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Potential and Kinetic Energy Problems. What is the potential energy of a rock that weighs 100 Newtons that is sitting on top of a hill 300 meters high?. Potential and Kinetic Energy Problems.
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Potential and Kinetic Energy Problems What is the potential energy of a rock that weighs 100 Newtons that is sitting on top of a hill 300 meters high?
Potential and Kinetic Energy Problems What is the potential energy of a rock that weighs 100 Newtons that is sitting on top of a hill 300 meters high? PE = mgh
Potential and Kinetic Energy Problems What is the potential energy of a rock that weighs 100 Newtons that is sitting on top of a hill 300 meters high? PE = mgh PE = 100N x 300m
Potential and Kinetic Energy Problems What is the potential energy of a rock that weighs 100 Newtons that is sitting on top of a hill 300 meters high? PE = mgh PE = 100N x 300m PE = 30,000N*m
Potential and Kinetic Energy Problems What is the potential energy of a rock that weighs 100 Newtons that is sitting on top of a hill 300 meters high? PE = mgh PE = 100N x 300m PE = 30,000N*m PE = 30,000J
Potential and Kinetic Energy Problems 2. What is the kinetic energy of a bicycle with a mass of 14 kg traveling at a velocity of 3 m/s?
Potential and Kinetic Energy Problems KE = ½ mv2 2. What is the kinetic energy of a bicycle with a mass of 14 kg traveling at a velocity of 3 m/s?
Potential and Kinetic Energy Problems KE = ½ mv2 KE= ½ (14kg)(3m/s)2 2. What is the kinetic energy of a bicycle with a mass of 14 kg traveling at a velocity of 3 m/s? Square 1st
Potential and Kinetic Energy Problems KE= ½ mv2 KE= ½ (14kg)(3m/s)2 KE = ½ (14kg) (9m2/s2) 2. What is the kinetic energy of a bicycle with a mass of 14 kg traveling at a velocity of 3 m/s? Multiply 2nd
Potential and Kinetic Energy Problems KE= ½ mv2 KE= ½ (14kg)(3m/s)2 KE = ½ (14kg) (9m2/s2) KE = ½ 126 kg*m2/s2 = 126 N*m 2. What is the kinetic energy of a bicycle with a mass of 14 kg traveling at a velocity of 3 m/s?
Potential and Kinetic Energy Problems KE= ½ mv2 KE= ½ (14kg)(3m/s)2 KE = ½ (14kg) (9m2/s2) KE = ½ 126 kg*m2/s2 = 126 N*m 2. What is the kinetic energy of a bicycle with a mass of 14 kg traveling at a velocity of 3 m/s? Kg*m/s2
Potential and Kinetic Energy Problems KE= ½ mv2 KE= ½ (14kg)(3m/s)2 KE = ½ (14kg) (9m2/s2) KE = ½ 126 kg*m2/s2 = 126 N*m 2. What is the kinetic energy of a bicycle with a mass of 14 kg traveling at a velocity of 3 m/s? Divide last
Potential and Kinetic Energy Problems KE= ½ mv2 KE= ½ (14kg)(3m/s)2 KE = ½ (14kg) (9m2/s2) KE = ½ 126 kg*m2/s2 = 126 N*m KE = 63 Joules 2. What is the kinetic energy of a bicycle with a mass of 14 kg traveling at a velocity of 3 m/s?
Potential and Kinetic Energy Problems 3. A 1-kilogram ball has a kinetic energy of 50 joules. What is the velocity of the ball?
