ v$ ka matlab hai velocity change hui chahe $∣ v ∣$ na badla ho.
Verdict. $Δ v \neq = 0 \Rightarrow$ ek net force act ki — inward string tension (centripetal). Figure mein yellow arrow dekho jo centre ki taraf point kar raha hai.

Verify: Start pe $∣ v ∣$ $= 0^{2} + 4^{2} = 4$ ; end pe $= (- 4)^{2} + 0^{2} = 4$ — equal speeds, phir bhi $∣Δ v ∣ = (- 4)^{2} + (- 4)^{2} = 32 \approx 5.66 m/s \neq = 0$ . Constant speed, changed velocity — first law sahi se ek force demand karti hai. Zyada Uniform circular motion mein.

Worked example 5 — Cell F: braking-bus illusion (non-inertial frame)

$10 m/s$ pe ek bus mein khade ho, driver brake lagata hai aur tum aage jhuk jaate ho. Ek passenger claim karta hai "ek force ne mujhe aage push kiya." Ground (inertial) frame se, kya tumpe koi real forward force hai?

Forecast: Tum definitely feel kiya aage throw hue — real force ya fake?

Honest frame pe switch karo. Ground se dekho, ek inertial frame jahan first law hold karta hai.
- Yeh step kyun? First law sirf inertial frame mein guaranteed hai. Bus, brake lagate waqt, accelerating hai — ek non-inertial frame — isliye hum wahan "apparent" motion pe trust nahi kar sakte.
Apne torso pe law apply karo. Koi horizontal force tumhara upper body forward push nahi karta; first law ke by woh $10 m/s$ pe chalti rehti hai.
- Yeh step kyun? Inertia: constant $v$ ko koi cause nahi chahiye. Tumhara torso bas continue karta hai.
"Lurch" explain karo. Bus (aur tumhare feet) slow hokar, maan lo, $6 m/s$ pe aate hain; tumhara torso abhi bhi $10$ pe ⇒ bus ke relative woh aage move karta hai. Koi real forward force nahi chahiye.

Verify: Torso ki bus ke w.r.t. relative velocity $= 10 - 6 = + 4 m/s$ forward — lurch se match, zero real forward force ke saath produce hua. Bus frame mein tum aage $ma$ size ka pseudo-force invent karte; first law precisely woh flag hai ki yeh frame non-inertial hai.

Worked example 6 — Cell G: degenerate "zero interval" (tablecloth)

Ek plate ke neeche se cloth khichi jaati hai. Cloth se plate pe friction force $F = 2 N$ sirf $Δ t = 0.05 s$ ke liye act karta hai; plate ka mass $0.4 kg$ hai. Plate ki velocity kitni change hoti hai? Effectively, kya plate move karti hai?

Forecast: Ek force act karti hai — toh kya plate slide off nahi honi chahiye?

Impulse velocity change karta hai. Velocity mein change $Δ v = \frac{F Δ t}{m}$ .
- Yeh step kyun? Jab ek force bahut thode time ke liye act karta hai, velocity change force $\times$ time divided by mass hai. Jaise $Δ t \to 0$ , $Δ v \to 0$ : matrix ka "no time to change" degenerate case.
Plug in karo. $Δ v = \frac{2 \times 0.05}{0.4} = 0.25 m/s$ .
- Yeh step kyun? "Tiny" ko number mein badlte hain prove karne ke liye ki woh negligible hai.
Interpret karo. Ek quarter metre per second, $0.05 s$ mein gain hua, plate ko lagbhag $\frac{1}{2} \cdot 0.25 \cdot 0.05 \approx 0.006 m$ — chhe millimetre — move karta hai. Woh wahan rehti hai.

Verify: $Δ v = 0.25 m/s$ ; pull ke dauran distance $\approx \frac{1}{2} Δ v Δ t = 0.00625 m$ . Jaise $Δ t \to 0$ plate ki motion $\to 0$ : woh limiting behaviour jo first law promise karta hai — essentially koi interval nahi, inertia jeetti hai aur kuch nahi badlta. Friction dekho.

Worked example 7 — Cell H: real-world word problem (truck mein crate)

Ek truck ek crate ( $m = 50 kg$ ) rough bed pe le jaata hai. Truck seedhi road pe $6 s$ mein $0$ se $12 m/s$ accelerate karta hai. Crate ke truck ke saath move karne ke liye (peeche slide na kare), friction ko kitna forward force supply karna chahiye, aur kya crate ka frame accelerate karte waqt inertial hai?

Forecast: Guess karo — kya crate ko force chahiye, aur woh kahan se aata hai?

