Lesson Notes By Weeks and Term v5 - Grade 12
Mechanics: work, energy and power – Week 6 focus
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Mechanics: work, energy and power – Week 6 focus
Download the Lessonotes Mobile South Africa app for faster lesson access on Android and iPhone.
Subject: Physical Sciences
Class: Grade 12
Term: 1st Term
Week: 6
Theme: General lesson support
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This week, we delve into the crucial concepts of work, energy, and power. Understanding these concepts is fundamental to understanding how things move and interact in our world. From cars accelerating on the N1 to the electricity powering our homes during load shedding, work, energy, and power are at play. This knowledge provides a powerful foundation for understanding more complex physical phenomena and has direct relevance to many careers, from engineering to renewable energy technologies, sectors of increasing importance in South Africa.
2.1 Work: Work (W) is done when a force causes a displacement. It is defined as the product of the force and the displacement in the direction of the force. Mathematically, W = F Δx * cos θ Where: W = Work done (in Joules, J) F = Magnitude of the force (in Newtons, N) Δx = Magnitude of the displacement (in meters, m) θ = Angle between the force vector and the displacement vector If the force and displacement are in the same direction (θ = 0°), cos θ = 1, and W = F Δx. If the force is perpendicular to the displacement (θ = 90°), cos θ = 0, and W =
0. Crucially, if the force opposes the motion (θ = 180°), then cos θ = -1 and W = -FΔx. This means that the work done is negative, indicating energy is being removed from the system (e.g., friction).
Example 1: A bakkie exerts a constant force of 500 N to pull a trailer a distance of 10 meters along a straight road. If the force is applied horizontally, how much work is done by the bakkie on the trailer? F = 500 N Δx = 10 m θ = 0° (force and displacement are in the same direction) W = F Δx cos θ = 500 N 10 m cos 0° = 5000 J Example 2: A worker pushes a crate with a force of 200 N at an angle of 30° to the horizontal. The crate moves 5 meters along the floor. How much work is done by the worker on the crate? F = 200 N Δx = 5 m θ = 30° W = F Δx cos θ = 200 N 5 m cos 30° = 200 N 5 m 0.866 = 866 J Example 3: A soccer ball rolls to a stop due to friction. If the frictional force is 2 N and the ball travels 3 meters before stopping, what is the work done by friction? F = 2 N Δx = 3 m θ = 180° (friction opposes the motion) W = F Δx cos θ = 2 N 3 m cos 180° = 2 N 3 m (-1) = -6
J. The negative sign indicates friction removes 6J of energy from the ball. 2.2 Energy: Energy is the capacity to do work. The SI unit of energy is the Joule (J), the same as for work.
Kinetic Energy (Ek): The energy an object possesses due to its motion. Ek = ½ mv² Where: Ek = Kinetic energy (in Joules, J) m = Mass of the object (in kilograms, kg) v = Velocity of the object (in meters per second, m/s)
Potential Energy (Ep): The energy an object possesses due to its position relative to a reference point. We will focus on gravitational potential energy. Ep = mgh Where: Ep = Gravitational potential energy (in Joules, J) m = Mass of the object (in kilograms, kg) g = Acceleration due to gravity (approximately 9.8 m/s²) h = Height of the object above the reference point (in meters, m)
Example 4: A taxi with a mass of 1500 kg is moving at a velocity of 20 m/s. What is its kinetic energy? Ek = ½ 1500 kg (20 m/s)² = ½ 1500 kg 400 m²/s² = 300,000 J = 300 kJ Example 5: A brick with a mass of 2 kg is held 5 meters above the ground. What is its gravitational potential energy relative to the ground? Ep = 2 kg 9.8 m/s² 5 m = 98 J 2.3 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. Wnet = ΔEk = Ek(final) - Ek(initial) = ½ mvf² - ½ mvi² Where: Wnet = Net work done on the object vf = Final velocity of the object vi = Initial velocity of the object This is a powerful tool because it directly links work done to changes in motion.
Example 6: A trolley of mass 5 kg is initially at rest. A force of 10 N is applied to the trolley over a distance of 2 meters. Assuming no friction, what is the final velocity of the trolley? Wnet = F Δx = 10 N 2 m = 20 J ΔEk = Wnet = 20 J ½ mvf² - ½ mvi² = 20 J ½ 5 kg vf² - ½ 5 kg (0 m/s)² = 20 J 5 kg * vf² = 20 J vf² = 20 J / 2.5 kg = 8 m²/s² vf = √(8 m²/s²) = 2.83 m/s 2.4 Power: Power (P) is the rate at which work is done or the rate at which energy is transferred. P = W / Δt = ΔE / Δt Where: P = Power (in Watts, W) W = Work done (in Joules, J) Δt = Time interval (in seconds, s) ΔE = Change in energy (in Joules, J) Also, since W = F Δx, we can write P = F Δx / Δt = F * v, where v is the average velocity.
Example 7: A pump lifts 500 kg of water to a height of 10 meters in 20 seconds. What is the power output of the pump? First, calculate the work done: W = mgh = 500 kg 9.8 m/s² 10 m = 49000 J Then, calculate the power: P = W / Δt = 49000 J / 20 s = 2450 W Example 8: A car engine delivers 50 kW of power to the wheels. If the car is moving at a constant velocity of 25 m/s, what is the magnitude of the force exerted by the engine? P = 50 kW = 50000 W P = F * v 50000 W = F * 25 m/s F = 50000 W / 25 m/s = 2000 N 2.5 Conservative and Non-Conservative Forces: Conservative Forces: Forces for which the work done is independent of the path taken. Examples include gravity and the force exerted by a spring. The work done by a conservative force in a closed loop is zero.
Non-Conservative Forces: Forces for which the work done depends on the path taken. Examples include friction and air resistance. The work done by a non-conservative force in a closed loop is not zero (energy is dissipated as heat or sound). 2.6 Conservation of Mechanical Energy: Mechanical energy (Em) is the sum of kinetic and potential energy.