Work, Energy and Power: Class 11 Physics Chapter 5 Hinglish mein

Physics Part 1 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 18 Oct 2026 · Updated 18 Oct 2026

100 min
28 practice questions
26 MCQs

Is chapter mein aap kya seekhenge

Do vectors ka scalar product unke magnitudes aur angle se, ya components se, calculate karna, aur usse angles aur projections nikalna
Constant aur variable forces dwara kiya gaya work uske sign ke saath calculate karna, aur use force–displacement graph ke neeche ke area ke roop mein padhna
Constant aur variable forces ke liye work-energy theorem ko state, derive aur apply karna
Conservative aur non-conservative forces mein fark karna, aur Earth ke paas gravity ke liye aur spring ke liye potential energy use karna
Girti cheezon, vertical circles aur springs par mechanical energy ka conservation, aur friction ke saath uska modified form, apply karna
Power ko define karna, P = F·v use karna, aur watts, horsepower aur kilowatt hours ke beech convert karna
Momentum aur kinetic energy ke conservation se one aur two dimensions mein elastic aur inelastic collisions ko analyse karna

NCERT Physics Class 11, Chapter 5, pp. 71–91 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

5.1 Introduction

Teen logon se poocho ki work ka matlab kya hai, to shayad tum Mumbai mein cycle chala kar lunch boxes pahunchane wale dabbawala ke baare mein sunoge, entrance test ki revision karti ek student ke baare mein, aur railway station par sir par trunk sambhaale ek coolie ke baare mein. Roz ki bolchaal mein teeno mehnat se kaam kar rahe hain. Physics zyada nakhre wali hai. Woh work tabhi ginti hai jab force us cheez ko move kare jis par woh lag raha hai. Desk par baithi student apni books par lagbhag koi work nahi karti, chahe woh kitni bhi thaki ho, aur coolie customer ka intezaar karte hue khada rehta hai to trunk par koi work nahi karta.

Energy aur power ke bhi roz ke matlab hain. Hum kehte hain ki jo bachcha kabhi tik kar nahi baithta woh energy se bhara hai, aur jo fast bowler ball $145 k m / h$ par phenkta hai woh power se bowling karta hai. Is chapter mein yeh teeno shabd units wali measurable quantities ban jaate hain. Work ko force aur displacement ke terms mein ek precise definition milti hai, energy work karne ki capacity banti hai, aur power naapti hai ki work kitni jaldi hota hai. Raaste mein kuch surprises ke liye taiyaar raho: ek bhaari bag ko bilkul sthir pakde rehna mehnat ka kaam lagta hai, phir bhi usmein koi work nahi hota.

Pehle thoda vector algebra. Work do vectors, force aur displacement, ko jodta hai, phir bhi result ki koi direction nahi hoti. Jo operation yeh karta hai woh scalar product hai.

5.1.1 Scalar product

Forces ko jodna Chapter 3 se jaana-pehchana hai. Work ko ek alag combination chahiye: force aur displacement ko multiply karke ek single number, jiski koi direction na ho. Vector algebra theek yahi deta hai, scalar product, jise dot product bhi kehte hain. (Vector product, jo vector deta hai, Chapter 6 mein aayega.) Do vectors $𝐀$ aur $𝐁$ ko tail to tail rakho, unke beech angle $θ$ ho ( $0 \leq θ \leq 180^{\circ}$ ). Unka scalar product hai

Scalar (dot) product

𝐀 \cdot 𝐁 = A B \cos θ

(5.1a)

Ise 'A dot B' padho. Right side ka har factor ek simple number hai, isliye product ka size aur sign hota hai lekin direction nahi. Factors ko do tarah group karne se iski ek picture milti hai (Fig. 5.1). $A (B \cos θ)$ ke roop mein, yeh $𝐀$ ki length guna $𝐀$ par $𝐁$ ka projection hai: woh shadow jo $𝐁$ , $𝐀$ ki line par daalta hai jab light us line par right angle se padti hai. $B (A \cos θ)$ ke roop mein, yeh $𝐁$ ki length guna $𝐁$ par $𝐀$ ki shadow hai. Dono groupings same number deti hain.

2/4 $𝐁$ ki tip se perpendicular $𝐀$ par uska projection $B \cos θ$ mark karta hai, isliye $𝐀 \cdot 𝐁 = A (B \cos θ)$ .

Fig. 5.1NCERT Fig. 5.1 · p. 72

Vector A, O se right ki taraf hai aur vector B angle theta par upar aur right ki taraf hai. Step 1: sirf vectors aur angle. Step 2: B ki tip se perpendicular A ke along ek green bar, B cos theta, mark karta hai. Step 3: A ki tip se perpendicular B ke along ek green bar, A cos theta, mark karta hai. Step 4: B 90 degrees se aage ghoomta hai aur uski shadow, negative dikhane ke liye vermillion mein, O ke peeche girti hai.

Dot product ke rules ordinary multiplication jaise hi hain. Kisi bhi vectors $𝐏$ , $𝐐$ , $𝐑$ aur real number $λ$ ke liye:

$𝐏 \cdot 𝐐 = 𝐐 \cdot 𝐏$ : order se fark nahi padta (commutative law).
$𝐏 \cdot (𝐐 + 𝐑) = 𝐏 \cdot 𝐐 + 𝐏 \cdot 𝐑$ : sum ke saath dot karne par dots ka sum milta hai (distributive law).
$𝐏 \cdot (λ 𝐐) = λ (𝐏 \cdot 𝐐)$ : number ko bahar nikala ja sakta hai.

DerivationComponents se dot product

Unit vectors $\hat{𝐢}$ , $\hat{𝐣}$ , $\hat{𝐤}$ ki length 1 hai aur yeh aapas mein perpendicular hain. Isliye ek unit vector ka khud se dot 1 deta hai ( $\hat{𝐢} \cdot \hat{𝐢} = 1$ , aur $\hat{𝐣}$ aur $\hat{𝐤}$ ke liye bhi yahi), jabki do alag unit vectors 0 dete hain ( $\hat{𝐢} \cdot \hat{𝐣} = \hat{𝐣} \cdot \hat{𝐤} = \hat{𝐤} \cdot \hat{𝐢} = 0$ ).

$𝐀 = A_{x} \hat{𝐢} + A_{y} \hat{𝐣} + A_{z} \hat{𝐤}$ likho, $𝐁$ ko bhi isi tarah likho, aur distributive law se multiply karo. Nau products mein se, jo chhe alag unit vectors ko jodte hain woh zero ho jaate hain, aur bachta hai

\begin{aligned} 𝐀 \cdot 𝐁 & = A_{x} B_{x} \\ + A_{y} B_{y} \\ + A_{z} B_{z} \end{aligned}

(5.1b)

$𝐁 = 𝐀$ (yaani $θ = 0$ ) lene par yeh ek length ka square deta hai, $A^{2} = 𝐀 \cdot 𝐀$ :

\begin{aligned} A^{2} & = A_{x}^{2} + A_{y}^{2} \\ + A_{z}^{2} \end{aligned}

(5.1c)

Aur do non-zero vectors theek tab perpendicular hote hain jab unka dot product zero ho: right angles ka ek test jismein protractor ki zaroorat nahi.

Example 5.1NCERT Example 5.1 · p. 72

Methane ka bond angle

Methane molecule ( $C H_{4}$ ) mein carbon atom ek cube ke centre mein hota hai aur chaar hydrogen atoms alternate corners par. Do C–H bonds $𝐚 = \hat{𝐢} + \hat{𝐣} + \hat{𝐤}$ aur $𝐛 = \hat{𝐢} - \hat{𝐣} - \hat{𝐤}$ ke along point karte hain (cube ke aadhe edge ki units mein). Nikalo (a) $𝐚 \cdot 𝐛$ , (b) bonds ke beech angle, aur (c) $𝐛$ par $𝐚$ ka projection.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

2 mein se 1

\hat{𝐣} \cdot (\hat{𝐢} + \hat{𝐣})

kya hai?

1

0

\sqrt{2}

\hat{𝐤}

Answer dekho

Answer (A) $1$

\hat{𝐣} \cdot \hat{𝐢} = 0

aur

\hat{𝐣} \cdot \hat{𝐣} = 1

, isliye result 1 hai. Dot product hamesha ek number hota hai, kabhi

\hat{𝐤}

jaisa vector nahi.

5.2 Work aur kinetic energy ke notions: work-energy theorem

Work aur kinetic energy ko sirf define karne ke bajaye discover kiya ja sakta hai. Chapter 2 ke ek result se shuru karo, jo constant acceleration $a$ wali seedhi line ki motion ke liye initial speed $u$ , final speed $v$ aur displacement $s$ ko jodta hai:

v^{2} - u^{2} = 2 a s

(5.2)

Dono sides ko $m / 2$ se multiply karo aur body par net force ke liye Newton ka second law, $F = m a$ , use karo:

\begin{aligned} \frac{1}{2} m v^{2} - \frac{1}{2} m u^{2} \\ = m a s = F s \end{aligned}

(5.2a)

Koi nayi physics nahi aayi: yeh second law ko hi rearrange kiya gaya hai. Three dimensions mein kinematics $v^{2} - u^{2} = 2 𝐚 \cdot 𝐝$ ban jaati hai, jahan $𝐝$ displacement hai, aur wahi step deta hai

\begin{aligned} \frac{1}{2} m v^{2} - \frac{1}{2} m u^{2} \\ = m 𝐚 \cdot 𝐝 = 𝐅 \cdot 𝐝 \end{aligned}

(5.2b)

Equation (5.2b) do definitions ki taraf ishara karti hai. $\frac{1}{2} m v^{2}$ ko body ki kinetic energy $K$ kaho, aur $𝐅 \cdot 𝐝$ ko force dwara kiya gaya work $W$ kaho. In naamon ke saath equation aisi padhi jaati hai

Work-energy theorem

K_{f} - K_{i} = W

(5.3)

jahan $K_{i}$ aur $K_{f}$ initial aur final kinetic energies hain (Fig. 5.2).

