Lesson Notes By Weeks and Term v5 - Grade 12
Revision and examination preparation – Week 6 focus
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Revision and examination preparation – Week 6 focus
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Subject: Physical Sciences
Class: Grade 12
Term: Term 4
Week: 6
Theme: General lesson support
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This week focuses on consolidating our understanding of key concepts covered thus far in Grade 12 Physical Sciences, specifically targeting areas known to be problematic in past examinations. This focused revision is crucial for improving your overall performance and confidence going into the final exams. Physical Sciences plays a vital role in various sectors in South Africa, from mining and manufacturing to energy production and environmental management. A strong foundation in these concepts is essential not only for academic success but also for future career opportunities and informed participation in scientific and technological advancements shaping our country.
2.1 Mechanics: Newton's Laws of Motion Newton's First Law (Law of Inertia): An object remains at rest or in uniform motion unless acted upon by a net force. This is fundamental to understanding why objects don't spontaneously start moving or stop moving.
Newton's Second Law: The net force acting on an object is equal to the rate of change of its momentum.
Mathematically: F_net = ma, where F_net is the net force, m is the mass, and a is the acceleration. This is the cornerstone of dynamics.
Newton's Third Law: For every action, there is an equal and opposite reaction. This means that when object A exerts a force on object B, object B simultaneously exerts an equal and opposite force on object
A. Crucially, these forces act on different objects.
Free Body Diagrams (FBDs): Essential for solving problems. Draw a diagram representing the object, and draw vectors representing all the forces acting on the object. Clearly label each force (e.g., F_g for gravitational force, F_N for normal force, F_f for frictional force, F_T for tension).
Friction: A force that opposes motion. Kinetic friction (f_k) acts on a moving object, and its magnitude is given by f_k = μ_k N, where μ_k is the coefficient of kinetic friction and N is the normal force. Static friction (f_s) prevents an object from starting to move, and its magnitude can range from 0 up to a maximum value of μ_s * N, where μ_s is the coefficient of static friction. μ_s is generally greater than μ_k.
Example 1: A 5 kg box is pulled along a horizontal surface by a force of 20 N applied at an angle of 30 degrees above the horizontal. The coefficient of kinetic friction between the box and the surface is 0.
2. Calculate the acceleration of the box.
Solution: Draw a Free Body Diagram: Draw the box. Show the applied force (F_A) at 30 degrees, gravitational force (F_g) downwards, normal force (F_N) upwards, and frictional force (F_f) to the left.
Resolve forces into components: F_A has horizontal component F_Ax = 20cos(30°) = 17.32 N and vertical component F_Ay = 20sin(30°) = 10
N. Apply Newton's Second Law in the vertical direction (y-axis): Since the box is not accelerating vertically, F_N + F_Ay - F_g =
0. Therefore, F_N = F_g - F_Ay = (5 kg)(9.8 m/s²) - 10 N = 39
N. Calculate the frictional force: F_f = μ_k F_N = 0.2 39 N = 7.8
N. Apply Newton's Second Law in the horizontal direction (x-axis): F_Ax - F_f = ma.
Therefore, a = (F_Ax - F_f) / m = (17.32 N - 7.8 N) / 5 kg = 1.90 m/s². 2.2 Mechanics: Work, Power, and Energy Work (W): The energy transferred to or from an object by a force acting on it. W = F d * cos(θ), where F is the force, d is the displacement, and θ is the angle between the force and the displacement.
Kinetic Energy (KE): The energy an object possesses due to its motion. KE = (1/2)mv², where m is the mass and v is the velocity.
Potential Energy (PE): Stored energy. Gravitational potential energy PE_g = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height above a reference point. Elastic potential energy PE_e = (1/2)kx², where k is the spring constant and x is the displacement from equilibrium.
Power (P): The rate at which work is done. P = W/t, where W is the work done and t is the time taken. Also, P = Fv, where F is the force and v is the velocity (when the force is constant and parallel to the velocity).
Work-Energy Theorem: The net work done on an object is equal to the change in its kinetic energy: W_net = ΔKE = KE_final - KE_initial.
Example 2: A motor lifts a 200 kg crate vertically at a constant speed of 2 m/s. Calculate the power output of the motor.
Solution: Calculate the force required to lift the crate: Since the crate moves at a constant speed, the net force on it is zero.
Therefore, the lifting force must be equal to the weight of the crate: F = mg = (200 kg)(9.8 m/s²) = 1960
N. Calculate the power: P = Fv = (1960 N)(2 m/s) = 3920 W. 2.3 Electricity and Magnetism: Electrostatics Coulomb's Law: The electrostatic force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them: F = k |q1 * q2| / r², where k is Coulomb's constant (8.99 x 10^9 Nm²/C²), q1 and q2 are the charges, and r is the distance between them.
Electric Field (E): The force per unit charge experienced by a small positive test charge placed at that point: E = F/q. For a point charge, E = k q / r². Electric field lines point away from positive charges and towards negative charges.
Electric Potential (V): The electric potential energy per unit charge: V = PE/q. For a point charge, V = k q / r. The potential difference (ΔV) between two points is the work done per unit charge to move a charge between those points.
Example 3: Two point charges, q1 = +4 μC and q2 = -6 μC, are separated by a distance of 0.3 m. Calculate the magnitude and direction of the electrostatic force between them.