Class 9 Physics Chapter 5 Complete Guide Book — Work, Energy and Power | PunjabTest.com

Class 9 Physics Chapter 5 — Work, Energy and Power
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📑 اس Chapter میں یہ Topics ہیں / اس باب میں یہ ٹاپکس ہیں

  1. 1. 🔨 Work — کام
  2. 2. 🏃 Kinetic Energy — حرکی توانائی
  3. 3. ⛰️ Potential Energy and Conservation of Energy — مخفی توانائی اور توانائی کی بقا
  4. 4. ⚙️ Power and Efficiency — طاقت اور کارکردگی
PROVENANCE / ماخذ کی وضاحت

This revised page was built from the existing Class 9 Physics library chapter (chapter-5-complete-guide-book.html) in the same series. No official Class 9 Physics book copy was on disk for a fresh page-by-page check, so no book page numbers are claimed and the exercise is carried over from the master chapter as given, not newly verified against the book. Answers follow the master chapter.

یہ نظرثانی شدہ صفحہ اسی سلسلے کے موجودہ جماعت نہم فزکس کے باب سے بنایا گیا ہے۔ کتاب کی کوئی مصدقہ نقل ڈسک پر موجود نہیں تھی، اس لیے کتاب کے صفحہ نمبرز کا دعویٰ نہیں کیا گیا، اور ایکسرسائز ماسٹر باب سے جوں کی توں لی گئی ہے۔

1. 🔨 Workکام
🏷 Tags: work
ENGLISH

Work is done when a force moves a body through a distance in the direction of the force: W = F × S. If the force acts at an angle θ to the displacement, W = FS cos θ. Unit: joule (J) = 1 N m. A joule can also be written as kg m2 s−2.

Work is zero when there is no displacement (a man pushing a fixed wall does no work, however tired he gets) or when force is perpendicular to motion (θ = 90°, cos 90° = 0 — e.g. a satellite's gravity does no work on it in a circular orbit). Work is maximum when θ = 0° (force along displacement). Work is a scalar.

اردو

کام تب ہوتا ہے جب فورس جسم کو اپنی سمت میں فاصلہ طے کرائے: W = F × S۔ اگر فورس زاویے θ پر لگے تو W = FS cos θ۔ اکائی جول (J) = 1 N m، یعنی kg m²/s²۔

کام صفر ہوتا ہے جب ڈسپلیسمنٹ نہ ہو (دیوار کو دھکیلنے پر کوئی کام نہیں، چاہے تھکاوٹ ہو) یا فورس حرکت کے عموداً ہو (θ = 90°)۔ کام زیادہ سے زیادہ تب ہوتا ہے جب θ = 0° ہو۔ کام اسکیلر ہے۔

BoxF →W = F × S (θ = 0°: maximum)W = FS cos θ • 1 J = 1 N m
Figure samjho: Work = force × distance in the direction of the force.
کام = فورس × فورس کی سمت میں طے شدہ فاصلہ۔
2. 🏃 Kinetic Energyحرکی توانائی
🏷 Tags: kinetic, energy
ENGLISH

Kinetic energy is the energy a body has because of its motion: Ek = ½mv2 (unit: joule). It is found from the work done to accelerate the body: W = F × S = ma × S, and with v2 = 2aS this gives W = ½mv2. Kinetic energy can never be negative (m > 0 and v2 ≥ 0).

Because of the v2, doubling speed makes kinetic energy four times. A slow car can have more kinetic energy than a fast motorcycle because its much larger mass outweighs the speed difference. If momentum doubles, kinetic energy (p2/2m) becomes four times. An object at velocity v has more kinetic energy than a twice-heavy object at v/2 (½mv2 vs ¼mv2). A car moving at constant speed around a curve keeps the same kinetic energy — speed, not direction, matters.