TheoremWork-energy theorem

Kisi bhi displacement mein, particle par lage net force dwara kiya gaya work particle ki kinetic energy ke increase ke barabar hota hai.

Yeh work-energy theorem yahan sirf constant force ke liye prove hua hai; Section 5.6 yeh restriction hata deta hai.

3/3Work $𝐅 \cdot 𝐝$ gain $\frac{1}{2} m v^{2} - \frac{1}{2} m u^{2}$ ke barabar hai.

Fig. 5.2

Track par mass m ki ek cart ko constant force right ki taraf kheenchta hai. Step 1: ek chhota arrow uski initial speed u dikhata hai. Step 2: cart kaafi right chali gayi hai, neeche displacement arrow hai, aur ek lamba arrow speed v dikhata hai. Step 3: do barabar bars aate hain, ek kiye gaye work ke liye aur ek kinetic energy ke gain ke liye.

Example 5.2NCERT Example 5.2 · p. 73

Ghat road par freewheeling

Ek cyclist (cycle ke saath total mass $80 k g$ ) pahadi road ke top par pedal chalana band karti hai aur uske along $800 m$ freewheel karti hai, height mein $40 m$ neeche aate hue. Uski speed $2.0 m / s$ se badhkar $12 m / s$ ho jaati hai. Friction aur air drag us par aise lagte hain jinki details hum nahi jaante. $g = 9.8 m / s^{2}$ lekar nikalo (a) gravity dwara kiya gaya work, (b) friction aur drag dwara kiya gaya total work, aur (c) road ke along average resistive force.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

2 mein se 1

2.0 k g

trolley ki speed

3.0 m / s

se badhkar

5.0 m / s

ho jaati hai. Net force ne us par kitna work kiya?

34 J

25 J

4.0 J

16 J

Answer dekho

Answer (D) $16 J$

Δ K = \frac{1}{2} (2.0) ({5.0}^{2} - {3.0}^{2}) = 16 J

. Lalchane wala

4.0 J

(5.0 - 3.0)^{2}

se aata hai: kinetic energy har speed ke square ke saath chalti hai, change ke square ke saath nahi.

5.3 Work

Ab sirf work ko dekho. Ek constant force $𝐅$ ek body par lagta hai jab body ek displacement $𝐝$ se move karti hai, force se angle $θ$ par. Force ka sirf woh part jo displacement ke along hai, $F \cos θ$ , motion ko tez ya dheema karta hai; uske across wala part dono mein se kuch nahi karta. Isliye work woh component guna chali gayi distance hai:

Constant force dwara work

\begin{aligned} W & = (F \cos θ) d \\ = 𝐅 \cdot 𝐝 \end{aligned}

(5.4)

Definition

Work (constant force dwara)

Displacement guna uske along force ka component, $W = 𝐅 \cdot 𝐝$ . Yeh scalar hai aur positive, negative ya zero ho sakta hai.

ExampleFloor se $60^{\circ}$ par $40 N$ force se suitcase ko $10 m$ kheenchne mein $200 J$ work hota hai.

1/4 $θ < 90^{\circ}$ ke liye component $F \cos θ$ displacement ke along point karta hai, isliye $W > 0$ .

Fig. 5.3

Ek box floor par right ki taraf slide karta hai. Step 1: force lagbhag 35 degrees par aage point karta hai, motion ke along ek green component ke saath, aur W 0 se bada hai. Step 2: force seedha upar, right angles par point karta hai, aur W 0 ke barabar hai. Step 3: force lagbhag 145 degrees par peeche point karta hai, ek vermillion backward component ke saath, aur W 0 se kam hai. Step 4: force friction ki tarah seedha peeche point karta hai, aur W minus F d ke barabar hai.

Yeh definition roz ki bolchaal se alag raasta pakadti hai. Ek tug-of-war socho jo barabari par khatam hota hai: dono teams poore ek minute zor lagati hain, rassi nahi hilti, aur koi bhi team us par koi work nahi karti, chahe sab kitne bhi thak jaayein. Unki muscles sirf tight rehne ke liye food energy jalati hain, lekin woh energy unke apne shareer mein heat ban jaati hai, rassi par work nahi.

Isliye teen situations mein force body par zero work karta hai (Fig. 5.4):

Koi displacement nahi. Platform par sir par $40 k g$ ka trunk liye sthir khada coolie trunk par koi work nahi karta, chahe woh kitni bhi der intezaar kare.
Motion ke along koi force nahi. Smooth ice par fisalte puck par koi horizontal force nahi lagta, phir bhi woh kaafi door jaata hai: zero force, zero work.
Displacement ke perpendicular force. Jab coolie level platform par steady speed se chalta hai, to trunk par uska upward push trunk ke displacement ke right angles par hota hai, aur $\cos 90^{\circ} = 0$ . Isi tarah pendulum ki string ka tension bob par koi work nahi karta, kyunki bob hamesha string ke right angles par move karta hai.

3/3Level zameen par chalte hue upward force displacement ke perpendicular hai.

Fig. 5.4

Step 1: ek coolie sir par box liye upward force F ke saath sthir khada hai; text kehta hai d 0 ke barabar hai, isliye W 0 ke barabar hai. Step 2: ek puck motion ke along bina kisi force ke ice par right ki taraf fisalta hai. Step 3: coolie box lekar right ki taraf chalta hai, uska upward force displacement ke right angles par hai, isliye W 0 ke barabar hai.

Work ka sign batata hai ki force madad kar raha hai ya rukawat daal raha hai. Jis force ka component motion ke along hai ( $θ < 90^{\circ}$ ) woh energy daalta hai aur positive work karta hai; jiska component motion ke against hai ( $θ > 90^{\circ}$ ) woh energy nikaalta hai aur negative work karta hai. Sliding body par kinetic friction seedha peeche point karta hai, isliye us body par uska work hamesha negative hota hai.

Eq. (5.4) se, work ke dimensions force guna length hain, $[M L^{2} T^{- 2}]$ , bilkul kinetic energy jaise. Iski SI unit joule (J) hai: jab $1 N$ ka force apne point of application ko force ke along $1 m$ move karta hai, tab kiya gaya work. Yeh unit James Prescott Joule (1818–1889) ke naam par hai, jinke experiments ne dikhaya ki diya gaya work hamesha utni hi heat paida karta hai. Energy ki purani aur specialist units aaj bhi use hoti hain (Table 5.1).

Table 5.1Work aur energy ki doosri unitsNCERT Table 5.1 · p. 74

Unit	Joules mein value	Yeh kahan milti hai
erg	$10^{- 7} J$	CGS system, purani books
electron volt (eV)	$1.6 \times 10^{- 19} J$	atoms, electrons aur nuclei
calorie (cal)	$4.186 J$	heat; food labels kcal use karte hain
kilowatt hour (kWh)	$3.6 \times 10^{6} J$	bijli ke bill par 'unit'

Example 5.3NCERT Example 5.3 · p. 74

Conveyor belt par parcel

Airport par ek $6.0 k g$ ka parcel ek conveyor belt par giraya jaata hai jo steady $2.0 m / s$ se chal rahi hai. Jab tak parcel belt ke saath nahi chalne lagta, unke beech $30 N$ ka kinetic friction lagta hai. (a) Fisalte waqt parcel kitna move karta hai, aur belt ka friction us par kitna work karta hai? (b) Utne time mein belt kitna move karti hai, aur parcel ka friction belt par kitna work karta hai? (c) Dono works ko jodo aur result ka matlab samjhao.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Dekho yeh example kya dikhata hai. Newton ka third law forces ko pair karta hai, works ko nahi: belt ka parcel par push, parcel ke belt par push ke barabar hai, lekin dono surfaces alag-alag distances move karti hain, isliye works alag hote hain. Do bodies ke beech forces ke liye $𝐅_{12} + 𝐅_{21} = 0$ hamesha hota hai, jabki $W_{12} + W_{21}$ sirf special cases mein zero hota hai, jaise jab dono surfaces ek doosre par fisalti nahi.

5.4 Kinetic energy

Section 5.2 se, velocity $𝐯$ se chalti mass $m$ ki body ki kinetic energy hai

Kinetic energy

\begin{aligned} K & = \frac{1}{2} m 𝐯 \cdot 𝐯 \\ = \frac{1}{2} m v^{2} \end{aligned}

(5.5)

Kyunki $𝐯 \cdot 𝐯 = v^{2}$ , kinetic energy sirf speed par depend karti hai, direction par nahi: yeh scalar hai, aur kabhi negative nahi hoti. Work-energy theorem ko ulta padhne se pata chalta hai ki yeh kya naapti hai. Speed $v$ se chalti body, jo bhi use rokta hai us par $\frac{1}{2} m v^{2}$ work karti hai. Hammer ka head isi tarah keel par work karke use lakdi mein ghusa deta hai, aur chalti hawa Tamil Nadu ke Muppandal ki wind turbines ke blades par work karti hai.

