RRB NTPC & Group D Physics: Master Work, Power and Energy - Complete Guide

Introduction to Work, Power and Energy for RRB Exams

Welcome to the ultimate guide on Work, Power and Energy tailored specifically for aspirants preparing for the Railway Recruitment Board (RRB) NTPC, Group D, and Technician examinations. Physics constitutes a major chunk of the General Science section in these exams, and mechanics forms the backbone of physics questions. Understanding how work is done, how energy transforms, and how fast power is delivered is essential for securing high marks.

In the world of competitive exams, conceptual clarity combined with quick application of formulas is the key to success. This comprehensive guide will take you from the basics of mechanical work to advanced numerical problems, ensuring you are fully equipped to tackle any question thrown at you by the RRB examiners.

Topic Weightage and Importance

When analyzing previous years' question papers for RRB NTPC and RRB Group D, General Science consistently contributes about 20-25% of the total questions in the Computer Based Test (CBT) Stage 1 and Stage 2. Within General Science, Physics questions make up roughly 30-35% of the weightage, and mechanics (specifically Work, Power, and Energy) accounts for 2 to 3 direct questions.

These questions can range from direct formula-based numericals to conceptual statements regarding conservation laws or the sign convention of work. Mastering this topic not only guarantees those 2-3 marks but also builds a strong foundation for related chapters like Gravitation and Motion.

Key Concepts and Formulas

Let us break down the core theoretical concepts and mathematical expressions you need to memorize for the exam.

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 force. Mathematically, work is the dot product of force and displacement vectors:

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

Where:

• $W$ = Work done (measured in Joules, J or Newton-meters, Nm)
• $F$ = Applied force (in Newtons, N)
• $s$ = Displacement of the object (in meters, m)
• $ heta$ = Angle between the direction of force and displacement

Special Cases of Work Done:

• Maximum Positive Work ($ heta = 0^{\circ}$): When force and displacement are in the same direction, $\text{cos}(0^{\circ}) = 1$, so $W = F \times s$. Example: Pushing a car forward.
• Zero Work ($ heta = 90^{\circ}$): When force and displacement are perpendicular to each other, $\text{cos}(90^{\circ}) = 0$, so $W = 0$. Example: A coolie carrying luggage on his head walking on a horizontal platform.
• Maximum Negative Work ($ heta = 180^{\circ}$): When force and displacement are in opposite directions, $\text{cos}(180^{\circ}) = -1$, so $W = -F \times s$. Example: Work done by frictional force.

2. Energy ($E$)

Energy is defined as the capacity to do work. Like work, it is a scalar quantity and its SI unit is Joules (J). The total mechanical energy of a system is the sum of its kinetic and potential energies.

• Kinetic Energy ($KE$): The energy possessed by an object by virtue of its motion. Formula: $KE = \frac{1}{2}mv^2$ (where $m$ is mass and $v$ is velocity).
• Potential Energy ($PE$): The energy possessed by an object by virtue of its position or configuration. Gravitational Potential Energy Formula: $PE = mgh$ (where $g$ is acceleration due to gravity and $h$ is height).
• Work-Energy Theorem: The work done by a net force on an object equals the change in its kinetic energy: $W = \Delta KE = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$.

3. Power ($P$)

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

$P = \frac{W}{t} = \frac{F \times s}{t} = F \times v$

Where:

• $P$ = Power (measured in Watts, W or Joules per second, J/s)
• $W$ = Work done
• $t$ = Time taken
• $v$ = Velocity

Commercial Unit of Electrical Energy: Kilowatt-hour (kWh) or Board of Trade (BOT) unit. $1 \text{ kWh} = 3.6 \times 10^6 \text{ Joules}$.

Solved Examples (Step-by-Step)

Example 1: Calculating Work Done with an Angle

Problem: A force of $50\text{ N}$ is applied to pull a toy car, making an angle of $60^{\circ}$ with the horizontal ground. If the car is displaced by $4\text{ meters}$, calculate the work done by the force. (Given $\text{cos}(60^{\circ}) = 0.5$)

Solution:

Step 1: Identify the given values: $F = 50\text{ N}$, $s = 4\text{ m}$, $\theta = 60^{\circ}$.

Step 2: Use the work formula: $W = F \times s \times \text{cos}\theta$.

Step 3: Substitute the values: $W = 50 \times 4 \times \text{cos}(60^{\circ})$.

Step 4: Calculate: $W = 200 \times 0.5 = 100\text{ Joules}$.

Answer: The work done is $100\text{ J}$.

Example 2: Kinetic Energy and Momentum Relation

Problem: If the linear momentum of a moving body is increased by $50\%$, find the percentage increase in its kinetic energy.

Solution:

Step 1: Recall the relation between Kinetic Energy ($KE$) and Momentum ($p$): $KE = \frac{p^2}{2m}$.

Step 2: Let initial momentum be $p_1 = p$ and initial $KE$ be $KE_1 = \frac{p^2}{2m}$.

Step 3: New momentum $p_2 = p + 0.50p = 1.5p$.

Step 4: New kinetic energy $KE_2 = \frac{(1.5p)^2}{2m} = \frac{2.25p^2}{2m} = 2.25 \times KE_1$.

