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Showing posts with label Class 11 Physics. Show all posts
Showing posts with label Class 11 Physics. Show all posts
Vectors Class 11 Physics Notes, MCQs, Assertion Reason, Case Study & CBSE Questions
Complete Class 11 Physics Vectors Notes with formulas, diagrams, MCQs, assertion-reason, case studies and CBSE exam questions.
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NEET Physics - Vectors Notes
NEET Physics Chapter : Vectors
Vectors are one of the most important topics in NEET Physics.
Almost every chapter uses vectors.
Therefore understanding vectors properly makes Mechanics very easy.
1. Equality of Vectors
Two vectors are called equal if
Magnitude is same.
Direction is same.
Position does NOT matter.
Even if vectors are shifted parallel,
they are still equal.
Equal vectors ⇒ Same Magnitude + Same Direction
2. Addition of Vectors
Vector addition means combining two vectors to get one resultant vector.
Triangle Law
Place the tail of second vector at the head of first vector.
Join the starting point to the final point.
That gives resultant.
Remember:
Head to Tail Rule
3. Parallelogram Law
If two vectors start from the same point,
complete a parallelogram.
Diagonal gives resultant vector.
Resultant = Diagonal of Parallelogram
4. Magnitude of Resultant
Suppose
First Vector = a
Second Vector = b
Angle between them = θ
R = √(a² + b² + 2ab cosθ)
This is one of the MOST IMPORTANT formulas for NEET.
Learn it perfectly.
5. Direction of Resultant
tanα =
(b sinθ)/(a + b cosθ)
α = angle made by resultant with first vector.
6. Special Cases
Angle
Magnitude
0°
a+b
180°
|a-b|
90°
√(a²+b²)
7. Example
Question:
Two vectors have equal magnitude A.
Angle between them is θ.
Find resultant.
Solution
R = √(A²+A²+2A²cosθ)
= √(2A²(1+cosθ))
Using
1+cosθ=2cos²(θ/2)
R = 2A cos(θ/2)
Resultant = 2A cos(θ/2)
Direction:
α = θ/2
The resultant bisects the angle between two equal vectors.
8. Memory Tricks
✔ Triangle Rule → Head to Tail
✔ Parallelogram Rule → Diagonal
✔ Equal Vectors → Same Magnitude + Same Direction
✔ 90° → Pythagoras
✔ 180° → Subtraction
✔ 0° → Addition
9. NEET Important Points
Magnitude is always positive.
Direction decides vector.
Vectors obey triangle law.
Resultant depends on angle.
Equal vectors can have different positions.
Parallelogram law is frequently asked in NEET.
10. Practice Questions
Define equal vectors.
State triangle law.
State parallelogram law.
Write magnitude formula.
Write direction formula.
Find resultant when angle is 90°.
Find resultant when angle is 180°.
Two vectors 10 N each make 60°. Find resultant.
Two vectors 5 N each make 120°. Find resultant.
Why does the resultant bisect equal vectors?
Summary
Equal vectors → Same magnitude + same direction
Triangle Law → Head to Tail
Parallelogram Law → Diagonal
Magnitude → √(a²+b²+2abcosθ)
Direction → tanα=(bsinθ)/(a+bcosθ)
Equal vectors → Resultant = 2Acos(θ/2)
Direction = θ/2
CBSE Class 11 Physics - Vectors Question Bank
CBSE Class 11 Physics
Chapter : Vectors Question Bank
1. Multiple Choice Questions (MCQs)
1. A vector quantity has
Only magnitude
Only direction
Magnitude and direction
None
Answer: C
2. Equal vectors have
Equal magnitude only
Equal direction only
Equal magnitude and direction
Different directions
Answer: C
3. The diagonal of a parallelogram represents
Difference of vectors
Resultant vector
Unit vector
Zero vector
Answer: B
2. Very Short Answer Questions (1 Mark)
Q1. Define a vector.
A quantity having both magnitude and direction is called a vector.
Q2. Give one example of a vector.
Velocity.
Q3. What is a zero vector?
A vector whose magnitude is zero.
3. Short Answer Questions (2-3 Marks)
Q1. Define equal vectors.
Two vectors having equal magnitude and same direction are called equal vectors.
Q2. State the triangle law of vector addition.
If two vectors are represented by two sides of a triangle taken in order, the third side taken in opposite order represents the resultant.
4. Long Answer Questions (5 Marks)
Q1. Explain the parallelogram law of vector addition with diagram.
If two vectors acting simultaneously are represented by two adjacent sides of a parallelogram, then the diagonal passing through the common point represents the resultant vector.
Magnitude:
R = √(A² + B² + 2AB cosθ)
Direction:
tanα = (B sinθ)/(A + B cosθ)
5. Assertion and Reason
Assertion:
Equal vectors may have different positions.