Definition

Kinetic energy

Body ki motion ki wajah se uske paas jo energy hai, $K = \frac{1}{2} m v^{2}$ : woh work jo woh use rokne wali cheez par kar sakti hai.

Example $40 m / s$ par $0.16 k g$ cricket ball ke paas $128 J$ hota hai.

Factor $v^{2}$ ki wajah se speed mass se kahin zyada important hai: mass double karne se $K$ double hota hai, lekin speed double karne se yeh chaar guna ho jaata hai. Fig. 5.5 dikhata hai ki kinetic energies kitni zabardast range mein hoti hain.

Fig. 5.5Chalti cheezon ki rough kinetic energies, logarithmic scale par (masses aur speeds typical values hain).NCERT Table 5.2 · p. 75

Logarithmic energy scale par ek bar chart. Udte mosquito ke paas lagbhag 10 to the minus 6 joule hai, tez phenki cricket ball ke paas 1.3 guna 10 squared, cyclist ke paas 10 cubed, bhari auto-rickshaw ke paas 4.3 guna 10 to the 4, bhare truck ke paas 3.2 guna 10 to the 6, express train ke paas 4.5 guna 10 to the 8, aur orbit mein Earth ke paas lagbhag 2.7 guna 10 to the 33 joules, jo ek toote bar se dikhaya gaya hai.

Example 5.4NCERT Example 5.4 · p. 75

Speed limits kyun zaroori hain

$40 k m / h$ par chalti $1200 k g$ ki car, brakes lagne ke baad, full braking mein $10 m$ mein ruk jaati hai. (a) Average braking force kya hai? (b) Same braking force ke saath, $80 k m / h$ se rukne mein kitni distance lagegi?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

2 mein se 1

Ek scooter ki speed

20 k m / h

se badhkar

60 k m / h

ho jaati hai. Uski kinetic energy kitne guna badhti hai?

36927

Answer dekho

Answer (C) 9

Speed teen guna hoti hai aur

K \propto v^{2}

, isliye

K

3^{2} = 9

guna badhta hai. 3 chunne ka matlab square bhool jaana.

5.5 Variable force dwara kiya gaya work

Zyadatar real forces cheezon ke move karne ke saath badalte hain. Spring ka pull use kheenchne par badhta hai, rocket ka thrust fuel jalne ke saath badalta hai, aur atki hui almirah par tumhara push har pal badalta hai. Tab hum work kaise nikalein?

$x$ -axis ke along ek force $F (x)$ ke under motion lo jo position ke saath badalta hai. $x_{i}$ se $x_{f}$ tak ke path ko $Δ x$ width ke chhote tukdon mein kaato. Ek chhote tukde par force mushkil se badalta hai, isliye uska work lagbhag $F (x) Δ x$ hai, yaani $F$ – $x$ graph ke neeche ek patle rectangle ka area (Fig. 5.6). Tukdon ko jodne par,

W \approx \sum_{x_{i}}^{x_{f}} F (x) Δ x

(5.6)

Strips ko aur patla karte jao, to unke ubad-khabad tops curve se aur kareeb chipakte jaate hain. Limit mein unka total theek curve ke neeche ka area ban jaata hai, jise calculus ek definite integral ke roop mein likhta hai:

Variable force dwara work

\begin{aligned} W & = \lim_{Δ x \to 0} \sum F Δ x \\ = \int_{x_{i}}^{x_{f}} F (x) d x \end{aligned}

(5.7)

To badalte force dwara kiya gaya work, start aur end ke beech uske force–displacement graph ke neeche ka area hai, aur $x$ -axis ke neeche ka area negative work ginta hai. (Textbook ka Appendix 3.1 zaroori integrals yaad dilata hai.)

2/3Patli strips curve se behtar fit hoti hain, aur unka total asli work ke aur kareeb aata hai.

Fig. 5.6NCERT Fig. 5.3 · p. 76

Ek curve x i aur x f ke beech position ke saath badalta force F dikhata hai. Step 1: curve ke neeche chaar chaude rectangles hain, ek par label delta W equals F of x delta x. Step 2: terah patle rectangles curve se zyada kareeb fit hote hain. Step 3: curve ke neeche ka exact shaded region dikhta hai, label W equals area.

Example 5.5NCERT Example 5.5 · p. 76

Lift ka shuru hona

Ek lift aur uske passengers, total mass $400 k g$ , ground floor par rest se shuru hote hain. Upar jaate hue cable ka tension pehle $4.0 m$ mein $4320 N$ se steadily girkar $3920 N$ ho jaata hai aur phir $10 m$ tak $3920 N$ par rehta hai. $g = 9.8 m / s^{2}$ lo aur friction ko ignore karo. $10 m$ mein (a) cable dwara aur (b) gravity dwara kiya gaya work nikalo, aur (c) $10 m$ mark par lift ki speed.

Fig. 5.7Example 5.5 ke liye height ke saath cable tension aur weight.NCERT Fig. 5.4 · p. 76

Force ko height ke against plot kiya gaya hai. Cable tension pehle 4 metres mein 4320 newtons se seedhi line mein girkar 3920 newtons ho jaata hai, phir 10 metres tak 3920 newtons par rehta hai; iske region par label plus 40 000 joules hai. Weight, poore raaste minus 3920 newtons, axis ke neeche ek dashed rectangle hai jis par label minus 39 200 joules hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

2 mein se 1

Ek particle par force

x = 0

par

0

x = 4.0 m

par

10 N

tak seedhi line mein badhta hai. In

4.0 m

mein yeh kitna work karta hai?

2.5 J

10 J

40 J

20 J

Answer dekho

Answer (D) $20 J$

Area ek triangle hai,

\frac{1}{2} \times 4.0 \times 10 = 20 J

. Poori distance ke liye final force use karne se galat

40 J

aata hai.

5.6 Variable force ke liye work-energy theorem

Ab hum constant force maane bina theorem prove kar sakte hain. One dimension mein raho aur dekho ki kinetic energy kitni jaldi badalti hai.

DerivationBadalte force ke saath work-energy theorem

$K = \frac{1}{2} m v^{2}$ ko time ke respect mein differentiate karo (chain rule), phir Newton ka second law, $m d v / d t = F$ , use karo:

\begin{aligned} \frac{d K}{d t} & = m v \frac{d v}{d t} \\ = F v = F \frac{d x}{d t} \end{aligned}

To $d K = F d x$ : ek chhote step mein kinetic energy theek utni badhti hai jitna us step mein work hua. Position $x_{i}$ , jahan kinetic energy $K_{i}$ hai, se $x_{f}$ , jahan yeh $K_{f}$ hai, tak steps ko jodo:

\begin{aligned} K_{f} - K_{i} \\ = \int_{x_{i}}^{x_{f}} F d x \end{aligned}

(5.8a)

Right side Eq. (5.7) wala work hai, isliye

K_{f} - K_{i} = W

(5.8b)

Yeh phir se work-energy theorem hai, ab kisi bhi position par depend karne wale force ke liye.

Newton ke second law ko ball-by-ball commentary samjho aur work-energy theorem ko final scorecard. Scorecard result, yaani final speed, jaldi de deta hai, kyunki force ka effect path ke poore hisse par jod diya gaya hai. Lekin isse tum yeh nahi padh sakte ki kuch kab hua, aur kyunki yeh vector nahi balki ek single number hai, yeh nahi bata sakta ki body aakhir mein kis taraf chal rahi hai. In sawaalon ke liye tum wapas $𝐅 = m 𝐚$ par jaate ho.

Example 5.6NCERT Example 5.6 · p. 77

Aur khurdura hota carpet strip

Mass $0.40 k g$ ki ek toy car $5.0 m / s$ par freely rolling karti hui carpet ke ek strip par aati hai jiska pile uski length ke along mota hota jaata hai. Car par drag strip par chali distance $x$ ke saath $F = - c x^{2}$ ki tarah badhta hai, jahan $c = 3.0 N / m^{2}$ . (a) Carpet ke $1.0 m$ ke baad uski speed nikalo. (b) Agar strip kaafi lamba ho, to car kitni door tak jaayegi?

Fig. 5.8Drag

F = - 3.0 x^{2}

aur pehle metre mein uska work.

Toy car par force ko carpet ke along distance x ke against plot kiya gaya hai. Yeh zero se shuru hota hai aur minus 3.0 x squared ki tarah neeche mudta hai, x equals 1.0 metre par minus 3.0 newtons tak pahunchta hai. 0 se 1.0 metre tak curve aur axis ke beech ka region shaded hai aur uska label W equals minus 1.0 joule hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

2 mein se 1

Work-energy theorem bata sakta hai ki body ko ek speed se doosri speed tak dheema hone mein kitna time lagta hai.

SahiGalat

Answer dekho

Answer Galat

Yeh work ke through speeds ko positions se jodta hai; time ko jod kar hata diya gaya hai. Durations nikalne ke liye Newton ka second law ya kinematics chahiye.

5.7 Potential energy ka concept

Kinetic energy motion ki hai, lekin energy intezaar mein bhi rakhi ja sakti hai. Gulel ki rubber peeche kheencho aur chhod do, to patthar tezi se ud jaata hai. Toy car mein chaabi bharo aur neeche rakho, to woh floor par daud padti hai. Bhakra jaise dam ke peeche ruka paani gates khulne par turbines se hote hue tezi se neeche aata hai. Har case mein energy parts ko rakhne ya arrange karne ke tareeke, yaani unke configuration, mein jama hui thi, aur baad mein kinetic energy ke roop mein wapas mili. Is tarah jama ki gayi energy ko potential energy kehte hain.

