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NDA Work, Energy & Power

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This page covers NDA Work, Energy & Power with complete concept notes, 3 graded practice MCQs, key points and exam-specific tips. Free to study.

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Concept Notes

Work, Energy & Power— Rules & Concept

1.Core Concept

Read this first — the foundation of the topic
ENERGY

Energy is the capacity to do work. It exists in many forms but two main types dominate exams: (1) Kinetic Energy - energy due to motion = (1/2)mv², (2) Potential Energy - stored energy = mgh (gravitational). Energy cannot be created or destroyed, only converted from one form to another (Law of Conservation of Energy). Unit is Joule (J)

POWER

Power is the rate of doing work or rate of energy consumption. Power = Work/Time = Energy/Time. Unit is Watt (W). 1 Watt = 1 Joule/second. Commercial unit is kilowatt-hour (kWh). 1 kWh = 3.6 × 10⁶ J.

2.Formula Block

Memorise — at least one formula appears in every paper
• Work = Force × Distance (W = F × s)
• Kinetic Energy = (1/2)mv²
• Potential Energy = mgh
• Power = Work/Time = Energy/Time
• 1 HP = 746 Watts

3.Exam Patterns

What examiners ask — read before attempting PYQs

RRB asks direct formula applications, unit conversions, and energy transformations. Common questions involve calculating work done against gravity, comparing kinetic energies, and power consumption problems.

4.Shortcuts

Use these to save 30–60 seconds per question
ENERGY COMPARISON TRICK

If mass doubles, KE doubles. If velocity doubles, KE becomes 4 times (v² relationship). 2

POWER TRICK

Remember 1 HP = 746 W, 1 kWh = 36 lakh Joules 3

WORK-ZERO TRICK

Circular motion, carrying load horizontally - work done is zero

5.Worked Example

Solve this step-by-step before moving on
1
Step 1

Initial KE = (1/2)mv² = (1/2) × 2 × 10² = (1/2) × 2 × 100 = 100 J

2
Step 2

At maximum height, all KE converts to PE

3
Step 3

Therefore, PE at maximum height = 100 J

4
Step 4

Using PE = mgh, 100 = 2 × 10 × h

5
Step 5

Height h = 100/20 = 5 meters WORKED EXAMPLE 2: A motor of 2 kW power lifts 100 kg mass to 10 m height in 5 seconds. Find efficiency. Solution:

1
Step 1

Work done against gravity = mgh = 100 × 10 × 10 = 10,000 J

2
Step 2

Power input = 2 kW = 2000 W

3
Step 3

Energy input in 5 sec = 2000 × 5 = 10,000 J

4
Step 4

Useful work output = 10,000 J

5
Step 5

Efficiency = (Output/Input) × 100 = (10,000/10,000) × 100 = 100%

6.Exam Traps

Common mistakes students make — avoid these

#1: Students often confuse work and energy units. Both have same unit (Joule) but power has different unit (Watt). Another major error is forgetting that work done in circular motion or perpendicular direction is zero.

Many students also mix up kinetic energy formula - remember it's (1/2)mv², not mv². REMEMBER: Energy transformations are everywhere - when you climb stairs, chemical energy converts to kinetic, then to potential energy. Exams love testing these real-life applications with numerical problems.

Key Points to Remember

  • Work = Force × Distance, measured in Joules (J)
  • Kinetic Energy = (1/2)mv², depends on velocity squared
  • Potential Energy = mgh, depends on height above reference point
  • Power = Work/Time = Energy/Time, measured in Watts (W)
  • 1 Horsepower (HP) = 746 Watts (key conversion for exams)
  • 1 kilowatt-hour (kWh) = 3.6 × 10⁶ Joules (commercial unit)
  • Energy cannot be created or destroyed, only transformed
  • Work done in circular motion or perpendicular direction is zero
  • If velocity doubles, kinetic energy becomes 4 times (v² relationship)
  • All energy forms ultimately convert between kinetic and potential energy

Exam-Specific Tips

  • 1 Horsepower = 746 Watts
  • 1 kilowatt-hour = 3.6 × 10⁶ Joules
  • SI unit of work, energy is Joule (J)
  • SI unit of power is Watt (W)
  • Kinetic energy formula: KE = (1/2)mv²
  • Gravitational potential energy formula: PE = mgh
  • Work formula when force and displacement are parallel: W = F × s
  • 1 Watt = 1 Joule per second
Practice MCQs

Work, Energy & Power — Practice Questions

3graded MCQs · easy to hard · full solution & trap analysis

All MCQs →
Practice 1easy

A man pushes a box horizontally across a frictionless floor with a force of 50 N. The box moves a distance of 8 m in the direction of the applied force. How much work is done on the box?

Practice 2medium

A man pushes a box of mass 20 kg across a horizontal floor with a force of 100 N applied at an angle of 60° below the horizontal. The box moves 5 m. Calculate the work done by the applied force. (cos 60° = 0.5)

Practice 3hard

A 2 kg block is lifted vertically from the ground to a height of 5 m. The same block is then pushed horizontally along a frictionless surface for a distance of 10 m. If g = 10 m/s², what is the total work done against gravity during both operations?

60-Second Revision — Work, Energy & Power

  • Remember: Work = F × s, Energy has same unit as work (Joule)
  • Formula: KE = (1/2)mv², PE = mgh, Power = Work/Time
  • Conversion: 1 HP = 746 W, 1 kWh = 36 lakh Joules
  • Trap: Work done in circular motion is zero despite force application
  • Key: If velocity doubles, KE becomes 4 times (v² relationship)
  • Pattern: Energy conservation problems are 90% sure to appear
  • Quick check: Power questions often involve motor efficiency calculations
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