1. Conservation of Mechanical Energy Mechanical Energy – The sum of Potential and Kinetic Energies ME=PE+KE The conservation of mechanical energy states that mechanical energy remains constant in the absence of air resistance or friction (non-conservative forces). Conservation of mechanical energy is path independent. Conservation of mechanical energy is path independent. PE=100 J KE=0 J ME=100 J PE=30 J KE=70 J ME=100 J PE=0 J KE=100 J ME = 100 J

2. Conservation of Mechanical Energy (Pendulum) Maximum potential Energy, zero kinetic energy Maximum kinetic energy, zero potential energy Maximum potential Energy, zero kinetic energy Maximum kinetic Energy, zero potential energy Maximum potential Energy, zero kinetic energy Maximum potential Energy, zero kinetic energy PE=10 J KE=0 J ME=10 J PE=0 J KE=10 J ME=10 J PE=10 J KE=0 J ME=10 J

3. Conservation of Mechanical Energy (Problem Solving) ME 1 =ME 2 ME 1 =ME 2 PE 1 +KE 1 =PE 2 +KE 2 PE 1 +KE 1 =PE 2 +KE 2 mgh 1 + ½ mv 1 2 =mgh 2 + ½ mv 2 2 mgh 1 + ½ mv 1 2 =mgh 2 + ½ mv 2 2 If v 1 =0 and h 2 =0 then If v 1 =0 and h 2 =0 then mgh 1 + 0 = 0+ ½ mv 2 2 mgh 1 + 0 = 0+ ½ mv 2 2 PE 1 =KE 2 (The initial potential energy is converted to the final kinetic energy. The work accomplished by gravity causes the change of potential to kinetic energy.) PE 1 =KE 2 (The initial potential energy is converted to the final kinetic energy. The work accomplished by gravity causes the change of potential to kinetic energy.) 1 2 h1h1 h2h2

4. Conservation of Mechanical Energy (Something to Consider) ME 1 =ME 2 ME 1 =ME 2 PE 1 +KE 1 =PE 2 +KE 2 PE 1 +KE 1 =PE 2 +KE 2 KE 2 -KE 1 =-(PE 2 -PE 1 ) ΔKE=-ΔPE What is the interpretation of this equation? Multiply by -1 ΔPE=-ΔKE (Interpret this equation) 1 2 h1h1 h2h2

5. Conservation of Mechanical Energy (More things to consider) Is mass a factor in determining height or velocity? Is mass a factor in determining height or velocity? ME 1 =ME 2 ME 1 =ME 2 PE 1 +KE 1 =PE 2 +KE 2 PE 1 +KE 1 =PE 2 +KE 2 mgh 1 + ½ mv 1 2 =mgh 2 + ½ mv 2 2 mgh 1 + ½ mv 1 2 =mgh 2 + ½ mv 2 2 gh 1 + ½ v 1 2 =gh 2 + ½ v 2 2 gh 1 + ½ v 1 2 =gh 2 + ½ v 2 2 Multiply by 2 Multiply by 2 2gh 1 +v 1 2 =2gh 2 +v 2 2 (Is this equation familiar?) 2gh 1 +v 1 2 =2gh 2 +v 2 2 (Is this equation familiar?) v 2 2 =v 1 2 +2gh 1 -2gh 2 (Is this equation familiar?) v 2 2 =v 1 2 +2gh 1 -2gh 2 (Is this equation familiar?) v 2 2 =v 1 2 +2g(h 1 -h 2 ) (Is this equation familiar?) v 2 2 =v 1 2 +2g(h 1 -h 2 ) (Is this equation familiar?) g=-9.8 m/s 2, d=h 1 -h 2 g=-9.8 m/s 2, d=h 1 -h 2 v 2 2 =v 1 2 +2gd (Is this equation familiar?) v 2 2 =v 1 2 +2gd (Is this equation familiar?)