Reason:
A vector depends only on magnitude and direction.
Answer:
Both Assertion and Reason are true, and Reason correctly explains Assertion.
Assertion:
Scalar quantities have direction.
Reason:
Scalars possess only magnitude.
Answer:
Assertion is false.
Reason is true.
6. Fill in the Blanks
Question
Answer
A vector has ______ and ______.
Magnitude, Direction
The diagonal of a parallelogram gives the ______.
Resultant
A quantity having only magnitude is called ______.
Scalar
The SI unit of displacement is ______.
metre
7. Match the Columns
Column A
Column B
Velocity
Vector
Mass
Scalar
Acceleration
Vector
Time
Scalar
Answers
Velocity → Vector
Mass → Scalar
Acceleration → Vector
Time → Scalar
8. Statement Based Questions
Statement I:
The resultant of two equal vectors bisects the angle between them.
Statement II:
The magnitudes of both vectors are equal.
Both statements are true.
Statement II explains Statement I.
9. Case Study Questions
Rahul pushes a box with force 20 N towards east.
Aman pushes the same box with force 20 N making an angle of 60°.
Answer the following.
Power Physics Notes PDF | NEET Work, Energy and Power Chapter
Power in Physics – Complete NEET Notes with Formulas, Units, Examples and Quick Revision
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Complete Class 11 Physics Notes
NEET Physics Formula Handbook
CBSE Class 11 Physics — Power: Complete Question Bank
Work, Energy and Power · Unit IV
Power — Complete Question Bank
A full CBSE Class 11 Physics practice set on Power: MCQs, assertion–reason, case studies, match-the-columns, numericals, and a quick answer key — click any answer to reveal it.
Class 11 · CBSEChapter: Work, Energy & Power100+ QuestionsAnswer Key Included
Section A Multiple Choice Questions
Tap "Show Answer" under any question to reveal the solution.
Level · Easy
Q1.Power is defined as:
Easy
(a) Total work done
(b) Rate of doing work
(c) Energy stored in a body
(d) Force applied per unit area
Show Answer
Answer: (b) Rate of doing work
Q2.The SI unit of power is:
Easy
(a) Joule
(b) Newton
(c) Watt
(d) Pascal
Show Answer
Answer: (c) Watt
Q3.1 horsepower is equal to:
Easy
(a) 500 W
(b) 746 W
(c) 1000 W
(d) 980 W
Show Answer
Answer: (b) 746 W
Q4.A machine does 1000 J of work in 10 s. Its power is:
Easy
(a) 10 W
(b) 100 W
(c) 1000 W
(d) 10000 W
Show Answer
Answer: (b) 100 W
Q5.The dimensional formula of power is:
Easy
(a) [ML²T⁻²]
(b) [ML²T⁻³]
(c) [MLT⁻²]
(d) [M²L²T⁻²]
Show Answer
Answer: (b) [ML²T⁻³]
Q6.Power is a:
Easy
(a) Vector quantity
(b) Scalar quantity
(c) Neither scalar nor vector
(d) Tensor quantity
Show Answer
Answer: (b) Scalar quantity
Q7.A pump lifts 200 kg of water to a height of 5 m in 10 s (g = 10 m/s²). The power of the pump is:
Easy
(a) 100 W
(b) 500 W
(c) 1000 W
(d) 2000 W
Show Answer
Answer: (c) 1000 W
Q8.If force and velocity are perpendicular to each other, the power is:
Easy
(a) Maximum
(b) Minimum
(c) Zero
(d) Infinite
Show Answer
Answer: (c) Zero
Level · Medium
Q9.A car engine exerts a force of 500 N while moving at a constant speed of 20 m/s. The power of the engine in hp is approximately:
Medium
(a) 10.7 hp
(b) 13.4 hp
(c) 15.2 hp
(d) 20.0 hp
Show Answer
Answer: (b) 13.4 hpP = Fv = 500 × 20 = 10000 W = 10000/746 ≈ 13.4 hp
Q10.The position of a particle is given by x = 3t² + 2t (x in m, t in s). A force of 6 N acts on it. The power at t = 2 s is:
Medium
(a) 72 W
(b) 84 W
(c) 96 W
(d) 108 W
Show Answer
Answer: (b) 84 Wv = dx/dt = 6t + 2. At t = 2s, v = 14 m/s. P = Fv = 6 × 14 = 84 W
Q11.A body of mass 2 kg is moved by a force of 10 N at constant velocity of 5 m/s at 60° to the direction of force. The power is:
Q15.A man of mass 60 kg climbs up a staircase carrying a load of 20 kg. If the total height gained is 10 m in 20 s, the average power is: (g = 10 m/s²)
Medium
(a) 200 W
(b) 300 W
