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CONSERVATION of

CONSERVATION of. MECHANICAL ENERGY. INTRODUCTION. We already learned the different forms of energy in the previous lessons and discussions. Now, our report will discuss why energy transformation is the unifying principle among these various forms of energy.

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CONSERVATION of

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  1. CONSERVATION of MECHANICAL ENERGY

  2. INTRODUCTION We already learned the different forms of energy in the previous lessons and discussions. Now, our report will discuss why energy transformation is the unifying principle among these various forms of energy. “When energy changes from one form to another, the amount of energy stays the same.” -Law of Conservation of Energy

  3. Law of Conservation of Energy A bulb changes electrical energy into light and heat energy. Telephones convert sound energy into electrical energy as sound moves back and forth. A swing changes the energy of the rider from potential energy into kinetic energy as the swings moves back and forth. From the given examples, you have seen that energy changes into different forms, but the total amount of energy stays the same. This is the application of this law which states that: Energy can neither be created nor destroyed, it can only be transformed from one form to another.

  4. Computations • Potential Energy: The Stored Energy PEg = W = F x d = mgh • Kinetic Energy: The Energy in Motion Anything that is moving has kinetic energy. KE = ½ mv2

  5. Law of Conservation of Mechanical Energy The sum of the kinetic energy and potential energy of a system is called Mechanical Energy. In a conservative system, the total mechanical energy is constant. In this system, only conservative forces are present and, therefore, a decrease in potential energy is equal to an increase in kinetic energy, and vice versa. This is now what the law states: The sum of the kinetic energy and potential energy in a conservative system is constant and equal to the total mechanical energy of the system.

  6. Sample Problem A 50 kg box falls from a bridge and lands in the water 20 m below. Find its (a) initial energy PE, (b) maximum KE, (c) KE and PE 15 m above water and (d) velocity upon reaching the water. Given: m = 50 kg h = 20 m g = 9.8 m/s2

  7. Solution • Initial PE is taken from the top. Therefore, • PEi = mgh • =(50kg)(9.8m/s2)(20m) • = 9800 J b. The maximum KE is equal to the total PE at the top. Therefore, KEmax= 9800 J c. At the height 15m above the water, PE = mgh = (50kg)(9.8m/s2)(15m) = 7350 J KE = TE – PE = 9800 J – 7350 J = 2450 J d. The velocity upon reaching the water is v = square root of 2 KE/ m = square root of 2(9800 J)/50kg • = square root of 19600 kg x m2 / s2 / 50kg = 19.80 m/s The KE used is 9800 J at the bottom; the KE is maximum.

  8. Thanks for paying attention to our report! Any Questions?  Reporters: Ruaya, Rhobie A. Paloma, Janey Anne M. Agcol, Jacques Matthew E. Salmayor, Jizelle M. Ursabia, Joan D. Musslewhite, Arthur M.

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