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Energy

Energy. Physics 11. Page 217. Multilab… Sources of Energy http://www.youtube.com/watch?v=iLXDirj4JUA&feature=related. Try giving an example of…. As many different types of energy and an example that goes with each. What is energy?. The measure of a system’s ability to do work.

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Energy

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  1. Energy Physics 11

  2. Page 217 • Multilab… Sources of Energy • http://www.youtube.com/watch?v=iLXDirj4JUA&feature=related

  3. Try giving an example of… • As many different types of energy and an example that goes with each

  4. What is energy? • The measure of a system’s ability to do work. • http://www.youtube.com/watch?v=yUpiV2I_IRI

  5. Types of Energy • There are 2 main classifications of energy: • 1) Potential Energy – The energy stored in a body or system as a consequence of its position, shape or form. • Example: An object being held up has potential energy because of its position (gravitational potential energy). • Example: A compressed spring has potential energy (potential to spring open). • 2) Kinetic Energy – The energy of motion • Example: When you walk across the classroom you have kinetic energy.

  6. http://www.youtube.com/watch?v=0ASLLiuejAo

  7. Work and Energy Relationship • There is not much difference between work and energy. • In order to do work, an object must have energy. • In order to have energy, an object must have work done on it.

  8. Energy Formulae • Ek = KE = ½ mv2 • Eg = PE = mgh • Units: Joules (J) • NOTE: h = height of the object measured from the reference level (measured in metres).

  9. What is the effect of doing work on an object? • You can give an object more kinetic energy by doing more work on it. • W = ΔKE • W = KEf – Kei • W = ½ mvf2 - ½ mvi2

  10. Example 1 • A 145g tennis ball is thrown at a speed of 25m/s. • A) What is the ball’s kinetic energy? • B) How much work was done to reach this speed assuming the ball started from rest.

  11. Answers • A) 45 J • B) W = ΔKE = 45 J

  12. Ex 2: Work on an moving object A 2kg object is moving at 10 m/s when a force is applied to it accelerating it to 20m/s over a distance of 5m. What is the work done on the object?

  13. Answers

  14. Ex 3: Potential Energy and Work What is the work done on a 12kg object to raise it from the ground to a height of 1.5m?

  15. Potential Energy and Work What is the work done on a 12kg object to raise it from the ground to a height of 1.5m?

  16. Ex 4: • A 1000kg car moves from point A to point B and then point C. The vertical distance between A and B is 10.m and between A and C is 15 m. • A) What is the PE at B and C relative to A? • B) What is the ΔPE (ΔPE = PEf – PEi) when it goes from B to C? • C) Repeat a) and b) but take the reference level at C (switch all letters).

  17. Answers • A) B: 98100 J  98000J, C: -147150J  -150 000J • B) a decrease of 245250 J 250000J • C) A: 150 000J • B: 245250 J  250 000J • Difference from A to B: increase of 100000J

  18. Assignment on Energy – Kinetic and Potential

  19. Comprehension Check • A truck pushes a car by exerting a horizontal force of 500. N on it. A frictional force of 300. N opposes the car’s motion as it moves 4.0m. • A) Calculate the work done on the car by the truck. • B) Calculate the work done on the car by friction. • C) Calculate the work done on the car overall (net work).

  20. Answers • A) W = Fd = 500 x 4 = 2000 N • = 2.00 x 103 J • B) W = Fd = -300 x 4 = -1200. J • C) Wnet = 2000 – 1200 = 800J

  21. Comprehension Check • Calculate the work done by a horse that exerts an applied force of 100. N on a sleigh, if the harness makes an angle of 30’ with the ground and the sleigh moves 30.m across a flat, level ice surface (ie, no friction).

  22. Answer • W = Fd cosΘ = (100)(30)cos(30) = 2.6 x 103 J

  23. Comprehension Check • A 50. kg crate is pulled 40. m along a horizontal floor by a constant force exerted by a person (100. N) which acts at an angle of 37’. The floor is rough and exerts a force of friction of 50.N. Determine the work done by EACH FORCE acting on the crate, and the net work done on the crate. • DRAW A DIAGRAM!!!

  24. WFg = FdcosΘ Work is 0J as the force is perpendicular to gravitational force. • WFN = 0J (same reason as above) • WFapp = Fdcos Θ =(100)(40)cos37’ = 3195J  S.F.  3200 J • WFf = Fd = -50(40) = -2000 J • -2.0 x 103J • Wnet = 3200 – 2000 = 1200 J

  25. Comprehension Check • Mrs. Evans is holding a 2.4kg textbook at a height of 3.4m above the floor. • a) What is the type of energy (potential or kinetic)? How do you know? • b) How much energy is there (use your equation)? • c) What is the velocity of the book at this point (ie, velocity initial)? • d) If Mrs. Evans drops the book, what is the final velocity assuming she doesn’t throw it (use your kinematics equations!)? • e) If Mrs. Evans drops the book as in d), what is the type of energy when the book hits the floor? • f) How much of this energy is there when it touches the floor? • g) Is there any time when there are both kinds of energy? If so, when? Explain.

  26. Answers • A) Potential: It is not moving, it has the potential to move (fall). • B) PE = mgh = 2.4x9.81x3.4 = 80.J • C) v = 0 (at rest, not thrown) • D) vf2 = vi2 + 2ad = 2(9.81)(3.4) = 66.708 Vf = 8.2m/s [down] • KE = ½ mv2 = ½ (2.4)(8.2)2 = 80.J • 80.J • When the object is falling, there is both PE and KE. When it falls 1.7m, there is equal PE and KE. Before this point (higher than 1.7m) there is more PE. After this (lower than 1.7m) there is more KE.

  27. “The net work done on an object is equal to its change in energy" If the object is experiencing KE: if Wnet is +ve, KE increases (moves in direction of force or speeds up) if Wnet is -ve, KE decreases (moves in direction of friction or slows down) If the object is experiencing PE: if Wnet is +ve, PE increases (is lifted) if Wnet is -ve, PE decreases (is lowered) Work-Energy Theorem

  28. The total energy of an object is the kinetic energy added to the potential energy. As an object is dropped, the kinetic energy changes to potential energy until there is 0 PE and only KE. ET = PE + KE Total Energy and Work-Energy Theorem

  29. Law of Conservation of Energy • Within a closed, isolated system, energy can change form, but the total amount of energy is constant • Closed - no objects enter or leave the system. • Isolated - no net external force is exerted on it. • Energy cannot be created or destroyed • E1 = E2 • KE1 + PE1 = KE2 + PE2

  30. Examples • Book Drop • Collision into a spring • Car coasting up a hill h v v x v v h

  31. Example 1: • A heavy object is dropped. If this object reaches the floor at a speed of 3.2 m/s from what height was it dropped?

  32. Answer • Etop = Ebottom • KE + PE = KE + PE • 0 + PE = KE + 0 • mgh = ½ mv2 • NOTE: The masses will cancel! • gh = ½ v2 • 9.81h = ½ (3.2)2 • h = 0.52 m

  33. Example 2: • A heavy box slides down a frictionless incline. The incline has a slope of 30° and the length of the incline is 12m. If the box starts from rest at the top of the incline what is the speed at the bottom?

  34. Answer • V = 10.8m/s

  35. Example 3: • A 4.0 x 104 kg roller coaster starts from rest at point A. Neglecting friction, calculate its potential energy relative to the ground, its kinetic energy and its speed at points B,C and D in the illustration above.

  36. Answer

  37. Page 287 • Questions 1, 2, 3, 4, 6, 8 • A 15.0 kg box slides down an incline. If the box starts from rest at the top of the incline and has a speed of 6.0m/s at the bottom, how much work was done to overcome friction? • NOTE: The incline is 5.0m high (vertically) and the incline that the box goes down is 8.0m long (hypotenuse). • Remember: W = ΔE

  38. Try This … • A skier is gliding along with a speed of 2.00m/s at the top of a ski hill. The hill is 40.0m high. The skier slides down the icy (frictionless) hill. • A) What will the skier’s speed be at a height of 25.0m?B) At what height will the skier have a speed of 10.0m/s? • HINT: Use similar triangles!

  39. Known: vi = 2.00m/s • h1 = 40.0m • h2 = 25.0m

  40. REMEMBER… • W = ΔE • That E can be potential or kinetic energy!

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