Weight The weight of an object is defined as the gravitational force acting on the object. Unit: Newton (N)


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1 Gravitational Field A gravitational field as a region in which an object experiences a force due to gravitational attraction Gravitational Field Strength The gravitational field strength at a point in the gravitational field is the gravitational force acting on a mass of 1 kg placed at that point. Unit: N/kg Symbol: g Gravitational Field Strength Formula Gravitational Acceleration The gravitational acceleration is the acceleration of an object due to the pull of the gravitational force. Unit: ms2 Symbol: g Important notes: Gravitational acceleration does not depend on the mass of the moving object. The magnitude of gravitational acceleration is taken to be 10ms2. Gravitational Field Strength vs. Gravitational Acceleration Both the gravitational field strength and gravitational acceleration have the symbol, g and the same value (10ms2) on the surface of the earth. When considering a body falling freely, the g is the gravitational acceleration. When considering objects at rest, g is the Earth s gravitational field strength acting on it. Weight The weight of an object is defined as the gravitational force acting on the object. Unit: Newton (N)
2 Differences between Weight and Mass Weight Depends on the gravitational field strength Vector quantity Unit Newton (N) Mass Independent from the gravitational field strength Scalar Quantity Unit: Kilogram (kg) Free Falling Free falling is a motion under force of gravity as the only force acting on the moving object. Practically, free falling can only take place in vacuum. Falling from high place
3 Acceleration = 10ms2 Initial velocity = 0 Displacement = high of the location Launching object upward Acceleration = 10ms2 Velocity at maximum height = 0 Vector and Scalar Quantity A scalar quantity is a quantity which can be fully described by magnitude only. A vector quantity is a quantity which is fully described by both magnitude and direction. Vector Diagram
4 The arrow shows the direction of the vector. The length representing the magnitude of the vector. Equal Vector Two vectors A and B may be defined to be equal if they have the same magnitude and point in the same direction. Vector Addition  Triangle Method Join the tail of the 2nd vector to the head of the 1st vector. Normally the resultant vector is marked with double arrow. Vector Addition  Parallelogram Method Join the tail of the 2nd vector to the tail of the 1st vector. Normally the resultant vector is marked with double arrow. Addition of 2 Perpendicular Vectors
5 If 2 vectors (a & b) are perpendicular to each others, the magnitude and direction of the resultant vector can be determined by the following equation. Example 1 Two forces, P and Q of magnitude 10N and 12N are perpendicular to each others. What is the magnitude of the resultant force if P and Q are acting on an object? 10 N F 12 N Use Theorem Pythagoras F 2 = F = F = 15.6 N Example 2
6 Diagram above shows that four forces of magnitude 2N, 4N, 5N and 8N are acting on point O. All the forces are perpendicular to each others. What is the magnitude of the resulatant force that acts on point O? The resultant force of the horizntal component = 52 = 3N to the right The resultant force of the vertical component = 84 = 4N acting downward. Therefore, the magtitude of these 2 force components, Vector Resolution A vector can be resolve into 2 components which is perpendicular to each others. Example 3
7 Diagram above shows a lorry pulling a log with an iron cable. If the tension of the cable is 3000N and the friction between the log and the ground is 500N, find the horizontal force that acting on the log. Horizontal component of the tension = 3000 cos30o =2598N Friction = 500N Resultant horizontal force = 2598N  500N =2098N Example 4 Diagram above shows two forces of magnitude 25N are acting on an object of mass 2kg. Find the acceleration of object P, in ms2. Horizontal component of the forces = 25cos45o + 25cos45o = 35.36N Vertical component of the forces = 25sin45o  25sin45o = 0N The acceleration of the object can be determined by the equation F = ma (35.36) = (2)a a = ms2 Inclined Plane
8 Weight component along the plane = Wsinθ. Weight component perpendicular to the plane = Wcosθ. Example 5 A block of mass 2 kg is pulling along a plane by a 20N force as shown in diagram above. Given that the fiction between block and the plane is 2N, find the magnitude of the resultant force parallel to the plane. First of all, let's examine all the forces or component of forces acting along the plane. The force pulling the block, F = 20N The frictional force Ffric = 2N The weight component along the plane = 20sin30o = 10N The resultant force along the plane = = 8N Vectors in Equilibrium
9 When 3 vectors are in equilibrium, the resultant vector = 0. After joining all the vectors tail to head, the head of the last vector will join to the tail of the first vector. Forces in equilibrium Forces are in equilibrium means the resultant force in all directions are zero. Example 6 Diagram above shows a load of mass 500g is hung on a string C, which is tied to 2 other strings A and B. Find the tension of string A.
