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Section 4 Conservation of Energy. I. Energy Transformations PE = mgh PE = (515 kg)(9.8 m/s 2 )(70.0 m) PE = 353,290 J At the top of the hill, PE is at.

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Presentation on theme: "Section 4 Conservation of Energy. I. Energy Transformations PE = mgh PE = (515 kg)(9.8 m/s 2 )(70.0 m) PE = 353,290 J At the top of the hill, PE is at."— Presentation transcript:

1 Section 4 Conservation of Energy

2 I. Energy Transformations PE = mgh PE = (515 kg)(9.8 m/s 2 )(70.0 m) PE = 353,290 J At the top of the hill, PE is at it’s maximum. What about KE? KE = 0.0 J

3 I. Energy Transformations KE = ½mv 2 KE = ½(515 kg)(37.1) 2 KE = 353,290 J At the bottom of the hill, KE is at it’s maximum. What about PE? PE = 0.0 J All of the PE was converted to KE.

4 As the car reaches the top of the 2 nd hill, KE = PE. KE = ½mv 2 KE = ½(515 kg)(26.2) 2 PE = mgh PE = (515 kg)(9.8 m/s 2 )(35.0 m) PE = 176,645 J KE = 176,758 J

5 II. The Law of Conservation of Energy A.Energy cannot be created or destroyed. 1. Energy changes from one form to another a.Roller coasters lose energy through: Sound Friction Air Resistance

6 III. Efficiency of Machines A.Not all work done by a machine is useful 1. Work Input is not equal to Work Output a.Energy is lost through: Sound Friction Air Resistance

7 Example: With this pulley system, we should be able to lift 100 pounds with only 25 pounds of force. However, friction and heat would cause some of the input force (25 lbs.) to be lost.

8 B. Efficiency Equation

9 Sample problems pg. 407 Questions 1 – 3


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