Work, Power & Energy is a high-weightage chapter in AMU. Expect 2-3 questions — mostly on work formula (cosθ), kinetic/potential energy, conservation of energy, and power. The Work-Energy Theorem is the bridge between Newton's Laws and Energy. Master the cosθ sign convention — that's where students lose marks! This expanded guide covers collisions, simple machines, 15+ PYQs, 15+ practice numericals, common mistakes, and quick revision.
1) Work — The Physics Definition
In everyday life, any physical or mental effort is called work. But in Physics, work has a very specific meaning: Work is done only when a force causes a displacement in the body, and the displacement has at least a component along the direction of the force. If there is no displacement, NO work is done — even if you push a wall all day or hold a 50 kg bag above your head for hours!
Mathematical definition:
SI unit:
CGS unit:
Conversion:
W = F × s × cosθ
W = F · s (dot product of force and displacement vectors)
SI unit:
Joule (J) = N mCGS unit:
erg = dyne × cmConversion:
1 J = 10⁷ erg1 J = 0.239 cal (calorie, in heat context)
θ (theta) is the angle between the force vector and the displacement vector. This angle is critical — it determines whether the work done is positive, negative, or zero.
1.1 Cases of Work Done Based on θ
| Condition | Angle θ | cosθ | Work sign | Real-life example |
|---|---|---|---|---|
| Force in same direction as motion | 0° | cos0° = 1 | W = +Fs (Maximum positive) | Pulling a cart forward, pushing a box on floor |
| Force at acute angle to motion | 0° < θ < 90° | cosθ > 0 | Positive (less than Fs) | Pulling a suitcase with handle at 60° |
| Force perpendicular to motion | 90° | cos90° = 0 | W = 0 (Zero work) | Carrying a bag horizontally; centripetal force in circular motion |
| Force at obtuse angle to motion | 90° < θ < 180° | cosθ < 0 | Negative | Pushing a stalled car while it rolls forward (force applied backward) |
| Force opposite to motion | 180° | cos180° = -1 | W = -Fs (Maximum negative) | Friction, braking force, air resistance |
🟡 Memory trick for cos values:
cos0° = 1 (same direction → max positive work)
cos90° = 0 (perpendicular → zero work)
cos180° = -1 (opposite direction → max negative work)
Just memorize these three values and you can handle any AMU question on work sign!
cos0° = 1 (same direction → max positive work)
cos90° = 0 (perpendicular → zero work)
cos180° = -1 (opposite direction → max negative work)
Just memorize these three values and you can handle any AMU question on work sign!
1.2 Work Done by Variable Force
Real forces are not always constant. When the force varies with displacement (like stretching a spring), we use the graphical method. The work done by a variable force is given by the area under the Force-Displacement (F-s) graph.
Area under F-s graph = Work done
If the F-s graph is a triangle: Area = ½ × base × height = ½ × s × F = work done
If the graph is a rectangle: Area = length × breadth = F × s = work done
W = ∫ F·ds (for variable force — integral = area under curve)If the F-s graph is a triangle: Area = ½ × base × height = ½ × s × F = work done
If the graph is a rectangle: Area = length × breadth = F × s = work done
🔥 AMU PYQ: Area under force-displacement graph represents:
(A) Force (B) Displacement (C) Work (D) Power
Ans: C (Work done by the force)
(A) Force (B) Displacement (C) Work (D) Power
Ans: C (Work done by the force)
1.3 Numericals on Work
Numerical 1: A force of 50 N acts on a body at an angle of 60° to the horizontal and displaces it by 10 m. Find the work done.
Solution:
W = F s cosθ = 50 × 10 × cos60°
cos60° = ½ → W = 50 × 10 × ½ = 250 J
Solution:
W = F s cosθ = 50 × 10 × cos60°
cos60° = ½ → W = 50 × 10 × ½ = 250 J
Numerical 2: A man lifts a 20 kg suitcase from the floor to a height of 1.5 m. Find work done against gravity (g = 10 m/s²).
Solution:
Work against gravity = mgh = 20 × 10 × 1.5 = 300 J
(Here θ = 0° between applied force and displacement → positive work)
Solution:
Work against gravity = mgh = 20 × 10 × 1.5 = 300 J
(Here θ = 0° between applied force and displacement → positive work)
Numerical 3: A coolie carries a 15 kg bag on his head and walks 100 m horizontally. Work done against gravity?
Solution:
Force of gravity is vertical downward, displacement is horizontal → θ = 90°
cos90° = 0 → W = 0 J (No work done against gravity!)
Solution:
Force of gravity is vertical downward, displacement is horizontal → θ = 90°
cos90° = 0 → W = 0 J (No work done against gravity!)
2) Energy — The Capacity to Do Work
Energy is defined as the capacity of a body to do work. A body that has energy can do work, and when it does work, it loses energy. The SI unit of energy is the same as work — the Joule (J). Energy is a scalar quantity.
2.1 Different Forms of Energy
Energy exists in many forms. Here are the ones you need to know for AMU:
| Form of Energy | Description | Examples |
|---|---|---|
| Kinetic Energy (KE) | Energy due to motion | Moving car, flowing water, wind |
| Potential Energy (PE) | Energy due to position or configuration | Stored water in dam, stretched spring, lifted weight |
| Chemical Energy | Energy stored in chemical bonds | Battery, food, petrol, coal, wood |
| Heat Energy (Thermal) | Energy due to temperature / molecular motion | Hot water, steam, sun's heat |
| Light Energy (Radiant) | Energy carried by light / electromagnetic waves | Sunlight, laser, bulb light |
| Sound Energy | Energy carried by sound waves | Speaker, musical instrument, thunder |
| Electrical Energy | Energy due to flow of electric charges | Current in wires, lightning |
| Nuclear Energy | Energy released from nuclear reactions (fission/fusion) | Atomic bomb, nuclear power plant, sun (fusion) |
| Magnetic Energy | Energy stored in magnetic fields | Magnet, electromagnet, MRI machine |
3) Kinetic Energy (KE) — Energy of Motion
Kinetic energy is the energy a body possesses because of its motion. A moving hammer drives a nail into wood — that's kinetic energy at work. The faster an object moves, the more kinetic energy it has.
