Work Work tells us how much a force or combination of forces changes the energy of a system. Work is the product of the force or its component in the direction of motion and its displacement. W = Fdcos F: force (N) d : displacement (m) : angle between force and displacement

Slide 4

Units of Work SI System: Joule (N m) 1 Joule of work is done when 1 N acts on a body moving it a distance of 1 meter British System: foot-pound (not used in Physics B) cgs System: erg (dyne-cm) (not used in Physics B) Atomic Level: electron-Volt (eV)

Slide 5

Force and direction of motion both matter in defining work! There is no work done by a force if it causes no displacement. Forces can do positive, negative, or zero work. When an box is pushed on a flat floor, for example… The normal force and gravity do no work, since they are perpendicular to the direction of motion. The person pushing the box does positive work, since she is pushing in the direction of motion. Friction does negative work, since it points opposite the direction of motion.

Slide 6

Question If a man holds a 50 kg box at arms length for 2 hours as he stands still, how much work does he do on the box? None, 0 J, No Movement.

Slide 7

Question If a man holds a 50 kg box at arms length for 2 hours as he walks 1 km forward, how much work does he do on the box? None, force and movement are perpendicular to each other (cos 90 = 0)

Slide 8

Question If a man lifts a 50 kg box 2.0 meters, how much work does he do on the box? W =fdcos W = (mg)dcos W = (50)(9.8)(2) cos 0 W = 980 J

Slide 9

Work and Energy Work changes mechanical energy! If an applied force does positive work on a system, it tries to increase mechanical energy. If an applied force does negative work, it tries to decrease mechanical energy. The two forms of mechanical energy are called potential and kinetic energy.

Slide 10

Sample problem Jane uses a vine wrapped around a pulley to lift a 70-kg Tarzan to a tree house 9.0 meters above the ground. a)How much work does Jane do when she lifts Tarzan? b) m = 70 kg, d = 9 m, g = 9.8 m/s 2, = 0 W = ? W = Fdcos = (mg)dcos a) W = ( 70)(9.8)(9)cos 0 b)How much work does gravity do when Jane lifts Tarzan? Same, but in the opposite direction.

Slide 11

Sample problem Joe pushes a 10-kg box and slides it across the floor at constant velocity of 3.0 m/s. The coefficient of kinetic friction between the box and floor is 0.50. a)How much work does Joe do if he pushes the box for 15 meters? m = 10 kg, = 0.5, d = 15 m, = 0, v = 15 m/s (constant) W = ? W = F app dcos, since constant velocity F app = F k W = ( mg)dcos W = (.5)(10)(9.8)(15)cos 0 b) How much work does friction do as Joe pushes the box? Same but opposite.

Slide 12

Sample problem A father pulls his child in a little red wagon with constant speed. If the father pulls with a force of 16 N for 10.0 m, and the handle of the wagon is inclined at an angle of 60 o above the horizontal, how much work does the father do on the wagon? F = 16 N, d = 10 m, = 60 W = ? W = Fdcos W = (16)(10)cos 60 W =

Slide 13

Kinetic Energy Energy due to motion K = ½ m v 2 K: Kinetic Energy m: mass in kg v: speed in m/s Unit: Joules

Slide 14

Sample problem A 10.0 g bullet has a speed of 1.2 km/s. a)What is the kinetic energy of the bullet? Do on your own, remember to convert units to proper units. a)What is the bullets kinetic energy if the speed is halved? b)What is the bullets kinetic energy if the speed is doubled?

Slide 15

The Work-Energy Theorem The net work due to all forces equals the change in the kinetic energy of a system. W net = K W net : work due to all forces acting on an object K: change in kinetic energy (K f – K i )

Slide 16

Sample problem An 8.0-g acorn falls from a tree and lands on the ground 10.0 m below with a speed of 11.0 m/s. a)What would the speed of the acorn have been if there had been no air resistance? b)Did air resistance do positive, negative or zero work on the acorn? Why?

Slide 17

Sample problem An 8.0-g acorn falls from a tree and lands on the ground 10.0 m below with a speed of 11.0 m/s. c)How much work was done by air resistance? d)What was the average force of air resistance?

