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# Energy 4

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### Transcript

• 1. ENERGY
• The Potential and Kinetic Chronicles
• (The Law of Conservation of Mechanical Energy)
• 2. What is Energy?
• Energy is the ability to do work
• Potential Energy (PE), and Kinetic Energy (KE).
• Energy is measured in Joules
• 3.
• Energy cannot be created or destroyed, but can only be converted into different types of energy
Law of Conservation of Energy Wind turbines transfer energy from the wind and convert it into electricity we can use.
• 4. Potential Energy
• Is energy that is stored
• Chemical energy
• Nuclear energy
• Mechanical PE
• Gravitational PE (GPE)
• 5. Example of a GPE Problem
• Gravitation PE problems have to do with an object and its position relative to the earths surface
• PE = mgh = (mass)(gravity)(height)
• 6. Example of a GPE Problem
• h=60m, g=9.81m/s 2 , m=2,000kg
• PE= (60m)(9.81m/s 2 )(2,000kg)= 1,177,200 Joules
• A roller coaster of 2 metric tons is sitting at the top of the track at a height of 60 meters. What is its PE?
• use the PE=mgh formula
• 7. Mechanical PE Problems
• have to do with stored energy within a spring
• PE= 1/2kx 2 = 1/2(spring constant)( in spring length from equilibrium)
• 8. Example of a MPE Problem
• A 4 gram rubber band with the spring constant of 4.3 F s /m is stretched a distance of 30 centimeters. What is the PE of the rubber band?
• 9. Example of a MPE Problem
• Plug your information in the equation 1/2kx 2 formula
• k= 4.3F s /m, x=30cm=.3m
• PE s = .5(4.3 F s /m)(.3m)=.645 Joules
• PE s =.645 Joules
• 10. Kinetic Energy
• is energy of motion
• Mechanical KE (MKE)
• Thermal energy
• Electrical energy
Kinetic Energy
• 11. Example of a MKE Problem
• Mechanical KE problems involve objects that are in motion.
• PE=KE or mgh=1/2mv 2 = 1/2(mass)(velocity 2 )
• If one knows the PE within a certain system, then one can find the variables such as v (velocity)
• 12. Example of a MKE problem
• A roller coaster of 2 metric tons is sitting at the top of the track at a height of 60 meters has a PE of 1,177,200 . What is the velocity when it reaches the ground?
• PE=KE, 1,177,200=1/2(2,000)(v 2 )
• sqrt(1,177,200/1,000)=sqrt(v 2 )
• v=34.3m/s
• 13. PE of a spring problem
• A 4 gram rubber band with the spring constant of 4.3 N/m is fired after being stretched a distance of 30 centimeters it has a PE is .645 Joules. What is the velocity when it is fired?
• PE s =KE, .645=.5(.004kg)(v 2 )
• sqrt(.645/.002)=sqrt(v 2 )
• v=18