KE = ½ m v²m = mass of the body (in kg)v = velocity of the body (in m/s)SI unit: Joule (J)
Dimension: [M¹L²T⁻²]
3.1 Derivation of KE = ½mv² (Using Work-Energy Theorem)
Consider a body of mass m initially at rest (u = 0). A constant force F acts on it, producing acceleration a, and displacing it by s. The final velocity becomes v.
- Using Newton's 2nd Law: F = ma
- Using 3rd equation of motion: v² = u² + 2as → v² = 0 + 2as → as = v²/2
- Work done: W = F × s = ma × s = m × (as) = m × (v²/2)
- KE = W = ½mv² (This work done to bring the body from rest to velocity v is stored as kinetic energy.)
3.2 Relation Between Kinetic Energy and Momentum
Momentum: p = mv
KE in terms of p:
KE in terms of p:
KE = ½mv² = ½ × m × v² = ½ × (mv)² / m = p² / (2m)KE = p² / 2mp = √(2m·KE)
🟡 Key relationships:
• KE ∝ m (for same v) — heavier objects have more KE
• KE ∝ v² (for same m) — doubling velocity quadruples KE
• KE ∝ p² (for same m) — if momentum doubles, KE becomes 4×
• p ∝ √(KE) — to double momentum, KE must quadruple
• KE ∝ m (for same v) — heavier objects have more KE
• KE ∝ v² (for same m) — doubling velocity quadruples KE
• KE ∝ p² (for same m) — if momentum doubles, KE becomes 4×
• p ∝ √(KE) — to double momentum, KE must quadruple
3.3 Numericals on KE
Numerical 1: A body of mass 10 kg moves with a velocity of 5 m/s. Find its kinetic energy.
Solution:
KE = ½ × 10 × (5)² = ½ × 10 × 25 = 125 J
Solution:
KE = ½ × 10 × (5)² = ½ × 10 × 25 = 125 J
Numerical 2: What is the kinetic energy of a 1500 kg car moving at 20 m/s?
Solution:
KE = ½ × 1500 × (20)² = ½ × 1500 × 400 = 750 × 400 = 3,00,000 J = 300 kJ
Solution:
KE = ½ × 1500 × (20)² = ½ × 1500 × 400 = 750 × 400 = 3,00,000 J = 300 kJ
Numerical 3: If a body's KE is 200 J and its mass is 4 kg, find its velocity.
Solution:
v = √(2KE / m) = √(2 × 200 / 4) = √(400 / 4) = √100 = 10 m/s
Solution:
v = √(2KE / m) = √(2 × 200 / 4) = √(400 / 4) = √100 = 10 m/s
Numerical 4: The momentum of a body is 40 kg m/s and its KE is 100 J. Find its mass.
Solution:
KE = p² / 2m → 100 = (40)² / (2m) = 1600 / (2m) = 800/m
m = 800 / 100 = 8 kg
Solution:
KE = p² / 2m → 100 = (40)² / (2m) = 1600 / (2m) = 800/m
m = 800 / 100 = 8 kg
4) Potential Energy (PE) — Stored Energy
Potential energy is the energy stored in a body due to its position (relative to a reference) or its configuration (shape, state, deformation). It is the energy that has the potential to do work.
4.1 Gravitational Potential Energy
When a body is raised to a height h above the ground, work is done against gravity. This work is stored as gravitational potential energy.
PE = mghm = mass (kg), g = acceleration due to gravity (m/s²), h = height above reference level (m)Derivation:
Force needed to lift body = mg (to balance gravity)
Displacement = h (upward)
Work done = Force × Displacement = mg × h = mgh
This work gets stored as PE = mgh
🟡 Important points about GPE:
• PE depends on the reference level (usually ground is taken as zero)
• PE is relative — PE at height h above ground = mgh; at ground = 0
• PE can be negative if the reference is above the object
• For AMU, take ground as zero level unless otherwise specified
• g = 10 m/s² (or 9.8 m/s² if stated)
• PE depends on the reference level (usually ground is taken as zero)
• PE is relative — PE at height h above ground = mgh; at ground = 0
• PE can be negative if the reference is above the object
• For AMU, take ground as zero level unless otherwise specified
• g = 10 m/s² (or 9.8 m/s² if stated)
4.2 Elastic Potential Energy (Spring)
When a spring is stretched or compressed from its natural (unstretched) length, work is done against the restoring force. This work is stored as elastic potential energy.
PE (elastic) = ½ k x²k = spring constant (stiffness) — N/mx = extension or compression from natural length (m)Derivation (basic idea):
For a spring, force F = kx (Hooke's Law — variable force as x changes)
Average force = (0 + kx)/2 = ½kx
Work = Average force × displacement = ½kx × x = ½kx²
This equals the elastic PE stored.
4.3 Numericals on PE
Numerical 1: A 5 kg object is raised to a height of 8 m. Find its PE (g = 10 m/s²).
Solution:
PE = mgh = 5 × 10 × 8 = 400 J
Solution:
PE = mgh = 5 × 10 × 8 = 400 J
Numerical 2: A spring of k = 200 N/m is stretched by 0.1 m. Find elastic PE stored.
Solution:
PE = ½ × 200 × (0.1)² = ½ × 200 × 0.01 = ½ × 2 = 1 J
Solution:
PE = ½ × 200 × (0.1)² = ½ × 200 × 0.01 = ½ × 2 = 1 J
Numerical 3: A 2 kg mass is at a height of 5 m. If g = 9.8 m/s², find PE and the work done by gravity if it falls to ground.
