Work, Energy and Power: CBSE Class 11 Physics NCERT Guide

Core Concepts: Work, Energy, and Power

Work (W)

In physics, work is done on an object when a force causes it to move through a distance. It is the scalar (dot) product of the force vector (F) and the displacement vector (d). The formula is W = F ⋅ d = Fd cos(θ), where θ is the angle between the force and displacement. Its SI unit is the Joule (J). Work can be positive (θ < 90°), negative (θ > 90°), or zero (θ = 90°).

Energy (E)

Energy is the capacity to do work. It is a scalar quantity and exists in various forms. In this chapter, we focus on mechanical energy, which has two main types: Kinetic Energy (KE), the energy of motion (KE = ½mv²), and Potential Energy (PE), the energy stored due to an object's position or configuration (e.g., Gravitational PE = mgh). The SI unit of energy is also the Joule (J).

Power (P)

Power is the rate at which work is done or energy is transferred. Average power is P_avg = W/t. Instantaneous power is the derivative of work with respect to time, P = dW/dt. It can also be expressed as the dot product of force and velocity, P = F ⋅ v. The SI unit of power is the Watt (W), where 1 Watt = 1 Joule/second.

The Work-Energy Theorem: Connecting Force and Motion

The Work-Energy Theorem is one of the most powerful and useful principles in mechanics. It provides a direct link between the work done on an object and the change in its kinetic energy. The theorem states: The net work done by all forces acting on an object is equal to the change in its kinetic energy.

Mathematically, this is expressed as: W_net = ΔKE = KE_f - KE_i = ½mv_f² - ½mv_i²

Where:

• W_net is the total work done by the net force.
• KE_f and KE_i are the final and initial kinetic energies.
• v_f and v_i are the final and initial velocities.

Intuition behind the proof: Let's consider a particle of mass 'm' moving along the x-axis under a net force 'F'. From Newton's second law, F = ma. The work done in a small displacement dx is dW = F dx = (ma) dx. We know that acceleration a = dv/dt. Using the chain rule, we can write a = (dv/dx)(dx/dt) = v(dv/dx). Substituting this into our work equation gives: dW = m(v dv/dx)dx = mvdv. To find the total work done as the velocity changes from v_i to v_f, we integrate this expression: W_net = ∫dW = ∫(from v_i to v_f) mvdv = m[v²/2] (from v_i to v_f) = ½mv_f² - ½mv_i². This elegant result connects the integral of force over distance (work) to the change in the state of motion (kinetic energy).

Worked Examples

• Example 1: Calculating Work Done A 10 kg block is pulled across a horizontal floor by a rope that exerts a force of 50 N at an angle of 30° above the horizontal. The block moves a distance of 5 m. Calculate the work done by the rope. Solution: Step 1: Identify the given values. Force, F = 50 N Displacement, d = 5 m Angle, θ = 30° Step 2: Use the formula for work. The formula for work is W = Fd cos(θ). Step 3: Substitute the values and calculate. W = (50 N) (5 m) cos(30°) We know that cos(30°) = √3 / 2 ≈ 0.866 W = 250 0.866 = 216.5 J Final Answer: The work done by the rope on the block is 216.5 Joules.
• Example 2: Applying the Work-Energy Theorem A car of mass 1000 kg is traveling at 10 m/s. The driver applies the brakes, and a net braking force of 2000 N brings the car to a stop. Find the distance the car travels before stopping. Solution: Step 1: Identify initial and final states. Mass, m = 1000 kg Initial velocity, v_i = 10 m/s Final velocity, v_f = 0 m/s (since it stops) Net Force, F_net = -2000 N (negative because it opposes motion) Step 2: Apply the Work-Energy Theorem: W_net = ΔKE. The work done by the net force is W_net = F_net d cos(180°) = -F_net d = -2000d. The change in kinetic energy is ΔKE = ½mv_f² - ½mv_i². Step 3: Calculate ΔKE. ΔKE = ½(1000)(0)² - ½(1000)(10)² ΔKE = 0 - ½(1000)(100) = -50,000 J * Step 4: Equate W_net and ΔKE to find the distance 'd'. -2000d = -50,000 d = -50,000 / -2000 = 25 m Final Answer: The car travels 25 meters before stopping.
• Example 3: Calculating Power An elevator of mass 500 kg is lifted upwards at a constant velocity of 0.4 m/s. Calculate the power delivered by the motor to lift the elevator. (Use g = 9.8 m/s²). Solution: Step 1: Analyze the forces. Since the elevator moves at a constant velocity, the net force is zero. This means the upward force (F) exerted by the motor must balance the downward force of gravity (mg). F = mg = 500 kg 9.8 m/s² = 4900 N. Step 2: Use the formula for power. Power can be calculated as P = F v, where F and v are in the same direction. Step 3: Substitute the values and calculate. P = (4900 N) (0.4 m/s) = 1960 W Final Answer: The power delivered by the motor is 1960 Watts or 1.96 kW.