10 Tension of string C, TC = weight of the load = 5N All forces in the system are in equilibrium, hence Vertical component of tension A (TA) = TC TAcos60o = TC TA = TC/cos60o TA = 5/cos60o = 10N Work Work done by a constant force is given by the product of the force and the distance moved in the direction of the force. Unit: Nm or Joule (J) Work is a scalar quantity. Formula of work Example 1
11 A force of 50 N acts on the block at the angle shown in the diagram. The block moves a horizontal distance of 3.0 m. Calculate the work being done by the force. Work done, W = F s cos θ W = cos30o = 129.9J Formula of work 2 When the direction of force and motion are same, θ = 0o, therefore cosθ = 1 Work done, W = F s Example 2 Diagram above shows a 10N force is pulling a metal. The friction between the block and the floor is 5N. If the distance travelled by the metal block is 2m, find a. the work done by the pulling force b. the work done by the frictional force Asnwer (a) The force is in the same direction of the motion. Work done by the pulling force, W = F s = (10)(2) = 20J (b) The force is not in the same direction of motion, work done by the frictional force W = F s cos180o= (5)(2)(1) = 10J Work Done Against the Force of Gravity
12 Example 3 Ranjit runs up a staircase of 35 steps. Each steps is 15cm in height. Given that Ranjit's mass is 45kg, find the work done by Ranjit to reach the top of the staircase. In this case, Ranjit does work to overcome the gravity. Ranjit's mass = 45kg Vertical height of the motion, h = Gravitational field strength, g = 10 ms2 Work done, W =? W = mgh = (45)(10)( ) = J Force  Displacement Graph In a ForceDisplacement graph, work done is equal to the area in between the graph and the horizontal axis. Example 4
13 The graph above shows the force acting on a trolley of 5 kg mass over a distance of 10 m. Find the work done by the force to move the trolley. In a ForceDisplacement graph, work done is equal to the area below the graph. Therefore, work done Energy Energy is defined as the capacity to do work. Work is done when energy is converted from one form to another. Unit: Nm or Joule(J) Kinetic Energy Kinetic energy is the energy of motion. Example 5 Determine the kinetic energy of a 2000kg bus that is moving with a speed of 35.0 m/s. : Kinetic Energy, E k = ½ mv 2
14 Gravitational Potential Energy Gravitational potential energy is the energy stored in an object as the result of its vertical position (i.e., height). Formula: Example 6 A ball of 1kg mass is droppped from a height of 4m. What is the maximum kinetic energy possessed by the ball before it reached the ground? According to the principle of conservation of energy, the amount of potential energy losses is equal to the amount of kinetic energy gain. Maximum kinetic energy = Maximum potentila energy losses = mgh = (1)(10)(4) = 40J Elastic Potential Energy Elastic potential energy is the energy stored in elastic materials as the result of their stretching or compressing. Formula:
15 Example 7 Diagram above shows a spring with a load of mass 0.5kg. The extention of the spring is 6cm, find the energy stored in the spring. The energy stored in the spring is the elastics potential energy. Conservation of Energy and Work Done During a conversion of energy, Amount of Work Done = Amount of Energy Converted Example 8 A trolley of 5 kg mass moving against friction of 5 N. Its velocity at A is 4ms1 and it stops at B after 4 seconds. What is the work done to overcome the friction? In this case, kinetic energy is converted into heat energy due to the friction. The work done to overcome the friction is equal to the amount of kinetic energy converted into heat energy, hence Power Power is the rate at which work is done, which means how fast a work is done. Formula:
16 Example 1 An electric motor takes 20 s to lift a box of mass 20kg to a height of 1.5 m. Find the amount of work done by the machine and hence find the power of the electric motor. Work done, W = mgh = (20)(10)(1.5) = 300J Power, P = Efficiency The efficiency of a device is defined as the percentage of the energy input that is transformed into useful energy. In the example above, the input power is 100J/s, the desire output power (useful energy) is only 75J/s, the remaining power is lost as undisire output. Therefore, the efficiency of this machine is 75/100 x 100% = 75%
17 Elasticity Elasticity is the ability of a substance to recover its original shape and size after distortion. Forces Between Atoms The intermolecular forces consist of an attractive force and a repulsive force. At the equilibrium distance d, the attractive force equal to the repulsive force. If the 2 atoms are brought closer, the repulsive force will dominate, produces a net repulsive force between the atoms. If the 2 atoms are brought furhter, the attractive force will dominate, produces a net attractive force between the atoms. Graph of Forces Between 2 atoms X 0 = Equilibrium Distance When the particles are compressed, x < x0, the repulsive force between the particles increases. When the particles are stressed, x > x0, the attractive force between the particles increases. If the distance x exceeds the elastic limit, the attractive force will decreases.
18 Hooke's Law Hooke's Law states that if a spring is not stretched beyond its elastic limit, the force that acts on it is directly proportional to the extension of the spring. Elastic Limit The elastic limit of a spring is defined as the maximum force that can be applied to a spring such that the spring will be able to be restored to its original length when the force is removed. Equation derived from Hooke's Law From Hook's Law, we can derived that Spring Constant Spring constant is defined as the ratio of the force applied on a spring to the extension of the spring. It is a measure of the stiffness of a spring or elastic object.
19 Graph of Streching Force  Extension Gradient = Spring constant Area below the graph = Work done Fx graph and spring constant The higher the gradient, the greater the spring constant and the harder (stiffer) spring. For example, the stiffness of spring A is greater than spring B. System of Spring
20 Arrangement in series: Extension = x number of spring Stiffness decreases Spring constant = k/number of spring Arrangement in parallel: Extension = x number of spring Stiffness increases Spring constant = k number of spring Factors Affecting the Stiffness of Spring Stiffer Less stiff Material type of spring Diameter of wire of spring Diameter of the spring Length of the string
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