Har force energy jama karne nahi deta. Khurdure floor par box ko dhakelo to tumhara work heat ban kar chala jaata hai; box ko koi wapas nahi dhakelta. Lekin box ko shelf par uthao, to work wapas mil sakta hai: box ko girne do aur gravity use kinetic energy ke roop mein lauta deti hai. Is tarah ke recoverable forces ko conservative kehte hain. Aisa force do points ke beech jo work karta hai woh sirf un do points par depend karta hai, unke beech ke raaste par nahi. Isse hum har position $x$ ke saath ek number $V (x)$ , potential energy, jod sakte hain, aisa ki $x_{i}$ se $x_{f}$ tak force ka work $V (x_{i}) - V (x_{f})$ ho. One dimension mein yeh kehne ke barabar hai ki

Potential energy se force

F (x) = - \frac{d V}{d x}

kyunki tab $\int_{x_{i}}^{x_{f}} F d x = - \int d V = V_{i} - V_{f}$ .

Zameen ke paas gravity iska standard example hai. Earth ke radius $R_{E}$ se bahut chhoti heights ke liye $g$ mushkil se badalta hai (Chapter 7 is change ko dekhta hai), isliye mass $m$ par neeche ki taraf constant pull $m g$ lagta hai. Height $h$ ko zameen se upar ki taraf naapo. Jab body zameen se $h$ tak uthti hai, gravity $- m g h$ work karti hai, isliye

Earth ke paas gravitational potential energy

V (h) = m g h

aur sach mein $- d V / d h = - m g$ , ek downward force. Body ko height $h$ se girao, to kinematics $v^{2} = 2 g h$ se uski landing speed deti hai, yaani $\frac{1}{2} m v^{2} = m g h$ : upar jaate waqt jama hua har joule neeche aate waqt wapas mil jaata hai.

Kyunki gravity ka work sirf height ke change par depend karta hai, height $h$ ki kisi bhi frictionless slide par chhoda gaya block neeche $\sqrt{2 g h}$ speed se pahunchta hai, slide ki slope ya shape chahe jo ho (Fig. 5.9). Friction yeh bigaad deta hai: uska work path ki length ke saath badhta hai aur motion par depend karta hai, isliye uski koi potential energy nahi hoti aur use non-conservative kehte hain.

3/4Ek curved frictionless slide bhi same speed deti hai: gravity ka work sirf drop $h$ par depend karta hai.

Fig. 5.9

Height h ke tower ke top se teen slides zameen tak jaati hain: steep, kam dhalan wali aur curved. Steps 1 se 3 mein baari-baari har frictionless slide highlight hoti hai, aur uske neeche ball ki speed same hai, root 2 g h. Step 4 mein kam dhalan wali slide khurduri hai, aur ball bahut chhote arrow ke saath pahunchti hai, yaani speed root 2 g h se kam.

Potential energy ke dimensions $[M L^{2} T^{- 2}]$ hain aur unit joule hai, work aur kinetic energy ki tarah. Conservative force ke liye, chhota displacement potential energy ko kiye gaye work ke minus jitna badalta hai:

Δ V = - F (x) Δ x

(5.9)

Definition

Potential energy

Conservative force ke under kisi body ki position ya configuration se jama energy; iska change us force dwara kiye gaye work ka minus hai, $Δ V = - F Δ x$ .

Example $2.0 k g$ ka bag $1.5 m$ upar shelf par rakhne se lagbhag $29 J$ jama hota hai.

Definition

Conservative force

Aisa force jiska body par work sirf start aur end points par depend karta hai, isliye kisi bhi closed path par uska work zero hota hai aur uski potential energy hoti hai.

ExampleGravity aur ideal spring ka force conservative hain; friction nahi.

Quick check

2 mein se 1

3.0 k g

ka box zameen se

4.0 m

upar ek landing tak le jaaya jaata hai, ek baar seedhiyon se aur ek baar lambe ramp se. Dono trips mein gravity ke work ko compare karo.

Ramp par zyada, kyunki woh lamba haiSeedhiyon par zyada, kyunki woh zyada steep hainBarabar, har baar lagbhag

- 118 J

Dono baar zero, kyunki box aakhir mein rest par hai

Answer dekho

Answer (C) Barabar, har baar lagbhag $- 118 J$

Gravity conservative hai, isliye uska work sirf upar uthne par depend karta hai: kisi bhi raaste ke liye

- m g h = - (3.0) (9.8) (4.0) \approx - 118 J

. Box ka aakhir mein rest par hona sirf yeh batata hai ki net work zero hai; gravity ka hissa phir bhi negative hai.

5.8 Mechanical energy ka conservation

Ball ko upar uchhalo aur dekho kaise woh dheemi hoti hai, rukti hai aur neeche aate waqt phir tez hoti hai. Kinetic energy gaayab hoti hai aur wapas aati hai. Potential energy batati hai ki beech mein woh kahan gayi.

DerivationK + V fixed kyun rehta hai

Ek body ko sirf ek conservative force $F (x)$ ke under chhote displacement $Δ x$ se move karne do. Work-energy theorem deta hai $Δ K = F (x) Δ x$ , aur Eq. (5.9) deta hai $Δ V = - F (x) Δ x$ . Jodne par,

\begin{aligned} Δ K + Δ V & = 0 \\ Δ (K + V) & = 0 \end{aligned}

(5.10)

Sum kisi bhi chhote step mein nahi badalta, isliye $x_{i}$ se $x_{f}$ tak ke poore path mein bhi nahi badalta:

\begin{aligned} K_{i} + V (x_{i}) \\ = K_{f} + V (x_{f}) \end{aligned}

(5.11)

Sum $E = K + V (x)$ body ki total mechanical energy hai. $K$ aur $V$ ko do jebon mein rakhe paise samjho: body ke move karne par har jeb ki rakam badalti hai, lekin dono ka total nahi. 'Conservative' naam yahi darj karta hai. Ab humare paas conservative force pehchanne ke teen barabar tareeke hain; koi bhi ek baaki do ko imply karta hai:

Yeh potential-energy graph ki slope ka minus hai, $F (x) = - d V / d x$ (Eq. (5.9)). Three dimensions ki motion ke liye is slope ka vector version chahiye, jo badi classes mein milega.
Do points ke beech iska work sirf end points se fix hota hai, $W = V (x_{i}) - V (x_{f})$ , raasta chahe jo ho.
Kisi bhi closed loop par yeh koi net work nahi karta: start aur finish ek hi jagah hain, isliye $V (x_{i}) - V (x_{f}) = 0$ .

PrincipleMechanical energy ka conservation

Agar kisi system par sirf conservative forces work karte hain, to uski total mechanical energy $K + V$ constant rehti hai.

Yeh principle ek nariyal par apply karo jo height $H$ par palm se toot kar freely girta hai (Fig. 5.10). Zameen par $V = 0$ lo aur air resistance ko ignore karo. Top par, neeche aate hue height $h$ par, aur zameen chhoone se theek pehle, uski mechanical energy hai

E_{H} = m g H

(5.11a)

E_{h} = m g h + \frac{1}{2} m v_{h}^{2}

(5.11b)

E_{0} = \frac{1}{2} m v_{f}^{2}

(5.11c)

Yahan gravity conservative hai, isliye teeno barabar hain. $E_{H} = E_{0}$ rakhne se landing speed $v_{f} = \sqrt{2 g H}$ milti hai, jaise Section 5.7 mein; $E_{H} = E_{h}$ rakhne se kisi bhi height par speed milti hai:

v_{h}^{2} = 2 g (H - h)

(5.11d)

Yeh kinematic result time ka zikr kiye bina nikal aaya. $15 m$ se girte $1.5 k g$ ke nariyal ke liye yeh lagbhag $17 m / s$ ki landing speed aur $220 J$ kinetic energy deta hai, isiliye koi samajhdaar insaan nariyal ke ped ke neeche nahi sota.

2/3 $h = H / 2$ par energy $V$ aur $K$ ke beech barabar bant jaati hai.

Fig. 5.10NCERT Fig. 5.5 · p. 78

Height H ke palm se ek nariyal girta hai; bars ek fixed total-energy line ke neeche uski potential energy V aur kinetic energy K dikhate hain. Step 1: top par V bhara hai aur K khaali. Step 2: aadhe raaste par, height h par, dono bars aadhe bhare hain. Step 3: zameen ke theek upar, V khaali hai aur K bhara.

3/4C par normal force abhi-abhi zero hua hai: sirf weight path ko modta hai, aur speed $\sqrt{g R}$ hai.

Fig. 5.11NCERT Fig. 5.6 · p. 79

Radius R ke vertical loop track ka centre O, bottom A, side point B aur top C hai. Step 1: A par block right ki taraf chalta hai, normal force upar O ki taraf point karta hai. Step 2: B par woh seedha upar chalta hai, normal force O ki taraf aur weight neeche point karta hai. Step 3: C par woh left ki taraf chalta hai, sirf weight lag raha hai, aur N 0 ke barabar hai. Step 4: track ka upar-left quarter hata diya gaya hai, aur block C se parabola par udta hai, A ke level par, 2R door girta hai.