(c) 400 W
(d) 800 W
Show Answer
Answer: (c) 400 WTotal mass = 80 kg. P = mgh/t = (80 × 10 × 10)/20 = 400 W
Level · Hard
Q16.A particle of mass m moves along a circular path of radius r with uniform speed v. The power delivered by the centripetal force is:
Hard
(a) mv²/r
(b) mv³/r
(c) Zero
(d) mv²r
Show Answer
Answer: (c) ZeroCentripetal force is always perpendicular to velocity (θ = 90°), so P = Fv cos90° = 0
Q17.The power delivered to a body moving in a straight line is given by P = 3t² + 2t (in watts). The work done in the first 2 seconds is:
Hard
(a) 10 J
(b) 12 J
(c) 14 J
(d) 16 J
Show Answer
Answer: (d) 16 JW = ∫P dt = ∫(3t² + 2t)dt = t³ + t². At t = 2: W = 8 + 8 = 16 J
Q18.A body of mass 1 kg is thrown vertically upward with initial velocity 20 m/s. The instantaneous power due to gravity at t = 1 s is: (g = 10 m/s²)
Hard
(a) 100 W
(b) −100 W
(c) 200 W
(d) −200 W
Show Answer
Answer: (b) −100 Wv = u − gt = 20 − 10 = 10 m/s (upward). F = mg = 10 N (downward). P = Fv cos180° = 10 × 10 × (−1) = −100 W
Q19.A pump motor is rated at 5 hp. How many kilograms of water can it raise in 1 minute through a height of 10 m? (g = 10 m/s², 1 hp = 746 W)
Hard
(a) 1492 kg
(b) 2238 kg
(c) 2984 kg
(d) 3730 kg
Show Answer
Answer: (b) 2238 kgP = 5 × 746 = 3730 W. W = P × t = 3730 × 60 = 223,800 J. m = W/(gh) = 223,800/(10×10) = 2238 kg
Q20.A vehicle of mass m accelerates uniformly from rest to velocity v in time t. The instantaneous power delivered by the engine at time t/2 is:
Hard
(a) mv²/2t
(b) mv²/4t
(c) mv²/t
(d) 3mv²/4t
Show Answer
Answer: (a) mv²/2ta = v/t (constant). At t/2, instantaneous velocity v′ = a(t/2) = v/2. F = ma = mv/t. P = F·v′ = (mv/t)(v/2) = mv²/2t.
Section B Very Short Answer Questions (1 Mark Each)
Q1.Define power.
Show Answer
Power is defined as the rate of doing work or the rate of energy transfer. Mathematically, P = W/t
Q2.Write the SI unit of power.
Show Answer
Watt (W), where 1 W = 1 J/s.
Q3.What is 1 horsepower in watts?
Show Answer
1 hp = 746 W.
Q4.Is power a scalar or vector quantity? Why?
Show Answer
Power is a scalar quantity because it has only magnitude and no direction. It is the ratio of two scalars (work and time).
Q5.Write the dimensional formula of power.
Show Answer
[M¹L²T⁻³]
Q6.A force acts perpendicular to the velocity of a body. What is the power?
Show Answer
Zero, because P = Fv cos90° = 0.
Q7.What is the relation between power, force, and velocity?
Show Answer
P = Fv (when force and velocity are in the same direction).
Q8.A machine does no work. Can it have power?
Show Answer
No. Since P = W/t, if W = 0, then P = 0.
Q9.What is the commercial unit of power?
Show Answer
Horsepower (hp).
Q10.Write the expression for instantaneous power.
Show Answer
P = dW/dt or P = F·v (dot product of force and instantaneous velocity).
Section C Short Answer Questions (2 Marks Each)
Q1.Distinguish between average power and instantaneous power.
Show Answer
Average Power
Instantaneous Power
Defined as total work divided by total time
Defined as power at a particular instant
Formula: Pavg = W/t
Formula: Pinst = F·v
Used when total work and time are given
Used when force or velocity varies with time
Q2.Derive the relation P = Fv.
Show Answer
We know power P = W/t. Work done W = F × s (for constant force in direction of displacement). Therefore, P = (F × s)/t = F × (s/t) = F × v. Hence, P = Fv
Q3.A pump delivers 1000 liters of water per minute to a tank at a height of 20 m. Find the power of the pump. (Density of water = 1000 kg/m³, g = 10 m/s²)
Show Answer
Volume per second = 1000/60 = 50/3 L/s = (50/3) × 10⁻³ m³/s. Mass per second = ρ × volume/s = 1000 × (50/3) × 10⁻³ = 50/3 kg/s. Power = (m/t) × gh = (50/3) × 10 × 20 = 10000/3 ≈ 3333.33 W ≈ 3.33 kW
Q4.Show that the power delivered by the centripetal force in uniform circular motion is zero.