Slide 18

Constant force and work The force shown is a constant force. W = F x can be used to calculate the work done by this force when it moves an object from x a to x b. The area under the curve from x a to x b can also be used to calculate the work done by the force when it moves an object from x a to x b F(x) x xaxa xbxb

Slide 19

Variable force and work The force shown is a variable force. W = F x CANNOT be used to calculate the work done by this force! The area under the curve from x a to x b can STILL be used to calculate the work done by the force when it moves an object from x a to x b F(x) x xaxa xbxb

Slide 20

Springs When a spring is stretched or compressed from its equilibrium position, it does negative work, since the spring pulls opposite the direction of motion. W s = - ½ k x 2 W s : work done by spring (J) k: force constant of spring (N/m) x: displacement from equilibrium (m) The force doing the stretching does positive work equal to the magnitude of the work done by the spring. W app = - W s = ½ k x 2

Slide 21

Springs: stretching mm x 0 F s = -kx (Hookes Law) 100 0 -100 -200 200 F(N) 012345 x (m) W s = negative area = - ½ kx 2 FsFs FsFs

Slide 22

Sample problem A spring with force constant 250 N/m is initially at its equilibrium length. a)How much work must you do to stretch the spring 0.050 m? Do on your own. a)How much work must you do to compress it 0.050 m?

Slide 23

Sample problem It takes 1000 J of work to compress a certain spring 0.10 m. a)What is the force constant of the spring? Do on your own a)To compress the spring an additional 0.10 m, does it take 1000 J, more than 1000 J, or less than 1000 J? Verify your answer with a calculation.

Slide 24

Sample Problem How much work is done by the force shown when it acts on an object and pushes it from x = 0.25 m to x = 0.75 m? Figure from Physics, James S. Walker, Prentice-Hall 2002

Slide 25

Sample Problem How much work is done by the force shown when it acts on an object and pushes it from x = 2.0 m to x = 4.0 m? Figure from Physics, James S. Walker, Prentice-Hall 2002

Slide 26

Power Power is the rate of which work is done. P = W/t W: work in Joules t: elapsed time in seconds When we run upstairs, t is small so P is big. When we walk upstairs, t is large so P is small.

Slide 27

Unit of Power SI unit for Power is the Watt. 1 Watt = 1 Joule/s Named after the Scottish engineer James Watt (1776-1819) who perfected the steam engine. British system horsepower 1 hp = 746 W

Slide 28

How We Buy Energy… The kilowatt-hour is a commonly used unit by the electrical power company. Power companies charge you by the kilowatt-hour (kWh), but this not power, it is really energy consumed. 1 kW = 1000 W 1 h = 3600 s 1 kWh = 1000 J/s 3600s = 3.6 x 10 6 J

Slide 29

Sample problem A man runs up the 1600 steps of the Empire State Building in 20 minutes. If the height gain of each step was 0.20 m, and the mans mass was 80.0 kg, what was his average power output during the climb? Give your answer in both watts and horsepower.

Slide 30

Sample problem Calculate the power output of a 0.10 g fly as it walks straight up a window pane at 2.0 cm/s.

Slide 31

Force types Forces acting on a system can be divided into two types according to how they affect potential energy. Conservative forces can be related to potential energy changes. Non-conservative forces cannot be related to potential energy changes. So, how exactly do we distinguish between these two types of forces?

Slide 32

Conservative forces Work is path independent. Work can be calculated from the starting and ending points only. The actual path is ignored in calculations. Work along a closed path is zero. If the starting and ending points are the same, no work is done by the force. Work changes potential energy. Examples: Gravity Spring force Conservation of mechanical energy holds!

Slide 33

Non-conservative forces Work is path dependent. Knowing the starting and ending points is not sufficient to calculate the work. Work along a closed path is NOT zero. Work changes mechanical energy. Examples: Friction Drag (air resistance) Conservation of mechanical energy does not hold!

