Chapter_8
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Chapter_8

Course Number: PHY107 55555, Spring 2009

College/University: SUNY Buffalo

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Chapter 8 Potential Energy and Conservation of Energy In this chapter we will introduce the following concepts: Potential Energy Conservative and non-conservative forces Mechanical Energy Conservation of Mechanical Energy The conservation of energy theorem will be used to solve a variety of problems As was done in Chapter 7 we use scalars such as work, kinetic energy, and mechanical energy rather than vectors....

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8 Potential Chapter Energy and Conservation of Energy In this chapter we will introduce the following concepts: Potential Energy Conservative and non-conservative forces Mechanical Energy Conservation of Mechanical Energy The conservation of energy theorem will be used to solve a variety of problems As was done in Chapter 7 we use scalars such as work, kinetic energy, and mechanical energy rather than vectors. Therefore the approach is mathematically simpler. (8-1) B h g Work and Potential Energy: A tomato of mass m is taken together with the earth as the system we study. o o v A v The tomato is thrown upwards with initial speed vo at point A. Under the gravitational force, it slows down and stops at point B. Then it falls back to point A and reaches the original speed vo. During the trip from A to B the gravitational force Fg does work U = W (8-2) S A o m B Fk Consider our system with the mass m attached to a spring of spring constant k. The mass has an initial speed vo at point A. Under the spring force FS m slows down and stops at point B (with a spring compression x). Then the mass reverses the direction of its motion and reaches point A. Its speed reaches the original value vo. v S o A F m B k v During the trip from A to B the spring force Fs does work U = W (8-3) Conservative and non-conservative forces. The gravitational force and the spring force are d A B called conservative because they can vo m transfer energy between kinetic energy and m x potential energy with no loss. Frictional and drag forces are called non-conservative for reasons that are explained below. Consider a block of mass m on the horizontal floor. The block starts to move with initial speed vo at point A. The coefficient of kinetic friction between the floor and the block is k. The block will slow down and stop at point B after traveling a distance d. During the trip from A to B the frictional force does work f f (8-4) (8-5) Path Independence of Conservative Forces How to decide whether a force is conservative or non-conservative? Closed-loop test: A force is conservative if the net work done on a particle during a round trip is always equal to zero (see fig.b). W =0 net In the examples of the tomato-earth and mass-spring system Wnet = Wa1b + Wb2a = 0 (because U = W Wa1b = U a U b , Wb 2 a = U b U a ) So gravitational force and spring force are conservative forces! Next, we prove the path independence of work done by a conservative force Wa1b = Wa 2b = U a U b Thus, the work done by a conservative force does not depend on the path taken but only on the initial and final points choose a simple path Path-independence of work done by conservative forces Work done by conservative forces (W = - U) is path-independent, it results in changes in the object's potential energy. Let's use gravity an example of a CONSERVATIVE FORCE. The block's final change in potential energy is the same whether it follows the path with the intermediate stops B, C and D or whether it is directly taken from A to E. Path-dependence of work done by nonconservative forces The work done by non-conservative forces does depend on the path. For example, a kinetic friction force is exerted on a sliding box moving from position 1 to 2 following two paths. Because Wf= -f d (assuming constant coefficient of kinetic friction) longer path requires more work. U = W Determining Potential Energy Values : . x How to determine the difference in potential ..x O energy U of a conservative force F between points x f and xi ? A conservative force F moves an object along the x-axis from an initial point xi F(x) to a final point x f . The work W that the force F does on the object is given by : xf W= F ( x)dx xi The corresponding change in potential energy U is: U = W Therefore the expression for U becomes: xf U = W = F ( x)dx xi (8-6) y . m y Gravitational Potential energy : Consider a particle of mass m moving vertically along the y -axis from yi to y f . dy mg . .O The gravitational force does work W on the particle which changes the potential energy U of the particle-earth system. U = F ( y )dy yi yf yf F = mg yf y U = ( mg ) dy = mg dy = mg [ y ] y f = mg ( y f yi ) = mg y. Since only changes in potential energy are physically menaingful, we set U i = 0 we assign y f = y, yi = 0 U ( y ) = mgy yi yi i U ( y ) = mgy y (8-7) Potential Energy of a spring : x O (a) x O (b) x O (c) Consider the block-mass system. The block