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Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
Work and Energy 2012
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Work and Energy 2012

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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. ExampleA 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 PowerW = Fs F = ma W = mas v2 – u2 = 2as 2 2 v u s 2a
  • 25. Energy and PowerAs 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 PowerA dart is fired out of a gun using a spring.
  • 43. Energy and PowerA 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 PowerA 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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