# Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Chapter 10. Energy Chapter Goal: To introduce the ideas of kinetic and.

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Chapter 10. Energy Chapter Goal: To introduce the ideas of kinetic and potential energy and to learn a new problem-solving strategy based on conservation of energy.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Ch. 10 – Student Learning Objectives To begin developing a concept of energy— what it is, how it is transformed, and how it is transferred. To introduce the concepts of kinetic and potential energy. To learn Hooke’s law for springs and the new idea of a restoring force.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Energy – The Big Picture Two basic types of mechanical energy: Kinetic energy (K) is an energy of motion. Gravitational potential energy (U g ) is an energy of position. Under some circumstances (e.g. freefall) these two kinds of energy can be transformed back and forth without loss from the system.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Kinetic and Potential Energy For some systems (e.g. systems in freefall):

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. ½ mv 2 f + mgy f = ½ mv 2 0 + mgy 0 where y is the height above an arbitrary zero, (not the displacement). This result is true for motion along any frictionless surface, regardless of the shape. This is a generalization of a relationship we already use for free fall: v 2 f = v 2 0 + 2(-g)(y f – y 0 )

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. A box slides along the frictionless surface shown in the figure. It is released from rest at the position shown. Is the highest point the box reaches on the other side at level a, at level b, or level c? A.At level a B.At level b C.At level c

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. A box slides along the frictionless surface shown in the figure. It is released from rest at the position shown. Is the highest point the box reaches on the other side at level a, at level b, or level c? A.At level a B.At level b C.At level c

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Energy Bar charts – a visual aid

EOC #4 a.What is the kinetic energy of a 1500 kg car traveling at 30 m/s? b.From what height would the car have to be dropped (!) to have the same amount of energy upon impact? c.Repeat parts a and b for a car of 3000 kg. By how much did the height change?

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. EOC#4 a.6.75 x 10 6 Joules b.Solve for the amount of initial potential energy needed to give the car a final kinetic energy equal to part a to get y = 46 m K 0 + U g0 = K f + U gf

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. EOC#4 a.6.75 x 10 6 Joules b.Solve for the amount of initial potential energy needed to give the car a final kinetic energy equal to part a to get y = 46 m. c.Same height K 0 + U g0 = K f + U gf

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The Zero of Potential Energy You can place the origin of your coordinate system, and thus the “zero of potential energy,” wherever you choose and be assured of getting the correct answer to a problem. The reason is that only ΔU has physical significance, not U g itself.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Conservation of Mechanical Energy The sum of the kinetic energy and the potential energy of a system is called the mechanical energy. the kinetic energy and the potential energy can change, as they are transformed back and forth into each other, but their sum remains constant.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Conservation of Mechanical Energy Under what conditions is E mech conserved? What happens to the energy when E mech is not conserved? Are there potential energies other than gravitational, and how do you calculate them?

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Hooke’s Law If you stretch a rubber band, a force appears that tries to pull the rubber band back to its equilibrium, or unstretched, length. A force that restores a system to an equilibrium position is called a restoring force. If s is the position of the end of a spring, and s e is the equilibrium position, we define Δs = s – s e. If (F sp ) s is the s-component of the restoring force, and k is the spring constant of the spring, then Hooke’s Law states that The minus sign is the mathematical indication of a restoring force.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The graph shows force versus displacement for three springs. Rank in order, from largest to smallest, the spring constants k 1, k 2, and k 3.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The graph shows force versus displacement for three springs. Rank in order, from largest to smallest, the spring constants k 1, k 2, and k 3. Answer: k 1 > k 2 > k 3

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Elastic Potential Energy U s = ½ k (∆s) 2 ∆s = (s-s e ) where s e is the equilibrium position of the end of the spring ∆U s = ½ k [(s f - s e ) 2 - (s 0 - se) 2 ]. Often s f or s 0 = s e as the spring begins or ends in equilibrium position. at s=s e, the ball will lose contact with the spring

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. EOC #21 A student places his 500 g physics book on a frictionless table and pushes it against a spring, compressing it by 4.0 cm. The spring constant is 1250 N/m. He then releases the book. What is the speed as it slides away from the spring? Draw a before and after picture:

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. EOC #21 Knowns m=.5 kg k = 1250 N/m x 1 = -.04 m v 1 = 0m/s Find: v 2, the speed of the book when it is released Draw an energy bar chart to determine what energy transformation takes place

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. EOC #21 The elastic potential energy of the spring is converted into the kinetic energy of the book. Solve by replacing each energy term in the bar chart with a value:

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. ½ mv 1 2 + ½ k(x 1 – x e ) 2 = ½ mv 2 2 + ½ k(x 2 – x e ) 2 This simplifies to: ½ k(x 1 ) 2 = ½ mv 2 2 and v 2 = = 2.0 m/s

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Problem with U g and U s (EOC #44) A 1000 kg safe is 2.0 m above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compresses it 50 cm. What is the spring constant of the spring? Draw a before and after picture and decide on a zero.

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Problem with U g and U s (EOC #44) Here the author decided on y e (equilibrium of the spring). Good choice, since we have no information on the length of the spring. The relevant information was given relative to the top of the spring. The safe has some velocity when it first hits the spring. Do we need to know that?

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Kinetic energy in the middle of the problem has no bearing and does not need to be calculated A 1000 kg safe is 2.0 m above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compresses it 50 cm. What is the spring constant of the spring?

Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. A 1000 kg safe is 2.0 m above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compresses it 50 cm. What is the spring constant of the spring? Using the non-zero terms from the energy bar chart, we get: mgy 0 = mgy 1 + ½ k (y 1 - y e ) 2 k= 196,000 N/m Note that mgy 1 is negative for the zero chosen.