Note: Slides are animations, so make sure to ’play’ them Homework Study examples at home for tomorrow’s special do now activity that has the potential to affect your grade Note: Slides are animations, so make sure to ’play’ them

Law of Conservation of Energy The total amount of energy in a closed system must remain constant – energy is never created or destroyed. Mechanical Energy = PE (both spring/gravitational) + KE Heat or Work done by friction = Q (Internal Energy)

Conservation of Energy http://www.physicsclassroom.com/Physics-Interactives/Work-and-Energy/Roller-Coaster-Model/Roller-Coaster-Model-Interactive

Example #1 – Cliff Diver A 500 newton cliff diver climbs to a height of 50 meters above the water. What is his total energy at the top of his climb? ET = PE + KE + Q ET = mgh + ½ mv2 + 0 ET = (500 N)(50 m) + 0 + 0 ET = 25000 J

Example #1 – Cliff Diver A 500 newton cliff diver climbs to a height of 50 meters above the water. What is his potential energy at the point where he is 40 meters above the water’s surface? ΔPE = mgΔh ΔPE = (500 N)(40 m) ΔPE = 20000 J

Example #1 – Cliff Diver A 500 newton cliff diver climbs to a height of 50 meters above the water. What is his kinetic energy at that point? ET = PE + KE + Q 25000 J = 20000 J + KE + 0 KE = 5000 J

PEtop = 25,000 J (relative to the water!) KEtop = 0 J 50 m PEmiddle = 12,500 J KEmiddle = 12,500 J 25 m 0 m PEbottom = 0 J KEbottom = 25,000 J

Example #2 – Carts and Hills ETOT = PE + KE + Q ETOT = 78.48 J + 50 J + 0 ETOT = 128.48 J KE = ½ mv2 KE = ½ (4.0 kg)(5 m/s)2 KE = 50 J KE = ½ mv2 128.48 J = ½ (4.0 kg)v2 v = 8.0 m/s Example #2 – Carts and Hills A 4.0 kilogram cart operating on a frictionless track has a speed of 5.0 meters per second while at the top of a 2.0 meter high hill. What is the cart’s total energy at the top of the hill? ET = PE + KE + Q ET = mgh + ½ mv2 + 0 (frictionless) ET = (4.0 kg)(9.81 m/s2)(2.0 m) + ½ (4.0 kg)(5.0 m/s)2 ET = 128.48 J

Example #2 – Carts and Hills A 4.0 kilogram cart operating on a frictionless track has a speed of 5.0 meters per second while at the top of a 2.0 meter high hill. What is the cart’s speed at the bottom of the hill? ET = PE + KE + Q 128.48 J = 0 + ½ mv2 + 0 128.48 J = ½ (4.0 kg)v2 v = 8.0 m/s

Example #3 – Spring Toy A spring toy with a mass of .002 kg has a spring with a spring constant of 120 newtons per meter. The toy is compressed .04 m. How much energy can be stored in the toy’s spring? PES = ½ kx2 PES = ½ (120 N/m)(0.04 m)2 PES = 0.096 J

Example #3 – Spring Toy A spring toy with a mass of 2.0 grams has a spring with a spring constant of 120 newtons per meter. The toy is compressed 4.0 centimeters. What maximum height will the spring toy reach if it is released? ΔPE = mgΔh 0.096 J = (0.002 kg)(9.81 m/s2) h h = 4.9 m

Label points where potential and kinetic energies are maximum/minimum. Example #4 – Pendulum Label points where potential and kinetic energies are maximum/minimum. PE max KE = 0 KE = 0 PE max KE max PE min

Walter Lewin MIT Demo https://www.youtube.com/watch?v=xXXF2C-vrQE