Introduction to Work, Power and Energy for RRB Exams

Welcome, future railway aspirants! Physics is a high-scoring section in RRB NTPC, Group D, and Technician examinations. Among various chapters, Work, Power and Energy holds immense importance due to its direct conceptual questions and numerical problems. Understanding this chapter helps you secure crucial marks in both General Science and basic engineering principles.

Topic Weightage and Importance

In the Computer Based Tests (CBT) for RRB NTPC and Group D, candidates can expect 2 to 4 questions from Work, Power and Energy. These questions generally test the direct application of formulas like $W = F s \cos \theta$, $P = \frac{W}{t}$, and the conservation of mechanical energy ($E = KE + PE$). Mastering this topic is essential for clearing the sectional cut-off.

Key Concepts and Formulas

Let us review the fundamental concepts and mathematical expressions required for solving railway exam problems:

  • Work ($W$): Work is done when a force produces displacement in the body. Formula: $W = F s \cos \theta$, where $F$ is force, $s$ is displacement, and $\theta$ is the angle between force and displacement vectors. Unit: Joule (J) or Newton-meter (N·m).
  • Power ($P$): Power is the rate of doing work or the rate of transfer of energy. Formula: $P = \frac{W}{t} = F v$, where $v$ is velocity. Unit: Watt (W) or Joule per second (J/s). Also, $1 \text{ horsepower (HP)} = 746 \text{ Watts}$.
  • Kinetic Energy ($KE$): Energy possessed by an object due to its motion. Formula: $KE = \frac{1}{2} m v^2$, where $m$ is mass and $v$ is velocity.
  • Potential Energy ($PE$): Energy possessed due to position or configuration. Gravitational Potential Energy Formula: $PE = m g h$, 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: $W = \Delta KE$.

Solved Examples (Step-by-Step)

Example 1: Calculating Work Done

Problem: A force of $20\text{ N}$ acts on an object and displaces it through $5\text{ m}$ in the direction of the force. Calculate the work done.

Solution:
Given: Force ($F$) = $20\text{ N}$, Displacement ($s$) = $5\text{ m}$, Angle ($\theta$) = $0^\circ$ (since displacement is in the direction of force).
Formula: $W = F s \cos \theta$
$W = 20 \times 5 \times \cos(0^\circ) = 100 \times 1 = 100\text{ Joules}$.
Answer: The work done is $100\text{ J}$.

Example 2: Kinetic Energy and Momentum Relation

Problem: If the momentum of an object of mass $2\text{ kg}$ is $10\text{ kg m/s}$, find its kinetic energy.

Solution:
Given: Mass ($m$) = $2\text{ kg}$, Momentum ($p$) = $10\text{ kg m/s}$.
Relation between Kinetic Energy ($KE$) and momentum ($p$): $KE = \frac{p^2}{2m}$.
$KE = \frac{10^2}{2 \times 2} = \frac{100}{4} = 25\text{ J}$.
Answer: The kinetic energy of the object is $25\text{ Joules}$.

Example 3: Power Calculation

Problem: An engine lifts a load of $500\text{ kg}$ to a height of $10\text{ m}$ in $5\text{ seconds}$. Find the power of the engine. (Take $g = 10\text{ m/s}^2$).

Solution:
Given: Mass ($m$) = $500\text{ kg}$, Height ($h$) = $10\text{ m}$, Time ($t$) = $5\text{ s}$, $g = 10\text{ m/s}^2$.
Work done ($W$) = Potential Energy gained ($m g h$) = $500 \times 10 \times 10 = 50,000\text{ J}$.
Power ($P$) = $\frac{W}{t} = \frac{50,000}{5} = 10,000\text{ W} = 10\text{ kW}$.
Answer: The power of the engine is $10,000\text{ Watts}$.

Common Mistakes to Avoid

  • Ignoring the angle $\theta$ in work formulas when force and displacement are not in the same direction.
  • Confusing units of Power (Watts) and Energy (Joules or kWh).
  • Forgetting to convert mass into kilograms or time into seconds while applying standard SI formulas.
  • Misinterpreting negative work (when force and displacement are in opposite directions, $\theta = 180^\circ$).

Practice Questions with Solutions

  1. Q1: A body of mass $10\text{ kg}$ is moving with a velocity of $4\text{ m/s}$. Find its kinetic energy.
    Solution: $KE = \frac{1}{2} m v^2 = \frac{1}{2} \times 10 \times (4)^2 = 5 \times 16 = 80\text{ J}$.
  2. Q2: What is the work done by the gravitational force on a satellite moving in a circular orbit around the Earth?
    Solution: $0\text{ J}$. The centripetal force and displacement are perpendicular ($\theta = 90^\circ$), so $\cos(90^\circ) = 0$.
  3. Q3: An electric bulb consumes $1000\text{ J}$ of electrical energy in $10\text{ seconds}$. What is its power?
    Solution: $P = \frac{W}{t} = \frac{1000}{10} = 100\text{ W}$.
  4. Q4: If the velocity of a car is doubled, what happens to its kinetic energy?
    Solution: $KE \propto v^2$. If velocity is doubled ($2v$), kinetic energy becomes $(2)^2 = 4$ times the original value.
  5. Q5: Find the potential energy of an object of mass $5\text{ kg}$ placed at a height of $6\text{ m}$. ($g = 9.8\text{ m/s}^2$).
    Solution: $PE = m g h = 5 \times 9.8 \times 6 = 294\text{ J}$.

Frequently Asked Questions (FAQs)

Q1: What are the SI units of Work, Power, and Energy?

Work and Energy are measured in Joules (J), while Power is measured in Watts (W).

Q2: Can work done be negative?

Yes, work done is negative when the force and displacement are in opposite directions (e.g., frictional force acting on a moving car).

Q3: How many Watts are there in 1 Horsepower?

There are exactly $746\text{ Watts}$ in $1\text{ Horsepower (HP)}$.

Conclusion and Final Tips

Mastering Work, Power and Energy requires clear understanding of formulas and rigorous practice of numerical problems. Revise your concepts regularly, practice unit conversions, and solve previous years' railway exam questions to boost your confidence. Stay consistent and success will follow!