Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Chapter 11. Work Chapter 11. Work Chapter Goal: To develop a more complete understanding of energy and its conservation.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Ch 11 Student Learning Objectives To introduce and use the basic energy model. To recognize transformations between kinetic, potential, and thermal energy. To define work and use the work-kinetic energy theorem. To develop a complete statement of the law of conservation of energy.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The Basic Energy Model W > 0: The environment does work on the system and the system’s energy increases. W < 0: The system does work on the environment and the system’s energy decreases.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Work and Kinetic Energy A force acts on a particle as the particle moves along the s- axis from s i to s f. The force component F s parallel to the s- axis causes the particle to speed up or slow down, thus transferring energy to or from the particle. We say that the force does work on the particle.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Work and Kinetic Energy A force acts on a particle as the particle moves along the s- axis from s i to s f. The force component F s parallel to the s- axis causes the particle to speed up or slow down, thus transferring energy to or from the particle. We say that the force does work on the particle. As the particle is moved by this single force, its kinetic energy changes as follows:

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Work-Kinetic Energy Theorem. Starting with Newton’s 2 nd Law: F s = ma s, it can be shown that: = ½ mv f 2 – ½ mv 0 2 or: This result is known as the Work-Kinetic Energy Theorem. Note that there is no mention of U, or potential energy, as of yet.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. A particle moving along the x-axis experiences the force shown in the graph. If the particle has 2.0 J of kinetic energy as it passes x = 0 m, what is its kinetic energy when it reaches x = 4 m?

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. A particle moving along the x-axis experiences the force shown in the graph. If the particle has 2.0 J of kinetic energy as it passes x = 0 m, what is its kinetic energy when it reaches x = 4 m? 6.0 J

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Work Done by a Constant Force A particle experiences a constant force which makes an angle θ with respect to the particle’s displacement, ∆r. The work done is Both F and θ are constant, so they can be taken outside the integral. Thus This as the dot product of the force vector and the displacement vector:

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Work done by a constant Force Therefore, for a constant force: W = |F∆r| cos θ

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Tactics: Calculating the work done by a constant force

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Tactics: Calculating the work done by a constant force

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Evaluate the dot products 1. 2.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. 1a)15.3 1b) c) 0 2a) b) -20 2c)

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. A. W g and W T are both zero. B. W g is negative and W T is negative. C. W g is negative and W T is positive. D. W g is positive and W T is positive. E. W g is positive and W T is negative. A crane lowers a steel girder into place at a construction site. The girder moves with constant speed. Consider the work W g done by gravity and the work W T done by the tension in the cable. Which of the following is correct?

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. A. W g and W T are both zero. B. W g is negative and W T is negative. C. W g is negative and W T is positive. D. W g is positive and W T is positive. E. W g is positive and W T is negative. A crane lowers a steel girder into place at a construction site. The girder moves with constant speed. Consider the work W g done by gravity and the work W T done by the tension in the cable. Which of the following is correct?

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Workbook Problem #7e

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Workbook Problem #7e

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The Work Done by a Variable Force To calculate the work done on an object by a force that either changes in magnitude or direction as the object moves, we use the following: We must evaluate the integral either geometrically, by finding the area under the curve, or by actually doing the integration (!).

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Workbook #13 A 1 kg particle moving along the x-axis experiences the force shown in the graph. The particle’s speed is 2 m/s at x = 0 m. What is its speed when it gets to x = 5 m?

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Workbook #13 A 1 kg particle moving along the x-axis experiences the force shown in the graph. The particle’s speed is 2 m/s at x = 0 m. What is its speed when it gets to x = 5 m? Ans: 4 m/s

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Conservation of Mechanical Energy revisited If no energy is added or removed from the system: K 0 + U g0 + U s0 = K + U g + U s If energy is added or removed via work, this becomes: K 0 + U g0 + U s0 + W = K + U g + U s

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Conservation of Mechanical Energy revisited K 0 + U g0 + U s0 + W = K + U g + U s It seems this would conflict with the Work- Kinetic Energy, W = K – K 0 This would imply that perhaps potential energy is a form of work. It can be shown that W = - ∆U But only for conservative forces. A conservative force is one that is path- independent.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The Work-Kinetic Energy Theorem when Nonconservative Forces Are Involved A force for which the work is not independent of the path is called a nonconservative force. It is not possible to define a potential energy for a nonconservative force. If W c is the work done by all conservative forces, and W nc is the work done by all nonconservative forces, then But the work done by the conservative forces is the negative of the change in potential energy, so the work- kinetic energy theorem becomes

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Energy Bar Charts We may express the conservation of energy concept as an energy equation. We may also represent this equation graphically with an energy par chart.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. EOC #36 A particle moves from A to D while experiencing a force of N. Calculate the work done for: a.Path ABD b.Path ACD c.Path AD d.Is this a conservative force?