Show Answer
In uniform circular motion, the centripetal force is always directed towards the center, while the velocity is always tangential. So θ = 90° always. P = Fv cosθ = Fv cos90° = 0.
Q5.The power of an engine is 5 kW. How much work can it do in 10 minutes?
Show Answer
Given: P = 5 kW = 5000 W, t = 10 min = 600 s. W = P × t = 5000 × 600 = 3,000,000 J = 3 × 10⁶ J
Q6.A car of mass 1000 kg moves up an incline of 1 in 20 at a constant speed of 10 m/s. Find the power of the engine. (g = 10 m/s², neglect friction)
Show Answer
Slope = 1/20, so sinθ = 1/20. Force required F = mg sinθ = 1000 × 10 × (1/20) = 500 N. Power P = Fv = 500 × 10 = 5000 W = 5 kW
Q7.Why is the power of a body moving with constant velocity on a frictionless horizontal surface zero?
Show Answer
On a frictionless horizontal surface, no external force is required to maintain constant velocity (Newton's first law). Since F = 0, power P = Fv = 0 × v = 0.
Q8.A 2 kW motor pump is used to pump water from a well 10 m deep. How much water can be pumped per minute? (g = 10 m/s²)
Show Answer
P = 2 kW = 2000 W, t = 60 s. W = P × t = 2000 × 60 = 120,000 J. m = W/(gh) = 120,000/(10 × 10) = 1200 kg
Section D Long Answer Questions (3–5 Marks Each)
Q1.(a) Define power and derive its SI unit. (2 marks) (b) A pump can throw 8000 kg of water per minute to a height of 15 m. Calculate the power of the pump. (g = 9.8 m/s²) (3 marks)
Show Answer
(a) Power is defined as the rate of doing work. If W is the work done in time t, then P = W/t. The SI unit of work is Joule (J) and time is second (s). Therefore, SI unit of power = J/s = Watt (W). 1 Watt = 1 Joule per second.
(b) Given: m = 8000 kg (per minute), h = 15 m, t = 60 s, g = 9.8 m/s². Work done per minute = mgh = 8000 × 9.8 × 15 = 1,176,000 J. Power P = W/t = 1,176,000/60 = 19,600 W = 19.6 kW
Q2.(a) Derive the expression for instantaneous power. (2 marks) (b) The position of a body of mass 2 kg is given by x = 2t³ + 3t² + 5, where x is in meters and t in seconds. A constant force of 12 N acts on the body in the direction of motion. Calculate the power delivered at t = 2 s. (3 marks)
Show Answer
(a) Consider a small amount of work dW done in a small time interval dt. Instantaneous power P = dW/dt. Since dW = F·ds, we get P = (F·ds)/dt = F·(ds/dt) = F·v. Therefore, P = F·v (dot product of force and instantaneous velocity).
(b) x = 2t³ + 3t² + 5 → v = dx/dt = 6t² + 6t. At t = 2 s: v = 6(4) + 6(2) = 24 + 12 = 36 m/s. Power P = F × v = 12 × 36 = 432 W
Q3.(a) Prove that power can also be expressed as the scalar product of force and velocity. (2 marks) (b) An engine of power 10 hp is used to pump water from a well 8 m deep. How many kilograms of water can be pumped in 1 hour? (1 hp = 746 W, g = 9.8 m/s²) (3 marks)
Show Answer
(a) Work done by a force F in displacing a body by ds is dW = F·ds. Power is rate of doing work: P = dW/dt = (F·ds)/dt = F·(ds/dt) = F·v. Hence, P = F·v.
(b) P = 10 hp = 10 × 746 = 7460 W. t = 1 hour = 3600 s. Total work W = P × t = 7460 × 3600 = 26,856,000 J. m = W/(gh) = 26,856,000/(9.8 × 8) = 26,856,000/78.4 ≈ 342,551 kg
Q4.(a) What is the difference between kilowatt and kilowatt-hour? (2 marks) (b) A family uses a 2 kW electric heater for 4 hours daily. Calculate the energy consumed in 30 days in kWh and joules. (3 marks)
Show Answer
(a)
Kilowatt (kW)
Kilowatt-hour (kWh)
Unit of power
Unit of energy
1 kW = 1000 W
1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J
Measures rate of energy consumption
Measures total energy consumed
(b) Power P = 2 kW, daily usage t = 4 h. Daily energy = 2 × 4 = 8 kWh. 30-day energy = 8 × 30 = 240 kWh. In joules: 240 kWh = 240 × 3.6 × 10⁶ = 8.64 × 10⁸ J
Section E Assertion and Reason Questions
Choose: (a) Both true, Reason correctly explains Assertion | (b) Both true, Reason does NOT explain Assertion | (c) Assertion true, Reason false | (d) Assertion false, Reason true
Q1.Assertion: Power is a scalar quantity. Reason: Power is the ratio of two scalar quantities, work and time.