اردو

حرکی توانائی حرکت کی وجہ سے ہوتی ہے: Ek = ½mv² (اکائی جول)۔ یہ اس کام سے معلوم ہوتی ہے جو جسم کو تیز کرنے میں ہوا۔ حرکی توانائی کبھی منفی نہیں ہو سکتی کیونکہ v² ہمیشہ مثبت ہوتا ہے۔

رفتار دوگنی ہو تو حرکی توانائی چار گنا ہو جاتی ہے۔ آہستہ کار کی حرکی توانائی تیز موٹر سائیکل سے زیادہ ہو سکتی ہے کیونکہ اس کی کمیت بہت زیادہ ہے۔

Ek = ½mv²Speed × 2→ Energy × 4Momentum × 2→ Energy × 4 (p²/2m)
Figure samjho: Kinetic energy depends on speed squared and on momentum squared.
حرکی توانائی رفتار کے مربع پر منحصر ہے۔
3. ⛰️ Potential Energy and Conservation of Energyمخفی توانائی اور توانائی کی بقا
🏷 Tags: potential, energy, conservation
ENGLISH

Potential energy is energy due to position: gravitational potential energy Ep = mgh for a body of mass m raised through height h. 'An object has one joule of potential energy' means it can do 1 J of work by virtue of its position (relative to a chosen reference level — potential energy is always relative).

Conservation of energy: energy cannot be created or destroyed, only transformed; total energy stays constant. A falling body converts Ep into Ek: at height h, Ep = mgh and Ek = 0; just before hitting the ground, Ep = 0 and Ek = mgh. A cyclist coasting down a hill gains speed (Ep → Ek) and can climb the next hill, but never quite as high, because some energy is lost to friction. That is also why a perpetual machine is impossible.

اردو

مخفی توانائی مقام کی وجہ سے ہوتی ہے: کشش ثقل کی مخفی توانائی Ep = mgh۔ 'کسی جسم کی مخفی توانائی ایک جول ہے' کا مطلب ہے کہ وہ اپنی پوزیشن کی وجہ سے 1 J کام کر سکتا ہے (یہ ہمیشہ کسی حوالہ سطح کے لحاظ سے ہوتی ہے)۔

توانائی کی بقا: توانائی پیدا یا ختم نہیں ہوتی، صرف شکل بدلتی ہے۔ گرتا جسم Ep کو Ek میں بدلتا ہے۔ اسی لیے پرپیچوئل مشین ناممکن ہے — رگڑ میں کچھ توانائی ہمیشہ ضائع ہوتی ہے۔

Top: Ep max, Ek = 0Bottom: Ek max, Ep = 0Ep → Ek as the body falls
Figure samjho: Energy transformation in a falling body and a coasting cyclist.
گرتے جسم میں مخفی توانائی حرکی توانائی میں بدلتی ہے۔
4. ⚙️ Power and Efficiencyطاقت اور کارکردگی
🏷 Tags: power, efficiency
ENGLISH

Power P = W ÷ t — the rate of doing work. Unit: watt (W); 1 W = 1 J s−1; 1 horsepower = 746 W. Two forces can do different work in different times: F1 doing 5 J in 10 s gives 0.5 W, while F2 doing 3 J in 5 s gives 0.6 W — so F2 delivers greater power despite less work.

Efficiency = (useful output energy ÷ input energy) × 100%. It can never be 100% because some energy is always wasted as heat through friction and air resistance. Example: a box pushed up a 15 m ramp with 80 N (input 1200 J) rising 5 m (useful 10 × 10 × 5 = 500 J) has efficiency 500 ÷ 1200 = 41.7%. There is a limit to a machine's efficiency for the same reason.

اردو

طاقت P = W ÷ t — کام کرنے کی شرح۔ اکائی واٹ (W)؛ 1 W = 1 J/s؛ 1 ہارس پاور = 746 W۔ دو فورسز میں سے وہ زیادہ طاقت دیتی ہے جس کی شرح زیادہ ہو، چاہے کام کم ہو: 5 J/10 s = 0.5 W بمقابلہ 3 J/5 s = 0.6 W۔

کارکردگی = (مفید آؤٹ پٹ ÷ ان پٹ) × 100%۔ یہ کبھی 100% نہیں ہو سکتی کیونکہ رگڑ میں کچھ توانائی ہمیشہ حرارت بن کر ضائع ہوتی ہے۔

P = W/t • Efficiency = (useful out / in) × 100%Input 1200 J → Useful 500 J → Efficiency 41.7%The rest is wasted as heat (friction) — never 100%
Figure samjho: Power is work per second; efficiency is always below 100%.
طاقت کام فی سیکنڈ ہے؛ کارکردگی ہمیشہ 100% سے کم ہوتی ہے۔
Section E: Numerical Problemsحصہ ہ: عددی مسائل
QUESTIONS + ANSWERS
5.1 A force of 20 N acting at an angle of 60° to the horizontal is used to pull a box through a distance of 3 m across a floor. How much work is done? (Book answer: 30 J)

