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Do Now for Wed., Feb. 12, 2014

Do Now for Wed., Feb. 12, 2014. 1. Write the definition and formula for potential energy. 2. Write the definition and formula for kinetic energy. Learning Intentions: Students will explain that energy is conserved & transformed from one type of energy to another.

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Do Now for Wed., Feb. 12, 2014

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  1. Do Now for Wed., Feb. 12, 2014 1. Write the definition and formula for potential energy. 2. Write the definition and formula for kinetic energy.

  2. Learning Intentions: • Students will explain that energy is conserved & transformed from one type of energy to another. • Students will apply the concept of conservation of energy to analyze the potential & kinetic energy of an object at various positions in its motion. Success Criteria: Students will apply the concept of conservation of energy to analyze the potential and kinetic energy of an object at various positions in its motion.

  3. Review of Energy Energyis the property of an object or system that allows it to do work. Energy is measured in Joules (J). Pushing a 1-kilogram object with a force of one newton for a distance of one meter uses one jouleof energy.

  4. Some Forms of Energy Mechanical energy is the energy possessed by an object due to its motion or its position. • Potential energy and kinetic energy are both forms of mechanical energy.

  5. CALCULATING POTENTIAL AND KINETIC ENERGY The formula for Potential Energy is: PE = mghwhere: PE = potential energy in Joules (J) m = mass in kilograms (kg) g = the acceleration of gravity on earth (g = 9.8 m/s2) The formula for Kinetic Energy is: KE = ½ mv2where: KE = kinetic energy in Joules (J) m = mass in kilograms (kg) v = speed in meters per second (m/s)

  6. PROBLEM 1-Calculate Potential Energy A 2 kg rock is at the edge of a cliff 20 meters above a lake. It becomes loose and falls toward the water below. Before the rock begins to fall, what is its potential energy? Looking for: PE Given: m = 2 kg; h = 20 m; g = 9.8 m/s2 Relationships: PE = mgh Solution: PE = 2 kg x 9.8 m/s2 x 20 m PE = 392 J

  7. PROBLEM 2 - Calculate Kinetic Energy Calculate the kinetic energy of a toy car with a mass of 2 kg that is moving at a speed of 2 m/s. Looking for: KE Given: m = 2 kg; v = 2 m/s Relationship: KE = ½ mv2; Solution: KE = ½ (2 kg)( 2 m/s)2 = ½ (2 kg)(4 m2/s2) KE = 4 J

  8. Conservation of Energy The idea that energy transforms from one form into another without a change in the total amount is called the Law of Conservation of Energy. The law of energy conservation says: - The total energy before a change equals the total energy after the change. - Energy cannot be created or destroyed. It can be transformed from one form into another, but the total amount of energy never changes.

  9. Conservation of Energy in a Pendulum A pendulum is a system in which a mass, called a bob, swings back & forth on a string.

  10. Conservation of Energy in a Pendulum When the bob is at its highest point, it briefly stops & all of its energy is PE,potential energy. When the bob is at its lowest point, all its energy is KE, kinetic energy.

  11. Conservation of Energy At each point in its travel, the sum of its potential & kinetic energy, PE + KE, is the same. The bob continues to travel until its energy is converted into heat by friction.

  12. Thurs. Feb. 13 Do Now • Calculate the potential energy of a person whose mass is 100 kg standing on a bridge 50 m above a river. • Calculate the kinetic energy of a horse of mass 1760 kg running at a speed of 10 m/s.

  13. Conservation of Energy in a Tossed Ball When you throw a ball in the air, the ball starts with kinetic energy. As the ball rises, some of its kinetic energy is transformed into potential energy. At the top of its travel, where it is momentarily stopped, the ball’s energy is all potential energy.

  14. Conservation of Energy in a Tossed Ball As the ball drops from its highest point, its potential energy is transformed back into kinetic energy.

  15. Let’s go back to Problem 1 & solve for PE & KE as the rock falls: A 2 kg rock is at the edge of a cliff 20 meters above a lake. It becomes loose and falls toward the water below. What is the potential and kinetic energy of the rock at the following positions: 1. Before the rock begins to fall (h= 20 m), 2. When the rock has fallen half way to the ground (h = 10 m), 3. Just before the rock hits the ground (h = 0 m).

  16. Solving Problems Looking for: PE& KE at h = 20 m, h = 10 m, & h = 0 m Given: mass = 2.0 kg; Starting: h = 20 m; v = 0 m/s Half way down: h = 10 m Before hitting the ground: h = 0 m Relationships: Assume the rock starts from rest (v = 0 at h = 20 m) PE =mgh; KE = ½ mv2 PE + KE is constant by the Law of Conservation of Energy.

  17. Solution: at the top of the cliff, h = 20 m m = 2 kg h = 20 m PE= (2 kg)(9.8 N/kg)(20 m) PE = 392 J h= 10 m; Before it drops, the rock is at rest so v = 0 and: KE= 0 J h = 0 m

  18. m = 2 kg Halfway down the cliff, where h = 10 m: • PE= (2 kg)(9.8 N/kg)(10 m) = 196 J By Conservation of Energy, PE + KE is the same at all heights so at h = 10 m: KE + 196 J = 392 J KE = 392J – 196J = 196 J Half of the potential energy (PE) has been transformed into kinetic energy, (KE). h= 10 m; h = 0 m

  19. m = 2 kg Just before the rock hits the ground, h = 0 m. PE = 2 kg x 9.8 m/s2 x 0 m = 0 J By Conservation of Energy, PE + KE is the same at all heights so at h = 0 m: KE = 392 J All of the potential energy from the top of the cliff has been transformed into kinetic energy as the rock is about to hit the ground! h = 0 m

  20. Calculate what happens when the car from Problem 2 rolls up a ramp: We calculated the kinetic energy of the toy car with a mass of 2 kg moving at a speed of 2 m/sto be KE = 4 J. If the car rolls up a ramp, how high will it travel before all of its kinetic energy is transformed into potential energy, causing the car to stop?

  21. When the car stops on the ramp, its kinetic energy will be 0 J & its potential energy will be 4 J (because all the KE has been transformed into PE).Looking for: hGiven: PE = 4 J, m = 2 kg, g = 9.8 m/s2Relationship: PE = mghSolution: 4 J = 2 kg x 9.8 m/s2 x h4 J = 19.6 kg m/s2 x h divide each side by 19.6 to solve for hh = 4/19.6 m = 0.20 m (20 cm)

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