RRB Group D Physics: Master Work, Energy, and Power with Formulas & Solved Questions

Introduction to Work, Energy, and Power for RRB Exams

Welcome, future railway employees! This comprehensive guide delves into the crucial physics concepts of Work, Energy, and Power, essential for acing the RRB Group D examination. These topics form a fundamental part of the physics syllabus and are frequently tested. Understanding these concepts not only helps in answering exam questions but also provides a solid foundation in classical mechanics. We will break down each concept, explore their relationships, and equip you with the knowledge and practice needed to tackle any question related to them.

Topic Weightage and Importance

The topics of Work, Energy, and Power are of significant importance in the RRB Group D Physics section. Typically, you can expect 2-4 questions from this area in the examination. These questions can range from straightforward application of formulas to more complex problems involving the interplay between these concepts. A thorough understanding is vital for maximizing your score in the Physics paper, which often includes a good number of objective-type questions.

Key Concepts and Formulas

Work

In physics, work is done when a force causes an object to move a certain distance. It is a scalar quantity.

Definition:

Work done (W) by a constant force (F) on an object is the product of the magnitude of the force and the distance (d) moved by the object in the direction of the force.

Formula:

If the force is applied in the same direction as the displacement:

W = F × d

If the force is applied at an angle (θ) to the displacement:

W = F × d × cos(θ)

Where:

• W = Work Done
• F = Magnitude of the force
• d = Magnitude of the displacement
• θ = Angle between the force vector and the displacement vector

Units of Work:

• SI Unit: Joule (J). 1 Joule is the work done when a force of 1 Newton moves an object by 1 meter in its direction.
• CGS Unit: Erg. 1 J = 10⁷ ergs.

Positive, Negative, and Zero Work:

• Positive Work: Done when the force is in the direction of displacement (0° ≤ θ < 90°). cos(θ) is positive.
• Negative Work: Done when the force is opposite to the direction of displacement (90° < θ ≤ 180°). cos(θ) is negative. Example: Friction acting on a moving object.
• Zero Work: Done when displacement is zero, force is zero, or the force is perpendicular to the displacement (θ = 90°). cos(90°) = 0. Example: A coolie carrying luggage on his head does no work on the luggage against gravity as the displacement is horizontal, while the force (gravity) is vertical.

Energy

Energy is the capacity to do work. It is a scalar quantity and exists in various forms, such as kinetic energy, potential energy, heat energy, chemical energy, etc. The law of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another.

Kinetic Energy (KE)

The energy possessed by an object due to its motion.

Formula:

KE = 1/2 × m × v²

Where:

• KE = Kinetic Energy
• m = mass of the object
• v = velocity of the object

Units of Kinetic Energy:

• SI Unit: Joule (J)

Potential Energy (PE)

The energy possessed by an object due to its position or configuration.

Gravitational Potential Energy:

Energy stored in an object due to its height above the ground.

Formula:

PE = m × g × h

Where:

• PE = Potential Energy
• m = mass of the object
• g = acceleration due to gravity (approximately 9.8 m/s² on Earth)
• h = height above the reference point

Units of Potential Energy:

• SI Unit: Joule (J)

Power

Power is the rate at which work is done or energy is transferred.

Definition:

Power is the work done per unit time.

Formula:

P = W / t

Alternatively, since W = F × d, and velocity v = d/t, we can also write:

P = (F × d) / t = F × (d/t) = F × v

Where:

• P = Power
• W = Work Done
• t = time taken
• F = Force
• v = velocity

Units of Power:

• SI Unit: Watt (W). 1 Watt is the power when 1 Joule of work is done in 1 second.
• Other units include Horsepower (HP). 1 HP = 746 Watts.

Work-Energy Theorem

The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy.

Formula:

W_net = ΔKE = KE_final - KE_initial

W_net = 1/2 × m × v_f² - 1/2 × m × v_i²

Solved Examples (Step-by-Step)

Example 1: Work Done Calculation

A force of 10 N is applied to a block, and it moves a distance of 5 meters in the same direction. Calculate the work done.