moves from xi to x f . The spring force does work W on the block which changes the potential energy of the block-spring system kxi2 U = W = F ( x)dx = kxdx = k xdx = 2 2 xi xi xi xf xf xf kx f 2 Since only changes in potential energy are physically meaningful, we set U i = 0 kx 2 we assign x f =x, xi = 0, U ( x) = 2 x kx 2 U= 2 x (8-8) Conservation of Mechanical Energy : Mechanical energy of an object is defined as the sum of potential and kinetic energies Emech = K + U work-kinetic-energy theorem K = Wnet equations 1 and 2 K = U (eqs.1) (eqs.2) If only conservative forces do work Wnet = Wc , U = Wc K f Ki = ( U f U i ) Ki + U i = K f + U f (8-9) This equation is known as the principle of conservation of mechanical energy. It can be summarized as: Emech = K + U = 0 If the forces are a mixture of conservative and non-conservative forces Wnet = Wc + Wnc = U + Wnc Emech = K + U = Wnc Here, Wnc is the work of all the non - conservative forces of the system Practice question 8.7.2. A rubber ball is dropped from rest from a height h. The ball bounces off the floor and reaches a height of 2h/3. How can we use the principle of the conservation of mechanical energy to interpret this observation? a) During the collision with the floor, the floor did not push hard enough on the ball for it to reach its original height. b) Some of the balls potential energy was lost in accelerating it toward the floor. c) The force of the earths gravity on the ball prevented it from returning to its original height. d) Work was done on the ball by the gravitational force that reduced the balls kinetic energy. e) Work was done on the ball by non-conservative forces that resulted in the ball having less total mechanical energy after the bounce. 8.7.2. A rubber ball is dropped from rest from a height h. The ball bounces off the floor and reaches a height of 2h/3. How can we use the principle of the conservation of mechanical energy to interpret this observation? a) During the collision with the floor, the floor did not push hard enough on the ball for it to reach its original height. b) Some of the balls potential energy was lost in accelerating it toward the floor. c) The force of the earths gravity on the ball prevented it from returning to its original height. d) Work was done on the ball by the gravitational force that reduced the balls kinetic energy. e) Work was done on the ball by non-conservative forces that resulted in the ball having less total mechanical energy after the bounce. Emec = mg (2h / 3 h) = mgh / 3 = Wnc Mechanical Energy Conservation for projectile motion K= m(v 2 + v 2 ) x y 2 2 + v0 y ) = 2 m(v0x + (voy gt) 2 ) 2 gt 2 = mg voy t 2 2 = K 0 mg(y y 0 ) = K 0 (U g U 0 ) K + Ug = K 0 + U0 2 m(v0x (8-10) An example of conservation of mechanical energy: a pendulum bob of mass m moving under the gravitational force The total mechanical energy of the bob-earth system is constant. L h U = mgh = mgL(1 cos ) K = mv 2 / 2 U max = K max As the pendulum swings, the energy E is transferred back and forth between kinetic energy K of the bob and gravitational potential energy U of the bob-earth system When is K at maximum? frame (a), (e) (U is at minimum there). When is U at maximum? frame (c), (g) (K is at minimum there) Mechanical Energy Conservation during a roller coaster ride ME=PE1 =50*9.8*4=1960J More on roller PE 2 =50*9.8*3=1470J KE 2 = ME PE 2 = 490J v 2 = 2KE 2 / m = 4.43m / s PE 3 =0J KE 3 = ME PE 3 = 1960J v3 = 2KE 3 / m = 8.85m / s KE 4 = mv 2 / 2 = 900J 4 PE 4 = ME KE 4 = 1060J h 4 = PE 4 / mg = 2.16m Energy Transformation for Downhill Skiing h d mv 2 mgh = 2 mv 2 0 = fd = k mgd 2 mgh = k mgd d = h / k Review of previous lecture Mechanical energy of an object is defined as the sum of potential and kinetic energies Emech = K + U Change in potential energy is defined in terms of the work done by a conservative force: U = Wc (eqs.2) work-kinetic-energy theorem K = Wnet =Wc +Wnc (eqs.1) If we compare equations 1 and 2 we have: K + U = Emech = Wnc If Wnc = 0, Emech = 0, which is known as the principle of conservation of m echanical energy. Example 1 vf vi d h 60. A cookie jar is moving up a 40 incline. At a point 55 cm from the bottom of the incline (measured along the incline), the jar has a speed of 1.4 m/s. The coefficient of friction kinetic between jar and incline is 0.15. (a) How much farther up the incline will the jar move? (b) How fast will it be going when it has slid back to the bottom of the incline? (c) Do the answers to (a) and (b) increase, decrease, or remain the same if we decrease the coefficient of kinetic friction (but do not change the given speed or location)? Example 1 vf vi d d 60. A cookie jar is moving up a 40 incline. At a point 55 cm from the bottom of the incline (measured along the incline), the jar has a speed of 1.4 m/s. The coefficient of kinetic friction between jar and incline is 0.15. (a) How much farther up the incline will the jar move? (b) How fast will it be going when it has slid back to the bottom of the incline? (c) Do the answers to (a) and (b) increase, decrease, or remain