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IBD Topic 2 Part 3

IBD Topic 2 Part 3

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- 1. Topic 2Work and Energy
- 2. Contents Work and Energy Kinetic Energy Potential Energy Elastic Potential Energy Conservation of Energy Power Centripetal Acceleration
- 3. Work and Energy Kinetic energy is defined as: Ek = ½ mv2 If a particle is moving freely with no unbalanced force acting on it: NI tells us that it will move with constant velocity. This means that kinetic energy will also be constant.
- 4. Work and Energy What happens however if an unbalanced force acts? A constant unbalanced force produces: a constant acceleration. Oneof the kinematic equations that can be used in this circumstance is:
- 5. Work and Energy v2 – u2 = 2as To find the K.E. multiply both sides by ½m. ½ mv2 - ½ mu2 = mas From NII, F = ma ½ mv2 - ½ mu2 = Fs
- 6. Work and Energy K.E. = Fs The term on the RHS of the equation is called WORK.
- 7. Work and Energy Thework done by a constant unbalanced force acting on a particle: which is moving in one dimension is given by, the product of the unbalanced force and, the displacement produced.W = Fs
- 8. Work and Energy This equation shows us that if an unbalanced force acts: there will always be a change in kinetic energy and, an amount of work done.A glider moving at constant velocity on an air track has: no unbalanced force acting on it.
- 9. Work and Energy However, if it is on a slope; there is an unbalanced force, of gravity (weight), acting on it and it will accelerate. This weight can be resolved into two components, parallel and perpendicular to the motion.
- 10. Work and Energy
- 11. Work and Energy The perpendicular component of the weight: is balanced by the reaction force, of the air track on the glider, air on the glider.
- 12. Work and Energy The unbalanced force is therefore the parallel component of the weight. This force: multiplied by the displacement along the track gives, the work done on the glider.
- 13. Work and Energy Whatpart does the angle of inclination play in calculating the work done?
- 14. ExampleA Woolworths supermarket trolley (that does move in the direction you push it), is pushed with a force of 200 N acting at an angle of 40o to the ground. Find the effective horizontal force pushing the trolley along.
- 15. Solution = 40o IFI = 200 N Draw vector diagram
- 16. Solution
- 17. Solution FH is the effective force pushing the trolley FH = F cos FH = 200 x cos 40o FH = 200 x 0.7660444 FH = 153 N Horizontally
- 18. Work and Energy Work can be determined by studying a force- displacement graph.
- 19. Work and Energy Area under graph = height x length Area under graph = Force x displacement Force x displacement = Work Area under graph = Work done Area under graph = 5 x 10 Area under graph = 50 J
- 20. Work and Energy Work is easy to calculate when the force is constant. What happens if the force is not constant? Use a F vs. disp. graph.
- 21. Work and Energy
- 22. Work and Energy Work = Area under a F vs. Disp. Graph Work = ½ (b x h) Work = ½ (5 x 10) Work = 25 J
- 23. Energy and Power Kinetic Energy Push an object and it can move. If an object moves: it is capable of doing work. The object has energy associated with its motion called: Kinetic Energy
- 24. Energy and PowerW = Fs F = ma W = mas v2 – u2 = 2as 2 2 v u s 2a
- 25. Energy and PowerAs W = mas 2 2 vu W ma 2a W = ½mv2 – ½mu2 W = ½mv2 The quantity ½mv2 is called: Kinetic Energy
- 26. Energy and Power Kinetic Energy, Ek, can be defined as: The product of half the object’s mass m, and the square of its speed v.
- 27. Energy and Power Potential Energy Kinetic energy is the ‘energy of motion’. We can develop an expression for the energy that is dependent on position; potential energy.
- 28. Energy and Power Consideran object that is dropped from a height above the floor, ht: where the floor is at height ho. Displacementis given by s = ht - ho. The unbalanced force is given by: the weight of the object mg.
- 29. Energy and Power As W = Fs W = mg(ht - ho) or W = mg h This gives the work done in terms of the objects position. This quantity mgh, is defined as the gravitational potential energy.
- 30. Energy and Power P.E. = mgh Work can also be defined as: the change in gravitational potential energy. When an object falls: it loses gravitational potential energy, and gains kinetic energy.
- 31. Energy and Power Work can be calculated by the change in either of these two terms. Generally, work is defined as the change in energy.
- 32. Energy and Power The relationship between Ek, Ep and work can be shown using a downhill skier.
- 33. Energy and Power Energytransformation can be shown using a roller coaster.
- 34. Energy and Power Elastic Potential Energy Consider a spring that has been compressed. When released for time t, the spring will return to, the uncompressed position.
- 35. Energy and Power This means there must be an unbalanced force acting. This force is given by Hooke’s Law. The restoring force in a spring is: proportional to its extension or compression. Graphically, it can be described as:
