Introduction: Mastering a Core Physics Topic for RRB Success
Welcome, future railway professionals! If you are gearing up for the competitive RRB NTPC, RRB Group D, or RRB Technician exams, you know that the General Science section holds significant weight. Within this section, Physics often poses a challenge with its blend of concepts and numerical problems. One of the most fundamental and frequently tested topics is Work, Energy, and Power. A solid understanding of these three interconnected concepts is crucial for scoring well. Questions from this chapter can be conceptual, formula-based, or involve unit conversions, making it a comprehensive test of your knowledge.
This detailed guide is designed to be your one-stop resource for mastering Work, Energy, and Power. We will break down each concept with simple explanations, provide all the necessary formulas, walk through solved examples, and test your understanding with a set of practice questions tailored to the RRB exam pattern. Let's build a strong foundation and turn this important topic into one of your scoring strengths!
Understanding Work (कार्य) in Physics
In our daily lives, we use the word 'work' to describe any activity that requires physical or mental effort. However, in physics, 'work' has a very specific and measurable definition. It’s not just about effort; it’s about accomplishing something tangible through force.
Scientific Definition of Work
Scientifically, work is said to be done when a force applied to an object causes a displacement of that object in the direction of the force.
Two conditions must be met for work to be done:
- A force must act on the object.
- The object must be displaced from its original position.
The Formula for Work
The amount of work done is calculated by the product of the magnitude of the force and the displacement of the object in the direction of the force.
The general formula for work is:
W = F × d × cos(θ)
- W is the Work done.
- F is the magnitude of the applied force.
- d is the magnitude of the displacement.
- θ (theta) is the angle between the direction of the force and the direction of the displacement.
When the force is applied in the same direction as the displacement, θ = 0°, and since cos(0°) = 1, the formula simplifies to W = F × d.
Types of Work: Positive, Negative, and Zero
Work can be categorized based on the angle (θ) between the force and displacement.
1. Positive Work (θ < 90°)
When the force and displacement are in the same general direction (the angle is acute), the work done is positive. This means the force is helping the motion.
Example: Pushing a box across the floor. The force you apply and the box's displacement are in the same direction.
2. Negative Work (θ > 90°)
When the force and displacement are in opposite general directions (the angle is obtuse), the work done is negative. This means the force is opposing the motion.
Example: The work done by friction. As a box slides to the right, the force of friction acts to the left, opposing the displacement. Work done by friction is always negative.
3. Zero Work (θ = 90° or d = 0)
Work done is zero in two main scenarios:
- When there is no displacement (d = 0): If you push against a solid wall, you might get tired, but since the wall doesn't move, the work done on the wall is zero.
- When the force is perpendicular to the displacement (θ = 90°): Since cos(90°) = 0, the work done is zero.
Example: A coolie carrying a heavy load on his head and walking horizontally on a platform. The force of gravity on the load acts downwards, while the displacement is horizontal. The angle is 90°, so the work done by gravity is zero. Similarly, the work done by the centripetal force on an object in a circular motion is zero because the force is always perpendicular to the direction of motion (the tangent).
Units of Work
- The SI unit of work is the Joule (J). One Joule is defined as the work done when a force of 1 Newton displaces an object by 1 meter in the direction of the force (1 J = 1 N·m).
- The CGS unit of work is the erg. 1 Joule = 107 erg.
Understanding Energy (ऊर्जा)
Energy is a fundamental concept that is closely related to work. In simple terms, energy is the capacity or ability to do work. If an object has energy, it can exert a force on another object to perform work.
The Law of Conservation of Energy is a critical principle: Energy can neither be created nor destroyed; it can only be transformed from one form to another. The total energy in an isolated system remains constant.
Units of Energy
Since energy is the capacity to do work, it has the same units as work.
- The SI unit of energy is the Joule (J).
- Other common units include:
- Calorie (cal): 1 cal ≈ 4.184 J.
- Kilowatt-hour (kWh): This is the commercial unit of electrical energy. 1 kWh = 3.6 × 106 J.
Types of Mechanical Energy
For RRB exams, the most important forms of energy to understand are the two types of mechanical energy: Kinetic and Potential Energy.
1. Kinetic Energy (KE) (गतिज ऊर्जा)
Kinetic energy is the energy possessed by an object due to its motion. The faster an object moves, the more kinetic energy it has.
The formula for kinetic energy is:
KE = ½ mv²
- KE is the Kinetic Energy.
- m is the mass of the object.
- v is the velocity of the object.
Key Insight: Kinetic energy is directly proportional to the mass (m) but proportional to the square of the velocity (v²). This means if you double the velocity of an object, its kinetic energy becomes four times greater!