Show Answer
Answer: (a) Both true and Reason correctly explains Assertion.
Q2.Assertion: When a body moves in a circular path with uniform speed, the power delivered by the centripetal force is zero. Reason: The centripetal force is always perpendicular to the velocity of the body.
Show Answer
Answer: (a) Both true and Reason correctly explains Assertion.
Q3.Assertion: A body moving with constant velocity on a frictionless horizontal surface has zero power. Reason: No force is required to maintain constant velocity on a frictionless surface.
Show Answer
Answer: (a) Both true and Reason correctly explains Assertion.
Q4.Assertion: 1 horsepower is equal to 1000 watts. Reason: Horsepower is the commercial unit of power.
Show Answer
Answer: (c) Assertion is false but Reason is true. (1 hp = 746 W, not 1000 W)
Q5.Assertion: The instantaneous power of a body can be negative. Reason: Power is always positive because it is the rate of doing work.
Show Answer
Answer: (c) Assertion is true but Reason is false. (Power can be negative when force opposes motion, e.g., friction doing negative work)
Q6.Assertion: Average power is always equal to instantaneous power. Reason: Average power is calculated over a time interval while instantaneous power is at a specific instant.
Show Answer
Answer: (d) Assertion is false but Reason is true.
Q7.Assertion: The power delivered by gravity to a body thrown vertically upward is negative. Reason: The gravitational force acts opposite to the direction of motion during upward journey.
Show Answer
Answer: (a) Both true and Reason correctly explains Assertion.
Q8.Assertion: A pump requires more power to lift the same amount of water to a greater height. Reason: Power is directly proportional to the height through which water is lifted.
Show Answer
Answer: (a) Both true and Reason correctly explains Assertion.
Section F Fill in the Blanks
Q1.Power is defined as the ______ of doing work.
Show Answer
rate
Q2.The SI unit of power is ______, named after the scientist ______.
Show Answer
Watt, James Watt
Q3.1 horsepower = ______ watts.
Show Answer
746
Q4.The dimensional formula of power is ______.
Show Answer
[M¹L²T⁻³]
Q5.When force and velocity are perpendicular to each other, the power is ______.
Show Answer
zero
Q6.The commercial unit of power is ______.
Show Answer
horsepower (hp)
Q7.For a body moving with constant velocity on a frictionless surface, the power is ______.
Show Answer
zero
Q8.Instantaneous power is given by the scalar product of ______ and ______.
Show Answer
force, velocity
Q9.1 kilowatt = ______ watts.
Show Answer
1000
Q10.The power of a pump lifting mass m through height h in time t is given by P = ______.
Show Answer
mgh/t
Section G Case Study Based Questions
Case Study 1 · Hydroelectric Power Plant
A hydroelectric power plant uses water falling from a height to generate electricity. Water from a reservoir at a height of 100 m flows down through penstocks to turn turbines. The plant has a capacity to process 5000 kg of water per second. (Take g = 10 m/s²)
Q1.What is the power generated by the falling water?
Show Answer
P = mgh/t = (5000 × 10 × 100)/1 = 5,000,000 W = 5 MW
Q2.If the efficiency of the turbine-generator system is 80%, what is the actual electrical power output?
Show Answer
Actual power = 80% of 5 MW = 0.8 × 5 = 4 MW
Q3.How much energy is produced in 1 hour at this actual power output?
Show Answer
E = P × t = 4 MW × 1 h = 4 MWh = 4 × 10⁶ Wh = 1.44 × 10¹⁰ J
Case Study 2 · Electric Vehicle
An electric car of mass 1500 kg accelerates from rest to a speed of 30 m/s in 10 seconds. The motor delivers constant power during this time.
Q1.What is the acceleration of the car?
Show Answer
a = (v − u)/t = (30 − 0)/10 = 3 m/s²
Q2.What is the average power delivered by the motor during acceleration?
Show Answer
Work done = ΔKE = ½mv² = ½ × 1500 × 30² = 675,000 J. Pavg = W/t = 675,000/10 = 67,500 W = 67.5 kW
Q3.If the car maintains a constant speed of 30 m/s on a level road with a frictional force of 500 N, what power is required to overcome friction?
Show Answer
P = Fv = 500 × 30 = 15,000 W = 15 kW
Case Study 3 · Human Power Output
A person of mass 70 kg climbs a staircase of 50 steps, each 20 cm high, in 20 seconds. (g = 9.8 m/s²)
Q1.What is the total height climbed?