Potential and Kinetic Energy Problems 3. A 1-kilogram ball has a kinetic energy of 50 joules. What is the velocity of the ball? KE = ½ mv2
Potential and Kinetic Energy Problems 3. A 1-kilogram ball has a kinetic energy of 50 joules. What is the velocity of the ball? KE= ½ mv2 50J = ½ (1kg)v2
Potential and Kinetic Energy Problems 3. A 1-kilogram ball has a kinetic energy of 50 joules. What is the velocity of the ball? KE= ½ mv2 50J = ½ (1kg)v2 2 x 50J = (1kg)v2x 2 2
Potential and Kinetic Energy Problems 3. A 1-kilogram ball has a kinetic energy of 50 joules. What is the velocity of the ball? KE= ½ mv2 50J = ½ (1kg)v2 2 x 50J = (1kg)v2x 2 2 100kg*m2/s2 = (1kg)v2 1 kg 1kg
Potential and Kinetic Energy Problems 3. A 1-kilogram ball has a kinetic energy of 50 joules. What is the velocity of the ball? KE= ½ mv2 50J = ½ (1kg)v2 2 x 50J = (1kg)v2x 2 2 100kg*m2/s2 = (1kg)v2 1 kg 1kg 100 m2/s2 = v2
Potential and Kinetic Energy Problems 3. A 1-kilogram ball has a kinetic energy of 50 joules. What is the velocity of the ball? KE= ½ mv2 50J = ½ (1kg)v2 2 x 50J = (1kg)v2x 2 2 100kg*m2/s2 = (1kg)v2 1 kg 1kg 100 m2/s2 = v2 10m/s = v
Potential and Kinetic Energy Problems 4. What is the potential energy of the rock?
Potential and Kinetic Energy Problems 4. What is the potential energy of the rock? Given in kg, not Newtons…must multiply mass times gravity
Potential and Kinetic Energy Problems 4. What is the potential energy of the rock? PE = mgh
Potential and Kinetic Energy Problems 4. What is the potential energy of the rock? PE = mgh PE = 95kg(9.8m/s2)(100m)
Potential and Kinetic Energy Problems 4. What is the potential energy of the rock? PE = mgh PE = 95kg(9.8m/s2)(100m) PE = 93,100 kg*m2/s2 kg*m/s2 = N *m
Potential and Kinetic Energy Problems 4. What is the potential energy of the rock? PE = mgh PE = 95kg(9.8m/s2)(100m) PE = 93,100 kg*m2/s2 kg*m/s2 = N *m PE = 93,100 N*m
Potential and Kinetic Energy Problems 4. What is the potential energy of the rock? PE = mgh PE = 95kg(9.8m/s2)(100m) PE = 93,100 kg*m2/s2 kg*m/s2 = N *m PE = 93,100 N*m PE = 93,100 J
Potential and Kinetic Energy Problems 5. What is the approximate difference in gravitational potential energy of the two shaded boxes?
Potential and Kinetic Energy Problems 5. What is the approximate difference in gravitational potential energy of the two shaded boxes? PE = mgh
Potential and Kinetic Energy Problems 5. What is the approximate difference in gravitational potential energy of the two shaded boxes? PE = mgh PE = 2.0kg(9.8m/s2)(3.0m)
Potential and Kinetic Energy Problems 5. What is the approximate difference in gravitational potential energy of the two shaded boxes? PE = mgh PE = 2.0kg(9.8m/s2)(3.0m) PE = 58.8 kg*m2/s2 or N*m
Potential and Kinetic Energy Problems 5. What is the approximate difference in gravitational potential energy of the two shaded boxes? PE = mgh PE = 2.0kg(9.8m/s2)(3.0m) PE = 58.8 kg*m2/s2 or N*m PE = 58.8 J
Potential and Kinetic Energy Problems 5. What is the approximate difference in gravitational potential energy of the two shaded boxes? PE = mgh
Potential and Kinetic Energy Problems 5. What is the approximate difference in gravitational potential energy of the two shaded boxes? PE = mgh PE = 2.0kg (9.8m/s2)(1.0m)
Potential and Kinetic Energy Problems 5. What is the approximate difference in gravitational potential energy of the two shaded boxes? PE = mgh PE = 2.0kg (9.8m/s2)(1.0m) PE = 19.6 kg*m/s2 or N*m
Potential and Kinetic Energy Problems 5. What is the approximate difference in gravitational potential energy of the two shaded boxes? PE = mgh PE = 2.0kg (9.8m/s2)(1.0m) PE = 19.6 kg*m/s2 or N*m PE = 19.6 J
Potential and Kinetic Energy Problems 5. What is the approximate difference in gravitational potential energy of the two shaded boxes? 58.8J – 19.6J = 39.2J