Solution:

Given:

• Force (F) = 10 N
• Distance (d) = 5 m
• The force is applied in the same direction as the displacement, so θ = 0°. cos(0°) = 1.

Using the formula W = F × d × cos(θ):

W = 10 N × 5 m × cos(0°)

W = 10 × 5 × 1

W = 50 Joules

Example 2: Kinetic Energy Calculation

A car of mass 1000 kg is moving with a velocity of 20 m/s. Calculate its kinetic energy.

Solution:

Given:

• Mass (m) = 1000 kg
• Velocity (v) = 20 m/s

Using the formula KE = 1/2 × m × v²:

KE = 1/2 × 1000 kg × (20 m/s)²

KE = 1/2 × 1000 × 400

KE = 500 × 400

KE = 200,000 Joules

Example 3: Power Calculation

A crane lifts a load of 500 kg to a height of 10 meters in 20 seconds. Calculate the power of the crane. (Assume g = 10 m/s²)

Solution:

First, calculate the work done in lifting the load. Work done against gravity is equal to the potential energy gained.

Given:

• Mass (m) = 500 kg
• Height (h) = 10 m
• Time (t) = 20 s
• Acceleration due to gravity (g) = 10 m/s²

Work Done (W) = PE = m × g × h

W = 500 kg × 10 m/s² × 10 m

W = 50000 Joules

Now, calculate the power using P = W / t:

P = 50000 J / 20 s

P = 2500 Watts

Example 4: Work-Energy Theorem

A ball of mass 0.5 kg is moving with an initial velocity of 10 m/s. A force acts on it, and its velocity increases to 20 m/s. Calculate the work done by the force.

Solution:

Given:

• Mass (m) = 0.5 kg
• Initial velocity (v_i) = 10 m/s
• Final velocity (v_f) = 20 m/s

Using the Work-Energy Theorem: W_net = 1/2 × m × v_f² - 1/2 × m × v_i²

W_net = 1/2 × 0.5 kg × (20 m/s)² - 1/2 × 0.5 kg × (10 m/s)²

W_net = 1/2 × 0.5 × 400 - 1/2 × 0.5 × 100

W_net = 0.25 × 400 - 0.25 × 100

W_net = 100 - 25

W_net = 75 Joules

Common Mistakes to Avoid

• Confusing Displacement with Distance: Work is calculated using displacement, not just the total distance covered.
• Ignoring the Angle (θ): Not considering the angle between force and displacement, especially when they are not in the same direction, leads to incorrect work calculations.
• Incorrectly Applying KE Formula: Forgetting to square the velocity in the kinetic energy formula.
• Units Mismatch: Using different units (e.g., km/h for velocity, grams for mass) without proper conversion to SI units (m/s, kg) leads to wrong answers.
• Confusing Energy and Power: Misunderstanding that power is the *rate* of doing work (work per unit time), not work itself.
• Assuming g = 10 m/s² always: While often used for simplicity, remember the standard value is 9.8 m/s² unless otherwise specified.
• Sign Errors in Work: Not correctly identifying when work done is positive, negative, or zero.

Practice Questions with Solutions

Question 1:

A body of mass 5 kg is at rest. A force of 20 N acts on it for 10 seconds. What is the work done by the force?

Question 2:

Calculate the kinetic energy of a particle of mass 10 g moving with a velocity of 5 m/s.

Question 3:

A motor pumps water at a rate of 3000 kg per minute from a well of depth 20 m. What is the power of the motor? (Assume g = 10 m/s²)

Question 4:

What is the work done by the force of gravity on a body of mass 1 kg dropped from a height of 10 m to the ground?

Question 5:

A train of mass 500 tonnes is moving at 72 km/h. Calculate its kinetic energy. (1 tonne = 1000 kg)

Question 6:

If a force of 5 N moves a body by 2 m at an angle of 60° to the direction of the force, what is the work done?

Question 7:

A body’s kinetic energy increases from 100 J to 400 J. If the mass remains constant, by what factor does its speed increase?

