Principles of Technology/Physics in Context (PT/PIC) Chapter 7 Work and Energy 2 Text p

Key Objectives At the conclusion of this chapter you’ll be able to: Solve problems involving power and work. State the equation for calculating kinetic energy, and solve problems using this equation. State the equation for calculating gravitational potential energy, and solve problems using this equation. Solve problems that relate changes in kinetic energy to changes in gravitational potential energy.

7.3 ENERGY When work is done on an object, the “energy” of the object is changed. Energy is a very broad term related to work, and it has a variety of forms. In this chapter we will consider two of these forms, kinetic and potential energy.

7.3 ENERGY Together the kinetic energy and the potential (gravitational and elastic) energies are called the mechanical energy of the object.

7.3 ENERGY Kinetic Energy An object is traveling along a frictionless horizontal surface. A constant force is applied to the object in the direction of its displacement. What is the result of the work done on the object?

7.3 ENERGY Kinetic Energy We know from Newton’s second law that F = ma, therefore W = ma d. If the force on the object is also constant, its acceleration is constant and we can write

7.3 ENERGY If we solve this relationship for a d, we find that

7.3 ENERGY We can substitute this solution for a d into our work relationship:

7.3 ENERGY The work done on the object changes a quantity called kinetic energy that is related to the mass and the square of the speed of the object.

7.3 ENERGY We define kinetic energy (KE) to be: Therefore, work is equal to the change in kinetic energy. W = KE f - KE i = ΔKE.

Assessment Question 1 All of the following statements are TRUE EXCEPT: A.When work is done on an object, the “energy” of the object does not change. B.Energy is a very broad term related to work, and it has a variety of forms. C.Together the kinetic energy and the potential (gravitational and elastic) energies are called the mechanical energy of the object. D.Work is equal to the change in kinetic energy.

7.3 ENERGY PROBLEM A 10-kilogram object subjected to a 20.-newton force moves across a horizontal, frictionless surface in the direction of the force. Before the force was applied, the speed of the object was 2.0 meters per second. When the force is removed, the object is traveling at 6.0 meters per second. Calculate the initial kinetic energy (KE i )

7.3 ENERGY SOLUTION

7.3 ENERGY PROBLEM A 10-kilogram object subjected to a 20.-newton force moves across a horizontal, frictionless surface in the direction of the force. Before the force was applied, the speed of the object was 2.0 meters per second. When the force is removed, the object is traveling at 6.0 meters per second. Calculate the final kinetic energy (KE f )

7.3 ENERGY SOLUTION

7.3 ENERGY PROBLEM A 10-kilogram object subjected to a 20.-newton force moves across a horizontal, frictionless surface in the direction of the force. Before the force was applied, the speed of the object was 2.0 meters per second. When the force is removed, the object is traveling at 6.0 meters per second. Calculate the change in kinetic energy (ΔKE)

7.3 ENERGY SOLUTION

Assessment Question 2 A 10 kg (m) object subjected to 10 N (F) force moves across a horizontal, frictionless surface in the direction of the force. Before the force was applied, the speed of the object was 2 m/s (v i ) When the force is removed, the object is traveling at 3 m/s (v f ) Calculate the change in kinetic energy (ΔKE) ΔKE = KE f - KE i = ½ (m v f 2 ) – ½ (m v i 2 ) = ΔKE = ½( 10 kg∙ (3 m/s) 2 ) - ½( 10 kg∙ (2 m/s) 2 ) = A.10 JC. 20 J B.15 JD. 25 J

7.3 ENERGY PROBLEM A 10-kilogram object subjected to a 20.-newton force moves across a horizontal, frictionless surface in the direction of the force. Before the force was applied, the speed of the object was 2.0 meters per second. When the force is removed, the object is traveling at 6.0 meters per second. Calculate work (W)

7.3 ENERGY SOLUTION

7.3 ENERGY PROBLEM A 10-kilogram object subjected to a 20.-newton force moves across a horizontal, frictionless surface in the direction of the force. Before the force was applied, the speed of the object was 2.0 meters per second. When the force is removed, the object is traveling at 6.0 meters per second. Calculate distance (d)

7.3 ENERGY SOLUTION

7.3 ENERGY Gravitational Potential Energy Darya lifts a textbook vertically off a desk and holds it above her head. It is clear that Darya has done work on the book because she has applied a force through a distance. However the change in the kinetic energy of the book is zero. What did Darya’s work accomplish?