Example 5.7NCERT Example 5.7 · p. 79

Loop ka chakkar

Mass $50 g$ ka ek chhota block ek ramp par bina friction ke fisalta hai jo radius $R = 0.20 m$ ke vertical circular loop mein jaata hai (Fig. 5.11). Ise loop ke bottom A se height $H$ par rest se chhoda jaata hai. B centre O ke level par hai aur C top hai. (a) Sabse kam $H$ nikalo jiske liye block C par track ke contact mein rahe. Is $H$ ke liye nikalo (b) A, B aur C par speeds aur ratio $K_{B} / K_{C}$ , aur (c) A par block par normal force. (d) Agar C par track mein gap hota to kya hota?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Teen ek jaise patthar cliff ke top se

10 m / s

par phenke jaate hain: ek seedha upar, ek horizontally aur ek seedha neeche. Air resistance ignore karke, neeche samudra tak sabse zyada speed se kaun pahunchta hai?

Upar phenka gaya, kyunki woh sabse upar jaata haiNeeche phenka gaya, kyunki woh neeche ki taraf shuru hota haiHorizontally phenka gaya, kyunki woh sabse door jaata haiTeeno same speed se pahunchte hain

Answer dekho

Answer (D) Teeno same speed se pahunchte hain

Har ek same

K

aur same

V

se shuru hota hai, isliye har ek same total energy ke saath, yaani same speed se pahunchta hai. Energy equation mein directions nahi aati; sirf flight times alag hote hain.

5.9 Spring ki potential energy

Ballpoint pen ki spring dabao to woh halke se wapas dhakelti hai; scooter ki suspension spring compress karo to woh zor se wapas dhakelti hai. Chhote deformations ke liye, ideal spring utna hi kheenchti ya dhakelti hai jitna uska free end $x$ relaxed position se hila ho, aur hamesha wapas usi position ki taraf:

Hooke ka law

F_{s} = - k x

Yeh Hooke ka law hai. Spring constant $k$ , $N / m$ mein, stiffness naapta hai: pen ki spring ka $k$ chhota hai, suspension spring ka bada. Fig. 5.12 mein smooth floor par ek block aisi spring se bandha hai: stretch karne par ( $x > 0$ ) spring force negative hai; compress karne par ( $x < 0$ ) positive hai. Zameen ke paas gravity ke ulat, yeh force position ke saath badalta hai, phir bhi yeh bhi conservative nikalta hai.

4/4Axis ke neeche shaded triangle stretch $x_{m}$ ke liye spring ka work, $- \frac{1}{2} k x_{m}^{2}$ , hai.

Fig. 5.12NCERT Fig. 5.7 · p. 80

Smooth floor par ek block spring se deewar se bandha hai; neeche spring force vs x ka graph origin se hokar neeche jhukti seedhi line hai, current state ke liye ek dot ke saath. Step 1: relaxed, dot origin par. Step 2: right ki taraf stretched, spring wapas kheenchti hai, dot positive x aur negative force par. Step 3: left ki taraf compressed, spring bahar dhakelti hai, dot negative x aur positive force par. Step 4: phir se stretched, axis ke neeche ka triangle spring ke work ke roop mein shaded.

DerivationSpring dwara kiya gaya work

Block ko $x_{i}$ se $x_{f}$ tak dheere move karne do. Spring force badalta hai, isliye use integrate karo (Eq. (5.7)):

\begin{aligned} W_{s} & = - \int_{x_{i}}^{x_{f}} k x d x \\ = \frac{1}{2} k x_{i}^{2} \\ - \frac{1}{2} k x_{f}^{2} \end{aligned}

(5.17)

Do special cases important hain. Relaxed length ( $x_{i} = 0$ ) se $x_{f} = x_{m}$ tak stretch karne par milta hai

W_{s} = - \frac{1}{2} k x_{m}^{2}

(5.15)

jo Fig. 5.12 mein shaded triangle ka area hai (base $x_{m}$ , height $k x_{m}$ ), negative kyunki yeh axis ke neeche hai: block bahar jaata hai jabki spring wapas kheenchti hai. Stretch karne wala haath har point par $+ k x$ se kheenchta hai, isliye woh ulta amount ka work karta hai:

W = + \frac{1}{2} k x_{m}^{2}

(5.16)

$x_{c} < 0$ tak compression same results deta hai, kyunki sirf $x_{c}^{2}$ aata hai. Aur ek round trip, $x_{i}$ se bahar aur wapas $x_{i}$ tak, deta hai

\begin{aligned} W_{s} & = \frac{1}{2} k x_{i}^{2} - \frac{1}{2} k x_{i}^{2} \\ = 0 \end{aligned}

(5.18)

To spring force sirf position par depend karta hai, uska work sirf end points par depend karta hai, aur closed cycle mein woh koi net work nahi karta: yeh conservative hai. Relaxed spring ke liye $V = 0$ chunne par, Eq. (5.17) kehta hai ki extension ya compression $x$ par uski potential energy hai

Spring ki potential energy

V (x) = \frac{1}{2} k x^{2}

(5.19)

Differentiate karne se confirm hota hai: $- d V / d x = - k x$ . Ab block ko $x = x_{m}$ par rest se chhodo. Friction na hone par uski kinetic plus potential energy wahi value $\frac{1}{2} k x_{m}^{2}$ rakhti hai jisse woh shuru hui thi:

\begin{aligned} \frac{1}{2} m v^{2} + \frac{1}{2} k x^{2} \\ = \frac{1}{2} k x_{m}^{2} \end{aligned}

To block relaxed position se guzarte waqt sabse tez chalta hai, jahan spring kuch jama nahi karti aur $\frac{1}{2} m v_{m}^{2} = \frac{1}{2} k x_{m}^{2}$ :

Sabse zyada speed

v_{m} = x_{m} \sqrt{\frac{k}{m}}

$200 N / m$ ki spring par $0.50 k g$ ka block, $10 c m$ bahar kheencha gaya, $v_{m} = 0.10 \times \sqrt{400} = 2.0 m / s$ tak pahunchta hai. Fig. 5.13 dikhata hai ki energy spring aur block ke beech aati-jaati hai jabki total wahi rehta hai, aur Fig. 5.14 wali lab mein tum ise chala sakte ho. Yeh motion tumhe Chapter 13 mein simple harmonic motion ke roop mein phir milegi.

3/3Har position par dono parts milkar same height $E$ tak pahunchte hain: energy ka form badalta hai, total nahi.

Fig. 5.13NCERT Fig. 5.8 · p. 81

Minus x m se plus x m tak position ke saath energy, total E ke liye ek dashed line. Step 1: ek purple upward parabola V dikhata hai, centre par zero aur ends par E. Step 2: ek dashed blue downward parabola K dikhata hai, centre par sabse bada. Step 3: ek position par stacked bar dikhata hai ki V aur K milkar theek E banate hain.

Fig. 5.14

k

m

aur stretch set karo, Play dabao aur

K

aur

V

ko jagah badalte dekho. Phir floor ko khurdura banao aur dekho energy kahan jaati hai.

Try karke dekho

Ghar par spring constant naapo

Tumhe chahiye

ek steel spring ya ek moti rubber band
handles wala kapde ya polythene ka bag
paani se bhari chaar 500 mL bottles
ek ruler aur ek hook ya keel

Pehle guess karo

Ek bhari bottle spring ko

4.0 c m

stretch karti hai. Agar spring Hooke ka law maanti hai, to teen bottles use kitna stretch karengi?

4.0 c m

8.0 c m

12 c m

36 c m

Spring ko hook se latkao aur uske neeche wale end se bag baandho. Ruler par note karo ki bag ka bottom kahan hai.
Ek bhari $500 m L$ bottle daalo (weight lagbhag $4.9 N$ ) aur extension $x$ naapo.
Ek-ek karke bottles daalo, har baar $x$ naapo.
Har row ke liye weight ÷ $x$ nikalo. Phir saari bottles ke saath jama energy, $\frac{1}{2} k x^{2}$ , nikalo.

Table 5.2

Bottles	Weight (N)	Extension x (m)	Weight ÷ x (N/m)
1	4.9
2	9.8
3	14.7
4	19.6

Tumhari entries apne aap save hoti hain.

Example 5.8NCERT Example 5.8 · p. 81

Spring par giraya gaya crate

$2.0 k g$ ka ek crate floor par khadi ek vertical spring ( $k = 2000 N / m$ ) ke top se $0.30 m$ upar se rest se giraya jaata hai. Air resistance aur spring ka mass ignore karke, spring ka sabse zyada compression aur uska sabse bada force nikalo. $g = 9.8 m / s^{2}$ lo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Teen baatein conservative forces ki picture poori karti hain.

Koi ghadi nahi. Energy conservation ne crate ka sabse zyada compression diya lekin yeh nahi ki dabne mein kitna time laga. Timing ke liye Newton ka second law chahiye, time ke function ke roop mein solve kiya hua.
Har force conservative nahi hota. Friction nahi hai, aur tab energy principle ko modify karna padta hai, jaise Example 5.9 dikhata hai.
Potential energy ka zero tum chun sakte ho. Humne relaxed spring ke liye aur zameen par $V = 0$ liya. Chapter 7 mein gravity ka zero infinitely door rakha jaayega. Koi bhi choice chalegi, lekin ek baar chun liya to poore problem mein wahi rakhna hai: beech mein zero badalna aisa hai jaise scorer innings ke beech mein scoreboard badal de.