Slide 34

Potential Energy Energy due to position or configuration Stored energy For gravity: U g = mgy m: mass g: acceleration due to gravity h: height above the zero point For springs: U s = ½ k x 2 k: spring force constant x: displacement from equilibrium position

Slide 35

Conservative forces and Potential energy W c = - U If a conservative force does positive work on a system, potential energy is lost. If a conservative force does negative work, potential energy is gained. For gravity W g = - U g = -(mgy – mgy o ) For springs W s = - U s = -(½ k x 2 – ½ k x o 2 )

Slide 36

More on paths and conservative forces. Q: Assume a conservative force moves an object along the various paths. Which two works are equal? A: W 2 = W 3 (path independence) Q: Which two works, when added together, give a sum of zero? A: W 1 + W 2 = 0 or W 1 + W 3 = 0 (work along a closed path is zero) Figure from Physics, James S. Walker, Prentice-Hall 2002

Slide 37

Sample problem Figure from Physics, James S. Walker, Prentice-Hall 2002 A box is moved in the closed path shown. a) How much work is done by gravity when the box is moved along the path A->B->C? b) How much work is done by gravity when the box is moved along the path A->B->C->D- >A?

Slide 38

Slide 39

Sample problem Figure from Physics, James S. Walker, Prentice-Hall 2002 A box is moved in the closed path shown. a) How much work would be done by friction if the box were moved along the path A->B->C? b) How much work is done by friction when the box is moved along the path A->B->C->D->A?

Slide 40

Slide 41

Sample problem A diver drops to the water from a height of 20.0 m, his gravitational potential energy decreases by 12,500 J. How much does the diver weigh?

Slide 42

Sample problem If 30.0 J of work are required to stretch a spring from a 2.00 cm elongation to a 4.00 cm elongation, how much work is needed to stretch it from a 4.00 cm elongation to a 6.00 cm elongation?

Slide 43

Law of Conservation of Energy In any isolated system, the total energy remains constant. Energy can neither be created nor destroyed, but can only be transformed from one type of energy to another.

Slide 44

Law of Conservation of Mechanical Energy E = K + U = Constant K: Kinetic Energy (1/2 mv 2 ) U: Potential Energy (gravity or spring) E = U + K = 0 K: Change in kinetic energy U: Change in gravitational/spring potential energy U – U o + K – K o = 0 U + K = U o + K o **** mgy + ½ mv 2 = mgy o + ½ mv o 2 ****

Slide 45

Roller Coaster Simulation Roller Coaster Physics Simulation (demonstrates gravitational potential energy and kinetic energy)

Slide 46

Pendulums and Energy Conservation Energy goes back and forth between K and U. At highest point, all energy is U. As it drops, U goes to K. At the bottom, energy is all K.

Slide 47

h Pendulum Energy ½mv max 2 = mgy For minimum and maximum points of swing K 1 + U 1 = K 2 + U 2 For any points 1 and 2.

Slide 48

Springs and Energy Conservation Transforms energy back and forth between K and U. When fully stretched or extended, all energy is U. When passing through equilibrium, all its energy is K. At other points in its cycle, the energy is a mixture of U and K.

Slide 49

Spring Energy mm -x m x 0 ½kx max 2 = ½mv max 2 For maximum and minimum displacements from equilibrium K 1 + U 1 = K 2 + U 2 = E For any two points 1 and 2 All U All K

Slide 50

Spring Simulation Spring Physics Simulation

Slide 51

Sample problem What is the speed of the pendulum bob at point B if it is released from rest at point A? 1.5 m A B 40 o

Slide 52

Slide 53

Sample problem Problem copyright Physics, James S. Walker, Prentice-Hall 2002 A 0.21 kg apple falls from a tree to the ground, 4.0 m below. Ignoring air resistance, determine the apples gravitational potential energy, U, kinetic energy, K, and total mechanical energy, E, when its height above the ground is each of the following: 4.0 m, 2.0 m, and 0.0 m. Take ground level to be the point of zero potential energy.

Slide 54

Sample problem Problem copyright Physics, James S. Walker, Prentice-Hall 2002 A 1.60 kg block slides with a speed of 0.950 m/s on a frictionless, horizontal surface until it encounters a spring with a force constant of 902 N/m. The block comes to rest after compressing the spring 4.00 cm. Find the spring potential energy, U, the kinetic energy of the block, K, and the total mechanical energy of the system, E, for the following compressions: 0 cm, 2.00 cm, 4.00 cm.