Show Answer
h = 50 × 0.20 = 10 m
Q2.What is the work done against gravity?
Show Answer
W = mgh = 70 × 9.8 × 10 = 6860 J
Q3.What is the average power output of the person?
Show Answer
P = W/t = 6860/20 = 343 W
Section H Statement Based Questions
Passage 1
"Power is the rate at which work is done. It is a scalar quantity. The SI unit of power is watt. When a force acts on a body in the direction of its motion, the power is given by P = Fv. If the force makes an angle θ with the velocity, then P = Fv cosθ."
Q1.Why is power called a scalar quantity?
Show Answer
Power is a scalar quantity because it has only magnitude and no direction. It is derived as the ratio of work (scalar) and time (scalar), and the dot product of two vectors (F·v) also yields a scalar.
Q2.A force of 20 N acts on a body moving with a velocity of 4 m/s. The force makes an angle of 60° with the direction of motion. Calculate the power.
Show Answer
P = Fv cosθ = 20 × 4 × cos60° = 80 × 0.5 = 40 W
Passage 2
"A pump is used to lift water from a well. The power of the pump depends on the mass of water lifted, the height to which it is lifted, and the time taken. The efficiency of the pump is defined as the ratio of useful power output to the total power input."
Q1.Write the expression for the power of a pump lifting water.
Show Answer
P = mgh/t
Q2.A pump of power 2 kW and efficiency 75% is used to lift water through 10 m. How much water can it lift in 1 minute? (g = 10 m/s²)
Show Answer
Useful power = 75% of 2 kW = 0.75 × 2000 = 1500 W. Work done in 1 min = 1500 × 60 = 90,000 J. m = W/(gh) = 90,000/(10 × 10) = 900 kg
Section I Match the Columns
Match the Following 1
Column A
Column B
(a) Unit of power
(p) [M¹L²T⁻³]
(b) Dimensional formula of power
(q) Joule
(c) Unit of work
(r) Watt
(d) 1 hp
(s) 746 W
Show Answer
(a) → (r) | (b) → (p) | (c) → (q) | (d) → (s)
Match the Following 2
Column A (Physical Quantity)
Column B (Expression)
(a) Average power
(p) F·v
(b) Instantaneous power
(q) W/t
(c) Power in lifting
(r) mgh/t
(d) Power when F ⊥ v
(s) Zero
Show Answer
(a) → (q) | (b) → (p) | (c) → (r) | (d) → (s)
Match the Following 3
Column A (Situation)
Column B (Power)
(a) Body moving with constant velocity on frictionless surface
(p) Positive
(b) Body thrown upward against gravity
(q) Negative
(c) Body falling freely under gravity
(r) Zero
(d) Body in uniform circular motion
(s) Variable
Show Answer
(a) → (r) Zero, as F = 0 | (b) → (q) Negative, gravity opposes motion | (c) → (p) Positive, gravity aids motion | (d) → (r) Zero, centripetal force ⊥ velocity
Match the Following 4
Column A
Column B
(a) James Watt
(p) Unit of energy
(b) Joule
(q) Commercial unit of power
(c) Horsepower
(r) SI unit of power
(d) Kilowatt-hour
(s) Improved steam engine
Show Answer
(a) → (s) | (b) → (p) | (c) → (q) | (d) → (p) — kWh is a unit of energy, like Joule
Match the Following 5
Column A (Value)
Column B (Equivalent)
(a) 1 kW
(p) 3.6 × 10⁶ J
(b) 1 kWh
(q) 1000 W
(c) 1 hp
(r) 746 W
(d) 1 W
(s) 1 J/s
Show Answer
(a) → (q) | (b) → (p) | (c) → (r) | (d) → (s)
Section J True or False
Q1.Power and work have the same dimensions.
Show Answer
False. [Work] = [ML²T⁻²], [Power] = [ML²T⁻³]
Q2.1 kilowatt-hour is a unit of power.
Show Answer
False. It is a unit of energy.
Q3.The power of a body can be negative.
Show Answer
True. When force opposes motion (e.g., friction), power is negative.
Q4.A body in uniform circular motion has zero power due to centripetal force.
Show Answer
True. Centripetal force is perpendicular to velocity.
Q5.The SI unit of power is named after James Prescott Joule.
Show Answer
False. It is named after James Watt.
Section K Numerical Problems (With Detailed Solutions)
Q1.A pump lifts 1200 kg of water per minute to a height of 25 m. Calculate the power of the pump in kW. (g = 10 m/s²)
Show Solution
m = 1200 kg, h = 25 m, t = 60 s, g = 10 m/s². P = mgh/t = (1200 × 10 × 25)/60 = 300,000/60 = 5000 W = 5 kW
Q2.A car of mass 1000 kg moves on a level road at a constant speed of 20 m/s against a resistance of 400 N. Find the power of the engine in horsepower.