2. Potential Energy (PE) (स्थितिज ऊर्जा)
Potential energy is the stored energy an object has due to its position or configuration. The most common type asked in exams is Gravitational Potential Energy.
Gravitational Potential Energy is the energy stored in an object as a result of its vertical position or height in a gravitational field.
The formula for potential energy is:
PE = mgh
- PE is the Potential Energy.
- m is the mass of the object.
- g is the acceleration due to gravity (approximately 9.8 m/s² or often taken as 10 m/s² for simpler calculations).
- h is the height of the object above a reference point.
Example: Water stored in a dam has potential energy. When released, this potential energy is converted into kinetic energy as the water flows, which then turns turbines to generate electricity.
Work-Energy Theorem
This important theorem states that the net work done on an object is equal to the change in its kinetic energy.
Wnet = ΔKE = KEfinal - KEinitial
Understanding Power (शक्ति)
Power is the measure of how quickly work is done or how quickly energy is transferred. Two people might do the same amount of work, but the one who does it faster is more powerful.
Definition and Formula of Power
Power is defined as the rate of doing work.
The formula for power is:
P = W / t or P = E / t
- P is the Power.
- W is the Work done.
- E is the Energy transferred.
- t is the time taken.
Units of Power
- The SI unit of power is the Watt (W), named after James Watt. One Watt is defined as one Joule of work done per second (1 W = 1 J/s).
- A larger unit of power is the Kilowatt (kW), where 1 kW = 1000 W.
- Another common unit is Horsepower (HP). It is often used for engines and motors. 1 HP ≈ 746 Watts. This is a very important conversion factor for RRB exams.
Relationship Between Kinetic Energy and Momentum
Momentum (p) is the product of an object's mass and velocity (p = mv). There is a direct mathematical relationship between kinetic energy (KE) and momentum (p) that is frequently tested.
We know KE = ½ mv². Let's manipulate this formula:
KE = (m²v²) / 2m
Since p = mv, then p² = m²v². Substituting this in the equation:
KE = p² / 2m
This formula is extremely useful for solving questions that relate changes in momentum to changes in kinetic energy.
For example, if the momentum of a body is doubled, its kinetic energy will become four times (as KE ∝ p²).
Solved Examples for RRB Exams
Let's apply these concepts to problems similar to what you'll find in your RRB exam.
Example 1: Calculating Work
Question: A boy pulls a toy car with a force of 10 N using a string that makes an angle of 60° with the horizontal. If the car moves a distance of 5 meters on the ground, calculate the work done by the boy.
Solution:
- Force (F) = 10 N
- Displacement (d) = 5 m
- Angle (θ) = 60°
- We use the formula: W = F × d × cos(θ)
- W = 10 × 5 × cos(60°)
- We know that cos(60°) = 0.5 or 1/2.
- W = 10 × 5 × 0.5
- W = 25 Joules
Answer: The work done by the boy is 25 J.
Example 2: Calculating Kinetic Energy
Question: A train of mass 10,000 kg is moving with a velocity of 20 m/s. What is its kinetic energy?
Solution:
- Mass (m) = 10,000 kg
- Velocity (v) = 20 m/s
- We use the formula: KE = ½ mv²
- KE = ½ × 10,000 × (20)²
- KE = 5,000 × 400
- KE = 2,000,000 J or 2 × 106 J
- This can also be expressed as 2000 kJ (kiloJoules) or 2 MJ (megaJoules).
Answer: The kinetic energy of the train is 2,000,000 J.
Example 3: Calculating Potential Energy
Question: An object of mass 15 kg is at a certain height. If the potential energy of the object is 450 J, find the height at which the object is with respect to the ground. (Take g = 10 m/s²)
Solution:
- Potential Energy (PE) = 450 J
- Mass (m) = 15 kg
- Acceleration due to gravity (g) = 10 m/s²
- We use the formula: PE = mgh
- 450 = 15 × 10 × h
- 450 = 150 × h
- h = 450 / 150
- h = 3 meters
Answer: The object is at a height of 3 meters.
Example 4: Calculating Power
Question: An electric motor lifts a 100 kg mass to a height of 10 meters in 5 seconds. Calculate the power of the motor. (Take g = 9.8 m/s²)
Solution:
First, we need to calculate the work done. The work done is equal to the potential energy gained by the mass.
- Work done (W) = PE = mgh
- W = 100 kg × 9.8 m/s² × 10 m
- W = 9800 Joules
- Time (t) = 5 seconds
- Now, we use the formula for power: P = W / t
- P = 9800 J / 5 s
- P = 1960 Watts or 1.96 kW
Answer: The power of the motor is 1960 Watts.
