Lesson 8: Work, Energy, Power (Sections ) Chapter 6 Conservation of Energy

Similar documents
Chapter 5 Circular Motion

Description Linear Angular position x displacement x rate of change of position v x x v average rate of change of position

Review. Physics 231 fall 2007

Physics 2A Chapter 11 - Universal Gravitation Fall 2017

Chapter 13 - Universal Gravitation

Dynamics of Rigid Bodies

LINEAR MOMENTUM. product of the mass m and the velocity v r of an object r r

Page 1. Clicker Question 9: Physics 131: Lecture 15. Today s Agenda. Clicker Question 9: Energy. Energy is Conserved.

Remember: When an object falls due to gravity its potential energy decreases.

PHYS 1443 Section 004 Lecture #12 Thursday, Oct. 2, 2014

Physics 131: Lecture 16. Today s Agenda

Objectives. Chapter 6. Learning Outcome. Newton's Laws in Action. Reflection: Reflection: 6.2 Gravitational Field

Rotational Kinematics. Rigid Object about a Fixed Axis Western HS AP Physics 1

VEKTORANALYS FLUX INTEGRAL LINE INTEGRAL. and. Kursvecka 2. Kapitel 4 5. Sidor 29 50

a v2 r a' (4v) 2 16 v2 mg mg (2.4kg)(9.8m / s 2 ) 23.52N 23.52N N

III. Electromechanical Energy Conversion

Spring Force and Power

2/24/2014. The point mass. Impulse for a single collision The impulse of a force is a vector. The Center of Mass. System of particles

Chapter 07: Kinetic Energy and Work

One-dimensional kinematics

Physics 207, Lecture 13, Oct. 15. Energy

PHY121 Formula Sheet

Energy in Closed Systems

Physics 111 Lecture 11

Momentum. Momentum. Impulse. Momentum and Collisions

EMU Physics Department

Physics 105: Mechanics Lecture 13

PHYS 1441 Section 002 Lecture #16

RE 6.d Electric and Rest Energy RE 6.e EP6, HW6: Ch 6 Pr s 58, 59, 91, 99(a-c), 105(a-c)

Chapter 7: Conservation of Energy

PHY126 Summer Session I, 2008

Physics 201 Lecture 4

How does the momentum before an elastic and an inelastic collision compare to the momentum after the collision?

PHYS 1441 Section 002 Lecture #15

Physics 2A Chapters 6 - Work & Energy Fall 2017

Physics 101 Lecture 9 Linear Momentum and Collisions

in state i at t i, Initial State E = E i

DYNAMICS VECTOR MECHANICS FOR ENGINEERS: Kinematics of Rigid Bodies in Three Dimensions. Seventh Edition CHAPTER

TEST-03 TOPIC: MAGNETISM AND MAGNETIC EFFECT OF CURRENT Q.1 Find the magnetic field intensity due to a thin wire carrying current I in the Fig.

1. A body will remain in a state of rest, or of uniform motion in a straight line unless it

24-2: Electric Potential Energy. 24-1: What is physics

Name: PHYS 110 Dr. McGovern Spring 2018 Exam 1. Multiple Choice: Circle the answer that best evaluates the statement or completes the statement.

EMU Physics Department.

You will analyze the motion of the block at different moments using the law of conservation of energy.

Period & Frequency. Work and Energy. Methods of Energy Transfer: Energy. Work-KE Theorem 3/4/16. Ranking: Which has the greatest kinetic energy?

Scalars and Vectors Scalar

total If no external forces act, the total linear momentum of the system is conserved. This occurs in collisions and explosions.

Modeling motion with VPython Every program that models the motion of physical objects has two main parts:

K = 100 J. [kg (m/s) ] K = mv = (0.15)(36.5) !!! Lethal energies. m [kg ] J s (Joule) Kinetic Energy (energy of motion) E or KE.

AP Physics Enosburg Falls High School Mr. Bushey. Week 6: Work, Energy, Power

Physics 1: Mechanics

Problem While being compressed, A) What is the work done on it by gravity? B) What is the work done on it by the spring force?

Lecture 16. Chapter 11. Energy Dissipation Linear Momentum. Physics I. Department of Physics and Applied Physics

Chapter 7. Potential Energy and Conservation of Energy

CHAPTER 8 Potential Energy and Conservation of Energy

Chapter 12 Equilibrium and Elasticity

Linear Momentum. Center of Mass.

