Work-energy theorem — derivation from Newton's second law: Exercises

1.3.3 · D4 · Physics › Work, Energy & Power › Work-energy theorem — derivation from Newton's second law

Numbers se pehle, ek picture dekho jo "work" aur " $Δ K$ " ka matlab dikhaye taaki koi bhi symbol unearned na lage. Neeche di gayi figure mein red arrow dekho: woh force $F$ hai jo kaam kar rahi hai jab block displacement $d$ (black double-arrow) se slide karta hai. Upar ke labels block ki kinetic energy shuruat mein ( $K_{i}$ , left mein) aur ant mein ( $K_{f}$ , right mein) dikhate hain. Poori page sirf woh statement hai jo us picture ke upar red mein likhi hai: red arrow jitna kaam karta hai woh $K_{i}$ se $K_{f}$ tak ki jump ke barabar hai.

Figure — Work-energy theorem — derivation from Newton's second law

Definition Har problem mein zaroori symbols

$m$ = mass = object mein kitna "stuff" hai, kilograms ( $kg$ ) mein measure hota hai. Bada $m$ matlab speed up karna mushkil.
$v$ = speed = object kitni tez move karta hai, metres per second ( $m/s$ ) mein. Initial speed ke liye $v_{i}$ aur final speed ke liye $v_{f}$ likhte hain.
$x$ = position = object apni line of travel mein kahan hai, metres ( $m$ ) mein. Jab koi force object ke move karne ke saath change hoti hai, tab hum ise function $F (x)$ likhte hain — "position $x$ par force".
$d$ = displacement (distance) = object kitna move hua, metres ( $m$ ) mein. Yeh woh length hai jiske through force kaam karti hai. (Jab $x$ start se end tak jaata hai, $d$ bas woh total gap hai.)
$W$ = work = force ko us distance $d$ se multiply karo jiske through woh motion ke saath saath push karta hai. Units: joule $J$ . Agar force aur motion ek hi direction mein point karein, $W > 0$ (speed up hota hai); opposite direction mein, $W < 0$ (slow down hota hai).
$K = \frac{1}{2} m v^{2}$ = kinetic energy = "motion ki energy", mass $m$ aur speed $v$ se bani. Speed squared par depend karti hai, isliye hamesha $\geq 0$ hoti hai aur direction ki kabhi parwah nahi karti. Dekho Kinetic Energy.
$Δ K = K_{f} - K_{i}$ = final minus initial. Positive matlab object speed up hua.
$F_{n e t}$ = net force = woh single force jo tumhe har force ko sign ke saath add karne ke baad milti hai (aage push karna $+$ hai, friction ya opposition $-$ hai). Theorem hamesha is net force ko use karti hai, kabhi akeli ek force ko nahi. Dekho Newton's Second Law.
$θ$ = angle = force ki direction aur motion ki direction ke beech ki tilting (degrees mein). Jab $θ = 0$ tab force seedha path ke saath khichti hai; bada $θ$ matlab force ka zyada hissa "sideways waste" ho raha hai.
$t$ = time = koi cheez kitni der chalti hai, seconds ( $s$ ) mein.
$a$ = acceleration = speed har second kitni tez change hoti hai, $m/s^{2}$ mein; Newton's Second Law se, $a = F_{n e t} / m$ .
$p$ = momentum = mass times velocity, $p = m v$ , $kg\cdotm/s$ mein. Iska change $Δ p$ woh "kick" hai jo ek force time ke saath deliver karta hai — work ka time-sibling. Dekho Impulse–Momentum Theorem.

Level 1 — Recognition

Goal: theorem tumhe kaunsi quantity deta hai yeh identify karo, bina kisi trap ke.

Recall Solution L1.1

KYA: hum $W_{n e t}$ chahte hain. THEOREM KYU: yeh seedha kehta hai net work $Δ K$ ke barabar hai — energies pata hone ke baad koi forces ya distances ki zaroorat nahi. $W_{n e t} = Δ K = K_{f} - K_{i} = 30 - 12 = 18 J$ $4 m$ ek distractor hai — theorem ne pehle se distance ko $K$ mein absorb kar liya hai.

Recall Solution L1.2

$W = F d$ YAH KYU: force constant aur motion ke parallel hai, yeh woh akela case hai jahan integral $\int F d x$ ek simple product mein collapse ho jaata hai. $W = F d = 8 \times 3 = 24 J \Rightarrow Δ K = 24 J$

Level 2 — Application

Goal: theorem use karke speed solve karo, ek force ek baar.

