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CLB 10102 PHYSICS CHAPTER 4 Work, Energy and Power Ahmad Azahari Hamzah [email protected] F2F WEEK TOPIC L T P O NON F2F DELIVERY SLT METHOD/ ASSESSMENT 7 *Quiz 1-2 Physical Quantities and Dimensions 1 2 4 2-3 Forces Acting at a Point 2 4 5 11 *Quiz 3-4 Newton’s Laws of Motion and Friction 2 4 7 13 5-6 Work, Energy and Power 2 4 7 13 *Quiz * TEST 1 *Quiz 6-7 Optics 1 4 7 *Quiz 8-9 1 2 2 Simple Machines 10 - 11 5 8 here. *Assignment 1 We are * TEST 2 3 6 9 18 *Quiz * Mini project 1 2 4 7 *Assignment 2 1 2 4 7 *Quiz The Effects of Forces on Materials 12 - 13 13 - 14 Heat Energy Thermal Expansion CLB 10102 Physics CHAPTER 4 Work, Energy & Power 4.1 Work Work is the product of the force in the direction of the motion and the displacement. W = Fd The work done by an agent exerting a constant force (F) and causing a displacement (d) equals the magnitude of the displacement, times the component of along the direction of. CLB 10102 Physics CHAPTER 4 Work, Energy & Power In Figure 4.1, the work done by is: W = Fd cos θ F θ F cos θ d Figure 4.1: Work CLB 10102 Physics CHAPTER 4 Work, Energy & Power Scenario A A force acts rightward upon an object as it is displaced rightward. In such an instance, the force vector and the displacement vector are in the same direction. Thus, the angle between F and d is 0o. F d θ = 0o CLB 10102 Physics CHAPTER 4 Work, Energy & Power Scenario B A force acts leftward upon an object which is displaced rightward. In such an instance, the force vector and the displacement vector are in the opposite direction. Thus, the angle between F and d is 180o. F d θ = 180o CLB 10102 Physics CHAPTER 4 Work, Energy & Power Scenario C A force acts upward upon an object as it is displaced rightward. In such an instance, the force vector and the displacement vector are at right angles to each other. Thus, the angle between F and d is 90o. F θ = 90o d CLB 10102 Physics CHAPTER 4 Work, Energy & Power If d = 0 ➱W = 0. (i.e: no work is done when holding a heavy box, or pushing against a wall). W = 0 if F ┴ d (i.e: no work is done by carrying a bucket of water horizontally). The sign of W depends on the direction of F relative to d: a) W > 0 when component of F along d is in the same direction as d. b) W < 0 when it is in the opposite direction. If F acts along the direction of d then W = Fd since cos θ = cos 0 = 1. The SI units of work are Joules (J) CLB 10102 Physics CHAPTER 4 Work, Energy & Power Examples: A teacher applies a force to a wall and becomes exhausted. ➦ This is not an example of work because the wall is not displaced. A book falls off a table and free falls to the ground. ➦ There is a force (gravity) which acts on the book which causes it to be displaced in a downward direction. CLB 10102 Physics CHAPTER 4 Work, Energy & Power A waiter carries a tray full of meals above his head by one arm across the room. ➦ This is not an example of work because there is a force ( the waiter pushes up the tray), and there is a displacement (the tray is moved horizontally across the room) but the force does not cause the displacement. CLB 10102 Physics CHAPTER 4 Work, Energy & Power Calculating of Work Done by Forces When a force acts to cause an object to be displaced, 3 quantities must be known in order to calculate the work: o Force o Displacement o Angle between the force and displacement The work is subsequently calculated as: Force x Displacement x Cosine θ where θ = angle between the force and the displacement vectors. CLB 10102 Physics CHAPTER 4 Work, Energy & Power Example: Diagram F = 100 N 15 kg F = 100 N 30o 15 kg F 15 kg Statement of the problem A 100 N force is applied to move a 15 kg object a horizontal distance of 5 m at constant speed. Solution W = Fd cos θ = (100 N)(5 m) cos 0o = 500 Nm = 500 J A 100 N force is applied at W = Fd cos θ an angle 30o to the = (100 N)(5 m) cos 30o horizontal to move a 15 kg object at a constant speed = 433 Nm for a horizontal distance of 5 = 433 J m. An upward force is applied to lift a 15 kg object to a height of 5 m at constant speed. W = Fd cos θ = (150 N)(5 m) cos 0o = 750 Nm = 750 J CLB 10102 Physics CHAPTER 4 Work, Energy & Power 4.2 Energy Energy is the capacity or ability to do work. It is a scalar quantity and has the units of Joules, (J). There are various types of energy, such as mechanical, thermal, electrical and chemical energy. Mechanical energy of a body or a system is due to its position, its motion or its internal structure. There are two kinds of mechanical energy: i) Potential energy, EP ii) Kinetic energy, EK CLB 10102 Physics CHAPTER 4 Work, Energy & Power 4.3 Potential Energy Potential energy is the stored energy of a body due to its internal characteristics or its position. Water behind a dam Hammer over head Food on the plate Potential energy: Gravitational potential energy Elastic potential energy CLB 10102 Physics CHAPTER 4 Work, Energy & Power The gravitational potential energy acquired by a weight, mg, in lifting it a distance, h, above the ground is equal to the work done in lifting the weight, i.e., mgh. EP = mgh Where m = mass g = gravitational force h = height CLB 10102 Physics CHAPTER 4 Work, Energy & Power Elastic