Exam Tip: Law of Conservation of Energy

A very common trap in exams involves the Law of Conservation of Energy. The law states that the total mechanical energy (KE + PE) of a system remains constant IF the work is done only by conservative forces (like gravity or ideal springs).

Common Mistake: Students often forget to account for non-conservative forces like friction or air resistance. When these forces are present, mechanical energy is NOT conserved; it is usually converted into heat.

The Correct Approach (The General Work-Energy Relation):
W_nc + W_c = ΔKE
Since the work done by conservative forces is equal to the negative change in potential energy (W_c = -ΔPE), we can rewrite this as:
W_nc - ΔPE = ΔKE

Rearranging gives the master equation: W_nc = ΔKE + ΔPE = ΔE_mechanical

This means the work done by non-conservative forces equals the change in the total mechanical energy of the system. If there are no non-conservative forces (W_nc = 0), then ΔKE + ΔPE = 0, which is the principle of conservation of mechanical energy.

Practice Questions with Solutions

• Q: A person pushes a 40 kg crate 5 m along a horizontal floor by a constant force of 120 N. If the coefficient of kinetic friction is 0.2, find the work done by (a) the applied force, (b) the frictional force, and (c) the net force. (Use g = 10 m/s²) A: (a) Work done by applied force: Step 1: Use the formula W_app = F_app d cos(0°) as the force is parallel to displacement. Step 2: W_app = 120 N 5 m 1 = 600 J. (b) Work done by frictional force: Step 1: Calculate the frictional force, f_k = μ_k N. The normal force N equals the weight mg, so N = 40 kg 10 m/s² = 400 N. Step 2: f_k = 0.2 400 N = 80 N. This force opposes motion, so the angle is 180°. Step 3: Calculate the work done by friction, W_f = f_k d cos(180°) = 80 N 5 m * (-1) = -400 J. (c) Work done by the net force: Step 1: The net work is the sum of the work done by all forces. W_net = W_app + W_f. Step 2: W_net = 600 J + (-400 J) = 200 J. Final answer: (a) 600 J, (b) -400 J, (c) 200 J.
• Q: A 2 kg object is dropped from a height of 10 m. Use the Work-Energy Theorem to find its speed just before it hits the ground. (Use g = 9.8 m/s²) A: Step 1: Identify the forces and energies. The only force doing work is gravity. The object is dropped, so initial velocity v_i = 0. Step 2: Calculate the work done by gravity. The force of gravity is F_g = mg = 2 kg 9.8 m/s² = 19.6 N. The displacement is d = 10 m. The angle is 0°. Step 3: W_g = F_g d = 19.6 N * 10 m = 196 J. This is the net work done. Step 4: Apply the Work-Energy Theorem, W_net = ΔKE = ½mv_f² - ½mv_i². Step 5: 196 J = ½(2 kg)v_f² - 0. So, 196 = v_f². Step 6: v_f = √196 = 14 m/s. Final answer: The speed just before hitting the ground is 14 m/s.
• Q: A bullet of mass 20 g moving at 500 m/s strikes a wooden block and comes to rest after penetrating 10 cm. What is the average resistive force exerted by the block on the bullet? A: Step 1: Convert all units to SI. Mass m = 20 g = 0.02 kg. Penetration distance d = 10 cm = 0.1 m. Initial velocity v_i = 500 m/s. Final velocity v_f = 0 m/s. Step 2: Calculate the change in kinetic energy (ΔKE). ΔKE = ½mv_f² - ½mv_i² = 0 - ½(0.02 kg)(500 m/s)² = -0.01 250000 = -2500 J. Step 3: Apply the Work-Energy Theorem. The work done by the resistive force is W_net = F_avg d cos(180°) = -F_avg d. Step 4: Equate work and change in KE: W_net = ΔKE. -F_avg * (0.1 m) = -2500 J. Step 5: Solve for F_avg. F_avg = 2500 / 0.1 = 25000 N. Final answer: The average resistive force is 25,000 N.
• Q: An engine pumps 3000 kg of water to a height of 10 m in 1 minute. Calculate the power of the engine, assuming no energy is wasted. (Use g = 10 m/s²) A: Step 1: Calculate the work done by the engine. The work is done against gravity to lift the water, so W = mgh. W = (3000 kg) (10 m/s²) (10 m) = 300,000 J. Step 2: Convert the time to seconds. Time t = 1 minute = 60 s. Step 3: Calculate the power using the formula P = W/t. P = 300,000 J / 60 s = 5000 W. Final answer: The power of the engine is 5000 W or 5 kW.

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