Study Guide For Exam Two

Capítulo. Three Dimensions

Chapter 3 and Chapter 4

PHYS 705: Classical Mechanics. Derivation of Lagrange Equations from D Alembert s Principle

4.4 Continuum Thermomechanics

Physics 11b Lecture #2. Electric Field Electric Flux Gauss s Law

10/2/2003 PHY Lecture 9 1

Chap13. Universal Gravitation

Physics 202, Lecture 2. Announcements

Chapter 8. Potential Energy and Conservation of Energy

Rigid Bodies: Equivalent Systems of Forces

Physics 207 Lecture 13. Lecture 13

Chapter 8. Linear Momentum, Impulse, and Collisions

Prof. Dr. I. Nasser T /16/2017

Chapter 8: Potential Energy and The Conservation of Total Energy

Linear Momentum. Equation 1

6. Introduction to Transistor Amplifiers: Concepts and Small-Signal Model

Physics Exam II Chapters 25-29

VECTOR MECHANICS FOR ENGINEERS: Vector Mechanics for Engineers: Dynamics. In the current chapter, you will study the motion of systems of particles.

GAUTENG DEPARTMENT OF EDUCATION SENIOR SECONDARY INTERVENTION PROGRAMME PHYSICAL SCIENCES GRADE 12 SESSION 1 (LEARNER NOTES)

Integral Vector Operations and Related Theorems Applications in Mechanics and E&M

COLLEGE OF FOUNDATION AND GENERAL STUDIES PUTRAJAYA CAMPUS FINAL EXAMINATION TRIMESTER /2017

Physics 2A Chapter 3 HW Solutions

Motion in One Dimension

DYNAMICS OF UNIFORM CIRCULAR MOTION

Uniform Circular Motion

9/19/2013. PHY 113 C General Physics I 11 AM-12:15 PM MWF Olin 101

Physics 207: Lecture 20. Today s Agenda Homework for Monday

Chapter 3: Vectors and Two-Dimensional Motion

Physics in Context Correlation with Arkansas Science Curriculum Framework (Revised 2005)

Final Exam. covering the entire semester. Extra time granted about 1 hour about 5 Problems about 30 Multiple Choice

Chapter 2. Pythagorean Theorem. Right Hand Rule. Position. Distance Formula

Recitation: Energy, Phys Energies. 1.2 Three stones. 1. Energy. 1. An acorn falling from an oak tree onto the sidewalk.

Motion in a Plane Uniform Circular Motion

Department of Physics, Korea University Page 1 of 5

10/23/2003 PHY Lecture 14R 1

PHYS 172: Modern Mechanics. Summer Lecture 4 The Momentum Principle & Predicting Motion Read

Contact, information, consultations

( ) α is determined to be a solution of the one-dimensional minimization problem: = 2. min = 2

c) (6) Assuming the tires do not skid, what coefficient of static friction between tires and pavement is needed?

2 dependence in the electrostatic force means that it is also

p p +... = p j + p Conservation Laws in Physics q Physical states, process, and state quantities: Physics 201, Lecture 14 Today s Topics

PHYS Week 5. Reading Journals today from tables. WebAssign due Wed nite

Transcription:

Lesson 8: Wok, negy, Powe (Sectons 6.-6.8) Chapte 6 Conseaton o negy Today we begn wth a ey useul concept negy. We wll encounte many amla tems that now hae ey specc dentons n physcs. Conseaton o enegy Wok Potental Powe In some cases, t can be agued that these tems hae a physcs denton that s smla to ts eeyday usage. The Law o Conseaton o negy The total enegy n the unese s unchanged by any physcal pocess: total enegy beoe = total enegy ate om page 9: In odnay language, conseng enegy means tyng not to waste useul enegy esouces. In the scentc meanng o conseaton, enegy s always conseed no matte what happens. Conseaton o enegy s one o the ew unesal pncples o physcs. No ecepton has ee been ound. It apples to physcal, chemcal, and bologcal systems. Also om page 9: Some poblems can be soled usng ethe enegy conseaton o Newton s second law, so t always pays to consde both methods. Usually the enegy method s ease. Usng Newton s second law noles ecto methods snce oces ae ecto quanttes. In most poblems, enegy noles scala quanttes, whch ae much ease to deal wth (and moe amla). When decdng whch o these two appoaches to use to sole a poblem, ty usng enegy conseaton st. Kangaoos ae mentoned at the begnnng o ths chapte. http://www.youtube.com/watch?=hjysrmy oms o negy Page 93: At the most undamental leel thee ae thee knds o enegy. Knetc enegy enegy due to moton. Potental enegy enegy due to nteacton a. Gatatonal potental enegy nteacton between the ath and a mass b. lastc potental enegy nteacton between a spng and a mass 3. Rest enegy ntenal enegy to a body Lesson 8, page