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. EOC #36 A particle moves from A to D while experiencing a force of N. Calculate the work done for: a.Path ABD: 50 J b.Path ACD: 50 J c.Path AD: 50 J d.Is this a conservative force? - Yes

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. EOC #36 -Answer Path ABD: 50 J Path ACD

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Conservation of Mechanical Energy, revisited K 0 + U g0 + U s0 + W nc = K + U g + U s can be rewritten as: ∆K + ∆U = W nc and ∆K + ∆U = ∆E mech Therefore mechanical energy is conserved if there are no non-conservative forces doing work on the system

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Finding Force from Potential Energy W = - ∆U: The work done by a conservative force is equal to the negative change in potential energy Therefore: F s = - dU/ds so F s = dW/ds

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. A function and its derivative F s = - dU/ds The derivative of a function shows the trend the change of that function. If the derivative is zero, the function is not changing (but not necessarily 0!). If the derivative is constant, the function is changing at a constant rate. If the derivative is increasing/decreasing the function is changing increasingly faster/slower. The negative sign is a bit of a challenge.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The top graph is of the gravitational force, mg which we model as constant. In our coordinate system it is negative. The gravitational potential energy curve is linear (changing at a constant rate), with a positive slope equal to the magnitude of mg.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. A particle moves along the x-axis with the potential energy shown. The force on the particle when it is at x = 4 m is

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. A particle moves along the x-axis with the potential energy shown. The force on the particle when it is at x = 4 m is –2 N.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Workbook #17 a,b,c The graph shows the potential- energy curve of a particle moving along the x-axis under the influence of a conservative force. a.At which intervals of x is the force on the particle to the right (positive)? b.At which intervals of x is the force on the particle to the left (negative)? c.At what value(s) of x is the force on the particle maximum? d. At what value(s) is the force on the particle zero?

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Workbook #17 a,b,c d. force is zero at: x=2m, x=5m, x=8m (where slope is zero).

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Dissipative Forces – Two perspectives Dissipative forces (e.g. friction and drag) can be modeled in two ways: 1.One can consider the dissipative force as an external, non-conservative force that is always negative and takes energy away from the system: K 0 + U g0 + U s0 + W nc – W dis = K + U g + U s

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Dissipative Forces – Two perspectives 2. One can consider the dissipative force as increasing the thermal energy of the system, as discussed in the text: K 0 + U g0 + U s0 + W nc = K + U g + U s + ∆E th This method is probably the more correct way to think about it. However ∆E th can only be computed as work (e.g. ∆E th due to friction is u k n ∆s), so in the end it doesn’t matter where you include it on the bar chart or the equation. Pick a method and stick with it

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Conservation of Energy

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. A child at the playground slides down a pole at constant speed. This is a situation in which A. U  E th. E mech is conserved. B. U  E th. E mech is not conserved but E sys is. C. U  W ext. Neither E mech nor E sys is conserved. D. U  K. E mech is not conserved but E sys is. E. K  E th. Emech is not conserved but E sys is.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. A child at the playground slides down a pole at constant speed. This is a situation in which A. U  E th. E mech is conserved. B. U  E th. E mech is not conserved but E sys is. C. U  W ext. Neither E mech nor E sys is conserved. D. U  K. E mech is not conserved but E sys is. E. K  E th. Emech is not conserved but E sys is.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. A child at the playground slides down a pole at constant speed. This is a situation in which A. U  Eth. Emech is conserved. B. U  Eth. Emech is not conserved but Esys is. C. U  Wext. Neither Emech nor Esys is conserved. D. U  K. Emech is not conserved but Esys is. E. K  Eth. Emech is not conserved but Esys is.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. EOC problems 26, 27,28

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. EOC #26 - Answer

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. EOC #27 - Answer

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. EOC #28 - Answer W ext = -1 J. This means energy was transferred from the system to the environment

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. EOC #37 A 100 g particle experiences only the one-dimensional, conservative force shown in the figure. a.Draw a graph of potential energy from x = 0m to x = 5 m. Assume U = 0 J at 0m. b.The particle is shot toward the right from x = 1.0 m with a speed of 25 m/s. What is the particle’s total mechanical energy?

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. EOC #37 a.see graph below b.Total E mech = E mech = (K + U)

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. EOC #48 A runaway truck ramp near Denver, CO slopes upward at 6.0° and has a large coefficent of rolling friction, u r = Find the length of a ramp that will stop a 15,000 kg truck that enters the ramp at 35.0 m/s. a. Draw an energy bar chart for this situation. b. Use work and energy and kinematics to solve.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. EOC #48: Answer = 124 meters A runaway truck ramp near Denver, CO slopes upward at 6.0° and has a large coefficent of rolling friction, u r = Find the length of a ramp that will stop a 15,000 kg truck that enters the ramp at 35.0 m/s. a. Draw an energy bar chart for this situation. b. Use work and energy and kinematics to solve.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. EOC #55 A 5.0 kg box slides down a 5.0 m high frictionless hill starting from rest. It slides across a 2-m long “rough” horizontal section with u k = At the end of the horizontal section the box hits a horizontal spring with k = 500 N/m. a.How fast is the box going before reaching the rough surface? b.How fast is it going just before hitting the spring? c.How far is the spring compressed? d.How many complete crossings can the box make before coming to rest?

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. EOC #55 - Answers A 5.0 kg box slides down a 5.0 m high frictionless hill starting from rest. It slides across a 2-m long “rough” horizontal section with u k = At the end of the horizontal section the box hits a horizontal spring with k = 500 N/m. a.How fast is the box going before reaching the rough surface? 9.9 m/s b.How fast is it going just before hitting the spring? 9.4 m/s c.How far is the spring compressed? 0.94 m d.How many complete crossings can the box make before coming to rest? 10 trips

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Final Review #1 (Impulse momentum A superball of mass 0.05 kg is dropped from a height of 10 cm above floor. It bounces and rebounds to a three-fourths of its original height. a. Sketch a quantitatively correct graph of the force of gravity on the ball during the time of impact. Use an appropriate number scale for the axes: b. Is the magnitude of the average force of the ball on the floor greater than, less than or equal to the magnitude of the average force of the floor on the ball? Explain your answer. c. Is the magnitude of the average force of the floor on the ball greater than, less than or equal to the force of gravity on the ball? Explain your answer. d. Determine the maximum force that the floor exerts on the ball during the collision. Assume that the impulse due to the weight is negligible.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Final Review #2 – Ch 8 - EOC #47 A 100 g ball on a