Crate ka required velocity change nikalo. Isse $0 \to 12 m/s$ jaana hai, isliye $Δ v = 12 m/s$ $6 s$ mein, yaani acceleration $a = 12/6 = 2 m/s^{2}$ .
- Yeh step kyun? Crate ki $v$ change hoti hai ⇒ first law kehti hai usse ek net force act karna chahiye. Hum quantify karte hain kitne change ki zarurat hai.
Force naam karo. Ek maatra forward horizontal agent truck bed se static friction hai, isliye friction $= ma = 50 \times 2 = 100 N$ forward.
- Yeh step kyun? First law culprit ko naam deta hai (kuch to $v$ change karta hai); value required change se aati hai. Enough friction ke bina crate apni purani velocity rakhta aur truck ke relative peeche slide karta — pure inertia.
Frame question. Crate ek accelerating truck pe ride karta hai — woh co-moving frame non-inertial hai; road frame mein (inertial) first law aur yeh force analysis hold karta hai.

Verify: $a = 2 m/s^{2}$ , friction $= 100 N$ . Sanity: agar bed frictionless hoti (Ex 1 ka Cell A horizontally apply hota), friction $= 0$ , crate $v = 0$ rakhta, aur truck uske neeche se nikal jaata — exactly wahi tablecloth logic bade scale pe. Consistent.

Worked example 8 — Cell I: exam twist (ek axis balanced, doosra nahi)

Ek box frictionless ramp se slide karta hai. Surface ke along (isse $x$ kaho) kuch bhi downhill pull ko balance nahi karta; surface ke perpendicular ( $y$ ) normal force gravity ke perpendicular part ko balance karta hai. Kya $F_{net} = 0$ hai? Kya box constant velocity rakhta hai?

Forecast: Ek direction balanced hai — kya woh first law ke liye "kaafi" hai?

Har axis alag check karo. Perpendicular ( $y$ ): normal force $=$ perpendicular gravity ⇒ net $y = 0$ . Along-ramp ( $x$ ): gravity ka downhill component ke paas cancel karne ke liye kuch nahi ⇒ net $x \neq = 0$ .
- Yeh step kyun? First law chahti hai ki poora vector $F_{net}$ vanish kare. Ek vector zero hota hai tabhi jab har component zero ho. Ek axis balance karna kaafi nahi.
Combine karo. Net $y = 0$ lekin net $x \neq = 0 \Rightarrow F_{net} \neq = 0$ .
- Yeh step kyun? Yahi classic exam trap hai: "normal force gravity ko balance karta hai, isliye net force zero hai." Woh sirf perpendicular part balance karta hai.
Verdict. $F_{net} \neq = 0 \Rightarrow v$ constant nahi — box slope ke neeche speed up karta hai.

Verify: Agar $F_{net}$ sach mein $0$ hota, box constant velocity pe glide karta; experience kehta hai woh downhill accelerate karta hai, isliye net nonzero hona chahiye. Sirf perpendicular axis balanced hai. First law ki demand — sab components zero — exactly yahi catch karti hai.

Recall Pure matrix pe quick self-check

Har cell ek sawaal pe reduce hoti hai ::: "Kya $v$ (magnitude ya direction) change hua?" — agar haan, toh net force act ki. "Bahut saari forces" phir bhi zero net force kyun mean kar sakti hain? ::: Kyunki first law sirf sum padhti hai; jo forces cancel hote hain (Ex 2) woh $v$ constant rehne dete hain. Ek axis balance karna kyun kaafi nahi? ::: $F_{net} = 0$ ke liye har component zero chahiye; Ex 8 ka ramp $y$ balance karta hai lekin $x$ nahi. Braking-bus "force" real kyun feel hoti hai lekin hai nahi? ::: Bus ek non-inertial frame hai; ground se dekho tumhara torso simply apni velocity rakhta hai (inertia), koi real forward force nahi.

Connections

Newton's second law — F=ma — "ek force act ki" verdict ko number mein badalta hai (Ex 3, 6, 7 mein use hua).
Newton's third law — action-reaction — crate pe friction (Ex 7) ka truck pe reaction hota hai.
Inertial vs non-inertial frames — Ex 5 aur Ex 7 is baat pe hinge karte hain ki kaun sa frame honest hai.
Friction — Ex 3 (uski absence), Ex 6, Ex 7 mein hidden force.
Uniform circular motion — Ex 4 mein reveal hui force.
Pseudo-forces — braking bus ke andar tum kya invent karte (Ex 5).
Mass vs weight — weight ek force ke roop mein (Ex 1, Ex 8) vs mass inertia ke roop mein.

Worked examples — Newton's first law — inertia, operational definition of force

Scenario matrix

Worked examples

Connections

Case Map