Slide 55

Law of Conservation of Energy E = U + K + E int = Constant E int is thermal energy. U + K + E int = 0 Mechanical energy may be converted to and from heat.

Slide 56

Work done by non- conservative forces W net = W c + W nc Net work is done by conservative and non-conservative forces W c = - U Potential energy is related to conservative forces only! W net = K Kinetic energy is related to net force (work-energy theorem) K = - U + W nc From substitution W nc = U + K = E Nonconservative forces change mechanical energy. If nonconservative work is negative, as it often is, the mechanical energy of the system will drop.

Slide 57

Sample problem Problem copyright Physics, James S. Walker, Prentice-Hall 2002 Catching a wave, a 72-kg surfer starts with a speed of 1.3 m/s, drops through a height of 1.75 m, and ends with a speed of 8.2 m/s. How much non-conservative work was done on the surfer?

Slide 58

Sample problem Problem copyright Physics, James S. Walker, Prentice-Hall 2002 A 1.75-kg rock is released from rest at the surface of a pond 1.00 m deep. As the rock falls, a constant upward force of 4.10 N is exerted on it by water resistance. Calculate the nonconservative work, W nc, done by the water resistance on the rock, the gravitational potential energy of the system, U, the kinetic energy of the rock, K, and the total mechanical energy of the system, E, for the following depths below the waters surface: d = 0.00 m, d = 0.500 m, d = 1.00 m. Let potential energy be zero at the bottom of the pond.

Slide 59

Pendulum lab Figure out how to demonstrate conservation of energy with a pendulum using the equipment provided. The photo-gates must be set up in gate mode this time. The width of the pendulum bob is an important number. To get it accurately, use the caliper.caliper Turn in just your data, calculations, and result. Clearly show the speed you predict for the pendulum bob from conservation of energy, the speed you measure using the caliper and photo-gate data, and a % difference for the two.

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Transcript

Slide 1

Slide 2

Work and Energy An Introduction

Slide 3

Work Work tells us how much a force or combination of forces changes the energy of a system. Work is the product of the force or its component in the direction of motion and its displacement. W = Fdcos F: force (N) d : displacement (m) : angle between force and displacement

Slide 4

Units of Work SI System: Joule (N m) 1 Joule of work is done when 1 N acts on a body moving it a distance of 1 meter British System: foot-pound (not used in Physics B) cgs System: erg (dyne-cm) (not used in Physics B) Atomic Level: electron-Volt (eV)

Slide 5

Force and direction of motion both matter in defining work! There is no work done by a force if it causes no displacement. Forces can do positive, negative, or zero work. When an box is pushed on a flat floor, for example… The normal force and gravity do no work, since they are perpendicular to the direction of motion. The person pushing the box does positive work, since she is pushing in the direction of motion. Friction does negative work, since it points opposite the direction of motion.

Slide 6

Question If a man holds a 50 kg box at arms length for 2 hours as he stands still, how much work does he do on the box? None, 0 J, No Movement.

Slide 7

Question If a man holds a 50 kg box at arms length for 2 hours as he walks 1 km forward, how much work does he do on the box? None, force and movement are perpendicular to each other (cos 90 = 0)

Slide 8

Question If a man lifts a 50 kg box 2.0 meters, how much work does he do on the box? W =fdcos W = (mg)dcos W = (50)(9.8)(2) cos 0 W = 980 J

Slide 9

Work and Energy Work changes mechanical energy! If an applied force does positive work on a system, it tries to increase mechanical energy. If an applied force does negative work, it tries to decrease mechanical energy. The two forms of mechanical energy are called potential and kinetic energy.

Slide 10

Sample problem Jane uses a vine wrapped around a pulley to lift a 70-kg Tarzan to a tree house 9.0 meters above the ground. a)How much work does Jane do when she lifts Tarzan? b) m = 70 kg, d = 9 m, g = 9.8 m/s 2, = 0 W = ? W = Fdcos = (mg)dcos a) W = ( 70)(9.8)(9)cos 0 b)How much work does gravity do when Jane lifts Tarzan? Same, but in the opposite direction.

