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

Introduction to Work, Energy, and Power for RRB Exams

Welcome to your ultimate preparation guide for Indian Railway Recruitment Board (RRB) examinations, including RRB NTPC, Group D, and Technician posts. General Science is a high-scoring section in all RRB exams, and Physics forms the backbone of this section. Among the core physics chapters, Work, Energy, and Power is one of the most conceptually rich and frequently tested topics. Whether you are solving direct formula-based numericals or answering conceptual questions about kinetic and potential energy, mastering this chapter will give you a significant edge over other aspirants. In this comprehensive guide, we will break down every single concept, formula, and shortcut needed to crack RRB exam questions effortlessly.

Topic Weightage and Importance

In both RRB NTPC (CBT 1 and CBT 2) and RRB Group D examinations, General Science contributes around 20 to 25 questions, depending on the specific exam phase. Out of these, Physics questions generally account for 7 to 10 questions. Candidates can safely expect 1 to 3 direct questions from Work, Energy, and Power in every shift. These questions typically include:

• Direct numerical problems calculating work done when force and displacement are at an angle.
• Inter-conversion of kinetic energy and potential energy.
• Conservation of mechanical energy principles.
• Calculations involving power, time, and rate of doing work.
• SI units and dimensional formulas of work, energy, and power.

Because the questions are often repetitive in pattern, mastering the formulas and understanding the sign conventions can guarantee full marks from this topic.

Key Concepts and Formulas

Let us break down the fundamental definitions, formulas, and units required for solving problems in this topic.

1. Work (W)

In physics, work is said to be done when a force applied on an object displaces it in the direction of the application of force. Mathematically, work is defined as the dot product of force vector and displacement vector.

Formula: $W = F \times s \times heta = Fs \text{ cos} heta$

Where:

• $F$ = Applied Force (in Newtons, N)
• $s$ = Displacement (in meters, m)
• $ heta$ = Angle between the direction of force and displacement

Special Cases of Work Done:

• When $ heta = 0^ \text{o}$ (Force and displacement in the same direction): $W = Fs \text{ cos}(0^ \text{o}) = Fs$ (Maximum positive work).
• When $ heta = 90^ \text{o}$ (Force perpendicular to displacement): $W = Fs \text{ cos}(90^ \text{o}) = 0$ (Zero work done, e.g., coolie carrying luggage horizontally while walking).
• When $ heta = 180^ \text{o}$ (Force opposite to displacement): $W = Fs \text{ cos}(180^ \text{o}) = -Fs$ (Maximum negative work, e.g., work done by frictional force).

Units of Work: SI unit is Joule (J) or Newton-meter ($ \text{N}\bullet \text{m}$). CGS unit is Erg ($1 \text{ Joule} = 10^7 \text{ Ergs}$). Work is a scalar quantity.

2. Energy (E)

Energy is defined as the capacity of a body to do work. Like work, energy is a scalar quantity and its SI unit is also Joule (J). Mechanical energy is broadly classified into two categories:

• Kinetic Energy (K.E.): The energy possessed by a body due to its motion. Formula: $ \text{K.E.} = \frac{1}{2}mv^2$, where $m$ is mass and $v$ is velocity. Alternatively, in terms of momentum ($p$), $ \text{K.E.} = \frac{p^2}{2m}$.
• Potential Energy (P.E.): The energy possessed by a body due to its position or configuration. Gravitational Potential Energy Formula: $ \text{P.E.} = mgh$, where $m$ is mass, $g$ is acceleration due to gravity, and $h$ is height.

3. Power (P)

Power is defined as the rate of doing work or the rate of transfer of energy.

Formula: $P = \frac{W}{t} = \frac{Fs}{t} = Fv$ (where $v$ is velocity).

Units of Power: SI unit is Watt (W) or Joule per second ($ \text{J/s}$). Commercial unit of electrical energy is Kilowatt-hour ($ \text{kWh}$), where $1 \text{ kWh} = 3.6 \times 10^6 \text{ Joules}$. Another common unit is Horsepower ($ \text{HP}$), where $1 \text{ HP} = 746 \text{ Watts}$.

Solved Examples (Step-by-Step)

Let us practice some standard problems frequently asked in RRB exams.

Example 1: Calculating Work Done at an Angle

Problem: A constant force of $50 \text{ N}$ acts on a body and displaces it by $10 \text{ m}$ in a direction making an angle of $60^ \text{o}$ with the force. Calculate the work done.

Step-by-Step Solution:

• Identify given parameters: $F = 50 \text{ N}$, $s = 10 \text{ m}$, $ heta = 60^ \text{o}$.
• Recall the work formula: $W = Fs \text{ cos} heta$.
• Substitute the values: $W = 50 \times 10 \times \text{ cos}(60^ \text{o})$.
• Since $ \text{cos}(60^ \text{o}) = 0.5$, we get: $W = 500 \times 0.5 = 250 \text{ Joules}$.
• Answer: The work done is $250 \text{ J}$.