Lesson 8: Wok, negy, Powe (Sectons 6.-6.8) negy s measued n Joules (J). Wok Suppose a oce causes an object to moe a dstance paallel to The wok done by a constant oce s dened as DO NOT MMORIZ! W Suppose a constant oce causes an object to moe along not paallel to Lesson 8, page

Lesson 8: Wok, negy, Powe (Sectons 6.-6.8) whee s the angle between and W cos. MMORIZ. Wok s a scala quantty and can be poste, negate, o zeo. Poste: between 0 o and 90 o Negate: between 90 o and 80 o Zeo: = 90 o o Usually tenson and nomal oce do no wok. The ecepton s (c) below. The wok done by seeal oces can be ound om the net oce W total W W W cos Wok and Knetc negy Choosng the as along net, (usng = cos ) We had an equaton om Chapte W net total net ma a N Lesson 8, page 3

Lesson 8: Wok, negy, Powe (Sectons 6.-6.8) a ( ) Substtutng nto Wtotal W total m( Snce the net oce s n the -decton, ay and az ae both zeo. Only the -component o the elocty changes ) ( y z ) ( y z ) and W total The tanslatonal knetc enegy s dened as The wok-knetc enegy theoem s m( K m W total K Whle ths epesson s oundatonal to ths chapte, do not memoze. We shall dee a moe useul om. ) Gatatonal Potental negy () The weght can do wok. Toss a ball up and t slows down. In ou new language, ts knetc enegy deceases. The knetc enegy s coneted nto anothe om o enegy we call gatatonal potental enegy. The change n gatatonal potental enegy U ga W ga In tems o poston U ga mgy Ths equaton s tue een the object does not tael n a staght lne. Lesson 8, page 4

Lesson 8: Wok, negy, Powe (Sectons 6.-6.8) The gatatonal potental enegy s U ga mgy The nal om o ou elaton s W nc K U Ths s t. You need to know t. We hae anothe enty nto ou cause and eect table. Anothe useul om s K U W W nc nc K U K K U U K U Wnc s the wok done by nonconseate oces. Nonconseate oces do not hae a potental enegy. A good eample s cton. The mechancal enegy s mech K U Conseaton o Mechancal negy When nonconseate oces do no wok, mechancal enegy s conseed: mech, mech, The zeo o potental enegy s abtay. Choose whatee s conenent, usually the gound. The wok done by a conseate oce s ndependent o the path taken. Poblem: A 0.0-kg ball s thown at 5.0 m/s om a 0 m towe. What s ts speed when t s 5.0 m aboe the gound? What s ts speed when t hts the gound? Soluton: Use the conseaton o mechancal enegy. mgy gy m ech, U K m U m ech, gy K mgy m Lesson 8, page 5

Lesson 8: Wok, negy, Powe (Sectons 6.-6.8) g( y y ) (5m/s) (9.8m/s )(0m 5m). m/s At the bottom, y = 0, = 4.9 m/s. Notce the decton o the thow s not mentoned. No matte whch way the ball s thown, t has the same speed at the same heght! Ths s ey had to poe usng Newton s second law. Gatatonal Potental negy () o objects a om the ath, U GmM Whle ths looks ey deent om mgy, the tet (p. 08) shows they ae equalent. ample 6.8 What s the escape elocty o the ath? Soluton: Use conseaton o enegy. We cannot use U = mgy o an object a om the eath! Gmm mech, U K U m mech, K Gmm m When an object escapes om the ath, ath s gaty s not actng on t and. I the object baely escapes the ath, = 0. Gmm m 0 Gm The statng poston s on the ath s suace and = R = 6.36 0 6 m. The mass o the ath s m = 5.97 0 4 kg. The escape elocty s Gm (6.670 Nm / kg )(5.970 6 (6.370 kg) 4 kg),00m/s Ths s Lesson 8, page 6

