3.2.3 Conservation of Energy Which Way Does it Flow? 3.2.3 Conservation of Energy
Law of Conservation of Energy The total amount of energy in a closed system must remain constant – energy is never created or destroyed.
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 – Skier A skier begins at rest at the top of a snowy, frictionless hill. He slides down the hill and after reaching the bottom slides across a long patch of unpacked snow. At which points is the total energy completely potential / kinetic / internal? PE max Q only Heat from Friction KE max Total Mechanical Energy = PE + KE
Example #3 – 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 #3 – 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 #3 – 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 – Carts and Hills PE at the top becomes KE. Some energy bounces back and forth between KE and PE. In the end, all of the PE becomes KE. PE max Some PE / Some KE KE max
Example #4 – 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. 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 #4 – 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 #5 – Pendulum Label points where potential and kinetic energies are maximum/minimum. PE max KE = 0 KE = 0 PE max KE max PE min
End of 3.2.3 - PRACTICE