Example 5.9NCERT Example 5.9 · p. 82

Khurdure floor par spring launcher

Ek $1.0 k g$ ka block khurdure floor ( $μ_{k} = 0.30$ ) par ek spring ( $k = 400 N / m$ ) se sata hai jo $0.15 m$ compressed hai. Block chhoda jaata hai aur jab spring apni natural length par aati hai tab woh spring chhod deta hai. Nikalo (a) spring chhodte waqt block ki speed, aur (b) release point se woh kul kitni distance fisalta hai. $g = 9.8 m / s^{2}$ lo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 5.9 ne general rule use kiya. Body par lage forces ko conservative, $F_{c}$ , aur non-conservative, $F_{n c}$ , mein baanto. Chhote displacement ke liye work-energy theorem deta hai $(F_{c} + F_{n c}) Δ x = Δ K$ , jabki $F_{c} Δ x = - Δ V$ . Isliye $Δ (K + V) = F_{n c} Δ x$ , aur poore path par

Non-conservative forces ke saath

E_{f} - E_{i} = W_{n c}

Mechanical energy theek utni badalti hai jitna non-conservative forces ne work kiya. Conservative force ke work ke ulat, $W_{n c}$ , $i$ se $f$ tak ke path par depend karta hai: block ko lambe raaste se fisalao to friction zyada le leta hai.

Quick check

2 mein se 1

Same force do springs ko stretch karta hai: P jiska

k = 200 N / m

aur Q jiska

k = 400 N / m

. Kaun zyada energy jama karti hai?

Dono same energy jama karti hainQ, kyunki woh zyada stiff haiKoi nahi, kyunki forces barabar hainP, kyunki woh zyada stretch hoti hai

Answer dekho

Answer (D) P, kyunki woh zyada stretch hoti hai

Force

F

ke liye

x = F / k

aur

V = \frac{1}{2} k x^{2} = F^{2} / 2 k

. Naram spring P double stretch hoti hai aur double energy jama karti hai. 'Stiff spring zyada jama karti hai' sirf barabar stretch ke liye sahi hai.

5.10 Power

Do students ek jaise bags same seedhiyon se upar le jaate hain. Ek ko 10 seconds lagte hain, doosre ko 20. Dono gravity ke against same work karte hain, phir bhi hum pehli ko zyada fit kahenge, kyunki usne yeh jaldi kiya (Fig. 5.15). Work kitni jaldi hota hai, ya energy kitni jaldi aage jaati hai, ise power naapti hai.

Agar koi force time $t$ mein work $W$ karta hai, to uski average power hai

Average power

P_{a v} = \frac{W}{t}

Time interval ko zero tak chhota karne par instantaneous power milti hai:

P = \frac{d W}{d t}

(5.20)

Chhote displacement $d 𝐫$ mein force $d W = 𝐅 \cdot d 𝐫$ work karta hai. $d t$ se divide karne par,

P = 𝐅 \cdot \frac{d 𝐫}{d t} = 𝐅 \cdot 𝐯

(5.21)

jahan $𝐯$ us pal ki velocity hai. Power scalar hai, dimensions $[M L^{2} T^{- 3}]$ ke saath. Iski SI unit watt (W) hai, $1 W = 1 J / s$ , jo James Watt (1736–1819) ke naam par hai, jinke steam engine ke sudhaaron ne Industrial Revolution ko aage badhaya. Vehicle brochures aaj bhi aksar engine output horsepower (hp) mein batate hain, jahan $1 h p = 746 W$ .

2/3A $10 s$ mein top par pahunchti hai jabki B aadhe raaste par hai: A ka work same, lekin B se double rate par.

Fig. 5.15

Ghadi ke saath ek seedhi. Climber A ek circle hai, climber B ek square. Step 1: dono 0 seconds par neeche. Step 2: 10 seconds par A top par aur B aadhe raaste par, aur A ki power W over 10 seconds hai. Step 3: 20 seconds par B bhi pahunchta hai, aur B ki power W over 20 seconds hai.

Bijli ke bill energy ke paise lete hain, power ke nahi. Bill par 'unit' kilowatt hour (kWh) hai, yaani woh energy jo $1 k W$ ka appliance ek ghante mein use karta hai:

\begin{aligned} 1 k W h & = (1000 J / s) \\ \times (3600 s) \\ = 3.6 \times 10^{6} J \end{aligned}

$1.5 k W$ ka geyser roz 20 minute, 30 din tak chalane par $1.5 \times \frac{1}{3} \times 30 = 15 k W h$ khata hai, jo ₹7 per unit ke hisaab se ₹105 ka padta hai. Kilowatt hour energy naapta hai; kilowatt power naapta hai.

Definition

Power

Per unit time kiya gaya work (ya aage di gayi energy), $P = d W / d t = 𝐅 \cdot 𝐯$ ; iski SI unit watt hai, $1 W = 1 J / s$ .

Example $50 k g$ ki bori ko $4.0 s$ mein $2.0 m$ uthane ke liye average $245 W$ chahiye.

Try karke dekho

Apni khud ki power naapo

Tumhe chahiye

kam se kam do floors ki seedhiyan
ek stopwatch ya phone timer
ek measuring tape
ek dost

Pehle guess karo

50 k g

ki ek student

10 s

mein kul

6.0 m

ki height chadhti hai. Gravity ke against uska power output lagbhag kitna hai?

30 W

300 W

3000 W

30 000 W

Ek step ki height naapo aur steps gino, taaki metres mein kul chadhai $h$ mil jaaye.
Jab tum tezi se chalkar upar jao (daudna nahi) tab dost se time lene ko kaho. Phir aaram ki speed se dohrao.
Apne mass se $m g h$ nikalo, phir watts mein power ke liye har time se divide karo, aur horsepower ke liye 746 se.

Table 5.3

Chadhai	Time (s)	Work mgh (J)	Power (W)
Tez
Aaram se

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Example 5.10NCERT Example 5.10 · p. 83

Pahadi road par car

$1200 k g$ ki ek car ek seedhi pahadi road par steady $54 k m / h$ se chadhti hai, jo uske along har $20 m$ chalne par $1 m$ upar uthti hai ( $\sin θ = 0.050$ ). Air drag aur rolling friction milkar $600 N$ se motion ko rokte hain. Nikalo (a) engine ko wheels tak kitni power deni padti hai, kW aur hp mein, aur (b) level road par same speed ke liye kitni power chahiye. $g = 9.8 m / s^{2}$ lo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

2 mein se 1

Inmein se kaun energy ki unit hai, power ki nahi?

kilowatthorsepowerkilowatt hourjoule per second

Answer dekho

Answer (C) kilowatt hour

Kilowatt hour power guna time hai, jo energy ki ek matra hai. Baaki teen ek rate naapte hain.

5.11 Collisions

Jab do bodies takraati hain, to unke beech forces bahut bade, chhote samay ke aur complicated hote hain: lagbhag ek millisecond ke contact mein cricket ball bat se saaf dikhne layak chapti ho jaati hai. Hum in forces ko kabhi detail mein track nahi kar sakte. Achhi baat yeh hai ki zaroorat bhi nahi, kyunki do conservation laws, momentum aur energy ke, inke bina hi result ka bahut kuch bata dete hain. Carrom, kanche aur billiards sab collisions ke khel hain.

Masses $m_{1}$ aur $m_{2}$ ki do bodies ke beech ek idealised collision lo. Body 1 velocity $𝐯_{1 i}$ (i yaani initial) se chalti hai aur body 2 rest par hai. Isse generality kam nahi hoti, kyunki hum hamesha body 2 ke saath chalte frame se collision dekh sakte hain. Baad mein bodies original line se angles $θ_{1}$ aur $θ_{2}$ par velocities $𝐯_{1 f}$ aur $𝐯_{2 f}$ se nikal jaati hain.

5.11.1 Elastic aur inelastic collisions

DerivationHar collision mein momentum conserve hota hai

Contact ke dauraan body 2, body 1 ko force $𝐅_{12} (t)$ se dhakelti hai aur body 1, body 2 ko $𝐅_{21} (t)$ se, dono tezi se badalte hue. Har momentum ka change contact time mein us par lage force ka impulse hai:

\begin{aligned} Δ 𝐩_{1} & = \int 𝐅_{12} d t \\ Δ 𝐩_{2} & = \int 𝐅_{21} d t \end{aligned}

Newton ka third law har pal $𝐅_{12} (t) = - 𝐅_{21} (t)$ banata hai, isliye dono integrals cancel ho jaate hain aur $Δ 𝐩_{1} + Δ 𝐩_{2} = 0$ , forces chahe kitne bhi complicated hon.

Kinetic energy ki kahani alag hai: takraati bodies dab jaati hain, aur daba hua material garam ho sakta hai, goonj sakta hai, toot sakta hai ya pichka reh sakta hai. Spring plunger wali trolleys se yeh samajhna aasaan hai (Fig. 5.16). Agar dabi spring poori wapas khul jaaye, to baad ki total kinetic energy pehle ke barabar hoti hai: collision elastic hai, bhale hi contact ke dauraan energy ka kuch hissa spring mein rehta hai. Agar trolleys aapas mein jud jaayein, to dabaav kabhi wapas nahi khulta aur woh ek saath aage badhti hain: collision completely inelastic hai. Zyadatar real collisions beech mein hote hain, jinmein kuch kinetic energy kho jaati hai; yeh inelastic collisions hain.

2/4Maximum compression par dono carts $u / 2$ se chalti hain, aur aadhi kinetic energy spring mein jama hai.