Step 5: Percentage increase = $\frac{KE_2 - KE_1}{KE_1} \times 100 = \frac{2.25KE_1 - KE_1}{KE_1} \times 100 = 1.25 \times 100 = 125\%$.

Answer: The kinetic energy increases by $125\%$.

Example 3: Power Calculation

Problem: An engine pumps $2000\text{ kg}$ of water up a height of $10\text{ meters}$ in $20\text{ seconds}$. Find the power of the engine. (Take $g = 10\text{ m/s}^2$)

Solution:

Step 1: Identify the given data: $m = 2000\text{ kg}$, $h = 10\text{ m}$, $t = 20\text{ s}$, $g = 10\text{ m/s}^2$.

Step 2: Work done by the engine equals the potential energy gained by the water: $W = mgh$.

Step 3: $W = 2000 \times 10 \times 10 = 2,00,000\text{ Joules}$.

Step 4: Power $P = \frac{W}{t} = \frac{200,000}{20} = 10,000\text{ Watts}$ or $10\text{ kW}$.

Answer: The power of the engine is $10,000\text{ W}$ ($10\text{ kW}$).

Common Mistakes to Avoid

• Ignoring Units: Always convert masses to kilograms (kg), distances to meters (m), and time to seconds (s) before applying formulas.
• Misinterpreting Angle $ heta$: Remember that $ heta$ is the angle between the force vector and displacement vector, not necessarily the angle given with the vertical or horizontal unless specified.
• Confusing Work and Power Units: Work is measured in Joules (J), whereas Power is measured in Watts (W) or Joules per second (J/s).
• Sign Confusion: Negative work does not mean scalar value is negative magnitude-wise in basic multiple-choice questions, but signifies opposition (like friction or gravity acting upwards while body moves down).

Practice Questions with Solutions

Q1. A body of mass $2\text{ kg}$ is dropped from a height of $5\text{ m}$. What is its kinetic energy just before striking the ground? ($g = 9.8\text{ m/s}^2$)

A) $49\text{ J}$
B) $98\text{ J}$
C) $196\text{ J}$
D) $24.5\text{ J}$

Q2. Which of the following quantities has the same unit as work?

A) Power
B) Torque
C) Momentum
D) Pressure

Q3. If the velocity of a moving object is doubled, what happens to its kinetic energy?

A) Doubled
B) Halved
C) Quadrupled
D) Remains unchanged

Q4. A pump is rated at $5\text{ kW}$. How much water can it lift to a height of $20\text{ m}$ in $1\text{ minute}$? ($g = 10\text{ m/s}^2$)

A) $1000\text{ kg}$
B) $1500\text{ kg}$
C) $2000\text{ kg}$
D) $2500\text{ kg}$

Q5. When a person carries a heavy suitcase and walks horizontally across a railway platform, the work done by gravity on the suitcase is:

A) Positive
B) Negative
C) Zero
D) Maximum

Solutions to Practice Questions

1. Solution (B): By the law of conservation of energy, Potential Energy at top = Kinetic Energy at bottom. $KE = mgh = 2 \times 9.8 \times 5 = 98\text{ Joules}$. Correct option is B.

2. Solution (B): Work = Force $\times$ Displacement ($F \times s$). Torque = Force $\times$ Perpendicular distance ($F \times d$). Both have the dimensional formula $[ML^2T^{-2}]$ and SI unit Joules/Newton-meter. Correct option is B.

3. Solution (C): Kinetic Energy formula is $KE = \frac{1}{2}mv^2$. If velocity becomes $2v$, $KE' = \frac{1}{2}m(2v)^2 = 4 \times (\frac{1}{2}mv^2) = 4 \times KE$. Correct option is C.

4. Solution (B): Power $P = 5\text{ kW} = 5000\text{ W}$, Time $t = 60\text{ s}$, height $h = 20\text{ m}$. Total work $W = P \times t = 5000 \times 60 = 300,000\text{ J}$. Equating to $mgh$: $300,000 = m \times 10 \times 20$, which gives $m = \frac{300,000}{200} = 1500\text{ kg}$. Correct option is B.

5. Solution (C): The force of gravity acts vertically downwards, while the displacement is horizontal ($\theta = 90^{\circ}$). Since $\text{cos}(90^{\circ}) = 0$, work done by gravity is zero. Correct option is C.

Frequently Asked Questions (FAQs)

Q1. Is work a scalar or vector quantity?

Work is a scalar quantity. Although it involves two vector quantities (force and displacement), their dot product results in a scalar value.

Q2. Can kinetic energy be negative?

No, kinetic energy can never be negative because mass is always positive and velocity is squared ($v^2$), which always yields a positive value.

Q3. What is 1 Horsepower (HP) in Watts?

1 Horsepower is equal to $746\text{ Watts}$, a standard conversion frequently asked in RRB and technical exams.

Conclusion and Final Tips

Mastering Work, Power and Energy is crucial for cracking the General Science section in RRB NTPC, Group D, and Technician examinations. Focus heavily on understanding the sign conventions of work, units of commercial energy, and relationship graphs between momentum and kinetic energy. Regular practice of numerical problems will build the speed and accuracy required to excel in CBT exams. Stay consistent, keep revising formulas, and success will be yours!