GIG Read the passage and mark your answers on your whiteboard. NOT ON THE PAPER. Questions 10-11 or 25-26.

Conservation of Energy

Conservation of Energy The law of conservation of energy states that the total energy in the universe never increases or decreases. Likewise, within a closed system, the total energy remains unchanged.

A Note about Relativity The law of conservation of energy and the law of conservation of matter were discovered independently. The theory of relativity showed that matter and energy are related and matter can be transformed into energy (such as nuclear energy holding an atom together). Matter and energy form a continuum through which both are conserved.

Conservation of Energy in an Ideal System In high school physics, we perform conservation of energy problems within an ideal system where all energy is conserved in either kinetic or potential form, which we call mechanical energy. Mechanical energy is the sum of the kinetic energy and potential energy of an object. It is represented by E.

Formula for Mechanical Energy E = PE + KE Because energy is conserved, this can also be written as: E = PEi + KEi = PEf + KEf Mechanical Energy = initial potential energy + initial kinetic energy = final potential energy + final kinetic energy

Finding Velocity One of the main uses of the mechanical energy formula is to find velocity of an object in free fall.

E = PEi + KEi = PEf + KEf PE = mgh (mass ✕ gravity ✕ height) If we treat the spot where the ball was hit as 0m PEi = 0 PEf = .15kg ✕ 9.81m/s2 ✕ 7.2m = 11J

E = PEi + KEi = PEf + KEf KE = .5m(v2) ½ mass ✕ velocity2 KEi = (.5)(.15kg)(36m/s)2 = 97J

E = PEi + KEi = PEf + KEf Put it all together to get KEf 0J + 97J = 11J + KEf KEf = 86J

E = PEi + KEi = PEf + KEf Now, solve for vf 86J = .5mvf2 86J = (.5)(.15kg)(vf2) 1146.7J/kg = vf2 Vf = 34m/s

Practice Problems Read the examples on pages 209 and 210. Then, try the LessonCheck problems on page 211.

Homework None.

Closure A 96.5kg stuntman jumps from a rooftop 7.4m above the ground, toward a pile of styrofoam cubes. What is his velocity when he is 1.0m above the ground?

Closure A 96.5kg stuntman jumps from a rooftop 7.40m above the ground, toward a pile of styrofoam cubes. What is his velocity when he is 1.00m above the ground? PEi = 96.5kg(9.81m/s2)(7.40m) = 7005J KEi = 0J PEf = 96.5kg(9.81m/s2)(1.00)m = 947J

Closure A 96.5kg stuntman jumps from a rooftop 7.40m above the ground, toward a pile of styrofoam cubes. What is his velocity when he is 1.00m above the ground? PEi + KEi = 7005J = PEf + KEf KEf = 7005J - 947J = 6058J

Closure A 96.5kg stuntman jumps from a rooftop 7.40m above the ground, toward a pile of styrofoam cubes. What is his velocity when he is 1.00m above the ground? KEf = 6058J = .5m(vf2) 6058J = .5(96.5kg)(vf2) vf2 = 125.6m2/s2 vf = 11.2 m/s

Closure A 96.5kg stuntman jumps from a rooftop 7.40m above the ground, toward a pile of styrofoam cubes. What is his velocity when he is 1.00m above the ground? Another way to solve this is to use the vertical position- time formula to solve for time, and then use the vertical velocity-time equations to find final velocity. Again, vf = 11.2m/s