Solution:
PE = 2 × 9.8 × 5 = 98 J
Work done by gravity = mgh = same = 98 J (positive, since gravity pulls downward)
(PE converts entirely to KE just before hitting ground)
Solution:
PE = 2 × 9.8 × 5 = 98 J
Work done by gravity = mgh = same = 98 J (positive, since gravity pulls downward)
(PE converts entirely to KE just before hitting ground)
| Energy Type | Formula | Depends on | SI Unit |
|---|---|---|---|
| Kinetic Energy | ½ mv² | Mass & speed | J |
| Gravitational PE | mgh | Mass, g, height | J |
| Elastic PE (Spring) | ½ kx² | Spring constant, stretch/compression | J |
5) Work-Energy Theorem — The Bridge
The Work-Energy Theorem connects Newton's Laws with Energy. It states: The net work done by all forces (conservative + non-conservative) acting on a body equals the change in its kinetic energy.
Statement:
W_net = ΔKE = KE_final − KE_initialW_net = ½ mv² − ½ mu²Where: u = initial velocity, v = final velocity
5.1 Proof of Work-Energy Theorem
Consider a body of mass m moving with initial velocity u. A constant net force F acts, producing acceleration a, over displacement s, changing velocity to v.
- F = ma (Newton's 2nd Law)
- v² = u² + 2as → as = (v² − u²)/2
- W_net = F × s = ma × s = m × (as) = m × (v² − u²)/2
- W_net = ½mv² − ½mu² = ΔKE
5.2 Applications of Work-Energy Theorem
- Stopping distance: Braking force × stopping distance = loss in KE
- Kinetic energy from work: If you push a block and do 50 J of work, its KE increases by 50 J
- Variable force: For variable forces, using W = ΔKE is often easier than integrating
Numerical — Stopping distance:
A car of mass 1000 kg moving at 20 m/s applies brakes. Friction force = 5000 N. Find stopping distance.
Solution:
KE = ½ × 1000 × 400 = 2,00,000 J
Work done by friction = −Fs (negative work, force opposite to motion)
W_net = ΔKE → −Fs = 0 − KE → −5000 × s = −2,00,000
s = 2,00,000 / 5000 = 40 m
A car of mass 1000 kg moving at 20 m/s applies brakes. Friction force = 5000 N. Find stopping distance.
Solution:
KE = ½ × 1000 × 400 = 2,00,000 J
Work done by friction = −Fs (negative work, force opposite to motion)
W_net = ΔKE → −Fs = 0 − KE → −5000 × s = −2,00,000
s = 2,00,000 / 5000 = 40 m
🔥 AMU PYQ: The kinetic energy of a body of mass 10 kg moving with velocity 4 m/s is:
(A) 20 J (B) 40 J (C) 80 J (D) 160 J
Ans: C (KE = ½ × 10 × 16 = 80 J)
🔥 AMU PYQ: Work-energy theorem states that work done by net force equals change in:
(A) Potential energy (B) Kinetic energy (C) Mechanical energy (D) Momentum
Ans: B
(A) 20 J (B) 40 J (C) 80 J (D) 160 J
Ans: C (KE = ½ × 10 × 16 = 80 J)
🔥 AMU PYQ: Work-energy theorem states that work done by net force equals change in:
(A) Potential energy (B) Kinetic energy (C) Mechanical energy (D) Momentum
Ans: B
6) Law of Conservation of Energy
Statement: Energy can neither be created nor destroyed — it can only be transformed from one form to another. The total energy of an isolated system always remains constant.
6.1 Conservation of Mechanical Energy (Special Case)
In the absence of non-conservative forces (friction, air resistance, viscous drag), the sum of kinetic energy and potential energy (mechanical energy) of a body remains constant.
E = KE + PE = constant½mv² + mgh = constantKE₁ + PE₁ = KE₂ + PE₂Energy can transform from KE ↔ PE, but total stays same.
6.2 Proof for a Freely Falling Body
Consider a body of mass m dropped from height h. Let's calculate total energy at three positions:
| Position | Height | Velocity | KE = ½mv² | PE = mgh | Total E = KE + PE |
|---|---|---|---|---|---|
| At top (just released) | h | 0 | 0 | mgh | mgh |
| At height x from ground | x | v = √[2g(h−x)] | ½m·2g(h−x) = mg(h−x) | mgx | mg(h−x) + mgx = mgh |
| Just before hitting ground | 0 | v = √(2gh) | ½m·2gh = mgh | 0 | mgh |
🟡 Key observation: Total mechanical energy = mgh at all points. Energy is just converted from PE to KE as the body falls. This is the Law of Conservation of Mechanical Energy in action!
6.3 Examples of Energy Conservation
- Simple Pendulum: At extremes, PE=max, KE=0; at mean position, KE=max, PE=min; total energy constant
- Roller Coaster: At top, PE=max, KE=min; at bottom, KE=max, PE=min; energy transforms continuously
- Bouncing Ball: In ideal case (no energy loss), ball returns to same height; in real case, some energy lost as heat/sound
- Hydroelectric Dam: PE of stored water → KE of flowing water → KE of turbine → Electrical energy
🔥 AMU PYQ: A body of mass 5 kg is thrown vertically upward with a speed of 10 m/s. Its kinetic energy at the highest point is:
(A) 250 J (B) 0 (C) 50 J (D) 125 J
Ans: B (0 — at highest point, v = 0, so KE = 0)
🔥 AMU PYQ: A ball of mass 2 kg is dropped from a height of 20 m. Its velocity just before hitting ground is (g = 10 m/s²):
Ans: Using energy conservation: mgh = ½mv² → v = √(2gh) = √(2×10×20) = √400 = 20 m/s
🔥 AMU PYQ: In a simple pendulum, the total mechanical energy is maximum at:
(A) Mean position (B) Extreme position (C) Same at all positions (D) Zero at extremes
Ans: C (conservation of mechanical energy — same everywhere)
(A) 250 J (B) 0 (C) 50 J (D) 125 J
Ans: B (0 — at highest point, v = 0, so KE = 0)
🔥 AMU PYQ: A ball of mass 2 kg is dropped from a height of 20 m. Its velocity just before hitting ground is (g = 10 m/s²):
Ans: Using energy conservation: mgh = ½mv² → v = √(2gh) = √(2×10×20) = √400 = 20 m/s
🔥 AMU PYQ: In a simple pendulum, the total mechanical energy is maximum at:
(A) Mean position (B) Extreme position (C) Same at all positions (D) Zero at extremes
Ans: C (conservation of mechanical energy — same everywhere)
7) Power — Rate of Doing Work
Power tells us how fast work is done. Two persons may do the same work, but the one who does it faster has more power. Formally: Power is the rate of doing work or the rate of energy transfer.
