4 - WorkPowerEnergy - Kinetic, Potential and Elastic Energy

University:
High School
Course:
High School Physics
Academic year:

2024
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194

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12
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Unit 5: Work, Power and Energy Gravitational Potential, Kinetic and Elastic Potential Energy 5.4 Gravitational Potential Energy: ◼ Energy stored in a raised object Eg = mgh Note: Same formula as W = Fd h = h final − hinitial Gravitational Potential Energy (J) Mass of object (kg) (m) Gravity (N/kg) Ex #1 How much potential energy does a 61.2 kg boy have if he is sitting on his bed, 0.500 m above the floor? m = 61.2kg g = 9.80N/kg ∆h = 0.500m Eg = mgh Eg = (61.2)(9.80)(0.500) E g = 3.00  10 J 2 Ex #2 A 0.0400kg rubber ball drops from a height of 5.00m to the ground & bounces back to a height of 3.00m. a) What is the change in potential energy of the ball from release to ground?E g = (0.04)(9.8)(−5) m = 0.0400kg g = 9.80N/kg Eg = mgh E g = −1.96J ∆h = -5.00 m b) How much potential energy does the ball regain on the trip back up? E = 1.18J g Eg = mgh Eg = (0.04)(9.8)(3) 5.5 Kinetic Energy: Energy of an object due to its motion ◼ Result of work having been done to the object 2 k ◼ 1 E = mv 2 Kinetic Energy (J) Mass of object (kg) Speed of the object (m/s) Ex #3 What is the kinetic energy of a 6.00kg curling stone sliding at 4.00m/s? m = 6 kg v = 4m/s 1 2 Ek = mv 2 1 2 Ek = (6)(4) 2 E k = 48.0 J Ex #4 What is the speed of a 5.44kg shotput if its kinetic energy is 68.0J? m = 5.44 kg Ek = 68J 1 2 Ek = mv 2 1 2 68 = (5.44)(v ) v = 5.00m / s 2 5.6 Elastic Potential Energy: ◼ Energy stored in an object due to a stretch or compression from equilibrium. 1 2 E e = k ( x ) 2 Elastic Potential Energy (J) Spring constant of object (N/m) Displacement from equilibrium position (m) Ex #5 What is the elastic potential energy stored in a spring with a spring constant of 60.0N/m after it has been stretched 25.0cm? k = 60.0N/m Δx = 0.250m 1 2 E e = k ( x ) 2 1 2 EE = (60.0)(0.25) E e = 1.88J 2 Ex #6 What is the spring constant of a spring that compresses 62.5 cm when 25.0 J of work is done on the spring? Ee = 25.0 J Δx = 0.625 m 1 2 E e = k ( x ) 2 1 2 25 = k (0.625) 2 k = 128N / m Ex #7 A 0.500kg rubber ball is thrown into the air. At a height of 20.0m, it is traveling at 15.0m/s. What is the ball’s Ek, Eg, and total energy? m = 0.5 kg v = 15m/s 1 Ek = mv 2 2 So ball has 154.3 J of total energy 1 Ek = (0.5)(15) 2 2 E k = 56.3J m = 0.5 kg g = 9.80N/kg h = 20 m Eg = mgh Eg = (0.5)(9.8)(20) E g = 98.0 J Quantum: HW: pg 341 #2, 4 pg 349 #1, 4 pg 373 #1, 5