Physics Chapter Work power & energy
WORK, POWER AND ENERGY
FSc Pre-Medical | Pakistan Armed Forces Nursing Service
Chapter Outline
- Work Done by a Force
- Kinetic and Potential Energy
- Law of Conservation of Energy
- Power
- Efficiency
- Simple Machines
- Practice MCQs (75 Questions)
- Short Practice Test
1:Work Done by a Force
In Physics, work is done only when a force causes a displacement in the direction of the force. Simply applying a force without movement = zero work.
1.1 Definition and Formula
KEY CONCEPT: Work W = F · d · cosθ, where F = force (N), d = displacement (m), θ = angle between force and displacement. SI unit: Joule (J) = N·m.
| Angle θ | Work Done | Situation |
| θ = 0° | W = Fd (maximum) | Force and displacement in same direction |
| θ = 90° | W = 0 (zero work) | Force perpendicular to displacement |
| θ = 180° | W = −Fd (negative) | Force opposite to displacement (friction) |
| 0° < θ < 90° | W = Fd cosθ (positive) | Force at angle to displacement |
EXAMPLE: A force 50 N at 60° to horizontal pulls a box 10 m. W = 50 × 10 × cos60° = 50 × 10 × 0.5 = 250 J.
QUICK TIP: Zero work examples: carrying a bag horizontally (weight perpendicular to motion), a satellite in circular orbit (gravity perpendicular to velocity).
1.2 Work Done Against Gravity
KEY CONCEPT: Work done to lift object = mgh, where m = mass, g = 9.8 m/s², h = height gained. This work is stored as gravitational PE.
EXAMPLE: Lifting a 5 kg box 3 m high. W = mgh = 5 × 10 × 3 = 150 J.
1.3 Work Done by Friction
Friction always acts opposite to displacement, so θ = 180°, cosθ = −1. Work done by friction is NEGATIVE (energy removed from system).
EXAMPLE: A block slides 4 m. Friction force = 20 N. Work by friction = 20 × 4 × cos180° = −80 J. (Energy lost as heat.)
- Kinetic and Potential Energy
2.1 Kinetic Energy (KE)
KEY CONCEPT: Kinetic Energy KE = ½mv². It is the energy of motion. SI unit: Joule (J). KE is always positive (scalar).
- KE increases when speed increases
- If speed doubles → KE quadruples (because of v²)
- KE = Work done to bring object to that speed from rest
EXAMPLE: A 2 kg ball moving at 5 m/s. KE = ½ × 2 × 5² = ½ × 2 × 25 = 25 J.
QUICK TIP: AFNS trap: If speed doubles, KE becomes 4 times (not 2 times). If speed triples → KE becomes 9 times. Always square the speed!
2.2 Work-Energy Theorem
KEY CONCEPT: Net work done on a body = change in kinetic energy. W_net = ΔKE = ½mv² − ½mu²
EXAMPLE: A car (mass 1000 kg) accelerates from 10 m/s to 20 m/s. ΔKE = ½×1000×400 − ½×1000×100 = 200,000 − 50,000 = 150,000 J = 150 kJ.
2.3 Gravitational Potential Energy (GPE)
KEY CONCEPT: GPE = mgh, where h is height above reference level. It is the energy stored due to position. SI unit: Joule (J).
- GPE increases when height increases
- GPE = 0 at reference level (usually ground)
- GPE can be negative if below reference level
EXAMPLE: A 3 kg book placed on a shelf 2 m high. GPE = mgh = 3 × 10 × 2 = 60 J.
2.4 Elastic Potential Energy
KEY CONCEPT: Elastic PE = ½kx², where k = spring constant (N/m), x = extension or compression. Energy stored in stretched/compressed spring.
- k is the stiffness of the spring (Hooke’s Law: F = kx)
- More extension → more energy stored (x² relationship)
EXAMPLE: Spring with k = 200 N/m compressed by 0.1 m. PE = ½ × 200 × 0.01 = 1 J.
- Law of Conservation of Energy
KEY CONCEPT: Energy cannot be created or destroyed. It can only be transformed from one form to another. Total energy of an isolated system remains constant.
3.1 Conservation in Mechanical Systems
For a falling body (no friction): Total Mechanical Energy = KE + PE = constant.
- As object falls: PE decreases, KE increases
- At highest point: all PE, zero KE
- Just before hitting ground: all KE, zero PE
- At any point: KE + PE = constant = initial total energy
EXAMPLE: Ball dropped from 20 m (m=1 kg, g=10). At 20 m: PE=200J, KE=0. At 10 m: PE=100J, KE=100J. At ground: PE=0, KE=200J. Total = 200J always.