potential energy is the energy stored in elastic materials as the result of their stretching or compressing. Can be stored in rubber bands, bungee chords, trampolines, springs, an arrow drawn into a bow, etc. The amount of elastic potential energy stored in such a device is related to the amount of stretch of the device - the more stretch, the more stored energy. A force is required to compress a spring; the more compression there is, the more force which is required to compress it further. CLB 10102 Physics CHAPTER 4 Work, Energy & Power For certain springs, the amount of force is directly proportional to the amount of stretch or compression (x); the constant of proportionality is known as the spring constant (k). Elastic potential energy: EP spring = ½kx2 Where k = spring constant x = amount of compression CLB 10102 Physics CHAPTER 4 Work, Energy & Power Example: A cart is loaded with a brick and pulled at constant speed along an inclined plane to the height of a seat-top. If the mass of the loaded cart is 4.0 kg and the height of the seat top is 0.55 meters, then what is the potential energy of the loaded cart at the height of the seat-top? EP = mgh = (4.0 kg)(9.8 ms-2)(0.55 m) = 21.56 J CLB 10102 Physics 4.4 Kinetic Energy Kinetic energy is the energy of motion. An object which has motion whether it be vertical or horizontal motion - has kinetic energy. CHAPTER 4 Work, Energy & Power CLB 10102 Physics CHAPTER 4 Work, Energy & Power There are many forms of kinetic energy: • Vibrational (the energy due to vibrational motion) • Rotational (the energy due to rotational motion) • Translational (the energy due to motion from one location to another). CLB 10102 Physics CHAPTER 4 Work, Energy & Power The amount of translational kinetic energy which an object has depends upon two variables: mass (m) speed (v) The following equation is used to represent the kinetic energy (EK) of an object. EK = ½mv2 This equation reveals that the kinetic energy of an object is directly proportional to the square of its speed. CLB 10102 Physics CHAPTER 4 Work, Energy & Power That means: for a twofold increase in speed, the kinetic energy will increase by a factor of 4; for a threefold increase in speed, the kinetic energy will increase by a factor of 9. The kinetic energy is dependent upon the square of the speed. Example: Determine the kinetic energy of a 1000 kg roller coaster car that is moving with a speed of 20.0 m/s. EK = ½mv2 = ½(1000 kg)(20.0 m/s)2 = 200 000 J = 200 kJ CLB 10102 Physics CHAPTER 4 Work, Energy & Power 4.5 WORK-ENERGY THEOREM • Forces can be categorized as: Internal forces External forces Internal Forces External Forces Gravitational Force, Fgrav Applied Forces, Fapp Spring Forces, Fspring Frictional Forces, Ffrict Air Resistance Forces, Fair Tensional Forces, Ftens Normal Forces, Fnorm CLB 10102 Physics CHAPTER 4 Work, Energy & Power • When work is done upon an object by an external force, the TME = (EK + EP) of that object is changed. • If the work is "positive work“ ( +ve) the object will gain energy. • If the work is "negative work“ ( -ve) the object will lose energy. • The gain or loss in energy can be in the form of potential energy, kinetic energy, or both. EKo + EPo + Wext = EK + EP CLB 10102 Physics CHAPTER 4 Work, Energy & Power • When work is done upon an object by an internal force, the TME = (EK + EP) of that object remains constant. • When the only forces doing work are internal forces, energy changes forms or transform from EK to EP (or vice versa); the sum of the EK + EP remain constant. the TME is conserved. • In these situations, the sum of the EK + EP is everywhere the same. EKo + EPo = EK + EP CLB 10102 Physics CHAPTER 4 Work, Energy & Power EXAMPLE: • As an object falls from rest, its EP is converted to EK. CLB 10102 Physics CHAPTER 4 Work, Energy & Power 4.6 Principle of Conservation of Energy Energy cannot be created or destroyed; it may be transformed from one form into another, but the total amount of energy never changes. At any point, the total mechanical energy is given by: E = Ek + Ep = ½mv2 + mgh CLB 10102 Physics CHAPTER 4 Work, Energy & Power Description of Motion EK to EP / EP to EK? 1. A ball falls from a height of 2 meters in the absence of air resistance. EP to EK 2. A skier glides from location A to location B across the friction free ice. EP to EK 3. A bungee cord begins to exert an upward force upon a falling bungee jumper. EK to EP 4. The spring of a dart gun exerts a force on a dart as it is launched from an initial rest position. EP to EK CLB 10102 Physics CHAPTER 4 Work, Energy & Power 4.7 Power Power is the rate of doing work. P = Work = Time W t The unit for power is Watt. CLB 10102 Physics CHAPTER 4 Work, Energy & Power Example: An escalator is used to move 20 passengers every minute from the 1st floor of a department store to the second. The 2nd floor is located 5 meters above the 1st floor. The average passenger's mass is 60 kg. Determine the power requirement of the escalator in order to move this number of passengers in this amount of time. Work done to lift 1 passenger = (60 kg)(9.8 ms-2)(5 m) = 2940 J Work done to lift 20 passenger = 20 x 2940 J = 58800 J Power = W/t = 58800 J = 980 Watts 60 s