Recall Solution L2.1

$W = F d$ YAH KYU: force constant hai aur motion ke saath point karti hai, isliye work integral ek plain product mein collapse ho jaata hai (L1.2 jaisi reasoning). $W_{n e t} = F d = 10 \times 5 = 50 J$ . Kyunki $K_{i} = \frac{1}{2} (2) (0)^{2} = 0$ : $\frac{1}{2} m v_{f}^{2} = 50 \Rightarrow v_{f}^{2} = \frac{2 \times 50}{2} = 50 \Rightarrow v_{f} = 50 \approx 7.07 m/s$ $F = ma$ SE BEHTAR KYU: humne kabhi acceleration ya time solve nahi kiya — force-over-distance seedha speed tak pahuncha.

Recall Solution L2.2

$K_{i} = \frac{1}{2} (1000) (20)^{2} = 200000 J$ , aur $K_{f} = 0$ . $W_{n e t} = Δ K = 0 - 200000 = - 200000 J$ NEGATIVE KYU: friction motion ke opposite point karti hai, isliye woh kinetic energy remove karti hai. Car ko decelerate karne wali opposing force ke liye Newton's Second Law dekho.

Recall Solution L2.3

$W_{n e t} = \frac{1}{2} m (v_{f}^{2} - v_{i}^{2}) = \frac{1}{2} (0.5) (1 0^{2} - 4^{2}) = 0.25 \times (100 - 16) = 0.25 \times 84 = 21 J$ Note karo ki humhe dono speeds chahiye theen — sirf $K$ ka change matter karta hai, koi bhi akela value nahi.

Level 3 — Analysis

Goal: variable forces (integrate karo), aur multiple forces (pehle net karo).

Recall Solution L3.1

INTEGRATE KYU: $F$ position $x$ ke saath change hoti hai, isliye $W = F d$ illegal hai; work $F$ – $x$ graph ke neeche ka area hai (figure mein red region), jo exactly $\int F d x$ hai. Dekho Work done by a variable force. $W = \int_{0}^{2} 6 x d x = [3 x^{2}]_{0}^{2} = 3 (4) - 0 = 12 J$ $\frac{1}{2} (3) v_{f}^{2} = 12 \Rightarrow v_{f}^{2} = 8 \Rightarrow v_{f} = 22 \approx 2.83 m/s$

Recall Solution L3.2

PEHLE NET KYU: theorem net force $F_{n e t}$ use karta hai jo panel mein define hai — har force ko sign ke saath add karo (forward push $+ 30$ , opposing friction $- 10$ ). $F_{n e t} = 30 - 10 = 20 N, W_{n e t} = 20 \times 5 = 100 J$ $\frac{1}{2} (4) v_{f}^{2} = 100 \Rightarrow v_{f}^{2} = 50 \Rightarrow v_{f} = 50 \approx 7.07 m/s$ Equivalently: tumhara work $30 \times 5 = 150 J$ plus friction ka work $- 10 \times 5 = - 50 J$ = $100 J$ . Same answer.

Recall Solution L3.3

$cos θ$ KYU: sirf force ka woh component jo displacement ke saath hai work karta hai — yeh dot product $F \cdot d r = F d cos θ$ hai, jahan $θ$ force aur motion ke beech ka angle hai (panel mein define kiya gaya). Vertical component zero work karta hai kyunki koi vertical motion nahi hai. $W = F d cos θ = 20 \times 6 \times 0.5 = 60 J$ $\frac{1}{2} (5) v_{f}^{2} = 60 \Rightarrow v_{f}^{2} = 24 \Rightarrow v_{f} = 24 = 26 \approx 4.90 m/s$

Level 4 — Synthesis

Goal: theorem ko gravity, inclines, aur energy conservation ke saath chain karo.

Neeche di gayi figure L4.1 aur L4.2 ke liye ramp dikhati hai. Red block upar baitha hai; woh vertical height jitni woh girata hai $h$ se marked hai. Note karo ki block slope ke saath travel karta hai (black arrow), lekin height label $h$ triangle ki vertical side hai — yahi distinction agle do solutions ka poora point hai.

Recall Solution L4.1

SIRF GRAVITY KA VERTICAL DROP KYU MATTER KARTA HAI: figure mein, normal force slope-arrow ke perpendicular hai (zero work karta hai), aur gravity ka work $m g h$ hai ramp angle se independent — yeh sirf picture mein marked vertical side $h$ par depend karta hai, slope length par nahi. $W_{n e t} = m g h = 2 \times 10 \times 3 = 60 J$ $\frac{1}{2} (2) v_{f}^{2} = 60 \Rightarrow v_{f}^{2} = 60 \Rightarrow v_{f} = 60 \approx 7.75 m/s$ Yeh exactly Conservation of Mechanical Energy disguise mein hai: gravity ka work kinetic energy ban gaya.