P = W / tP = F × v (for constant force — derived: P = W/t = Fs/t = Fv)P = F v cosθ (general form, when F and v have angle θ between them)SI unit:
Watt (W) = J/sDimension:
[M¹L²T⁻³]
7.1 Average vs Instantaneous Power
Average Power = Total work done / Total time taken. P_avg = W_total / t_total
Instantaneous Power = Power at a given instant. P_inst = F · v (where v is instantaneous velocity)
7.2 Units of Power and Energy (Commercial)
| Unit | Conversion | Used for |
|---|---|---|
| 1 Watt (W) | 1 J/s | SI unit of power |
| 1 Horsepower (hp) | 746 W | Engine power, motors |
| 1 Kilowatt (kW) | 1000 W | Electrical appliances |
| 1 Megawatt (MW) | 10⁶ W | Power plants, large generators |
| 1 kilowatt-hour (kWh) | 3.6 × 10⁶ J | Commercial unit of energy (electricity bills) |
⚠️ Common Mistake: kWh is a unit of energy, NOT power! Don't confuse kW (power) with kWh (energy).
kW = kilowatt (power unit), kWh = kilowatt-hour (energy unit)
1 kWh = 1 kW × 1 h = 1000 W × 3600 s = 3.6 × 10⁶ J
kW = kilowatt (power unit), kWh = kilowatt-hour (energy unit)
1 kWh = 1 kW × 1 h = 1000 W × 3600 s = 3.6 × 10⁶ J
7.3 Electricity Bill Calculations
Energy consumed (in kWh) = Power (in kW) × Time (in hours)
Bill amount = Units (kWh) × Rate per unit
Example: A 100 W bulb is used for 10 hours daily. In 30 days, how many units (kWh) are consumed?
Power = 100 W = 0.1 kW, Time = 10 h/day × 30 days = 300 h
Energy = 0.1 × 300 = 30 kWh (30 units)
If rate is Rs 5/unit → bill = 30 × 5 = Rs 150
Bill amount = Units (kWh) × Rate per unit
Example: A 100 W bulb is used for 10 hours daily. In 30 days, how many units (kWh) are consumed?
Power = 100 W = 0.1 kW, Time = 10 h/day × 30 days = 300 h
Energy = 0.1 × 300 = 30 kWh (30 units)
If rate is Rs 5/unit → bill = 30 × 5 = Rs 150
7.4 Numericals on Power
Numerical 1: A motor does 5000 J of work in 25 s. Find power.
Solution:
P = W/t = 5000/25 = 200 W
Solution:
P = W/t = 5000/25 = 200 W
Numerical 2: A 60 kg person climbs stairs of total height 10 m in 20 s (g = 10 m/s²). Find power.
Solution:
Work = mgh = 60 × 10 × 10 = 6000 J
P = 6000/20 = 300 W
Solution:
Work = mgh = 60 × 10 × 10 = 6000 J
P = 6000/20 = 300 W
Numerical 3: A car engine provides a force of 500 N when the car moves at 30 m/s. Find power output in hp.
Solution:
P = Fv = 500 × 30 = 15000 W = 15 kW
In hp: 15000/746 ≈ 20.1 hp
Solution:
P = Fv = 500 × 30 = 15000 W = 15 kW
In hp: 15000/746 ≈ 20.1 hp
Numerical 4: An electric pump lifts 2000 kg of water to a height of 20 m in 10 s. Find power (g = 10). Express in hp.
Solution:
Work = mgh = 2000 × 10 × 20 = 4,00,000 J
P = 4,00,000/10 = 40,000 W = 40 kW
In hp: 40000/746 ≈ 53.6 hp
Solution:
Work = mgh = 2000 × 10 × 20 = 4,00,000 J
P = 4,00,000/10 = 40,000 W = 40 kW
In hp: 40000/746 ≈ 53.6 hp
🔥 AMU PYQ: The SI unit of power is:
(A) Joule (B) Newton (C) Watt (D) Pascal
Ans: C
🔥 AMU PYQ: 1 kWh = ? J
(A) 3.6 × 10⁵ (B) 3.6 × 10⁶ (C) 3.6 × 10⁷ (D) 3.6 × 10⁴
Ans: B
🔥 AMU PYQ: A 2 kW motor works for 5 hours. The energy consumed in kWh is:
(A) 2 (B) 5 (C) 10 (D) 20
Ans: C (Energy = 2 × 5 = 10 kWh)
🔥 AMU PYQ: Which of the following is NOT a unit of power?
(A) Watt (B) J/s (C) kWh (D) hp
Ans: C (kWh is energy, not power)
(A) Joule (B) Newton (C) Watt (D) Pascal
Ans: C
🔥 AMU PYQ: 1 kWh = ? J
(A) 3.6 × 10⁵ (B) 3.6 × 10⁶ (C) 3.6 × 10⁷ (D) 3.6 × 10⁴
Ans: B
🔥 AMU PYQ: A 2 kW motor works for 5 hours. The energy consumed in kWh is:
(A) 2 (B) 5 (C) 10 (D) 20
Ans: C (Energy = 2 × 5 = 10 kWh)
🔥 AMU PYQ: Which of the following is NOT a unit of power?
(A) Watt (B) J/s (C) kWh (D) hp
Ans: C (kWh is energy, not power)
8) Collisions — Elastic & Inelastic
When two objects hit each other, it's called a collision. In all collisions, momentum is always conserved (if no external force acts). The difference lies in whether kinetic energy is also conserved.