7.3 ENERGY Gravitational Potential Energy In this case, the work overcame the attraction of the gravitational field, and as a result the position of the book with respect to the Earth changed. We relate this change to a quantity we call gravitational potential energy.

7.3 ENERGY Gravitational Potential Energy To calculate the change in the gravitational potential energy (PE) of the object, we measure the work done on the object.

7.3 ENERGY Gravitational Potential Energy The force needed to overcome gravity is simply F g, which is equal to mg. Therefore, since W = F g ∙d, we define the change in the gravitational potential energy (ΔPE) as:

7.3 ENERGY Gravitational Potential Energy Here we use Δh, rather than d, to represent the change in vertical displacement above the Earth.

7.3 ENERGY Gravitational Potential Energy Since we are dealing with a scalar quantity, we will not consider the algebraic signs of the quantities involved.

7.3 ENERGY Gravitational Potential Energy We will simply agree that an object decreases its gravitational potential energy as it moves closer to the Earth (and vice versa).

Assessment Question 3 All of the following statements are TRUE EXCEPT: A.To increase the gravitational potential energy of an object work must be performed on it. B.You have greater gravitational potential energy in class and less when you are flying in an airplane. C.Gravitational Potential Energy is a scalar quantity. D.An object decreases its gravitational potential energy as it moves closer to the Earth

7.3 ENERGY Gravitational Potential Energy PROBLEM A 2.00-kilogram mass is lifted to a height of 10.0 meters above the surface of the Earth. Calculate the change in the gravitational potential energy of the object.

7.3 ENERGY Gravitational Potential Energy SOLUTION Since the object has moved away from the Earth’s surface, its gravitational potential energy has increased by 196 joules.

Assessment Question 4 A 12 kg (m) mass is lifted to a height of 15 m (h) above the surface of the Earth. g = 9.8 m/s 2 Calculate the change in the gravitational potential energy (ΔPE) of the object. ΔPE = mgh = 12 kg∙ 9.8 m/s 2 ∙ 15 m = A.520 J B.1800 J C.2500 J D.3200 J

7.3 ENERGY Gravitational Potential Energy For a change in gravitational energy to occur, there must be a change in the vertical displacement of an object; If it is moved only horizontally, its gravitational potential energy change is zero.

7.3 ENERGY Gravitational Potential Energy If an object is moved up an inclined plane, its potential energy change is measured by calculating only its vertical displacement; the horizontal part of its displacement does not change its potential energy.

Assessment Question 5 All of the following statements are TRUE EXCEPT: A.For a change in gravitational energy to occur, there must be a change in the vertical displacement of an object; B.If an object is moved only horizontally, its gravitational potential energy change is zero. C.If an object is moved up an inclined plane, its potential energy change is measured by calculating only its vertical displacement; D.The horizontal component of an object’s displacement is needed to calculate its potential energy.

7.3 ENERGY Interaction of Gravitational Potential and Kinetic Energy (Conservation of Mechanical Energy) Ketan tosses an object upward, and it returns to the Earth. Let’s analyze the motion of this object using an “energy” point of view. For simplicity we will ignore air resistance.

7.3 ENERGY Interaction of Gravitational Potential and Kinetic Energy (Conservation of Mechanical Energy) Ketan’s hand does work on the object. The work is transformed into kinetic energy. As the object rises, we observe that its speed decreases to zero. As a result, the kinetic energy of the object is decreasing while its potential energy is increasing. This represents a transformation of energy from kinetic to potential.

7.3 ENERGY Interaction of Gravitational Potential and Kinetic Energy (Conservation of Mechanical Energy) On the downward trip, the speed of the object increases. As the potential energy of the object decreases, its kinetic energy increases. This represents a transformation from potential to kinetic energy.

7.3 ENERGY Interaction of Gravitational Potential and Kinetic Energy (Conservation of Mechanical Energy) In the system, the sum of potential energy and kinetic energy (the total mechanical energy) has been conserved (i.e., is constant); a change in one is accompanied by an opposite change in the other.