Practice Questions with Solutions
Now it's time to test your knowledge. Try to solve these questions on your own before looking at the solutions.
Q1. When a body falls freely towards the earth, its total energy:
- Increases
- Decreases
- Remains constant
- First increases and then decreases
Q2. In which of the following cases is the work done equal to zero?
- A person pushing a stationary wall.
- A satellite revolving around the Earth.
- A man carrying a suitcase on his head and walking on a level road.
- All of the above.
Q3. A body of mass 5 kg is moving with a velocity of 10 m/s. If its velocity is doubled, what will be the new ratio of its kinetic energy?
- 1:2
- 1:4
- 2:1
- 4:1
Q4. The commercial unit of energy is:
- Watt
- Joule
- Kilowatt-hour (kWh)
- Newton
Q5. A force of 20 N displaces an object through 2 m and does work of 20 J. The angle between the force and displacement is:
- 30°
- 60°
- 90°
- 0°
Q6. One horsepower (HP) is equal to approximately:
- 746 Watts
- 1000 Watts
- 764 Watts
- 500 Watts
Q7. If the momentum of an object is increased by 100% (doubled), its kinetic energy will increase by:
- 100%
- 200%
- 300%
- 400%
Q8. An electric bulb of 60 W is used for 10 hours a day. The energy consumed in 30 days is:
- 18 kWh
- 1.8 kWh
- 180 kWh
- 1800 kWh
Q9. A spring is compressed. The potential energy of the spring:
- Increases
- Decreases
- Remains unchanged
- Becomes zero
Q10. Work done by the force of friction is always:
- Positive
- Negative
- Zero
- Either positive or negative
Solutions to Practice Questions
Ans 1: (c) Remains constant. According to the law of conservation of energy, the total energy of the system remains constant. As the body falls, its potential energy converts into kinetic energy, but the sum (total energy) stays the same.
Ans 2: (d) All of the above. In (a), displacement is zero. In (b), the centripetal force is perpendicular to the displacement. In (c), the gravitational force is perpendicular to the horizontal displacement. In all cases, work done is zero.
Ans 3: (b) 1:4. Kinetic Energy (KE) is proportional to v². Let initial KE be KE₁ = ½ mv². When velocity is doubled (2v), the new KE₂ = ½ m(2v)² = ½ m(4v²) = 4(½ mv²) = 4KE₁. The question asks for the ratio of initial to new, which is KE₁ : KE₂ = 1:4.
Ans 4: (c) Kilowatt-hour (kWh). This is the unit used in electricity bills for measuring energy consumption.
Ans 5: (b) 60°. Using W = Fd cos(θ). We have 20 = 20 × 2 × cos(θ). So, 20 = 40 cos(θ). This gives cos(θ) = 20/40 = 0.5. The angle whose cosine is 0.5 is 60°.
Ans 6: (a) 746 Watts. This is a standard conversion factor that you should memorize.
Ans 7: (c) 300%. We know KE = p²/2m. Let the initial momentum be 'p'. The initial KE is proportional to p². When momentum is increased by 100%, the new momentum becomes p + 100% of p = 2p. The new KE is proportional to (2p)² = 4p². The new KE is 4 times the original KE. The increase in KE is New KE - Old KE = 4KE - KE = 3KE. The percentage increase is (Increase / Original) × 100 = (3KE / KE) × 100 = 300%.
Ans 8: (a) 18 kWh. Energy consumed per day = Power × Time = 60 W × 10 h = 600 Wh = 0.6 kWh. Energy consumed in 30 days = 0.6 kWh/day × 30 days = 18 kWh.
Ans 9: (a) Increases. When a spring is compressed or stretched, work is done on it, which gets stored as elastic potential energy. Hence, its potential energy increases.
Ans 10: (b) Negative. The force of friction always acts in the direction opposite to the motion (displacement). Therefore, the work done by friction is always negative.
Conclusion: Your Path to Mastery
Work, Energy, and Power are not just abstract physics concepts; they are principles that govern the world around us and are a staple of the RRB General Science syllabus. By thoroughly understanding the definitions, mastering the formulas (W=Fdcos(θ), KE=½mv², PE=mgh, P=W/t), and recognizing the different scenarios for positive, negative, and zero work, you can confidently tackle any question from this topic.
The key to success is consistent practice. Revisit the solved examples and practice questions. Try to find more problems from previous years' RRB papers. The more you practice, the faster and more accurate you will become. Keep up the hard work, and you will surely achieve your goal of securing a position in the Indian Railways. All the best!