Show Solution
At constant speed, engine force = resistance = 400 N. P = Fv = 400 × 20 = 8000 W. In hp: 8000/746 ≈ 10.7 hp
Q3.The power of an engine is 5 kW. How much time will it take to lift a load of 500 kg to a height of 40 m? (g = 10 m/s²)
Show Solution
W = mgh = 500 × 10 × 40 = 200,000 J. P = 5000 W. t = W/P = 200,000/5000 = 40 s
Q4.A force F = (3i + 4j) N acts on a body moving with velocity v = (4i − 3j) m/s. Calculate the power.
Show Solution
P = F·v = (3)(4) + (4)(−3) = 12 − 12 = 0 W
Q5.A motor pump is rated at 3 hp. Calculate the maximum mass of water it can lift in 2 minutes through a height of 15 m. (1 hp = 746 W, g = 9.8 m/s²)
Show Solution
P = 3 × 746 = 2238 W. t = 120 s. W = P × t = 2238 × 120 = 268,560 J. m = W/(gh) = 268,560/(9.8 × 15) = 268,560/147 ≈ 1827 kg
A moving car compresses a spring, demonstrating conservation of mechanical energy and work-energy theorem.
Internal Links
NCERT Class 11 Physics Chapter 5 Notes
Work, Energy and Power Formulas
Conservation of Mechanical Energy
Work-Energy Theorem Explained
Hooke's Law and Spring Force
Potential Energy and Kinetic Energy
NCERT Physics Example 5.1–5.7 Solutions
NCERT Physics Chapter 5 Exercise Solutions
JEE Physics Work Energy Questions
NEET Physics Chapter-wise MCQs
Energy Conservation Numerical Problems
Class 11 Physics Formula Sheet
NCERT Physics Class 11 - Examples 5.8 & 5.9
NCERT Physics Class 11
Chapter 5 - Work, Energy and Power
Example 5.8 & Example 5.9
Question 1
A car of mass 1000 kg is moving with a speed of
18 km h-1 on a smooth horizontal road and
collides with a spring of spring constant
5.25 × 103 N m-1.
Find the maximum compression of the spring.
Answer
Given
Mass (m) = 1000 kg
Speed (v) = 18 km h-1 = 5 m s-1
Spring constant (k) = 5.25 × 103 N m-1
Formula
At maximum compression,
Kinetic Energy = Spring Potential Energy
½mv² = ½kx²
Step 1 : Calculate Kinetic Energy
K = ½ × 1000 × 5²
K = 12500 J
Step 2 : Calculate Compression
12500 = ½ × 5.25 × 10³ × x²
12500 = 2625x²
x² = 12500 / 2625
x² = 4.76
x = √4.76
x ≈ 2.18 m
Maximum Compression = 2.0 m (Approx.)
Question 2
Using Example 5.8, if the coefficient of friction is
0.5, calculate the maximum compression of the spring.
Answer
Given
Mass = 1000 kg
Speed = 5 m s-1
Spring constant = 5.25 × 10³ N m-1
Coefficient of friction = 0.5
g = 10 m s-2
Formula
Work-Energy Theorem
ΔK = W
½mv² = ½kx² + μmgx
Step 1
12500 = 2625x² + 5000x
Step 2
2625x² + 5000x − 12500 = 0
Step 3
Using quadratic formula,
x = 1.35 m
Maximum Compression = 1.35 m
Practice Questions
Practice Question 1
A spring of spring constant 400 N m-1 is compressed
by 0.5 m. Calculate the elastic potential energy stored.
Formula
U = ½kx²
U = ½ × 400 × (0.5)²
U = 50 J
Answer = 50 J
Practice Question 2
A body of mass 2 kg moves with speed 10 m s-1.
Find its kinetic energy.
Formula
K = ½mv²
K = ½ × 2 × 10²
K = 100 J
Answer = 100 J
Practice Question 3
A spring stores 100 J of energy.
Its spring constant is 500 N m-1.
Find the compression.
100 = ½ × 500 × x²
100 = 250x²
x² = 0.4
x = 0.63 m
Answer = 0.63 m
Practice Question 4
A 500 kg car moving at 10 m s-1 hits a spring
of spring constant 10000 N m-1.
Find the maximum compression.
K = ½mv²
K = ½ × 500 × 10²
K = 25000 J
25000 = ½ × 10000 × x²
25000 = 5000x²
x² = 5
x = 2.24 m
Answer = 2.24 m
Important Formulae
Formula
Expression
Kinetic Energy
K = ½mv²
Spring Potential Energy
U = ½kx²
Work-Energy Theorem
ΔK = W
Mechanical Energy
KE + PE = Constant (Without Friction)
Important Viva Questions
Why does the car stop at maximum compression?