7.3 ENERGY Interaction of Gravitational Potential and Kinetic Energy (Conservation of Mechanical Energy) In the system, the sum of potential energy and kinetic energy (the total mechanical energy) has been conserved (i.e., is constant);

7.3 ENERGY Interaction of Gravitational Potential and Kinetic Energy (Conservation of Mechanical Energy) This can also be expressed as the law of conservation of energy, which states that energy can neither be created nor destroyed.

7.3 ENERGY Interaction of Gravitational Potential and Kinetic Energy (Conservation of Mechanical Energy) In an ideal mechanical system (a closed system upon which no friction or other external forces act) the total mechanical energy is constant.

Assessment Question 6 All of the following statements are TRUE EXCEPT: A.According to the law of conservation of energy, friction forces can be ignored because energy is conserved. B.According to the law of conservation of energy, a change in potential or kinetic energy is accompanied by an opposite change in the other. C.According to the law of conservation of energy, the sum of potential energy and kinetic energy (total mechanical energy) in a system is conserved D.According to the law of conservation of energy, energy can neither be created nor destroyed.

7.3 ENERGY Interaction of Gravitational Potential and Kinetic Energy (Conservation of Mechanical Energy) PROBLEM A 0.50-kilogram ball is projected vertically and rises to a height of 2.0 meters above the ground. Calculate: (a) the increase in the ball’s potential energy

7.3 ENERGY Interaction of Gravitational Potential and Kinetic Energy (Conservation of Mechanical Energy) SOLUTION

7.3 ENERGY Interaction of Gravitational Potential and Kinetic Energy (Conservation of Mechanical Energy) PROBLEM A 0.50-kilogram ball is projected vertically and rises to a height of 2.0 meters above the ground. Calculate: (b) the decrease in the ball’s kinetic energy

7.3 ENERGY Interaction of Gravitational Potential and Kinetic Energy (Conservation of Mechanical Energy) SOLUTION

7.3 ENERGY Interaction of Gravitational Potential and Kinetic Energy (Conservation of Mechanical Energy) PROBLEM A 0.50-kilogram ball is projected vertically and rises to a height of 2.0 meters above the ground. Calculate: (c) the initial kinetic energy, and

7.3 ENERGY Interaction of Gravitational Potential and Kinetic Energy (Conservation of Mechanical Energy) SOLUTION (c) At the highest point, the speed of the ball is zero; therefore its kinetic energy is zero. As a result, the initial kinetic energy represents the change in the kinetic energy of the object.

7.3 ENERGY Interaction of Gravitational Potential and Kinetic Energy (Conservation of Mechanical Energy) PROBLEM A 0.50-kilogram ball is projected vertically and rises to a height of 2.0 meters above the ground. Calculate: (d) the initial speed of the ball.

7.3 ENERGY Interaction of Gravitational Potential and Kinetic Energy (Conservation of Mechanical Energy) SOLUTION (c)

7.3 ENERGY Interaction of Gravitational Potential and Kinetic Energy (Conservation of Mechanical Energy) SOLUTION (d)

Assessment Question 7 A 2 kg (m) ball is thrown vertically to a height of 10 m (h) above the surface of the Earth. g = 9.8 m/s 2 Calculate the initial speed (v i ) of the ball. ΔPE = ΔKE = mgh = ½ (m v i 2 ) v i = √(2gh) = √(2 ∙ 9.8 m/s 2 ∙ 10 m) = A.14 m/s B.22 m/s C.35 m/s D.49 m/s

7.3 ENERGY Interaction of Gravitational Potential and Kinetic Energy (Conservation of Mechanical Energy) An amusement-park roller coaster is an example of the interchange of the kinetic and potential energies of the coaster car.

7.3 ENERGY Interaction of Gravitational Potential and Kinetic Energy (Conservation of Mechanical Energy) As the car falls, its kinetic energy increases; as it rises, its kinetic energy decreases, as shown in the diagram.

7.3 ENERGY Interaction of Gravitational Potential and Kinetic Energy (Conservation of Mechanical Energy) Subsequent hills are made shorter and shorter so that the car will continue to have kinetic energy as it moves along the track.