the same if we decrease the coefficient of kinetic friction (but do not change the given speed or location)? Finding the Force F ( x) from the potential energy U ( x) . O A x . F . x x+ B x Consider an object that moves along the x-axis under the influence of an unknown conservative force F given its potential energy U ( x). The object moves from point A (at x) to point B (at x + x). The force F does work W on the object: W = F x (eqs.1) The potential energy U of the system is changed by U = W (eqs.2) If we combine equations 1 and 2 we get: U x We take the limit as x 0 and we get the following equation: F = dU ( x) F ( x) = dx U = F ( x)dx xi xf (8-11) The potential Energy Curve If we plot the potential energy U versus x for a force F that acts along the x-axis, we can get a lot of information about the motion of a particle on which F is acting. x2 x3 x4 We can determine the force F(x) using the equation: F ( x) = dU ( x) dx An example is given in the figures above. In fig.a we plot U(x) versus x. In fig.b we plot F(x) versus x. At x2 , x3 and x4 the slope of the U(x) vs x curve is zero, thus F = 0. The slope dU/dx between x3 and x4 is negative F > 0 for this interval. The slope dU/dx between x2 and x3 is positive F < 0 for this interval. (8-13) Turning Points : The total mechanical energy is Emec = K ( x) + U ( x) Emec is constant (equal to 5 J in the figure) and is thus represented by a horizontal line. We can solve this equation for K ( x): allowed x1 forbidden K ( x) = Emec U ( x) mv 2 Because kinetic energy K = 0, we can determine which regions 2 of the x-axis motion is allowed: K ( x) = Emec U ( x) 0 If K = Emech U ( x) 0 U ( x) Emec Motion is allow ed If K = Emech U ( x) < 0 U ( x) > Emec Motion is forbidden The points at which Emec = U ( x) are known as turning points for the motion. For example x1 is the turning point for the U versus x plot above. At the turning point K = 0. (8-14) Given the U(x) versus x curve the turning points and the regions for which motion is allowed depends on the value of the mechanical energy Emec 1 5 In the above figure, if Emec = 4 J: x x The turning points (Emec = U ) occur at x1 and x > x5. Motion is allowed (Emec U ) for x > x1 If we reduce Emec to 3 J or 1 J the turning points and regions of allowed motion change accordingly. Equilibrium Point: a position at which the slope dU/dx = 0 (and thus F = 0) stable equilibrium: positions of minima in the U versus x curve unstable equilibrium: positions of maxima in the U versus x curve neutral equilibrium: a region for which F = 0. For example, if Emec = 4 J, the kinetic energy K = 0 if x > x5 and any particle moving in this region will be stationary (8-15) dU ( x) F ( x) = dx Positions of Stable Equilibrium. An example is point x4 where U has a minimum. If Emec = 1 J then K = 0 at point x4. A particle with Emec = 1 J is stationary at x4. If we displace slightly the particle either to the right or to the left of x4, the force will bring it back to the equilibrium position this equilibrium is stable. Positions of Unstable Equilibrium. An example is point x3 where U has a maximum. If Emec = 3 J then K = 0 at point x3. A particle with Emec = 3 J is stationary at x3. If we displace slightly the particle either to the right or to the left of x3, the force will take it further away from the equilibrium position this equilibrium is unstable x3 x4 Quiz of Ch8 A small, initially stationary block is released on a frictionless ramp at a height of 3.0 m. Hill heights along the ramp are as shown. The hills have identical circular tops, and the block does not fly off any hill. For which hill is the speed of the block greatest? D) C) B) A) A small, initially stationary block is released on a frictionless ramp at a height of 3.0 m. Hill heights along the ramp are as shown. The hills have identical circular tops, and the block does not fly off any hill. For which hill is the speed of the block greatest? D) C) B) A) mv 2 mgH = mgh + 2 v = 2 g ( H h) 21. The string in Fig. 8-37 is long, has a ball attached to one end, and is fixed at its other end. The distance d from the fixed end to a fixed peg at point P is 75.0 cm. When the initially stationary ball is released with the string horizontal as shown, it will swing along the dashed arc. What is its speed when it reaches (a) its lowest point and (b) its highest point after the string catches on the peg? Example 1 (a) vb (b) yb=2r check mv 2 / r mg b 39. Figure 8-50 shows a plot of potential energy U versus position x of a 0.90 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) Three values are , , and . The particle is released at with an initial speed of 7.0 m/s, headed in the negative x direction. (a) If the particle can reach , what is its speed there, and if it cannot, what is its turning point? What are the (b) magnitude and (c) direction of the force on the particle as it begins to move to the left of ? Suppose, instead, the particle is headed in the positive x direction when it is released at at speed 7.0 m/s. (d) If the particle can reach , what is its speed there, and if it