Slide 11

Sample problem Joe pushes a 10-kg box and slides it across the floor at constant velocity of 3.0 m/s. The coefficient of kinetic friction between the box and floor is 0.50. a)How much work does Joe do if he pushes the box for 15 meters? m = 10 kg, = 0.5, d = 15 m, = 0, v = 15 m/s (constant) W = ? W = F app dcos, since constant velocity F app = F k W = ( mg)dcos W = (.5)(10)(9.8)(15)cos 0 b) How much work does friction do as Joe pushes the box? Same but opposite.

Slide 12

Sample problem A father pulls his child in a little red wagon with constant speed. If the father pulls with a force of 16 N for 10.0 m, and the handle of the wagon is inclined at an angle of 60 o above the horizontal, how much work does the father do on the wagon? F = 16 N, d = 10 m, = 60 W = ? W = Fdcos W = (16)(10)cos 60 W =

Slide 13

Kinetic Energy Energy due to motion K = ½ m v 2 K: Kinetic Energy m: mass in kg v: speed in m/s Unit: Joules

Slide 14

Sample problem A 10.0 g bullet has a speed of 1.2 km/s. a)What is the kinetic energy of the bullet? Do on your own, remember to convert units to proper units. a)What is the bullets kinetic energy if the speed is halved? b)What is the bullets kinetic energy if the speed is doubled?

Slide 15

The Work-Energy Theorem The net work due to all forces equals the change in the kinetic energy of a system. W net = K W net : work due to all forces acting on an object K: change in kinetic energy (K f – K i )

Slide 16

Sample problem An 8.0-g acorn falls from a tree and lands on the ground 10.0 m below with a speed of 11.0 m/s. a)What would the speed of the acorn have been if there had been no air resistance? b)Did air resistance do positive, negative or zero work on the acorn? Why?

Slide 17

Sample problem An 8.0-g acorn falls from a tree and lands on the ground 10.0 m below with a speed of 11.0 m/s. c)How much work was done by air resistance? d)What was the average force of air resistance?

Slide 18

Constant force and work The force shown is a constant force. W = F x can be used to calculate the work done by this force when it moves an object from x a to x b. The area under the curve from x a to x b can also be used to calculate the work done by the force when it moves an object from x a to x b F(x) x xaxa xbxb

Slide 19

Variable force and work The force shown is a variable force. W = F x CANNOT be used to calculate the work done by this force! The area under the curve from x a to x b can STILL be used to calculate the work done by the force when it moves an object from x a to x b F(x) x xaxa xbxb

Slide 20

Springs When a spring is stretched or compressed from its equilibrium position, it does negative work, since the spring pulls opposite the direction of motion. W s = - ½ k x 2 W s : work done by spring (J) k: force constant of spring (N/m) x: displacement from equilibrium (m) The force doing the stretching does positive work equal to the magnitude of the work done by the spring. W app = - W s = ½ k x 2

Slide 21

Springs: stretching mm x 0 F s = -kx (Hookes Law) 100 0 -100 -200 200 F(N) 012345 x (m) W s = negative area = - ½ kx 2 FsFs FsFs

Slide 22

Sample problem A spring with force constant 250 N/m is initially at its equilibrium length. a)How much work must you do to stretch the spring 0.050 m? Do on your own. a)How much work must you do to compress it 0.050 m?

Slide 23

Sample problem It takes 1000 J of work to compress a certain spring 0.10 m. a)What is the force constant of the spring? Do on your own a)To compress the spring an additional 0.10 m, does it take 1000 J, more than 1000 J, or less than 1000 J? Verify your answer with a calculation.

Slide 24

Sample Problem How much work is done by the force shown when it acts on an object and pushes it from x = 0.25 m to x = 0.75 m? Figure from Physics, James S. Walker, Prentice-Hall 2002

Slide 25

Sample Problem How much work is done by the force shown when it acts on an object and pushes it from x = 2.0 m to x = 4.0 m? Figure from Physics, James S. Walker, Prentice-Hall 2002

Slide 26

Power Power is the rate of which work is done. P = W/t W: work in Joules t: elapsed time in seconds When we run upstairs, t is small so P is big. When we walk upstairs, t is large so P is small.