7.3 ENERGY Interaction of Gravitational Potential and Kinetic Energy (Conservation of Mechanical Energy) A simple pendulum, shown in the following diagram, is another device that illustrates the transformation between kinetic and potential energies.

7.3 ENERGY Interaction of Gravitational Potential and Kinetic Energy (Conservation of Mechanical Energy) In the absence of friction, the swing of a pendulum back and forth will go on continuously. This type of motion, known as simple harmonic motion (SHM), occurs often in nature. For example, the oscillation of a spring and the vibration of a tuning fork are examples of SHM.

Assessment Question 8 All of the following statements are TRUE EXCEPT: A.An amusement-park roller coaster car falls, its kinetic energy increases; as it rises, its kinetic energy decreases, B.The final hill on a roller coaster track is always the highest hill on the track and the first hill is the shortest. C.In the absence of friction, the swing of a pendulum back and forth will go on continuously. D.A swinging pendulum, the oscillation of a spring and the vibration of a tuning fork and other examples of simple harmonic motion (SHM), occurs often in nature.

7.3 ENERGY Interaction of Gravitational Potential and Kinetic Energy (Conservation of Mechanical Energy) The time needed to complete one full swing of the pendulum is known as its period (T).

7.3 ENERGY Interaction of Gravitational Potential and Kinetic Energy (Conservation of Mechanical Energy) The period of a pendulum (for small angles) is related to the length of the pendulum ( l ) and to gravitational acceleration (g) according to the relationship Note that the period of a simple pendulum is independent of the mass of the bob.

Assessment Question 9 Calculate the period (T) of a pendulum with a length of 1.5 m ( l ). g = 9.8 m/s 2 T = 2π√( l /g) = T = 2∙3.14√(1.5 m/ 9.8 m/s 2 ) = A.2.5 s B.7.9 s C.14 s D.29 s

7.3 ENERGY Interaction of Gravitational Potential and Kinetic Energy (Conservation of Mechanical Energy) PROBLEM A pendulum whose bob weighs 12 newtons is lifted a vertical height of 0.40 meter from its equilibrium position. Calculate: (a) the change in potential energy between maximum height and equilibrium height

7.3 ENERGY Interaction of Gravitational Potential and Kinetic Energy (Conservation of Mechanical Energy) SOLUTION (a) We take the potential energy at the lowest point to be zero. Then:

7.3 ENERGY Interaction of Gravitational Potential and Kinetic Energy (Conservation of Mechanical Energy) PROBLEM A pendulum whose bob weighs 12 newtons is lifted a vertical height of 0.40 meter from its equilibrium position. Calculate: (b) the gain in kinetic energy

7.3 ENERGY Interaction of Gravitational Potential and Kinetic Energy (Conservation of Mechanical Energy) SOLUTION A pendulum whose bob weighs 12 newtons is lifted a vertical height of 0.40 meter from its equilibrium position. Calculate: (b) the gain in kinetic energy

7.3 ENERGY Interaction of Gravitational Potential and Kinetic Energy (Conservation of Mechanical Energy) PROBLEM A pendulum whose bob weighs 12 newtons is lifted a vertical height of 0.40 meter from its equilibrium position. Calculate: (c) the velocity at the equilibrium point.

7.3 ENERGY Interaction of Gravitational Potential and Kinetic Energy (Conservation of Mechanical Energy) SOLUTION (c) First we calculate the mass of the bob:

7.3 ENERGY Interaction of Gravitational Potential and Kinetic Energy (Conservation of Mechanical Energy) SOLUTION (c) Then we calculate the velocity:

Conclusion If work is done on an object, the work may be used to change the object’s kinetic energy (the energy associated with its motion), its potential energy (the energy associated with its position), or its internal energy (the energy associated with its atoms and molecules).

Assessment Question 10 A pendulum whose bob has a mass of 1 kg (m) is hangs 7 m (h) from its equilibrium point. g = 9.8 m/s 2 Calculate the change in potential energy (ΔPE) between maximum height and equilibrium height ΔPE = PE f -PE i = mgh - 0 = ΔPE = 1 kg ∙ 7m∙ 9.8 m/s 2 = A.25 J B.48 J C.69 J D.94 J