8.1 Types of Collisions
| Feature | Elastic Collision | Inelastic Collision | Perfectly Inelastic Collision |
|---|---|---|---|
| Kinetic Energy conserved? | Yes ✓ (100%) | No ✗ (some KE → heat, sound, deformation) | No ✗ (maximum KE loss) |
| Momentum conserved? | Yes ✓ | Yes ✓ | Yes ✓ |
| Bodies separate after? | Yes (bounce off) | Yes (but deformed) | No (they stick together) |
| Coefficient of restitution (e) | e = 1 | 0 < e < 1 | e = 0 |
| Example | Two billiard balls colliding | Car crash (some damage, bounce apart) | Bullet hitting and embedding in a block |
1D Elastic Collision Formulas (Class 10 level):
Special cases (head-on, equal masses, elastic):
If m₁ = m₂, then they exchange velocities:
Momentum conservation: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂KE conservation: ½m₁u₁² + ½m₂u₂² = ½m₁v₁² + ½m₂v₂²Special cases (head-on, equal masses, elastic):
If m₁ = m₂, then they exchange velocities:
v₁ = u₂ and v₂ = u₁
Perfectly Inelastic Collision (1D):
KE lost = ½m₁u₁² + ½m₂u₂² − ½(m₁ + m₂)v_common²
m₁u₁ + m₂u₂ = (m₁ + m₂) × v_commonv_common = (m₁u₁ + m₂u₂) / (m₁ + m₂)KE lost = ½m₁u₁² + ½m₂u₂² − ½(m₁ + m₂)v_common²
🔥 AMU PYQ: In an elastic collision, which is conserved?
(A) Only momentum (B) Only KE (C) Both momentum and KE (D) Neither
Ans: C
🔥 AMU PYQ: In an elastic collision between two equal masses, after head-on collision they:
(A) Stick together (B) Exchange velocities (C) Stop (D) Move in same direction
Ans: B
🔥 AMU PYQ: A bullet of mass 50 g moving at 200 m/s hits a 2 kg block and gets embedded. The common velocity after collision is:
Ans: m₁u₁ + m₂u₂ = (m₁+m₂)v → 0.05×200 + 0 = 2.05v → v = 10/2.05 ≈ 4.88 m/s
(A) Only momentum (B) Only KE (C) Both momentum and KE (D) Neither
Ans: C
🔥 AMU PYQ: In an elastic collision between two equal masses, after head-on collision they:
(A) Stick together (B) Exchange velocities (C) Stop (D) Move in same direction
Ans: B
🔥 AMU PYQ: A bullet of mass 50 g moving at 200 m/s hits a 2 kg block and gets embedded. The common velocity after collision is:
Ans: m₁u₁ + m₂u₂ = (m₁+m₂)v → 0.05×200 + 0 = 2.05v → v = 10/2.05 ≈ 4.88 m/s
9) Simple Machines — MA & Efficiency
Simple machines make work easier by multiplying force or changing its direction. For AMU, you need to know lever, pulley, and inclined plane.
9.1 Lever
A lever is a rigid bar that pivots about a fulcrum. Based on the relative positions of fulcrum (F), load (L), and effort (E), there are three classes.
Mechanical Advantage (MA) = Load / Effort = Effort arm / Load armVelocity Ratio (VR) = Distance moved by effort / Distance moved by loadEfficiency = (MA / VR) × 100%For an ideal machine (no friction): MA = VR, Efficiency = 100%
| Type | F-L-E arrangement | Example |
|---|---|---|
| Class 1 | Fulcrum between Load and Effort | See-saw, scissors, crowbar |
| Class 2 | Load between Fulcrum and Effort | Wheelbarrow, nutcracker |
| Class 3 | Effort between Fulcrum and Load | Tweezers, fishing rod, broom |
9.2 Pulley
Single fixed pulley:
MA = 1, VR = 1 — only changes direction of effort (pulling down instead of lifting up)
Single movable pulley:
MA = 2, VR = 2 — load is supported by 2 rope segments
Block and tackle system (multiple pulleys):
MA = Number of rope segments supporting the load
MA = 1, VR = 1 — only changes direction of effort (pulling down instead of lifting up)
Single movable pulley:
MA = 2, VR = 2 — load is supported by 2 rope segments
Block and tackle system (multiple pulleys):
MA = Number of rope segments supporting the load
9.3 Inclined Plane
MA = Length of incline (L) / Height of incline (h)Effort required = (Load × h) / LExample: To lift a 100 kg load to height 1 m using a 5 m ramp:
Effort = (1000 N × 1) / 5 = 200 N (5× less effort!)
MA = 5
10) PYQ Battle — AMU Past Questions
Q1. The work done in lifting a body of mass 10 kg to a height of 5 m (g = 10 m/s²) is:
(A) 50 J (B) 100 J (C) 500 J (D) 1000 J
Ans: C (W = mgh = 10×10×5 = 500 J)
(A) 50 J (B) 100 J (C) 500 J (D) 1000 J
Ans: C (W = mgh = 10×10×5 = 500 J)
Q2. A body of mass 2 kg moves with velocity 3 m/s. Its kinetic energy is:
(A) 3 J (B) 6 J (C) 9 J (D) 18 J
Ans: C (KE = ½×2×9 = 9 J)
(A) 3 J (B) 6 J (C) 9 J (D) 18 J
Ans: C (KE = ½×2×9 = 9 J)
Q3. A force of 20 N acts on a body and displaces it by 5 m at 60° to the force. Work done is:
(A) 50 J (B) 100 J (C) 50√3 J (D) 25 J
Ans: A (W = 20×5×cos60° = 20×5×½ = 50 J)
(A) 50 J (B) 100 J (C) 50√3 J (D) 25 J
Ans: A (W = 20×5×cos60° = 20×5×½ = 50 J)
Q4. The power of a motor that does 6000 J of work in 2 minutes is:
(A) 50 W (B) 100 W (C) 3000 W (D) 12000 W
Ans: A (P = 6000/120 = 50 W)
(A) 50 W (B) 100 W (C) 3000 W (D) 12000 W
Ans: A (P = 6000/120 = 50 W)
Q5. A body of mass 5 kg is raised to a height of 10 m. The potential energy gained is (g = 10 m/s²):
(A) 50 J (B) 100 J (C) 250 J (D) 500 J
Ans: D (PE = 5×10×10 = 500 J)
(A) 50 J (B) 100 J (C) 250 J (D) 500 J
Ans: D (PE = 5×10×10 = 500 J)
Q6. If the speed of a body is doubled, its kinetic energy becomes:
(A) Half (B) Double (C) Four times (D) Eight times
Ans: C (KE ∝ v²)
(A) Half (B) Double (C) Four times (D) Eight times
Ans: C (KE ∝ v²)
Q7. A body of mass 50 kg has a momentum of 200 kg m/s. Its KE is:
(A) 200 J (B) 400 J (C) 600 J (D) 800 J
Ans: B (KE = p²/2m = 40000/100 = 400 J)
(A) 200 J (B) 400 J (C) 600 J (D) 800 J
Ans: B (KE = p²/2m = 40000/100 = 400 J)
Q8. A boy of mass 40 kg runs up stairs of 20 steps each 15 cm high in 10 s. Power developed (g = 10 m/s²):
(A) 120 W (B) 240 W (C) 60 W (D) 30 W
Ans: A (h = 20×0.15 = 3 m, W = mgh = 40×10×3 = 1200 J, P = 1200/10 = 120 W)
(A) 120 W (B) 240 W (C) 60 W (D) 30 W
Ans: A (h = 20×0.15 = 3 m, W = mgh = 40×10×3 = 1200 J, P = 1200/10 = 120 W)
Q9. When a body falls freely, its total mechanical energy:
(A) Increases (B) Decreases (C) Remains constant (D) Becomes zero
Ans: C (conservation of mechanical energy)
(A) Increases (B) Decreases (C) Remains constant (D) Becomes zero
Ans: C (conservation of mechanical energy)
Q10. Which of the following is NOT a unit of energy?