Slide 27

Unit of Power SI unit for Power is the Watt. 1 Watt = 1 Joule/s Named after the Scottish engineer James Watt (1776-1819) who perfected the steam engine. British system horsepower 1 hp = 746 W

Slide 28

How We Buy Energy… The kilowatt-hour is a commonly used unit by the electrical power company. Power companies charge you by the kilowatt-hour (kWh), but this not power, it is really energy consumed. 1 kW = 1000 W 1 h = 3600 s 1 kWh = 1000 J/s 3600s = 3.6 x 10 6 J

Slide 29

Sample problem A man runs up the 1600 steps of the Empire State Building in 20 minutes. If the height gain of each step was 0.20 m, and the mans mass was 80.0 kg, what was his average power output during the climb? Give your answer in both watts and horsepower.

Slide 30

Sample problem Calculate the power output of a 0.10 g fly as it walks straight up a window pane at 2.0 cm/s.

Slide 31

Force types Forces acting on a system can be divided into two types according to how they affect potential energy. Conservative forces can be related to potential energy changes. Non-conservative forces cannot be related to potential energy changes. So, how exactly do we distinguish between these two types of forces?

Slide 32

Conservative forces Work is path independent. Work can be calculated from the starting and ending points only. The actual path is ignored in calculations. Work along a closed path is zero. If the starting and ending points are the same, no work is done by the force. Work changes potential energy. Examples: Gravity Spring force Conservation of mechanical energy holds!

Slide 33

Non-conservative forces Work is path dependent. Knowing the starting and ending points is not sufficient to calculate the work. Work along a closed path is NOT zero. Work changes mechanical energy. Examples: Friction Drag (air resistance) Conservation of mechanical energy does not hold!

Slide 34

Potential Energy Energy due to position or configuration Stored energy For gravity: U g = mgy m: mass g: acceleration due to gravity h: height above the zero point For springs: U s = ½ k x 2 k: spring force constant x: displacement from equilibrium position

Slide 35

Conservative forces and Potential energy W c = - U If a conservative force does positive work on a system, potential energy is lost. If a conservative force does negative work, potential energy is gained. For gravity W g = - U g = -(mgy – mgy o ) For springs W s = - U s = -(½ k x 2 – ½ k x o 2 )

Slide 36

More on paths and conservative forces. Q: Assume a conservative force moves an object along the various paths. Which two works are equal? A: W 2 = W 3 (path independence) Q: Which two works, when added together, give a sum of zero? A: W 1 + W 2 = 0 or W 1 + W 3 = 0 (work along a closed path is zero) Figure from Physics, James S. Walker, Prentice-Hall 2002

Slide 37

Sample problem Figure from Physics, James S. Walker, Prentice-Hall 2002 A box is moved in the closed path shown. a) How much work is done by gravity when the box is moved along the path A->B->C? b) How much work is done by gravity when the box is moved along the path A->B->C->D- >A?

Slide 38

Slide 39

Sample problem Figure from Physics, James S. Walker, Prentice-Hall 2002 A box is moved in the closed path shown. a) How much work would be done by friction if the box were moved along the path A->B->C? b) How much work is done by friction when the box is moved along the path A->B->C->D->A?

Slide 40

Slide 41

Sample problem A diver drops to the water from a height of 20.0 m, his gravitational potential energy decreases by 12,500 J. How much does the diver weigh?

Slide 42

Sample problem If 30.0 J of work are required to stretch a spring from a 2.00 cm elongation to a 4.00 cm elongation, how much work is needed to stretch it from a 4.00 cm elongation to a 6.00 cm elongation?

Slide 43

Law of Conservation of Energy In any isolated system, the total energy remains constant. Energy can neither be created nor destroyed, but can only be transformed from one type of energy to another.

