Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Define work, energy, power, and the work-energy theorem.
• Apply the work-energy theorem to solve problems.
• Calculate work, kinetic energy, potential energy, and power.
• Explain the relationship between work, energy, and power.
• Solve problems involving the conservation of mechanical energy.

Lesson summary

This week, we delve into the fundamental concepts of work, energy, and power within the broader field of mechanics. These concepts are crucial for understanding how forces cause motion and how energy is transferred and transformed. Understanding work, energy, and power isn't just about passing exams; it's about understanding the physical world around us. From the design of safer cars on our roads to the development of sustainable energy solutions for our communities, these principles play a vital role.

Lesson notes

2.1 Work Definition: Work (W) is done when a force (F) causes a displacement (Δx) of an object. It is a scalar quantity, meaning it has magnitude but no direction.

Formula: W = FΔx cos θ Where: W is the work done (in Joules, J) F is the magnitude of the force (in Newtons, N) Δx is the magnitude of the displacement (in meters, m) θ is the angle between the force vector and the displacement vector.

Important Considerations: If the force and displacement are in the same direction (θ = 0°), then cos θ = 1, and W = FΔx (Maximum work) If the force and displacement are perpendicular (θ = 90°), then cos θ = 0, and W = 0 (No work done) If the force and displacement are in opposite directions (θ = 180°), then cos θ = -1, and W = -FΔx (Negative work done – the force is acting against the motion). This often occurs with friction.

Example 1: A worker pushes a box across a factory floor with a force of 50 N. The box moves 10 m horizontally. The force is applied in the direction of the motion. Calculate the work done by the worker.

Solution: F = 50 N Δx = 10 m θ = 0° (force and displacement are in the same direction) W = FΔx cos θ = (50 N)(10 m)(cos 0°) = 500 J Therefore, the worker does 500 J of work.

Example 2: A learner lifts a bag of potatoes with a mass of 5 kg to a height of 1.5 m. Calculate the work done by the learner.

Solution: The learner is applying a force to overcome the gravitational force on the bag.

Therefore, the applied force is equal to the weight of the bag, F = mg, where g is the acceleration due to gravity (9.8 m/s²). m = 5 kg g = 9.8 m/s² F = mg = (5 kg)(9.8 m/s²) = 49 N Δx = 1.5 m θ = 0° (force and displacement are in the same direction) W = FΔx cos θ = (49 N)(1.5 m)(cos 0°) = 73.5 J Therefore, the learner does 73.5 J of work.

Example 3: A car is moving along a road. A frictional force of 200 N acts on the car in the opposite direction to its motion. If the car travels 50 m, calculate the work done by the frictional force.

Solution: F = 200 N Δx = 50 m θ = 180° (force and displacement are in opposite directions) W = FΔx cos θ = (200 N)(50 m)(cos 180°) = -10000 J The work done by friction is -10000

J. This negative work represents energy being dissipated as heat. 2.2 Energy Definition: Energy is the ability to do work. It is also a scalar quantity, measured in Joules (J).

Types of Energy: Kinetic Energy (Ek): Energy possessed by an object due to its motion.

Gravitational Potential Energy (Ep): Energy possessed by an object due to its position relative to a reference point (usually the ground).

Elastic Potential Energy (Ee): Energy stored in a deformed elastic object, such as a spring or a rubber band.

Formulas: Kinetic Energy: Ek = ½mv² Where: m is the mass (kg) and v is the velocity (m/s)

Gravitational Potential Energy: Ep = mgh Where: m is the mass (kg), g is the acceleration due to gravity (9.8 m/s²), and h is the height above the reference point (m)

Elastic Potential Energy: Ee = ½kx² Where: k is the spring constant (N/m) and x is the extension or compression of the spring (m)

Example 1: A taxi with a mass of 1500 kg is travelling at a speed of 20 m/s. Calculate its kinetic energy.

Solution: m = 1500 kg v = 20 m/s Ek = ½mv² = ½(1500 kg)(20 m/s)² = 300000 J = 300 kJ The kinetic energy of the taxi is 300 k

J. Example 2: A rock with a mass of 2 kg is held 5 m above the ground. Calculate its gravitational potential energy.

Solution: m = 2 kg g = 9.8 m/s² h = 5 m Ep = mgh = (2 kg)(9.8 m/s²)(5 m) = 98 J The gravitational potential energy of the rock is 98

J. Example 3: A spring with a spring constant of 100 N/m is compressed by 0.1 m. Calculate the elastic potential energy stored in the spring.

Solution: k = 100 N/m x = 0.1 m Ee = ½kx² = ½(100 N/m)(0.1 m)² = 0.5 J The elastic potential energy stored in the spring is 0.5 J. 2.3 Work-Energy Theorem Statement: The net work done on an object is equal to the change in its kinetic energy.

Formula: Wnet = ΔEk = Ek(final) - Ek(initial) = ½mvf² - ½mvi² Where: Wnet is the net work done on the object vf is the final velocity of the object vi is the initial velocity of the object

Example: A shopper pushes a trolley with a mass of 20 kg, initially at rest (vi = 0 m/s), with a force of 50 N over a distance of 2 m. There is a frictional force of 10 N opposing the motion. Calculate the final velocity of the trolley.

Solution: Calculate the work done by the applied force: Fapplied = 50 N Δx = 2 m θ = 0° Wapplied = FappliedΔx cos θ = (50 N)(2 m)(cos 0°) = 100 J Calculate the work done by the frictional force: Ffriction = 10 N Δx = 2 m θ = 180° Wfriction = FfrictionΔx cos θ = (10 N)(2 m)(cos 180°) = -20 J Calculate the net work done: Wnet = Wapplied + Wfriction = 100 J - 20 J = 80 J Apply the work-energy theorem: Wnet = ΔEk = ½mvf² - ½mvi² 80 J = ½(20 kg)vf² - ½(20 kg)(0 m/s)² 80 J = (10 kg)vf² vf² = 8 m²/s² vf = √8 m/s = 2.83 m/s Therefore, the final velocity of the trolley is 2.83 m/s.