(A) Joule (B) kWh (C) Watt (D) erg
Ans: C (Watt is unit of power, not energy)
(A) Joule (B) kWh (C) Watt (D) erg
Ans: C (Watt is unit of power, not energy)
Q11. Work done by centripetal force in circular motion is always:
(A) Positive (B) Negative (C) Zero (D) Depends on speed
Ans: C (force ⟂ displacement → cos90°=0)
(A) Positive (B) Negative (C) Zero (D) Depends on speed
Ans: C (force ⟂ displacement → cos90°=0)
Q12. A coolie carries a load on his head and walks horizontally. Work done against gravity is:
(A) mgh (B) Fs (C) 0 (D) mgh/2
Ans: C (θ = 90° between gravity and horizontal displacement)
(A) mgh (B) Fs (C) 0 (D) mgh/2
Ans: C (θ = 90° between gravity and horizontal displacement)
Q13. A stone of mass 1 kg is thrown with KE = 200 J. Its momentum is:
(A) 10 kg m/s (B) 20 kg m/s (C) 200 kg m/s (D) 400 kg m/s
Ans: B (p = √(2m·KE) = √(2×1×200) = √400 = 20 kg m/s)
(A) 10 kg m/s (B) 20 kg m/s (C) 200 kg m/s (D) 400 kg m/s
Ans: B (p = √(2m·KE) = √(2×1×200) = √400 = 20 kg m/s)
Q14. If the KE of a body becomes 4 times, its momentum becomes:
(A) 2 times (B) 4 times (C) 8 times (D) 16 times
Ans: A (KE ∝ p² → if KE quadruples, p doubles)
(A) 2 times (B) 4 times (C) 8 times (D) 16 times
Ans: A (KE ∝ p² → if KE quadruples, p doubles)
Q15. A 100 W bulb is used for 10 h. Energy consumed in kWh:
(A) 0.1 (B) 1 (C) 10 (D) 100
Ans: B (Energy = 0.1 kW × 10 h = 1 kWh)
(A) 0.1 (B) 1 (C) 10 (D) 100
Ans: B (Energy = 0.1 kW × 10 h = 1 kWh)
Q16. Which simple machine has MA always greater than 1?
(A) Single fixed pulley (B) Single movable pulley (C) Class 3 lever (D) None
Ans: B (MA = 2)
(A) Single fixed pulley (B) Single movable pulley (C) Class 3 lever (D) None
Ans: B (MA = 2)
11) Practice Problems — 15+ Numericals with Solutions
Practice 1: A force of 100 N acts on a body at 30° to the horizontal and displaces it 20 m. Find work done.
Solution: W = Fs cosθ = 100 × 20 × cos30° = 2000 × (√3/2) = 2000 × 0.866 = 1732 J
Solution: W = Fs cosθ = 100 × 20 × cos30° = 2000 × (√3/2) = 2000 × 0.866 = 1732 J
Practice 2: A 2 kg block is pulled with 50 N force at 60° for 4 m on a frictionless surface. Find work done by applied force and final velocity.
Solution: W = 50 × 4 × cos60° = 200 × 0.5 = 100 J. By WET: 100 = ½×2×v² → v² = 100 → v = 10 m/s
Solution: W = 50 × 4 × cos60° = 200 × 0.5 = 100 J. By WET: 100 = ½×2×v² → v² = 100 → v = 10 m/s
Practice 3: How much work is done by a person lifting a 30 kg suitcase 2 m high (g = 9.8 m/s²)?
Solution: W = mgh = 30 × 9.8 × 2 = 588 J
Solution: W = mgh = 30 × 9.8 × 2 = 588 J
Practice 4: A car of 1200 kg moving at 54 km/h (15 m/s) brakes to stop. If braking force is 6000 N, find stopping distance.
Solution: KE = ½×1200×225 = 1,35,000 J. Work = Fs → 6000 × s = 1,35,000 → s = 22.5 m
Solution: KE = ½×1200×225 = 1,35,000 J. Work = Fs → 6000 × s = 1,35,000 → s = 22.5 m
Practice 5: A spring of k = 500 N/m is compressed by 0.05 m. Find elastic PE stored. If released against a 0.2 kg mass, find max speed of mass.