Recall Solution L4.2

Gravity phir bhi $m g h = 60 J$ karta hai (figure mein vertical $h$ ). Friction, however, slope arrow ki length $5 m$ ke saath act karti hai, isliye uska negative work woh slope distance use karta hai: $W_{f r i c} = - 4 \times 5 = - 20 J, W_{n e t} = 60 - 20 = 40 J$ $\frac{1}{2} (2) v_{f}^{2} = 40 \Rightarrow v_{f}^{2} = 40 \Rightarrow v_{f} = 40 \approx 6.32 m/s$ Frictionless $7.75 m/s$ se slower — friction ne $20 J$ chura liye.

Recall Solution L4.3

$W = \int_{0}^{3} 4 x d x = [2 x^{2}]_{0}^{3} = 18 J$ $\frac{1}{2} (1) v_{f}^{2} = 18 \Rightarrow v_{f}^{2} = 36 \Rightarrow v_{f} = 6 m/s$ Coasting ka matlab hai aage koi net work nahi, isliye speed $6 m/s$ rehti hai.

Level 5 — Mastery

Goal: power, momentum ke saath combine karo, aur limiting/edge cases ke baare mein reason karo.

Recall Solution L5.1

AB POWER KYU: power kaam karne ki rate hai, average par $P = \frac{W}{t}$ , panel mein defined time $t$ use karke. Dekho Power. $P_{a v g} = \frac{W}{t} = \frac{18}{2} = 9 W$ Theorem ne work diya; power bas time se divide karta hai.

Recall Solution L5.2

$6 m/s$ par $K$ : $\frac{1}{2} (3) (6)^{2} = 54 J$ . Direction matter nahi karta — $K$ speed squared use karta hai.

(a) Rokna: $W_{a} = 0 - 54 = - 54 J$ .
(b) Re-speeding: $W_{b} = 54 - 0 = + 54 J$ .
(c) Poora trip: start speed $6$ , end speed $6$ , toh $Δ K = 54 - 54 = 0 J$ . Insight: object doosri taraf move karte hue end karta hai phir bhi net work zero hai — kyunki Kinetic Energy left aur right mein farq nahi kar sakti.

Recall Solution L5.3

Time route (impulse): $\int F d t = Δ p$ , Impulse–Momentum Theorem, jahan $p = m v$ momentum hai jo panel mein define kiya gaya. $F t = m v_{f} \Rightarrow v_{f} = \frac{12 \times 2}{3} = 8 m/s$ $Δ K = \frac{1}{2} (3) (8)^{2} - 0 = 96 J$ . Distance route (work): acceleration $a = F / m = 4 m/s^{2}$ ke saath (panel mein defined $a$ ), hume covered distance $d$ chahiye. $d = \frac{1}{2} a t^{2}$ KYU: rest se constant acceleration ke under shuru karke, distance ek straight velocity–time line ke neeche ka area hai — base $t$ aur height $a t$ ka triangle, area $\frac{1}{2} (t) (a t) = \frac{1}{2} a t^{2}$ deta hai. Yeh standard constant-acceleration kinematic result hai: $d = \frac{1}{2} a t^{2} = \frac{1}{2} (4) (4) = 8 m$ $W = F d = 12 \times 8 = 96 J = Δ K ✓$ Dono siblings agree karte hain: time-integral momentum deta hai, distance-integral energy deta hai, aur yahan woh consistently land karte hain.

Recall Solution L5.4

Constant velocity ⇒ zero acceleration $a = 0$ ⇒ zero net force ⇒ friction $= 50 N$ opposing.

(a) $W_{a ppl i e d} = 50 \times 10 = + 500 J$ .
(b) $W_{f r i c} = - 50 \times 10 = - 500 J$ .
(c) $W_{n e t} = 500 - 500 = 0 J$ .
(d) $Δ K = 0 J$ — consistent, kyunki speed kabhi change nahi hui. Edge lesson: tum $500 J$ mehnat dal sakte ho aur kinetic energy kuch bhi change nahi kar sakte. Theorem sirf net track karta hai.

Recall checkpoints

Recall Answer key (sab values)

L1.1 $18 J$ · L1.2 $24 J$ · L2.1 $7.07 m/s$ · L2.2 $- 200000 J$ · L2.3 $21 J$ · L3.1 $12 J, 2.83 m/s$ · L3.2 $7.07 m/s$ · L3.3 $60 J, 4.90 m/s$ · L4.1 $7.75 m/s$ · L4.2 $6.32 m/s$ · L4.3 $6 m/s$ · L5.1 $9 W$ · L5.2 $- 54, + 54, 0 J$ · L5.3 $8 m/s, 96 J$ · L5.4 $+ 500, - 500, 0, 0$ .

Connections

Parent topic — woh derivation jise yeh exercises drill karti hain.
Newton's Second Law · Kinetic Energy · Work done by a variable force · Conservation of Mechanical Energy · Power · Impulse–Momentum Theorem.