Physics basics

Physics Basics – Haryana Police Constable (G.A. & Current Affairs)

Purpose: Quick, clear, and exam‑friendly notes on the fundamentals of physics that you are likely to encounter in the General Awareness & Current Affairs paper.

1. Simple Explanation

Physics is the science that studies matter, energy, motion, and the forces that act on them. In the police syllabus you only need the conceptual and practical bits:

2. Key Points (bullet‑style)

2.1 Kinematics (Motion)

• Displacement (s) – change in position (m).
• Velocity (v = \frac{s}{t}) – rate of displacement (m/s).
• Acceleration (a = \frac{\Delta v}{t}) – rate of change of velocity (m/s²).
• Uniformly Accelerated Motion (UAM) equations
  1. (v = u + at)
  2. (s = ut + \frac{1}{2}at^2)
  3. (v^2 = u^2 + 2as)

2.2 Newton’s Laws

2.3 Work, Energy & Power

• Work (W = F \cdot d \cdot \cos\theta) (J).
• Kinetic Energy (KE = \frac{1}{2}mv^2).
• Potential Energy (gravitational) (PE = mgh).
• Power (P = \frac{W}{t}) (W).
• Conservation of Mechanical Energy (ignoring friction).

2.4 Simple Harmonic Motion (SHM)

• Amplitude – maximum displacement.
• Period (T = 2\pi \sqrt{\frac{m}{k}}) (s).
• Frequency (f = \frac{1}{T}) (Hz).
• Real‑world: Police radio oscillators, clock pendulums.

2.5 Waves

• Speed (v = f\lambda).
• Types: Transverse (light, radio), Longitudinal (sound).
• Reflection, Refraction – relevant for radar, sonar.

2.6 Electricity & Magnetism

• Ohm’s Law (V = IR).
• Power in electrical circuits (P = VI = I^2R = \frac{V^2}{R}).
• Magnetic Field (B) – causes force on moving charges: (F = qvB).
• Electromagnetic Induction – basis of police radio, generators.

2.7 Thermodynamics (Brief)

• Temperature, Heat, Thermal Energy – heat in engines, fire safety.
• First Law: (Q = \Delta U + W).

3. Worked Example

Question:
A police vehicle of mass (m = 1500;\text{kg}) accelerates uniformly from rest to (v = 20;\text{m/s}) in (t = 10;\text{s}).

1. Find the acceleration.
2. Determine the force applied (ignoring friction).
3. Compute the work done by the engine.

Solution

1. Acceleration
   [ a = \frac{\Delta v}{t} = \frac{20-0}{10} = 2.0;\text{m/s}^2 ]

2. Force (Newton’s 2nd law)
   [ F = ma = 1500 \times 2.0 = 3000;\text{N} ]

3. Work Done
   Work = Force × Displacement (since force is constant and parallel to motion).
   First find displacement using (s = ut + \frac{1}{2}at^2) (with (u=0)): [ s = 0 + \frac{1}{2}\times 2.0 \times 10^2 = 100;\text{m} ] [ W = F \times s = 3000 \times 100 = 300,000;\text{J} ]

Answer:

• Acceleration = 2.0 m/s²
• Force = 3,000 N
• Work done = 3 × 10⁵ J

Why it matters: Knowing how to compute these helps in understanding vehicle dynamics, battery life for portable devices, and safety calculations.

4. Quick Revision Summary