Example 2: Relationship Between Kinetic Energy and Momentum

Problem: If the momentum of an object of mass $2 \text{ kg}$ is increased by $100%$, what will be the percentage increase in its kinetic energy?

Step-by-Step Solution:

• Recall formula: $ \text{K.E.} = \frac{p^2}{2m}$. This means $ \text{K.E.} ∘ p^2$.
• Let initial momentum $p_1 = 100$, so initial $ \text{K.E.}_1$ is proportional to $(100)^2 = 10000$.
• Since momentum increases by $100%$, new momentum $p_2 = 200$.
• New $ \text{K.E.}_2$ is proportional to $(200)^2 = 40000$.
• Percentage increase in $ \text{K.E.} = \frac{40000 - 10000}{10000} \times 100 = \frac{30000}{10000} \times 100 = 300%$.
• Answer: The kinetic energy increases by $300%$.

Example 3: Power Calculation

Problem: An electric pump can raise $2000 \text{ kg}$ of water to a height of $20 \text{ m}$ in $10 \text{ seconds}$. Find the power of the pump. (Take $g = 10 \text{ m/s}^2$).

Step-by-Step Solution:

• Mass ($m$) = $2000 \text{ kg}$, Height ($h$) = $20 \text{ m}$, Time ($t$) = $10 \text{ s}$, $g = 10 \text{ m/s}^2$.
• Work done by the pump equals potential energy gained by water: $W = mgh$.
• $W = 2000 \times 10 \times 20 = 4,00,000 \text{ Joules}$.
• Power ($P$) = $\frac{W}{t} = \frac{4,00,000}{10} = 40,000 \text{ Watts}$ or $40 \text{ kW}$.
• Answer: The power of the pump is $40,000 \text{ W}$.

Common Mistakes to Avoid

Aspirants often lose easy marks due to silly calculation errors or conceptual misconceptions. Keep these points in mind during the exam:

• Ignoring Angle $ heta$: Assuming work is always $W = Fs$ without checking if the force and displacement are collinear. If displacement is perpendicular to force, work done is zero.
• Confusion in Units: Forgetting to convert grams to kilograms or centimeters to meters before applying formulas. Always work in SI units unless specified otherwise.
• Confusing Power and Energy Units: Kilowatt-hour ($ \text{kWh}$) is a unit of energy, not power, even though it contains the word 'watt'.
• Sign Conventions of Work: Forgetting that frictional work is negative because it opposes motion ($ heta = 180^ \text{o}$).

Practice Questions with Solutions

1. Q1: A body of mass $5 \text{ kg}$ is thrown vertically upwards with a velocity of $10 \text{ m/s}$. Find its initial kinetic energy. ($g = 10 \text{ m/s}^2$)
2. Q2: If a force of $10 \text{ N}$ causes a displacement of $5 \text{ m}$ at an angle of $0^ \text{o}$, what is the work done?
3. Q3: What is the commercial unit of electrical energy, and how many Joules does it contain?
4. Q4: A machine does $400 \text{ J}$ of work in $20 \text{ seconds}$. What is the power of the machine?
5. Q5: If the velocity of a moving object is doubled, what happens to its kinetic energy?

Solutions to Practice Questions:

• Ans 1: $ \text{K.E.} = \frac{1}{2}mv^2 = \frac{1}{2} \times 5 \times (10)^2 = 0.5 \times 5 \times 100 = 250 \text{ J}$.
• Ans 2: $W = Fs \text{ cos}(0^ \text{o}) = 10 \times 5 \times 1 = 50 \text{ J}$.
• Ans 3: Kilowatt-hour ($ \text{kWh}$), which equals $3.6 \times 10^6 \text{ Joules}$.
• Ans 4: $P = \frac{W}{t} = \frac{400}{20} = 20 \text{ Watts}$.
• Ans 5: $ \text{K.E.} ∘ v^2$. If velocity is doubled ($2v$), kinetic energy becomes $(2)^2 = 4$ times the original value.

Frequently Asked Questions (FAQs)

Q1: Is work a vector or scalar quantity?

Work is a scalar quantity because it is the dot product of two vector quantities (force and displacement), resulting in a magnitude with no specific direction.

Q2: Can work done by a force be negative?

Yes, work done can be negative when the force and displacement are in opposite directions ($ heta = 180^ \text{o}$), such as the work done by friction or air resistance.

Q3: What is the power rating of standard household appliances in India?

Household electrical consumption is billed in Kilowatt-hours ($ \text{kWh}$), commonly known as 'units' of electricity.

Conclusion and Final Tips

Mastering Work, Energy, and Power is essential for scoring high in the General Science section of RRB NTPC, Group D, and Technician exams. Focus heavily on understanding the relationship between momentum and kinetic energy, practice unit conversions, and remember standard trigonometric values for angle-based work problems. Consistent practice of numericals will build your speed and accuracy. Stay motivated, keep revising daily, and success in your Indian Railways career dream is within your reach!