Solution: PE = ½×500×(0.05)² = 0.625 J. By COE: 0.625 = ½×0.2×v² → v² = 6.25 → v = 2.5 m/s
Solution: PE = ½×500×(0.05)² = 0.625 J. By COE: 0.625 = ½×0.2×v² → v² = 6.25 → v = 2.5 m/s
Practice 6: A body of mass 500 g has KE of 100 J. Find its velocity and momentum.
Solution: m = 0.5 kg. v = √(2KE/m) = √(200/0.5) = √400 = 20 m/s. p = mv = 0.5 × 20 = 10 kg m/s
Solution: m = 0.5 kg. v = √(2KE/m) = √(200/0.5) = √400 = 20 m/s. p = mv = 0.5 × 20 = 10 kg m/s
Practice 7: A 60 W fan runs for 12 hours daily. Find energy consumed in 30 days in kWh and J.
Solution: Power = 0.06 kW. Daily energy = 0.06 × 12 = 0.72 kWh. 30 days = 0.72 × 30 = 21.6 kWh. In J: 21.6 × 3.6×10⁶ = 7.78 × 10⁷ J
Solution: Power = 0.06 kW. Daily energy = 0.06 × 12 = 0.72 kWh. 30 days = 0.72 × 30 = 21.6 kWh. In J: 21.6 × 3.6×10⁶ = 7.78 × 10⁷ J
Practice 8: A pump lifts 500 kg of water to height 15 m in 30 s. Power (g = 10).
Solution: Work = 500×10×15 = 75000 J. P = 75000/30 = 2500 W = 2.5 kW ≈ 3.35 hp
Solution: Work = 500×10×15 = 75000 J. P = 75000/30 = 2500 W = 2.5 kW ≈ 3.35 hp
Practice 9: A bullet of 20 g moving at 300 m/s hits a 3 kg block at rest and gets embedded. Find common velocity and KE lost.
Solution: v = (0.02×300)/(3.02) = 6/3.02 = 1.987 m/s. KE_initial = ½×0.02×90000 = 900 J. KE_final = ½×3.02×(1.987)² ≈ 5.96 J. KE lost = 894 J
Solution: v = (0.02×300)/(3.02) = 6/3.02 = 1.987 m/s. KE_initial = ½×0.02×90000 = 900 J. KE_final = ½×3.02×(1.987)² ≈ 5.96 J. KE lost = 894 J
Practice 10: Two identical balls of mass 0.2 kg moving toward each other at 5 m/s and 3 m/s collide elastically head-on. Find final velocities.
Solution: For elastic equal mass collision, they exchange velocities. Ball 1 (5 m/s → 3 m/s), Ball 2 (3 m/s → 5 m/s) but opposite directions. Final: v₁ = 3 m/s (opposite direction), v₂ = 5 m/s (opposite direction).
Solution: For elastic equal mass collision, they exchange velocities. Ball 1 (5 m/s → 3 m/s), Ball 2 (3 m/s → 5 m/s) but opposite directions. Final: v₁ = 3 m/s (opposite direction), v₂ = 5 m/s (opposite direction).
Practice 11: A 10 kg object is at height 20 m. If it falls freely, find KE and PE at height 5 m (g = 10).
Solution: At h=5m: PE = 10×10×5 = 500 J. Total E = mgh_initial = 10×10×20 = 2000 J. KE = 2000 − 500 = 1500 J. v = √(2×1500/10) = √300 ≈ 17.32 m/s
Solution: At h=5m: PE = 10×10×5 = 500 J. Total E = mgh_initial = 10×10×20 = 2000 J. KE = 2000 − 500 = 1500 J. v = √(2×1500/10) = √300 ≈ 17.32 m/s
Practice 12: A man pushes a 50 kg box up a 30° incline of length 10 m with force parallel to incline. Find work done against gravity and total work if friction is absent.
Solution: Height = 10 × sin30° = 5 m. Work against gravity = mgh = 50×10×5 = 2500 J. Net work = same = 2500 J (stored as PE).
Solution: Height = 10 × sin30° = 5 m. Work against gravity = mgh = 50×10×5 = 2500 J. Net work = same = 2500 J (stored as PE).
Practice 13: A 1.5 kW motor runs for 2 hours. Calculate energy consumed in kWh and J. If rate is Rs 8 per kWh, find bill.
Solution: Energy = 1.5 × 2 = 3 kWh. In J: 3 × 3.6×10⁶ = 1.08×10⁷ J. Bill = 3 × 8 = Rs 24
Solution: Energy = 1.5 × 2 = 3 kWh. In J: 3 × 3.6×10⁶ = 1.08×10⁷ J. Bill = 3 × 8 = Rs 24
Practice 14: A body of mass 4 kg has 72 J KE. Find its speed. If speed is halved, find new KE.
Solution: v = √(2×72/4) = √36 = 6 m/s. New v = 3 m/s. New KE = ½×4×9 = 18 J (KE becomes ¼ when speed halves).
Solution: v = √(2×72/4) = √36 = 6 m/s. New v = 3 m/s. New KE = ½×4×9 = 18 J (KE becomes ¼ when speed halves).
Practice 15: A block is dragged 5 m on a horizontal surface with F = 40 N at θ = 0°. Friction = 15 N. Find work done by applied force, friction, and net work. Also find final KE if initial KE = 0.
Solution: W_applied = 40×5×1 = 200 J. W_friction = −15×5 = −75 J. W_net = 200 − 75 = 125 J. By WET: ΔKE = 125 J, so final KE = 125 J
Solution: W_applied = 40×5×1 = 200 J. W_friction = −15×5 = −75 J. W_net = 200 − 75 = 125 J. By WET: ΔKE = 125 J, so final KE = 125 J
Practice 16: Using an inclined plane of length 4 m and height 1 m, a load of 200 kg is raised. Find MA, effort required, and VR (g = 10).
Solution: MA = L/h = 4/1 = 4. Load = 2000 N. Effort = Load/MA = 2000/4 = 500 N. VR = 4 (same as MA if ideal).
Solution: MA = L/h = 4/1 = 4. Load = 2000 N. Effort = Load/MA = 2000/4 = 500 N. VR = 4 (same as MA if ideal).