Solutions to Practice Questions:

Solution 1:

Given: m = 5 kg, initial velocity (u) = 0 m/s (at rest), F = 20 N, t = 10 s.

First, find acceleration: a = F/m = 20 N / 5 kg = 4 m/s².

Next, find the distance covered: d = ut + 1/2 at² = (0)(10) + 1/2 (4)(10)² = 0 + 2(100) = 200 m.

Work done: W = F × d = 20 N × 200 m = 4000 Joules.

Solution 2:

Given: m = 10 g = 0.01 kg, v = 5 m/s.

KE = 1/2 mv² = 1/2 × 0.01 kg × (5 m/s)² = 1/2 × 0.01 × 25 = 0.005 × 25 = 0.125 Joules.

Solution 3:

Rate of pumping water = 3000 kg/min = 3000/60 kg/s = 50 kg/s. This is the mass 'm' lifted per second 't'.

Depth (h) = 20 m, g = 10 m/s².

Work done per second (Power) = (mass lifted per second) × g × h

P = 50 kg/s × 10 m/s² × 20 m = 10000 Watts.

Solution 4:

The force of gravity acts downwards, and the displacement is also downwards. So, θ = 0°.

F = mg = 1 kg × 9.8 m/s² = 9.8 N.

d = 10 m.

W = F × d × cos(0°) = 9.8 N × 10 m × 1 = 98 Joules.

Solution 5:

Mass m = 500 tonnes = 500 × 1000 kg = 500,000 kg.

Velocity v = 72 km/h. Convert to m/s: v = 72 × (5/18) m/s = 4 × 5 = 20 m/s.

KE = 1/2 mv² = 1/2 × 500,000 kg × (20 m/s)² = 1/2 × 500,000 × 400 = 250,000 × 400 = 100,000,000 Joules = 1 × 10⁸ Joules.

Solution 6:

F = 5 N, d = 2 m, θ = 60°.

W = F × d × cos(θ) = 5 N × 2 m × cos(60°) = 10 × (1/2) = 5 Joules.

Solution 7:

KE_initial = 1/2 mv_i² = 100 J

KE_final = 1/2 mv_f² = 400 J

Divide the second equation by the first:

(1/2 mv_f²) / (1/2 mv_i²) = 400 / 100

v_f² / v_i² = 4

(v_f / v_i)² = 4

v_f / v_i = √4 = 2

So, the speed doubles.

Frequently Asked Questions (FAQs)

Q1: What is the difference between work and energy?

Answer: Work is the transfer of energy that occurs when a force moves an object over a distance. Energy is the capacity to do work. Work is a process, while energy is a state or quantity.

Q2: Can negative work be done? Give an example.

Answer: Yes, negative work is done when the force acts in the opposite direction to the displacement. For example, when a moving object experiences friction, the frictional force does negative work on the object, opposing its motion.

Q3: What is the SI unit of power, and how is it related to horsepower?

Answer: The SI unit of power is the Watt (W). 1 Watt is defined as 1 Joule per second. 1 Horsepower (HP) is approximately equal to 746 Watts.

Q4: Does a satellite orbiting the Earth do work?

Answer: No, a satellite orbiting the Earth does no work against the Earth's gravitational force. This is because the gravitational force acts towards the center of the Earth (centripetal force), while the satellite's instantaneous velocity is tangential to its orbit. The angle between the force and the displacement is 90°, and cos(90°) = 0, hence the work done is zero.

Conclusion and Final Tips

Mastering Work, Energy, and Power is a significant step towards securing a good score in the RRB Group D Physics section. Remember the core definitions, formulas, and their interrelationships. Pay close attention to the units and ensure consistency. Practice the solved examples and the provided questions to solidify your understanding. Always analyze the direction of force and displacement carefully, especially for work calculations. By applying these principles diligently and practicing regularly, you can confidently tackle any question related to these fundamental physics concepts. Keep revising, stay motivated, and best of luck with your preparation!