QUICK TIP: To find speed at any point: ½mv² = mgh₁ − mgh₂ → v = √(2g(h₁−h₂)). This uses conservation of energy directly.
3.2 Forms of Energy
| Form | Description | Example |
| Kinetic | Energy of motion | Moving car, flowing water |
| Gravitational PE | Energy due to height | Water in dam, raised hammer |
| Elastic PE | Energy in deformed elastic | Stretched spring, bent bow |
| Chemical | Energy in chemical bonds | Food, fuel, batteries |
| Thermal | Energy due to temperature | Hot water, steam |
| Nuclear | Energy in atomic nucleus | Nuclear power plant |
| Electrical | Energy of electric charges | Current in wire |
| Light/Radiant | Energy of electromagnetic waves | Sunlight, X-rays |
- Power
KEY CONCEPT: Power P = Work done / Time = W/t. SI unit: Watt (W) = J/s. Power measures how FAST work is done.
4.1 Formula and Units
- P = W/t = Fd cosθ / t
- Also: P = F × v (power = force × velocity, when force is in direction of motion)
- 1 Watt = 1 Joule per second
- 1 kilowatt (kW) = 1000 W
- 1 Megawatt (MW) = 10⁶ W
- 1 horsepower (hp) = 746 W ≈ 750 W
EXAMPLE: A motor lifts 200 kg box by 5 m in 10 s (g=10). W = mgh = 200×10×5 = 10,000 J. P = W/t = 10,000/10 = 1000 W = 1 kW.
QUICK TIP: AFNS shortcut: P = Fv is very useful. If a car moves at constant speed v against friction F, then P = Fv. No need to find work or time separately.
4.2 Kilowatt-Hour (kWh)
KEY CONCEPT: 1 kWh = energy used by a 1 kW device in 1 hour = 1000 × 3600 = 3.6 × 10⁶ J = 3.6 MJ.
kWh is a unit of energy (not power). It is used by electricity companies to measure electrical energy consumption.
EXAMPLE: A 2 kW heater runs for 3 hours. Energy used = Power × time = 2 × 3 = 6 kWh = 6 × 3.6 × 10⁶ = 2.16 × 10⁷ J.
Q50. A girl of mass 50 kg climbs 3 m stairs in 6 s (g=10). Power =
- (A) 100 W
- (B) 150 W
- (C) 250 W
- (D) 300 W
Answer: (C) 250 W
Explanation: W = mgh = 50×10×3=1500 J. P = W/t = 1500/6 = 250 W.
Q52. Which uses more power: lift 20 kg by 5 m in 10 s OR 10 kg by 5 m in 5 s? (g=10)
- (A) First case
- (B) Second case
- (C) Equal
- (D) Cannot determine
Answer: (C) Equal
Explanation: P₁ = (20×10×5)/10 = 100 W. P₂ = (10×10×5)/5 = 100 W. Equal!
Q53. Average power when 1000 J work done in 25 s:
- (A) 25,000 W
- (B) 40 W
- (C) 25 W
- (D) 4 W
Answer: (B) 40 W
Explanation: P = W/t = 1000/25 = 40 W.
Q54. Energy efficiency can be expressed as:
- (A) Output/Input × 100
- (B) Input/Output × 100
- (C) (Input−Output)/Input
- (D) Output × Input
Answer: (A) Output/Input × 100
Explanation: Efficiency = (useful output / total input) × 100%.
Q55. A machine has MA=3 and VR=4. Efficiency =
- (A) 75%
- (B) 133%
- (C) 50%
- (D) 25%
Answer: (A) 75%
Explanation: Efficiency = (MA/VR) × 100 = (3/4) × 100 = 75%.
Section D: Energy Conservation & Machines (Q56–75)
Q56. The law of conservation of energy states:
- (A) Energy can be created
- (B) Energy can be destroyed
- (C) Energy transforms but total remains constant
- (D) KE equals PE always
Answer: (C) Energy transforms but total remains constant
Explanation: Energy cannot be created or destroyed — only converted from one form to another.
Q57. At the midpoint of a falling body, which is true?
- (A) KE > PE
- (B) KE < PE
- (C) KE = PE
- (D) KE = 0
Answer: (C) KE = PE
Explanation: At midpoint of fall: height = h/2, so PE = mgh/2. KE gained = mgh/2. KE = PE.
Q58. A pendulum at the highest point has:
- (A) maximum KE, zero PE
- (B) zero KE, maximum PE
- (C) equal KE and PE
- (D) maximum of both
Answer: (B) zero KE, maximum PE
Explanation: At highest point of pendulum, velocity = 0 (KE=0) and height is maximum (PE=max).