Fig. 5.16

Track par do barabar carts, cart 2 par spring bumper, aur neeche total kinetic energy ka bar. Step 1: cart 1 u se rest par khadi cart 2 ki taraf chalti hai; bar bhara hai. Step 2: spring dabi hai, dono u over 2 se chalti hain, aur bar ka aadha spring mein jama hai. Step 3: carts alag ho gayi hain, cart 1 rest par aur cart 2 u par; bar phir bhara hai. Step 4: carts judkar u over 2 par hain aur aadha bar kho gaya.

5.11.2 One dimension mein collisions

Ek line ke along completely inelastic collision se shuru karo ( $θ_{1} = θ_{2} = 0$ ). Bodies chipak jaati hain aur common velocity $v_{f}$ se aage badhti hain, isliye sirf momentum conservation se result tay ho jaata hai.

DerivationCompletely inelastic collision

Momentum conservation, $m_{1} v_{1 i} = (m_{1} + m_{2}) v_{f}$ , se milta hai

v_{f} = \frac{m_{1}}{m_{1} + m_{2}} v_{1 i}

(5.22)

Khoyi kinetic energy $Δ K = \frac{1}{2} m_{1} v_{1 i}^{2} - \frac{1}{2} (m_{1} + m_{2}) v_{f}^{2}$ hai. Eq. (5.22) se doosra term $\frac{1}{2} m_{1} v_{1 i}^{2} \cdot m_{1} / (m_{1} + m_{2})$ hai, aur $1 - m_{1} / (m_{1} + m_{2}) = m_{2} / (m_{1} + m_{2})$ , isliye

\begin{aligned} Δ K \\ = \frac{m_{1} m_{2} v_{1 i}^{2}}{2 (m_{1} + m_{2})} \end{aligned}

Yeh hamesha positive hai: chipakne wali bodies hamesha kuch kinetic energy barbaad karti hain.

Ab one dimension mein elastic collision. Ab momentum aur kinetic energy dono conserve hote hain, jisse do anjaani final velocities ke liye do equations milti hain.

DerivationOne-dimensional elastic collision ke baad velocities

\begin{aligned} m_{1} v_{1 i} & = m_{1} v_{1 f} \\ + m_{2} v_{2 f} \end{aligned}

(5.23)

\begin{aligned} m_{1} v_{1 i}^{2} & = m_{1} v_{1 f}^{2} \\ + m_{2} v_{2 f}^{2} \end{aligned}

(5.24)

(doosri mein $\frac{1}{2}$ ke factors cancel ho jaate hain). Body 1 ke terms left mein ikattha karo: $m_{1} (v_{1 i} - v_{1 f}) = m_{2} v_{2 f}$ aur $m_{1} (v_{1 i}^{2} - v_{1 f}^{2}) = m_{2} v_{2 f}^{2}$ .

$v_{1 i}^{2} - v_{1 f}^{2} = (v_{1 i} - v_{1 f}) (v_{1 i} + v_{1 f})$ factorise karo aur doosri line ko pehli se divide karo, jo allowed hai kyunki collision $v_{1}$ ko badalta hai:

v_{2 f} = v_{1 i} + v_{1 f}

(5.25)

Shabdon mein, bodies utni hi tezi se alag hoti hain jitni tezi se paas aayi thi. Wapas Eq. (5.23) mein daalne par milta hai

v_{1 f} = \frac{m_{1} - m_{2}}{m_{1} + m_{2}} v_{1 i}

(5.26)

v_{2 f} = \frac{2 m_{1}}{m_{1} + m_{2}} v_{1 i}

(5.27)

Ab final velocities poori tarah $m_{1}$ , $m_{2}$ aur $v_{1 i}$ se fix hain. Teen special cases yaad rakhne layak hain (Fig. 5.17):

Barabar masses, $m_{1} = m_{2}$ : $v_{1 f} = 0$ aur $v_{2 f} = v_{1 i}$ . Chalti body ekdum ruk jaati hai aur target uski velocity le leta hai, jaise jab ek carrom coin ek jaise coin se seedha takraata hai.
Bahut bhaari target, $m_{2} ≫ m_{1}$ : $v_{1 f} \approx - v_{1 i}$ aur $v_{2 f} \approx 0$ . Halki body lagbhag poori speed se seedhi wapas uchhalti hai, jaise deewar par phenki ball.
Bahut halka target, $m_{1} ≫ m_{2}$ : $v_{1 f} \approx v_{1 i}$ aur $v_{2 f} \approx 2 v_{1 i}$ . Bhaari body ko mushkil se pata chalta hai, aur halki body lagbhag double speed se nikal jaati hai.

1/3Barabar masses: ball 1 ruk jaati hai aur ball 2 velocity $u$ se nikalti hai.

Fig. 5.17

Har step mein ek pehle wali row aur ek baad wali row hai. Step 1, barabar balls: u par ball 1 rest par ball 2 se takraati hai, phir ball 1 ruk jaati hai aur ball 2 u se chalti hai. Step 2, chhoti ball badi ball se takraati hai: woh minus u se wapas uchhalti hai jabki badi ball mushkil se hilti hai. Step 3, badi ball chhoti se takraati hai: badi ball lagbhag u rakhti hai aur chhoti lagbhag 2u se nikalti hai.

Fig. 5.18Masses aur velocities set karo, collision ka type chuno aur Play dabao. Table pehle aur baad ke momentum aur kinetic energy ko compare karti hai.

Example 5.11NCERT Example 5.11 · p. 85

Kaun sa target sabse zyada energy leta hai?

$20 m / s$ se chalti $0.16 k g$ ki cricket ball rest par khadi ball se head-on elastic collision karti hai. Nikalo ki woh apni kinetic energy ka kitna fraction de deti hai agar target (a) $0.057 k g$ ki tennis ball, (b) ek aur $0.16 k g$ ki cricket ball, (c) $5.0 k g$ ka shot-put ho. Pehle dikhao ki rest par mass $m_{2}$ ke target ke liye fraction $4 m_{1} m_{2} / (m_{1} + m_{2})^{2}$ hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Ab tak sab kuch yeh maan kar tha ki motion ek hi line par rehti hai: head-on ya one-dimensional collision. Do smooth balls ke saath iske liye perfect nishaana chahiye, chalti ball doosri ke centre ki taraf seedhi jaaye. Nishaana thoda chooke to balls angles par alag ud jaati hain, two-dimensional collision mein.

Quick check

2 mein se 1

6.0 m / s

se chalti

2.0 k g

ki ball rest par

4.0 k g

ki ball se head-on elastic collision karti hai. Baad mein

2.0 k g

wali ball ki velocity kya hai?

- 2.0 m / s

+ 2.0 m / s

0

+ 4.0 m / s

Answer dekho

Answer (A) $- 2.0 m / s$

v_{1 f} = \frac{2.0 - 4.0}{6.0} (6.0) = - 2.0 m / s

: yeh wapas uchhalti hai.

4.0 k g

wali ball

+ 4.0 m / s

se nikalti hai; check karo ki momentum (

12 k g m / s

) aur kinetic energy (

36 J

) nahi badle.

5.11.3 Two dimensions mein collisions

Ab body 1 ko body 2 se off-centre takraane do, taaki dono original line ke dono taraf angles $θ_{1}$ aur $θ_{2}$ par alag ud jaayein (Fig. 5.19). Momentum vector hai, isliye uska conservation har direction ke liye ek equation deta hai. $x$ -axis ko $𝐯_{1 i}$ ke along lo aur $x$ – $y$ plane aisa lo ki usmein dono final velocities hon; tab $z$ -equation confirm karti hai ki kuch bhi is plane se bahar nahi jaata, aur baaki do aisi hain

\begin{aligned} m_{1} v_{1 i} \\ = m_{1} v_{1 f} \cos θ_{1} \\ + m_{2} v_{2 f} \cos θ_{2} \end{aligned}

(5.28)

\begin{aligned} 0 & = m_{1} v_{1 f} \sin θ_{1} \\ - m_{2} v_{2 f} \sin θ_{2} \end{aligned}

(5.29)

( $θ_{1} = θ_{2} = 0$ ke saath Eq. (5.28) bas Eq. (5.23) hai.) Agar collision elastic hai, to kinetic energy ek teesri equation jodti hai:

\begin{aligned} \frac{1}{2} m_{1} v_{1 i}^{2} & = \frac{1}{2} m_{1} v_{1 f}^{2} \\ + \frac{1}{2} m_{2} v_{2 f}^{2} \end{aligned}

(5.30)

Unknowns gino: $m_{1}$ , $m_{2}$ aur $v_{1 i}$ diye hone par bhi humein $v_{1 f}$ , $v_{2 f}$ , $θ_{1}$ aur $θ_{2}$ chahiye, zyada se zyada teen equations se chaar numbers. Sirf conservation laws result tay nahi kar sakte; yeh is par bhi depend karta hai ki bodies theek kahan chhooti hain, jaise carrom player coin ka nishaana lagata hai. Particle experiments mein ek angle (maan lo $θ_{1}$ ) target ke chaaron taraf chune hue angles par rakhe detector se naapa jaata hai, aur phir Eqs. (5.28) se (5.30) baaki de dete hain.

3/3Final momenta milkar initial momentum banate hain; elastic collision mein barabar masses ke liye triangle mein ek right angle hota hai.

