RRB NTPC & Group D Physics: Master Work, Energy, and Power – Complete Guide with Solved Problems

Introduction to Work, Energy, and Power for RRB Exams

In classical mechanics, Work, Energy, and Power form the absolute cornerstone of physics. For aspirants preparing for Indian Railway Recruitment Board (RRB) exams—such as RRB NTPC, RRB Group D, and RRB Technician (Grade I & III)—this chapter is one of the most high-yielding areas in the General Science section.

While the terms 'work', 'energy', and 'power' are used interchangeably in daily life, they have precise, mathematically defined definitions in physics. This comprehensive guide will walk you through the fundamental principles, essential formulas, derivation shortcuts, and typical numerical questions designed to help you score 100% accuracy in your upcoming RRB examinations.

Topic Weightage and Importance

Understanding the weightage of this topic helps prioritize your preparation strategy. Let's analyze how questions from Work, Energy, and Power are distributed across different RRB exams:

• RRB Group D & Technician Grade III: These exams place a very heavy emphasis on General Science. You can expect 2 to 3 questions directly from this chapter, often containing numerical problems based on kinetic energy, potential energy, and commercial consumption of electricity.
• RRB NTPC (CBT-1 & CBT-2): Included under the General Awareness umbrella, you can expect 1 to 2 questions, which generally lean toward conceptual, theoretical applications and simple direct numericals.
• RRB Technician Grade I (Signal): This exam requires a deep technical understanding. You can expect 3 to 4 analytical questions requiring application-based knowledge of conservation of mechanical energy and work-energy theorem.

Key Concepts and Formulas

To master this topic, you must understand the definitions, scientific conditions, SI units, and relationships between Work, Energy, and Power.

1. Work (W)

In physics, work is said to be done only when a force applied to an object causes it to undergo displacement. If a force F acts on a body and displaces it by a distance s, the work done is given by the scalar product of force and displacement:

W = F × s × cos(θ)

Where:

• F = Applied Force (in Newtons, N)
• s = Displacement of the object (in meters, m)
• θ (Theta) = Angle between the direction of force and the direction of displacement.

SI Unit: Joule (J). 1 Joule = 1 Newton-meter (N·m).
Dimensional Formula: [ML²T^-2]

Special Cases of Work Done:

2. Energy (E)

Energy is defined as the capacity of a body to do work. Since it represents capacity to do work, its unit and dimensions are identical to those of work.

SI Unit: Joule (J).
Note: Energy is a scalar quantity.

Mechanical Energy

Mechanical energy is the sum of Kinetic Energy and Potential Energy. In RRB exams, mechanical energy is classified into two main types:

• Kinetic Energy (K.E.): The energy possessed by a body by virtue of its motion.
  Formula: K.E. = ½ mv²
  Where m = mass of the body (kg) and v = velocity of the body (m/s).
• Potential Energy (P.E.): The energy possessed by a body by virtue of its position, shape, or configuration. The most common form is Gravitational Potential Energy.
  Formula: P.E. = mgh
  Where m = mass (kg), g = acceleration due to gravity (approx. 9.8 m/s² or 10 m/s²), and h = height above reference level (m).

Important Relations & Theorems:

1. Work-Energy Theorem: The work done by a net force on a body is equal to the change in its Kinetic Energy.
   W = ΔK.E. = ½ m(v² - u²) (where u is initial velocity, and v is final velocity).
2. Relation between Kinetic Energy (K.E.) and Linear Momentum (p):
   Momentum, p = mv. Therefore:
   K.E. = p² / 2m
   (Trick: If momentum is doubled, Kinetic Energy becomes 4 times!)
3. Law of Conservation of Energy: Energy can neither be created nor destroyed; it can only be transformed from one form to another. Total Mechanical Energy (K.E. + P.E.) of an isolated system remains constant.

3. Power (P)

Power is defined as the rate of doing work or the rate at which energy is consumed or transferred.

Power (P) = Work Done (W) / Time taken (t) = Energy (E) / Time (t)

Since Work = Force × Displacement, we can write:
P = (F × s) / t = F × (s / t). Since (s/t) is velocity (v):

P = F × v

SI Unit: Watt (W). 1 Watt = 1 Joule per second (J/s).
Other Units:

• 1 Kilowatt (kW) = 1000 W
• 1 Horsepower (HP) = 746 W (Extremely important for RRB Technician/Group D)

Commercial Unit of Electrical Energy

The standard commercial unit of electrical energy is the Kilowatt-hour (kWh), which is commonly referred to as a "Unit".

1 kWh = 1 Unit = 3.6 × 10⁶ Joules (or 3.6 MJ)

Solved Examples (Step-by-Step)

Let's look at standard, high-probability numerical problems patterned exactly on past RRB exams.

Example 1: Work Calculation at an Angle

Question: A porter lifts a suitcase of mass 15 kg from the ground and puts it on his head 1.5 m above the ground. Calculate the work done by him on the suitcase. (Take g = 10 m/s²).

Solution:
1. Identify Given Data:
Mass (m) = 15 kg
Displacement (h) = 1.5 m
Acceleration due to gravity (g) = 10 m/s²
Angle (θ) = 0° (The force applied by the porter is upwards, and displacement of the suitcase is also upwards).

2. Formula:
Work Done (W) = F × s = mgh

3. Calculation:
W = 15 kg × 10 m/s² × 1.5 m
W = 150 × 1.5 = 225 J

Answer: The work done by the porter on the suitcase is 225 Joules.