Lesson 8: Wok, negy, Powe (Sectons 6.-6.8) m m,00 7.0 s 609m Wok done by a aable oce. The adantage o enegy methods s seen when dealng wth a aable oce. Suppose the oce changes wth dsplacement. An eample o such a dece s shown on the net page. How do we calculate the wok done? We dde the oeall dsplacement nto a sees o small dsplacements,. Oe each o the smalle dsplacements, the oce s almost constant. The wok done o a small dsplacement s W m s The total wok s the aea unde the cue shown aboe. Ths pocedue was used to nd the gatatonal potental enegy o an object a om the ath GMm U Hooke s Law and Ideal Spngs The oce eeted by the ache nceases as the bowstng s dawn back. Robet Hooke poposed an deal spng whee the oce s popotonal to the dsplacement k The dsplacement o the spng om the elaed poston s. The constant k, s called the spng constant. It s measued n N/m and t ges the stength o the spng. The lage k s, the ste the spng. The mnus sgn ndcates that the spng s stetched to the ght, the spng pulls back to the let and ce esa. Lesson 8, page 7

Lesson 8: Wok, negy, Powe (Sectons 6.-6.8) lastc Potental negy As the spng s pulled (o pushed) om ts elaed poston, wok s done on t. The wok done s ndependent o the path taken and accodngly, a potental enegy can be dened. The elastc potental enegy s ound to be U elastc k Note that U = 0 when = 0. Poblem 09. A spng gun (k = 8 N/m) s used to shoot a 56-g ball hozontally. Intally the spng s compessed by 8 cm. The ball loses contact wth the spng and leaes the gun when the spng s stll compessed by cm. What s the speed o the ball when t hts the gound.4 m below the gun? Soluton: Ths appeas to be a pojectle poblem. It s an enegy poblem wth two potental eneges. Take the ntal poston to be at the top and the nal poston just beoe t hts the gound, Lesson 8, page 8

Lesson 8: Wok, negy, Powe (Sectons 6.-6.8) 0 k mgy m k 0 K U K U gy (9.8m/s )(.4m) (8N/m)((0.8m) 6.04m/s k( )/ m (0.m) )/(0.056kg) Powe Oten the ate o enegy coneson s mpotant. We use the tem powe to ee to the ate o enegy coneson. Oe an etended tme, the aeage powe s P a t Powe s measued n Joules/second o watts (W). Be caeul wth W o wok and W o watts. Remembe that wok changes the mechancal enegy o the system. P t cos t cos Poblem 95. A 500-kg ca coasts n neutal down a.0º hll. The ca attans a temnal speed o 0.0 m/s. (a) How much powe must the engne dele to de the ca on a leel oad at 0.0 m/s? (b) I the mamum useul powe that can be deleed by the engne s 40.0 kw, what s the steepest hll the ca can clmb at 0.0 m/s? Soluton: At the temnal speed, the -component o the weght o the ca s opposed by a esstance. When coastng down the hll, the ee body dagam s o the -component (along the nclne) Lesson 8, page 9

Lesson 8: Wok, negy, Powe (Sectons 6.-6.8) mgsn a a ma 0 mgsn (500kg)(9.8m/s )sn 53N (a) Now the ca s mong along a hozontal oad at constant elocty. om the BD ound at the top o the net page: moto a moto ma 0 a The powe deleed by the engne when the ca moes at 0 m/s (notce the angle below s not º!) s Pmoto moto cos a cos0 (53N)(0m/s)cos0 0,300W (b) Agan a ee body dagam s helpul. Lesson 8, page 0

Lesson 8: Wok, negy, Powe (Sectons 6.-6.8) Clmbng wth constant speed, a = 0, moto mgsn ma a moto 0 a mgsn The mamum powe suppled by the engne s 40.0 kw. The powe s Ths coesponds to a 5.8º angle. P m oto cos ( a mgsn ) cos0 a mgsn P a sn mg 3 40.0 0 W (53N)(0m/s) (500kg)(9.8m/s )(0m/s) 0.0 Lesson 8, page