Slide 44

Law of Conservation of Mechanical Energy E = K + U = Constant K: Kinetic Energy (1/2 mv 2 ) U: Potential Energy (gravity or spring) E = U + K = 0 K: Change in kinetic energy U: Change in gravitational/spring potential energy U – U o + K – K o = 0 U + K = U o + K o **** mgy + ½ mv 2 = mgy o + ½ mv o 2 ****

Slide 45

Roller Coaster Simulation Roller Coaster Physics Simulation (demonstrates gravitational potential energy and kinetic energy)

Slide 46

Pendulums and Energy Conservation Energy goes back and forth between K and U. At highest point, all energy is U. As it drops, U goes to K. At the bottom, energy is all K.

Slide 47

h Pendulum Energy ½mv max 2 = mgy For minimum and maximum points of swing K 1 + U 1 = K 2 + U 2 For any points 1 and 2.

Slide 48

Springs and Energy Conservation Transforms energy back and forth between K and U. When fully stretched or extended, all energy is U. When passing through equilibrium, all its energy is K. At other points in its cycle, the energy is a mixture of U and K.

Slide 49

Spring Energy mm -x m x 0 ½kx max 2 = ½mv max 2 For maximum and minimum displacements from equilibrium K 1 + U 1 = K 2 + U 2 = E For any two points 1 and 2 All U All K

Slide 50

Spring Simulation Spring Physics Simulation

Slide 51

Sample problem What is the speed of the pendulum bob at point B if it is released from rest at point A? 1.5 m A B 40 o

Slide 52

Slide 53

Sample problem Problem copyright Physics, James S. Walker, Prentice-Hall 2002 A 0.21 kg apple falls from a tree to the ground, 4.0 m below. Ignoring air resistance, determine the apples gravitational potential energy, U, kinetic energy, K, and total mechanical energy, E, when its height above the ground is each of the following: 4.0 m, 2.0 m, and 0.0 m. Take ground level to be the point of zero potential energy.

Slide 54

Sample problem Problem copyright Physics, James S. Walker, Prentice-Hall 2002 A 1.60 kg block slides with a speed of 0.950 m/s on a frictionless, horizontal surface until it encounters a spring with a force constant of 902 N/m. The block comes to rest after compressing the spring 4.00 cm. Find the spring potential energy, U, the kinetic energy of the block, K, and the total mechanical energy of the system, E, for the following compressions: 0 cm, 2.00 cm, 4.00 cm.

Slide 55

Law of Conservation of Energy E = U + K + E int = Constant E int is thermal energy. U + K + E int = 0 Mechanical energy may be converted to and from heat.

Slide 56

Work done by non- conservative forces W net = W c + W nc Net work is done by conservative and non-conservative forces W c = - U Potential energy is related to conservative forces only! W net = K Kinetic energy is related to net force (work-energy theorem) K = - U + W nc From substitution W nc = U + K = E Nonconservative forces change mechanical energy. If nonconservative work is negative, as it often is, the mechanical energy of the system will drop.

Slide 57

Sample problem Problem copyright Physics, James S. Walker, Prentice-Hall 2002 Catching a wave, a 72-kg surfer starts with a speed of 1.3 m/s, drops through a height of 1.75 m, and ends with a speed of 8.2 m/s. How much non-conservative work was done on the surfer?

Slide 58

Sample problem Problem copyright Physics, James S. Walker, Prentice-Hall 2002 A 1.75-kg rock is released from rest at the surface of a pond 1.00 m deep. As the rock falls, a constant upward force of 4.10 N is exerted on it by water resistance. Calculate the nonconservative work, W nc, done by the water resistance on the rock, the gravitational potential energy of the system, U, the kinetic energy of the rock, K, and the total mechanical energy of the system, E, for the following depths below the waters surface: d = 0.00 m, d = 0.500 m, d = 1.00 m. Let potential energy be zero at the bottom of the pond.

Slide 59

Pendulum lab Figure out how to demonstrate conservation of energy with a pendulum using the equipment provided. The photo-gates must be set up in gate mode this time. The width of the pendulum bob is an important number. To get it accurately, use the caliper.caliper Turn in just your data, calculations, and result. Clearly show the speed you predict for the pendulum bob from conservation of energy, the speed you measure using the caliper and photo-gate data, and a % difference for the two.