Because all kinetic energy is converted into spring potential energy.
Why is conservation of mechanical energy not applicable when friction is present?
Because friction is a non-conservative force and converts mechanical energy into heat.
What is the formula of spring potential energy?
U = ½kx²
What is the formula of kinetic energy?
K = ½mv²
NEET Physics Notes - Work, Energy and Power
NEET Physics Notes
Chapter: Work, Energy and Power
Topics Covered
Spring Force (Hooke's Law)
Spring Potential Energy
Kinetic Energy
Conservation of Mechanical Energy
Work-Energy Theorem
Friction
NCERT Example 5.8
NCERT Example 5.9
Important Formulae
NEET MCQs
1. Spring Force (Hooke's Law)
When a spring is stretched or compressed, it tries to return to its original length.
This restoring force is called Spring Force.
Formula
F = -kx
Symbol
Meaning
F
Spring Force (N)
k
Spring Constant (N/m)
x
Compression or Extension (m)
Remember
Negative sign shows restoring force.
Force acts opposite to displacement.
Larger k means stronger spring.
Smaller k means softer spring.
2. Spring Potential Energy
Energy stored inside a compressed or stretched spring.
Formula
U = ½ kx²
Symbol
Meaning
U
Potential Energy
k
Spring Constant
x
Compression
3. Kinetic Energy
Energy possessed by a moving object.
Formula
K = ½ mv²
Symbol
Meaning
m
Mass
v
Velocity
4. Conservation of Mechanical Energy
When there is no friction,
the total mechanical energy remains constant.
½mv² = ½kx²
Used to calculate maximum compression of a spring.
Physics Examples
Example 5.8
To simulate car accidents, auto manufacturers study the collisions of moving cars with mounted springs and other constraints. Consider a typical simulation with a car of mass 1000 kg moving at a speed of 18.0 km/h on a smooth road and colliding with a horizontal mounted spring of spring constant 5.25 × 10³ N m⁻¹. What is the maximum compression of the spring?
Example 5.9
Consider Example 5.8 taking the coefficient of friction, μ = 0.5, and calculate the maximum compression of the spring.
NCERT Example 5.8
Question
Mass = 1000 kg
Speed = 18 km/h
Spring Constant = 5.25 × 10³ N/m
Find the maximum compression.
Solution
Step 1
18 km/h = 5 m/s
Kinetic Energy
K = ½ ×1000×5²
K = 12500 J
Using Conservation of Energy
½kx² = 12500
x = 2.0 m
Answer:
Maximum Compression = 2.0 m
5. Work-Energy Theorem
The work done by all forces acting on a body equals the change in kinetic energy.
Formula
W = ΔK
6. Friction
Friction always opposes motion.
Formula
F = μN
On horizontal surface,
N = mg
Therefore,
F = μmg
NCERT Example 5.9
Question
Mass = 1000 kg
Speed = 5 m/s
k = 5.25 ×10³ N/m
μ = 0.5
Solution
Apply Work-Energy Theorem
−½mv² = −½kx² − μmgx
Rearranging,
kx² + 2μmgx − mv² = 0
x = 1.35 m
Answer:
Maximum Compression = 1.35 m
Important Formula Sheet
Formula
Equation
Kinetic Energy
½mv²
Spring Force
F = -kx
Spring Potential Energy
½kx²
Friction
μmg
Work-Energy Theorem
W = ΔK
Conservation of Energy
½mv² = ½kx²
NEET Important Points
Hooke's law is valid only within elastic limit.
Spring force is a restoring force.
Potential energy is always positive.
Friction is a non-conservative force.
Mechanical energy is conserved only without friction.
Work done by friction is negative.
Spring energy depends on x².
Kinetic energy depends on velocity².
NEET Practice MCQs
Q1.
The force exerted by a spring is
A. Constant
B. Proportional to displacement
C. Inversely proportional
D. Zero
Answer: B
Q2.
Potential energy stored in a spring is
A. kx
B. kx²
C. ½kx²
D. ½kx
Answer: C
Q3.
Which force is non-conservative?
A. Gravity
B. Spring
C. Electrostatic
D. Friction
Answer: D
Q4.
Mechanical energy remains constant when
A. Friction acts
B. Air resistance acts
C. Only conservative forces act
D. External work is done
Answer: C
Quick Revision
Hooke's Law → F = -kx
Spring Energy → ½kx²
Kinetic Energy → ½mv²
Work-Energy Theorem → W = ΔK
Without friction → Energy conserved.
With friction → Use Work-Energy Theorem.
Maximum compression occurs when kinetic energy becomes zero.