Example 2: Change in Kinetic Energy

Question: An object of mass 20 kg is moving with a uniform velocity of 5 m/s. What is the kinetic energy possessed by the object? If its velocity is doubled, what will be its new kinetic energy?

Solution:
1. Case 1: Finding Initial Kinetic Energy (K.E.₁)
Given: Mass (m) = 20 kg, Velocity (v₁) = 5 m/s
Formula: K.E. = ½ mv²
K.E.₁ = ½ × 20 × (5)²
K.E.₁ = 10 × 25 = 250 J

2. Case 2: When Velocity is Doubled (v₂ = 10 m/s)
K.E.₂ = ½ × 20 × (10)²
K.E.₂ = 10 × 100 = 1000 J

Shortcut Trick: Since K.E. is proportional to v², doubling the velocity increases the kinetic energy by a factor of 2² = 4 times.
New K.E. = 4 × 250 J = 1000 J.

Answer: The initial kinetic energy is 250 J and the new kinetic energy is 1000 J.

Example 3: Calculating Mechanical Power

Question: An electric motor of power rating 2 kW is used to pump water to a tank at a height of 10 m. How much mass of water can it lift in 1 minute? (Take g = 10 m/s²).

Solution:
1. Identify Given Data:
Power (P) = 2 kW = 2000 W
Height (h) = 10 m
Time (t) = 1 minute = 60 seconds
Acceleration due to gravity (g) = 10 m/s²

2. Formula:
Power = Work Done / Time = (mgh) / t
Therefore, m = (P × t) / (g × h)

3. Calculation:
m = (2000 × 60) / (10 × 10)
m = 120,000 / 100 = 1200 kg

Answer: The motor can lift 1200 kg of water in one minute.

Common Mistakes to Avoid

• Ignoring Unit Conversions: Time must always be converted to seconds (s), mass to kilograms (kg), and power to Watts (W) before starting calculations. A common error is calculating work with mass in grams or time in minutes.
• Confusion in Work Signs: Remember that gravity does negative work when you lift an object up (since force of gravity is downwards but displacement is upwards). However, the lifting force does positive work. Read the question carefully to see whose work done is being asked.
• The "Doubling" Trap: If momentum is doubled, many students think Kinetic Energy also doubles. Remember that $K.E. = p^2 / 2m$. Doubling the momentum increases K.E. by four times ($2^2 = 4$).
• Perpendicular Motion Work: A person carrying a suitcase walking horizontally does zero work against gravity because the force exerted is upwards (θ = 90° to displacement). Do not blindly use $W = F × s$ here.

Practice Questions with Solutions

Try to solve these questions on your own before checking the step-by-step solutions below.

Practice Questions

Q1. A body of mass 10 kg is moving with a constant speed of 4 m/s. Find its kinetic energy.

Q2. A force of 50 N acts on an object, displacing it by 3 meters in a direction making an angle of 60° with the force. Find the work done by the force. (Given cos(60°) = 0.5)

Q3. An electrical appliance of 500 W is used for 6 hours daily. Calculate the number of "units" of electrical energy consumed by it in 30 days.

Q4. If the kinetic energy of a body becomes 9 times its initial value, how does its linear momentum change?

Q5. A crane lifts a load of 2000 kg through a vertical height of 15 m in 10 seconds. Calculate the power of the crane. (Take g = 10 m/s²)

Detailed Solutions

Solution 1:
Given: m = 10 kg, v = 4 m/s.
Formula: K.E. = ½ mv²
Calculation: K.E. = ½ × 10 × 4 × 4 = 5 × 16 = 80 Joules.

Solution 2:
Given: F = 50 N, s = 3 m, θ = 60°, cos(60°) = 0.5.
Formula: W = F × s × cos(θ)
Calculation: W = 50 × 3 × 0.5 = 150 × 0.5 = 75 Joules.

Solution 3:
Given: Power (P) = 500 W = 0.5 kW. Daily time (t) = 6 hours.
Energy consumed in 1 day = Power × Time = 0.5 kW × 6 h = 3 kWh (Units).
Energy consumed in 30 days = 3 × 30 = 90 Units (kWh).

Solution 4:
We know that K.E. = p² / 2m. Therefore, Momentum p is proportional to √K.E.
If K.E. becomes 9 times, then the new momentum p' ∝ √9 = 3 times.
Thus, the linear momentum becomes 3 times its initial value.

Solution 5:
Given: m = 2000 kg, h = 15 m, t = 10 s, g = 10 m/s².
Work Done (W) = mgh = 2000 × 10 × 15 = 300,000 J.
Power (P) = W / t = 300,000 / 10 = 30,000 W = 30 kW (or 40.2 HP).

Frequently Asked Questions (FAQs)

Conclusion and Final Tips

Mastering Work, Energy, and Power is guaranteed to help you secure a top score in the science segment of your RRB NTPC, Group D, or Technician exams. Remember, RRB questions heavily feature the relation between momentum and kinetic energy, along with work done under gravitational forces.

Preparation Tip: Always pay close attention to structural calculations. Keep practicing physical unit conversions (such as converting minutes to seconds, and grams to kilograms) to eliminate avoidable errors. Keep reviewing these concepts, solve mock tests regularly, and stay focused on your dream of securing a career in the Indian Railways!