Q59. Mechanical advantage of a lever with effort arm 60 cm and load arm 20 cm:
- (A) 1/3
- (B) 1
- (C) 3
- (D) 6
Answer: (C) 3
Explanation: MA = effort arm / load arm = 60/20 = 3. (Longer effort arm → more force multiplication.)
Q60. Class 2 lever has load:
- (A) between effort and fulcrum
- (B) at the fulcrum
- (C) beyond the effort
- (D) between fulcrum and effort
Answer: (D) between fulcrum and effort
Explanation: Class 2: fulcrum at one end, effort at other end, load in between. Example: wheelbarrow.
Q61. A single fixed pulley has MA of:
- (A) 0
- (B) 1
- (C) 2
- (D) 4
Answer: (B) 1
Explanation: Single fixed pulley: only changes direction of force, does not multiply force. MA = 1.
Q62. Nuclear energy is converted to electrical energy in:
- (A) solar panels
- (B) windmills
- (C) nuclear power plants
- (D) hydroelectric dams
Answer: (C) nuclear power plants
Explanation: Nuclear → heat → steam → turbine → generator → electrical energy.
Q63. In a hydroelectric plant, energy conversion is:
- (A) Chemical → Electrical
- (B) GPE → Kinetic → Electrical
- (C) Nuclear → Electrical
- (D) Solar → Electrical
Answer: (B) GPE → Kinetic → Electrical
Explanation: Water behind dam has GPE → falls (KE) → spins turbine → generator produces electricity.
Q64. A 60% efficient machine needs 500 J input. Useful output =
- (A) 200 J
- (B) 300 J
- (C) 400 J
- (D) 833 J
Answer: (B) 300 J
Explanation: Output = efficiency × input = 0.60 × 500 = 300 J.
Q65. Hooke’s Law: F = kx means:
- (A) force is independent of extension
- (B) force is proportional to extension
- (C) force equals extension
- (D) force decreases with extension
Answer: (B) force is proportional to extension
Explanation: Hooke’s Law: extension is directly proportional to applied force (within elastic limit).
Q66. Energy in food is primarily:
- (A) kinetic
- (B) chemical
- (C) nuclear
- (D) electrical
Answer: (B) chemical
Explanation: Food stores energy in chemical bonds (glucose, proteins, fats) — released during metabolism.
Q67. An inclined plane of length L and height h has VR =
- (A) h/L
- (B) L/h
- (C) Lh
- (D) L−h
Answer: (B) L/h
Explanation: VR of inclined plane = length/height = L/h. Longer ramp → smaller effort needed.
Q68. A machine with VR=5 and efficiency 60% has MA =
- (A) 1
- (B) 3
- (C) 5
- (D) 8
Answer: (B) 3
Explanation: MA = efficiency × VR = 0.60 × 5 = 3.
Q69. Energy crisis means:
- (A) excess energy production
- (B) shortage of energy resources
- (C) perfect energy efficiency
- (D) zero friction
Answer: (B) shortage of energy resources
Explanation: Energy crisis refers to shortage of conventional energy sources (oil, gas, coal).
Q70. Solar energy is converted to electrical energy using:
- (A) turbines
- (B) generators
- (C) photovoltaic cells
- (D) condensers
Answer: (C) photovoltaic cells
Explanation: Solar panels (photovoltaic cells) convert light energy directly into electrical energy.
Q71. Which energy transformation occurs in a battery-powered torch?
- (A) Chemical→Light→Electrical
- (B) Chemical→Electrical→Light
- (C) Electrical→Chemical→Light
- (D) Light→Chemical→Electrical
Answer: (B) Chemical→Electrical→Light
Explanation: Battery: chemical energy → electrical energy → bulb → light (and some heat).
Q72. A car brakes and stops. Kinetic energy is converted to:
- (A) gravitational PE
- (B) chemical energy
- (C) heat
- (D) nuclear energy
Answer: (C) heat
Explanation: Braking uses friction. KE converts to heat in brakes and tyres. That is why brakes get hot.
Q73. Class 3 lever example is:
- (A) wheelbarrow
- (B) scissors
- (C) tweezers/forceps
- (D) see-saw
Answer: (C) tweezers/forceps
Explanation: Class 3: effort between fulcrum and load. Tweezers: pivot at one end, effort in middle, load at other end.
Q74. The device that converts mechanical energy to electrical energy:
- (A) motor
- (B) battery
- (C) generator
- (D) transformer
Answer: (C) generator
Explanation: Generator converts mechanical energy (rotation) to electrical energy. Motor does the reverse.
Q75. Total mechanical energy of an ideal pendulum is:
- (A) always increasing
- (B) always decreasing
- (C) zero at highest point
- (D) constant throughout
Answer: (D) constant throughout
Explanation: In an ideal (frictionless) pendulum: KE + PE = constant. Energy just transfers between forms.