- 36. Energy and Power Force Extension
- 37. Energy and Power Mathematically, it can be described as:F = -kx Where k is the slope of the graph.
- 38. Energy and Power The elastic potential energy can also be calculated. E.P.E. = ½kx2. This suggests that as a spring is compressed or extended: the energy increases.
- 39. Energy and Power Conservation of Energy Consider a ball thrown vertically into the air. It begins its motion with kinetic energy. As it reaches it’s highest point: The Ek is zero.
- 40. Energy and Power At the same time, the G.P.E. has: increased. The loss of one type of energy: is balanced by the gain in another. Total Energy = mgh + ½mv2. If a glass of whisky is pushed along a bar to a waiting gunslinger: is energy conserved?
- 41. Energy and Power In this case, the G.P.E. has not increased: when the K.E. has decreased. Thishowever is not an isolated system. Energy has been lost to friction. The total energy in any isolated system: is constant.
- 42. Energy and PowerA dart is fired out of a gun using a spring.
- 43. Energy and PowerA 3 kg cart moves down the hill. Calculate the Ep lost and Ek gained.
- 44. Energy and Power Ep = mgh Ep = 3 x 9.8 x (0.40 – 0.05) Ep = 10.3 J Ek = ½ mv2 Ek = ½ x 3 x 2.622 Ek = 10.3 J Energy is conserved.
- 45. Energy and Power Energy can be expended to perform a useful function. A device that turns energy into some useful form of work is called a: Machine
- 46. Energy and Power Machines cannot turn all the energy used to run the machine into useful work. In any machine, some energy goes to: atomic or molecular kinetic energy. This makes the machine warmer. Energy is dissipated as heat.
- 47. Energy and Power The amount of energy converted into: useful work by the machine is called, The efficiency. An example of a simple machine is: A pulley system. We can do 100 J of work.
- 48. Energy and Power Friction turn the pulleys which in turn rub on the axles. This may dissipate 40 J of energy as heat. The system is 60% efficient.
- 49. Energy and Power Efficiency can be expressed mathematically: useful work output efficiency total work input
- 50. Energy and Power Power Power is defined as: the rate at which work is done. W P t Units: Js-1 or Watts.
- 51. Energy and Power The work in this equation could be: the change in kinetic energy or, the work done on a mass that has been lifted. Itdoes not matter what form the energy takes: it is just the rate at which work is done.
- 52. Energy and PowerA 100W light globe produces 100 J of energy every second. To give an idea of the size of 1 W, a jumping flea produces 10-4 W, a person walking 300 W and, a small car 40 000 W.
- 53. Uniform Circular Motion Centripetal Acceleration A particle undergoing uniform circular motion is continually changing velocity: acceleration is changing.
- 54. Uniform Circular Motion v a v1 -va vb vc
- 55. Uniform Circular Motion v1 = vb - va v2 = vc - vb and so on The magnitude of v1 = v2 The direction is always to the centre of the circle. v a v1 -va vb vc
- 56. Uniform Circular Motion The acceleration, which produces these velocity changes in a direction which is, always towards the centre of the circular motion, is called: centripetal (centre seeking) acceleration.
- 57. Uniform Circular Motion Newton’s 2nd law tells us that: a centripetal acceleration can only happen if, there is an unbalanced force.
- 58. Uniform Circular Motion Any particle undergoing uniform circular motion is acted upon by: an unbalanced force which is, constant in magnitude and, directed towards the centre of the circle. This is called Centripetal Force.
- 59. Uniform Circular Motion
- 60. Uniform Circular Motion Witha centripetal force, the object moves in a circular path.
- 61. Uniform Circular Motion When the unbalanced force is released: the object moves along a tangential path, at a constant velocity.
- 62. Uniform Circular Motion Examplesinclude: Moon revolving around the Earth: Gravitational Force, Directed towards the centre of the Earth, Holds the moon in a near circular orbit.
- 63. Uniform Circular Motion Electrons revolve around the nucleus: Electric Force, Directed to centre of the nucleus, Holds electrons in circular orbit
- 64. Uniform Circular Motion Car rounding a corner: Sideways frictional force, Directed towards centre of turn, Force between car tyre and road. If force not great enough: Car skids.
- 65. Uniform Circular Motion The force acts on the passenger in the car if they do not have their seat belt on. Note: it is an European car
- 66. Uniform Circular Motion Washing Machine tub on spin cycle: Tub rotates at high speed, Inner wall exerts inwards force on clothes. Holes in tub allow water to follow a straight line. Water escapes. Force acts on clothes: not water.
- 67. Uniform Circular Motion Extreme Example TOK Centripetal force is real; centrifugal force is fictitious yet real to the person experiencing it. On what basis can we make this distinction?

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