Answer: W = FS cos θ = 20 × 3 × cos 60° = 60 × 0.5 = 30 J. جواب: 30 J۔

5.2 A body moves a distance of 5 metres in a straight line under the action of a force of 8 newtons. If the work done is 20 joules, find the angle which the force makes with the direction of motion of the body. (Book answer: 60°)

Answer: W = FS cos θ → 20 = 8 × 5 × cos θ → cos θ = 0.5 → θ = 60°. جواب: 60°۔

5.3 An engine raises 100 kg of water through a height of 80 m in 25 s. What is the power of the engine? (Book answer: 3200 W)

Answer: Work = mgh = 100 × 10 × 80 = 80,000 J. Power = 80,000 ÷ 25 = 3200 W. جواب: 3200 W۔

5.4 A body of mass 20 kg is at rest. A 40 N force acts on it for 5 seconds. What is the kinetic energy of the body at the end of this time? (Book answer: 1000 J)

Answer: a = 40 ÷ 20 = 2 m s−2; v = 2 × 5 = 10 m s−1. Ek = ½ × 20 × 100 = 1000 J. جواب: 1000 J۔

5.5 A ball of mass 160 g is thrown vertically upward. The ball reaches a height of 20 m. Find the potential energy gained by the ball at this height. (Book answer: 32 J)

Answer: Ep = mgh = 0.16 × 10 × 20 = 32 J. جواب: 32 J۔

5.6 A 0.14 kg ball is thrown vertically upward with an initial velocity of 35 m s−1. Find the maximum height reached by the ball. (Book answer: 61.25 m)

Answer: h = v2 ÷ 2g = 352 ÷ 20 = 1225 ÷ 20 = 61.25 m. جواب: 61.25 m۔

5.7 A girl is swinging on a swing. At the lowest point of her swing, she is 1.2 m from the ground, and at the highest point she is 2.0 m from the ground. What is her maximum velocity and where? (Book answer: 4 m s−1, at the lowest position)

Answer: Height difference = 0.8 m. ½mv2 = mgh → v = √(2 × 10 × 0.8) = √16 = 4 m s−1, at the lowest point (where kinetic energy is maximum). جواب: 4 m/s، سب سے نچلی پوزیشن پر۔

5.8 A person pushes a lawn mower with a force of 50 N making an angle of 45° with the horizontal. If the mower is moved through a distance of 20 m, how much work is done? (Book answer: 707 J)

Answer: W = 50 × 20 × cos 45° = 1000 × 0.707 = 707 J. جواب: 707 J۔

5.9 Calculate the work done in (i) Pushing a 5 kg box up a frictionless inclined plane 10 m long that makes an angle of 30° with the horizontal. (ii) Lifting the box vertically up from the ground to the top of the inclined plane. (Book answers: 250 J, 250 J)

Answer: (i) Height gained = 10 × sin 30° = 5 m. Work = mgh = 5 × 10 × 5 = 250 J. (ii) Lifting vertically to the same height: work = mgh = 250 J. Same — work against gravity depends only on height, not path. جواب: دونوں 250 J — کشش ثقل کے خلاف کام صرف اونچائی پر منحصر ہے۔

5.10 A box of mass 10 kg is pushed up along a ramp 15 m long with a force of 80 N. If the box rises up a height of 5 m, what is the efficiency of the system? (Book answer: 41.7%)

Answer: Input = 80 × 15 = 1200 J. Useful output = 10 × 10 × 5 = 500 J. Efficiency = 500 ÷ 1200 × 100% = 41.7%. جواب: 41.7%۔

5.11 A force of 600 N acts on a box to push it 5 m in 15 s. Calculate the power. (Book answer: 200 W)

Answer: Work = 600 × 5 = 3000 J. Power = 3000 ÷ 15 = 200 W. جواب: 200 W۔

5.12 A 40 kg boy runs up-stair 10 m high in 8 s. What power he developed. (Book answer: 500 W)

Answer: Work = 40 × 10 × 10 = 4000 J. Power = 4000 ÷ 8 = 500 W. جواب: 500 W۔

5.13 A force F acts through a distance L on a body. The force is then increased to 2F that further acts through 2L. Sketch a force-displacement graph and calculate the total work done. (Book answer: 5FL or 5 units)

Answer: Force–displacement graph: a step graph — height F for distance L (area FL), then height 2F for distance 2L (area 4FL). Total work = area under graph = FL + 4FL = 5FL. جواب: 5FL — گراف کے نیچے کا رقبہ۔