cannot, what is its turning point? What are the (e) magnitude and (f) direction of the force on the particle as it begins to move to the right of ? Example 2 Figure 8-50 shows a plot of potential energy U versus position x of a 0.90 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) Three values are , , and . The particle is released at with an initial speed of 7.0 m/s, headed in the negative x direction. (a) If the particle can reach , what is its speed there, and if it cannot, what is its turning point? What are the (b) magnitude and (c) direction of the force on the particle as it begins to move to the left of ? Suppose, instead, the particle is headed in the positive x direction when it is released at at speed 7.0 m/s. (d) If the particle can reach , what is its speed there, and if it cannot, what is its turning point? What are the (e) magnitude and (f) direction of the force on the particle as it begins to move to the right of ? Example 2 slope=(45-15)/(6-5)=30 > Figure 8-50 shows a plot of potential energy U versus position x of a 0.90 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) Three values are , , and . The particle is released at with an initial speed of 7.0 m/s, headed in the negative x direction. (a) If the particle can reach , what is its speed there, and if it cannot, what is its turning point? What are the (b) magnitude and (c) direction of the force on the particle as it begins to move to the left of ? Suppose, instead, the particle is headed in the positive x direction when it is released at at speed 7.0 m/s. (d) If the particle can reach , what is its speed there, and if it cannot, what is its turning point? What are the (e) magnitude and (f) direction of the force on the particle as it begins to move to the right of ? Example 2 133. A massless rigid rod of length L has a ball of mass m attached to one end. The other end is pivoted in such a way that the ball will move in a vertical circle. First, assume that there is no friction at the pivot. The system is launched downward from the horizontal position A with initial speed . The ball just barely reaches point D and then stops. (a) Derive an expression for in terms of L, m, and g. (b) What is the tension in the rod when the ball passes through B? (c) A little grit is placed on the pivot to increase the friction there. Then the ball just barely reaches C when launched from A with the same speed as before. What is the decrease in the mechanical energy during this motion? (d) What is the decrease in the mechanical energy by the time the ball finally comes to rest at B after several oscillations? 2 mvo / 2 = mgL 2 mvo / 2 + mgL = mv 2 / 2 Example 3 E=EC-EA=0- = -mgL E=EB-EA=0 ( +mgL) = -2mgL Review of sample test II y x

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1THE NATURE OF LIGHT- Light is also known as electromagnetic radiation. - Light is emitted by oscillating charges. - Oscillating charges create oscillating electric and magnetic fields.+-- Light is very peculiar in that it is a particle and a wave at
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1ENERGY AND ITS DIFFERENT FORMSWhat is Energy - good question! - capacity to do work or to supply heat - best to learn what is energy through examples Kinetic and Potential Energy Kinetic Energy - energy of movement - K.E. = mv2 m mass v velocity - exam
Skyline College - CHEM - 220
1THE GASEOUS STATEKinetic Theory of Gases an excellent model of gases 1. Gas molecules have negligible volume compared with space between molecules. - molecules are spread out 2. Gas molecules move. When they hit side of container they contribute to pre
Skyline College - ACTG - 100
Solutions to Suggested Exercises &amp; Problems in Unit 9E143. 1. 2. 3. 4. Cable T.V. Company (no gross profit; high property &amp; equipment) Accounting firm (high receivables; no gross profit) Retail jewelry store (high inventory; high gross profit) Grocery st
Skyline College - ACTG - 100
Solutions to Suggested Exercises &amp; Problems in Unit 8E131.O NA F O I F O F F O 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. Depreciation and amortization Cash collections from customers Dividends paid [Change in] Inventory Payments to acquire property and equipment R
Skyline College - ACTG - 100
Solutions to Suggested Exercises &amp; Problems in Unit 7E118. Stockholders Equity Contributed capital: Preferred stock, 6%, par $10, authorized 50,000 shares, issued and outstanding, 15,000 shares. Common stock, par $1, authorized 200,000 shares, issued and
Skyline College - ACTG - 100
Solutions to Suggested Exercises &amp; Problems in Unit 6E81.Hasbro, Inc.Excerpts from Balance Sheet (in millions) ASSETS Current Assets Cash and cash equivalents Accounts receivable (net of allowance for doubtful accounts, $39) Inventories Prepaid expense
Skyline College - ACTG - 100
Solutions to Suggested Exercises &amp; Problems in Unit 4E62. Sales revenue ($1,000 + $5,000 +$3,000). Less: Sales returns and allowances ($1,000 from T). Less: Sales discounts ($5,000 collected from S x 3%). Less: Credit card discounts ($1,000 from R x 2%).
