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## 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