12) Common Mistakes & Tricks
⚠️ Top common mistakes in AMU exam:
Mistake 1: Forgetting cosθ in work formula
→ Many students use W = Fs (which is only true when θ = 0°). Always check the angle!
Mistake 2: Work sign confusion
→ Work done by a force (like gravity) is different from work done against a force. "By gravity" = +mgh (downward). "Against gravity" = +mgh (upward). These have opposite signs.
Mistake 3: Double speed → double KE?
→ NO! KE ∝ v², so double speed = 4× KE. Triple speed = 9× KE!
Mistake 4: Confusing kW vs kWh
→ kW is power (rate), kWh is energy (quantity). Electricity bills charge for kWh (energy), not kW.
Mistake 5: Power numerical time conversion
→ Always convert minutes → seconds before using P = W/t. For example, 2 minutes = 120 s, not 2 s.
Mistake 6: Conservation of energy with friction
→ Energy is ALWAYS conserved (total). But mechanical energy is NOT conserved if friction is present (some converts to heat).
Mistake 1: Forgetting cosθ in work formula
→ Many students use W = Fs (which is only true when θ = 0°). Always check the angle!
Mistake 2: Work sign confusion
→ Work done by a force (like gravity) is different from work done against a force. "By gravity" = +mgh (downward). "Against gravity" = +mgh (upward). These have opposite signs.
Mistake 3: Double speed → double KE?
→ NO! KE ∝ v², so double speed = 4× KE. Triple speed = 9× KE!
Mistake 4: Confusing kW vs kWh
→ kW is power (rate), kWh is energy (quantity). Electricity bills charge for kWh (energy), not kW.
Mistake 5: Power numerical time conversion
→ Always convert minutes → seconds before using P = W/t. For example, 2 minutes = 120 s, not 2 s.
Mistake 6: Conservation of energy with friction
→ Energy is ALWAYS conserved (total). But mechanical energy is NOT conserved if friction is present (some converts to heat).
✅ Tricks to score full marks:
• cos0° = 1, cos90° = 0, cos180° = −1 — write these at the top of your rough sheet
• "Constant speed" or "no acceleration" → net force = 0 → net work = 0
• For "just before hitting ground" problems, use v = √(2gh) directly
• For pendulum: total energy = mgl(1−cosθ) at angle θ from vertical
• For spring: max KE = max elastic PE = ½kA² where A = amplitude
• In collision problems: always write momentum conservation equation FIRST
• 1 hp ≈ 746 W — remember 746, not 750!
• cos0° = 1, cos90° = 0, cos180° = −1 — write these at the top of your rough sheet
• "Constant speed" or "no acceleration" → net force = 0 → net work = 0
• For "just before hitting ground" problems, use v = √(2gh) directly
• For pendulum: total energy = mgl(1−cosθ) at angle θ from vertical
• For spring: max KE = max elastic PE = ½kA² where A = amplitude
• In collision problems: always write momentum conservation equation FIRST
• 1 hp ≈ 746 W — remember 746, not 750!
13) Important Formulae — At a Glance
| Quantity | Formula | SI Unit |
|---|---|---|
| Work | W = F s cosθ | J (N m) |
| Kinetic Energy | KE = ½ mv² | J |
| KE in terms of momentum | KE = p² / 2m | J |
| Gravitational PE | PE = mgh | J |
| Elastic PE (spring) | PE = ½ kx² | J |
| Work-Energy Theorem | W_net = ΔKE = ½mv² − ½mu² | J |
| Conservation of Energy | KE₁ + PE₁ = KE₂ + PE₂ | — |
| Power | P = W/t = Fv | W (J/s) |
| 1 HP | = 746 W | — |
| 1 kWh | = 3.6 × 10⁶ J | — |
| 1 J in CGS | 1 J = 10⁷ erg | — |
| Momentum | p = mv | kg m/s |
| Elastic collision (equal masses) | v₁ = u₂, v₂ = u₁ | — |
| Perfectly inelastic (common v) | v = (m₁u₁+m₂u₂)/(m₁+m₂) | m/s |
| Simple machine MA | MA = Load / Effort | — |
| Lever MA | MA = Effort arm / Load arm | — |
| Inclined plane MA | MA = L / h | — |
14) Quick Revision — Power-packed One-liners
- Work = F·s·cosθ. Unit: Joule (J). Scalar quantity.
- Positive work: θ between 0° and 90° (cosθ > 0)
- Zero work: θ = 90° (perpendicular) or s = 0 (no displacement)
- Negative work: θ between 90° and 180° (cosθ < 0)
- KE = ½mv². If v doubles → KE quadruples. If v triples → KE becomes 9×.
- KE and momentum: KE = p²/2m, p = √(2m·KE)
- Gravitational PE = mgh (depends on reference level)
- Elastic PE (spring) = ½kx² (x = extension/compression)
- Work-Energy Theorem: Net work = change in KE
- Law of Conservation of Energy: Energy neither created nor destroyed; only transformed
- Conservation of Mechanical Energy: KE + PE = constant (no non-conservative forces)
- Falling body: v = √(2gh), KE just before ground = mgh
- Power = W/t = Fv. Unit: Watt (W)
- 1 HP = 746 W. 1 kW = 1000 W. 1 MW = 10⁶ W
- 1 kWh = 3.6 × 10⁶ J (commercial unit of energy — used in electricity bills)
- Elastic collision: KE + momentum conserved. e = 1.
- Inelastic collision: Only momentum conserved. Some KE lost.
- Perfectly inelastic: bodies stick together, maximum KE loss.
- Equal mass elastic collision head-on → velocities exchange
- 1 J = 10⁷ erg
- Single movable pulley: MA = 2. Inclined plane: MA = L/h.
- Efficiency = (MA/VR) × 100%. Ideal machine: efficiency = 100%.
✅ Work, Power & Energy — COMPLETE. Remember: Energy is never created or destroyed — it only transforms. Master the cosθ sign convention, the Work-Energy Theorem, and the energy conservation equation. These three will solve 90% of AMU questions from this chapter! Next up: Pressure & Floatation!