Skyline College - ACTG - 100
Solutions to Suggested Exercises &amp; Problems in Unit 3E31.TERMF E G I D C M K J L E33. (1) Expenses (2) Gains (3) Revenue principle (4) Cash basis accounting (5) Unearned revenue (6) Operating cycle (7) Accrual basis accounting (8) Prepaid expenses (9)
Skyline College - ACTG - 100
Solutions to Suggested Exercises &amp; Problems in Unit 2E21. (1) E; (6) A; (2) F; (7) L; (3) B; (8) N; (4) O; (9) M; (5) J; (10) D.E22. Req. 1 Received (a) (b) (c) (d) (e) (f) (g) Equipment (A) Equipment (A) Cash (A) Building (A) Intangibles(A) [or Constru
Skyline College - ACTG - 100
Solutions to Suggested Exercises &amp; Problems in Unit 1E11. K G I E A D J F C L H B N M Term or Abbreviation (1) SEC (2) Audit (3) Sole proprietorship (4) Corporation (5) Accounting (6) Accounting entity (7) Audit report (8) Cost principle (9) Partnership
SUNY Stony Brook - ARCH - 362
Long Island Terminal ArchaicAnthropology 362Prehistoric Chronology for Long Island PeriodLate Woodland Middle Woodland Early Woodland Terminal Archaic Late Archaic Middle Archaic Early Archaic Paleo-IndianDatesA.D. 1000 1500 A.D. 0 1000 700 B.C A.D.
SUNY Stony Brook - ARCH - 362
The Peopling of Long IslandAnthropology 362Long IslandDutch TraderPrehistoric Chronology for Long Island PeriodLate Woodland Middle Woodland Early Woodland Terminal Archaic Late Archaic Middle Archaic Early Archaic Paleo-IndianDatesA.D. 1000 1500 A
SUNY Stony Brook - ARCH - 362
Long Island Early and Middle ArchaicAnthropology 362Prehistoric Chronology for Long Island PeriodLate Woodland Middle Woodland Early Woodland Terminal Archaic Late Archaic Middle Archaic Early Archaic Paleo-IndianDatesA.D. 1000 1500 A.D. 0 1000 700 B
SUNY Stony Brook - ARCH - 362
Lecture 3What is TechnologyTechnologyProvince of engineers practical applications often developed by trial &amp; error expand the realm of human possibility materialistMan the tool builder &amp; user Techne: Greek for skill Created implements since before we
SUNY Stony Brook - ARCH - 362
Lecture 2 What is Science?Science is what scientists do Theory experiment mathematics (models) Observation and manipulation Biology Chemistry PhysicsTremendous growth in science Numbers of practitioners: 150,000 in 1950 2,700,000 in 2000 Number of field
SUNY Stony Brook - ARCH - 362
Introduction to Science, Technology, and Society StudiesEST 202 David J. TonjesClass DataM-W-F 10:40-11:35 SBS S-218 (M-W) Old Chemistry 226 (F) Each student will be in either the A or B group (Fr) Names from A-Karnis are A Names from Kleigl-Z are BIn
PNCA - ACC - 12
RES342 WEEK THREE QUIZ DUE ON DAY 7 OF WEEK THREE - SEVEN QUESTIONS WITH EQUAL WEIGHT The mean weight of newborn infants at a community hospital is 6.6 pounds. A sample of seven infants is randomly selected and their weights at birth are recorded as 9.0,
Indian Institute Of Management, Ahmedabad - ME - 9421
STRUCTURE OF SOLIDS STRUCTURE Types of solids based on structure Types of solids based on bondingUNIVERSEHYPERBOLIC EUCLIDEAN SPHERICALSTRONG WEAKELECTROMAGNETICSPACEnD + tENERGYPARTICLESGRAVITYFIELDS NON-ATOMICMETAL SEMI-METALSEMI-CONDUCTOR