Fig. 5.19NCERT Fig. 5.10 · p. 83

Step 1: coin 1 ek line ke along right ki taraf chalta hai jo rest par khade coin 2 ke centre se theek upar se guzarti hai. Step 2: collision ke baad coin 1 original line ke upar theta 1 par aur coin 2 uske neeche theta 2 par nikalta hai. Step 3: ek vector triangle initial momentum ko dono final momenta ke sum ke roop mein dikhata hai, jo right angle par milte hain.

Example 5.12NCERT Example 5.12 · p. 85

Proton se tirchha takraata proton

Ek particle detector mein $3.0 \times 10^{6} m / s$ se chalta proton rest par khade proton se elastically takraata hai. Detector dikhata hai ki aane wala proton apni original direction se $25^{\circ}$ mud gaya. Nikalo (a) takraaye gaye proton ki direction, aur (b) collision ke baad dono protons ki speeds.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Bodies bina chhuye bhi 'takraa' sakti hain. Voyager 2 1979 mein Jupiter ke paas se ghoom kar nayi speed aur direction ke saath nikla, planet ko kabhi chhuye bina. Doori se lagne wale force ke through aisi mulaqat ko scattering kehte hain. Kisi bhi collision ki tarah momentum conserve hota hai, aur result initial velocities aur force ke nature par depend karta hai. Class 12 mein tumhe woh experiment milega jismein alpha particles ki scattering ne atomic nucleus ko dhoondh nikala.

Quick check

2 mein se 1

Ek billiard ball rest par khadi ek jaisi ball se off-centre, elastic collision mein takraati hai. Baad mein pehli ball apni original line se

35^{\circ}

par chalti hai. Doosri ball us line se kis angle par chalti hai?

90^{\circ}

, doosri taraf

35^{\circ}

, usi taraf

145^{\circ}

, doosri taraf

55^{\circ}

, doosri taraf

Answer dekho

Answer (D) $55^{\circ}$ , doosri taraf

Barabar masses, ek rest par, elastic aur head-on nahi: final velocities perpendicular hain, isliye dono angles milkar

90^{\circ}

hote hain, line ke opposite sides par.

Recap

Summary aur key points

Scalar product $𝐀 \cdot 𝐁 = A B \cos θ = A_{x} B_{x} + A_{y} B_{y} + A_{z} B_{z}$ ek scalar hai; yeh commutative aur distributive hai, aur perpendicular vectors ke liye zero hai (Section 5.1.1).
Work-energy theorem: body par saare forces dwara kiya gaya net work uski kinetic energy ke increase ke barabar hai, $K_{f} - K_{i} = W$ (Section 5.2, Section 5.6).
Constant force $W = 𝐅 \cdot 𝐝 = F d \cos θ$ work karta hai, jo positive, negative ya zero ho sakta hai; iski SI unit joule hai (Section 5.3).
Kinetic energy $K = \frac{1}{2} m v^{2}$ ek scalar hai jo kabhi negative nahi hoti (Section 5.4).
Variable force dwara kiya gaya work uske force–displacement graph ke neeche ka area hai, $W = \int F d x$ (Section 5.5).
Force conservative hai agar uska work sirf end points par depend karta hai (yaani har closed path par zero hai); tab $F (x) = - d V / d x$ . Earth ke paas $V = m g h$ (Section 5.7).
Agar sirf conservative forces work karte hain, to mechanical energy $K + V$ conserve hoti hai; non-conservative forces ise $W_{n c}$ se badalte hain (Section 5.8).
Ideal spring $F_{s} = - k x$ maanti hai aur potential energy $V = \frac{1}{2} k x^{2}$ jama karti hai (Section 5.9).
Power work karne ki rate hai, $P = d W / d t = 𝐅 \cdot 𝐯$ , watts mein naapi jaati hai; kilowatt hour energy ki unit hai (Section 5.10).
Har collision mein momentum conserve hota hai; kinetic energy sirf elastic collisions mein conserve hoti hai, aur bodies ke chipakne par kuch hamesha kho jaati hai (Section 5.11.1).
Body 2 rest par ho to 1D elastic collision mein $v_{1 f} = \frac{m_{1} - m_{2}}{m_{1} + m_{2}} v_{1 i}$ aur $v_{2 f} = \frac{2 m_{1}}{m_{1} + m_{2}} v_{1 i}$ (Section 5.11.2).
Two dimensions mein conservation laws ek quantity ko tay nahi karte; elastic tirchhe collision mein barabar masses right angles par nikalte hain (Section 5.11.3).

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercises 5.1	Practice Q1
NCERT Exercises 5.2	Practice Q2
NCERT Exercises 5.3	Practice Q3
NCERT Exercises 5.4	Practice Q4
NCERT Exercises 5.5	Practice Q5
NCERT Exercises 5.6	Practice Q6
NCERT Exercises 5.7	Practice Q7
NCERT Exercises 5.8	Practice Q8
NCERT Exercises 5.9	Practice Q9
NCERT Exercises 5.10	Practice Q10
NCERT Exercises 5.11	Practice Q11
NCERT Exercises 5.12	Practice Q12
NCERT Exercises 5.13	Practice Q13
NCERT Exercises 5.14	Practice Q14
NCERT Exercises 5.15	Practice Q15
NCERT Exercises 5.16	Practice Q16
NCERT Exercises 5.17	Practice Q17
NCERT Exercises 5.18	Practice Q18
NCERT Exercises 5.19	Practice Q19
NCERT Exercises 5.20	Practice Q20
NCERT Exercises 5.21	Practice Q21
NCERT Exercises 5.22	Practice Q22
NCERT Exercises 5.23	Practice Q23

Quick check: MCQs

5.1.1 Scalar product2 questions
5.2 Work aur kinetic energy ke notions: work-energy theorem2 questions
5.3 Work3 questions
5.4 Kinetic energy2 questions
5.5 Variable force dwara kiya gaya work2 questions
5.6 Variable force ke liye work-energy theorem2 questions
5.7 Potential energy ka concept2 questions
5.8 Mechanical energy ka conservation3 questions
5.9 Spring ki potential energy2 questions
5.10 Power2 questions
5.11.2 One dimension mein collisions2 questions
5.11.3 Two dimensions mein collisions2 questions

Glossary · 13

Key terms (English + Hindi)

conservative forceसंरक्षी बलsanrakshi bal: Aisa force jiska work sirf start aur end points par depend karta hai, aur kisi bhi closed path par zero hota hai.Chapter mein pehli baar yahan aaya →

elastic collisionप्रत्यास्थ संघट्टpratyasth sanghatt: Aisa collision jiske baad total kinetic energy pehle jitni hi hoti hai; momentum bhi conserve hota hai.Chapter mein pehli baar yahan aaya →

inelastic collisionअप्रत्यास्थ संघट्टapratyasth sanghatt: Aisa collision jismein kuch kinetic energy kho jaati hai; agar bodies chipak jaayein to yeh completely inelastic hai.Chapter mein pehli baar yahan aaya →

jouleजूलjool: Work aur energy ki SI unit: jab 1 N ka force apne point of application ko force ke along 1 m move karta hai, tab kiya gaya work.Chapter mein pehli baar yahan aaya →

kinetic energyगतिज ऊर्जाgatij oorja: Motion ki energy, $K = \frac{1}{2} m v^{2}$ .Chapter mein pehli baar yahan aaya →

mechanical energyयांत्रिक ऊर्जाyaantrik oorja: Body ki kinetic aur potential energies ka sum $K + V$ .Chapter mein pehli baar yahan aaya →

potential energyस्थितिज ऊर्जाsthitij oorja: Conservative force ke under position ya configuration se jama energy; $Δ V = - F Δ x$ .Chapter mein pehli baar yahan aaya →
powerशक्तिshakti: Work karne ki rate, $P = d W / d t = 𝐅 \cdot 𝐯$ , watts mein naapi jaati hai.Chapter mein pehli baar yahan aaya →

scalar productअदिश गुणनफलadish gunanphal: Do vectors ka product $𝐀 \cdot 𝐁 = A B \cos θ$ ; result ek scalar hota hai.Chapter mein pehli baar yahan aaya →
spring constantस्प्रिंग नियतांकspring niyatank: Hooke ke law $F_{s} = - k x$ mein constant $k$ ; yeh stiffness naapta hai, N/m mein.Chapter mein pehli baar yahan aaya →

wattवाटvaat: Power ki SI unit: ek joule per second.Chapter mein pehli baar yahan aaya →
workकार्यkaarya: Displacement guna uske along force ka component, $W = 𝐅 \cdot 𝐝$ ; joules mein naapa jaane wala scalar.Chapter mein pehli baar yahan aaya →
work-energy theoremकार्य-ऊर्जा प्रमेयkaarya-oorja pramey: Body par saare forces dwara kiya gaya net work uski kinetic energy ke increase ke barabar hota hai.Chapter mein pehli baar yahan aaya →

Work, Energy and Power: Class 11 Physics Chapter 5 Hinglish mein

Is chapter mein aap kya seekhenge

5.1 Introduction

5.1.1 Scalar product

5.2 Work aur kinetic energy ke notions: work-energy theorem

5.3 Work

5.4 Kinetic energy

5.5 Variable force dwara kiya gaya work

5.6 Variable force ke liye work-energy theorem

5.7 Potential energy ka concept

5.8 Mechanical energy ka conservation

5.9 Spring ki potential energy

5.10 Power

5.11 Collisions

5.11.1 Elastic aur inelastic collisions

5.11.2 One dimension mein collisions

5.11.3 Two dimensions mein collisions

Summary aur key points

Practice questions answers ke saath (NCERT “Exercises”)

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye