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Aim: How can we explain the Law of Conservation of Energy?

Aim: How can we explain the Law of Conservation of Energy?. Do Now: Homework Review. Conservation of Energy. The total energy (E T ) of an object or system is constant Energy cannot be created or destroyed Energy can be changed from one form to another. A Dropped Sphere.

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Aim: How can we explain the Law of Conservation of Energy?

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  1. Aim: How can we explain the Law of Conservation of Energy? Do Now: Homework Review

  2. Conservation of Energy • The total energy (ET) of an object or system is constant • Energy cannot be created or destroyed • Energy can be changed from one form to another.

  3. A Dropped Sphere • What energy does it have when held above the ground? Potential Energy • What happens to this PE as the ball drops? It becomes smaller and smaller • Is the energy just disappearing? No!! It is being converted into KE (the object is speeding up)

  4. At the top: PE = 9,800 J (PE = mgh) KE = 0 J (at rest) ET = 9,800 J (PE + KE = ET) m = 10 kg If PE = 4,900 J, what is KE? KE = 4,900 J 100 m PE = 3,000 J, what is KE? KE = 6,800 J At the bottom: PE = 0 J (no height) ET = 9,800 J (total energy is constant!) KE = 9,800 J (PE + KE = ET)

  5. As an object falls, PE is being converted into KE ΔPE = ΔKE mgΔh = ½mv2 Problem A 2 kg mass is dropped from a height of 10 m. What is the KE as it strikes the ground? ΔKE = ΔPE ΔKE = mgΔh ΔKE = (2 kg)(9.8 m/s2)(10 m) ΔKE = 196 J

  6. How fast is the object moving when it strikes the ground? KE = ½mv2 196 J = .5(2kg)v2 196 J/kg = v2 v = 14 m/s

  7. After the mass falls 5 m, what is its KE? ΔKE = ΔPE ΔKE = mgΔh ΔKE = (2 kg)(9.8 m/s2)(5 m) ΔKE = 98 J What is its PE? Solution 1 PE = mgh PE = (2 kg)(9.8 m/s2)(5 m) PE = 98 J Solution 2 PE + KE = ET PE + 98 J = 196 J PE = 98 J

  8. Sometimes energy is lost due to heat or friction When this happens: PE + KE + Q = ET Q = energy lost due to friction It is not used in every problem!

  9. m = 2 kg How much energy was lost due to friction? 10 m KE = 190 J • This means solve for Q • At the top: • No KE • Has not moved, so no frictional loss • The only energy is PE • PE at the top equals ET

  10. PE = mgh • PE = (2 kg)(9.8 m/s2)(10 m) • PE = 196 J • At the bottom, there is no height (no PE) • KE should equal PE at the top, unless there is frictional loss • PEtop = 196 J • KEbottom = 190 J • PE + KE + Q = ET • 0 J + 190 J + Q = 196 J • Q = 6 J

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