An Introduction. Work

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1 Work and Energy An Introduction Work Work tells us how much a force or combination of forces changes the energy of a system. Work is the bridge between force (a vector) and energy (a scalar). W = F Dr cos F: force (N) Dr : displacement (m) : angle between force and displacement SI Unit: Joule (N m) 1

2 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. 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? Zero work is done because there is zero displacement 2

3 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? Zero work is done because force is perpendicular to movement Question If a man lifts a 50 kg box 2.0 meters, how much work does he do on the box? W Fd 2 F mg (50 kg)(10 m / s ) 500N W (500 N)(2.0 m) 1000N m 3

4 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. 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. How much work does Jane do when she lifts Tarzan? W Fd 2 F mg (70 kg)(10 m / s ) 700N W (700 N)(9.0 m) 6300N m 4

5 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 a) How much work does Joe do if he pushes the box for 15 meters? W Fd F push f k 2 fk FN (0.50)(10 kg)(10 m / s ) 50N W (50 N)(15 m) 750N m b) How much work does friction do as Joe pushes the box? W 750N m 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? W Fd cos W (16 N)(10 m)cos60 o 80N m 5

6 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 Sample problem A 10.0 g bullet has a speed of 1.2 km/s. a) What is the kinetic energy of the bullet? 1 (0.01 )(1200 / ) 2 KE kg m s 2 KE 7200J b) What is the bullet s kinetic energy if the speed is halved? 7200J KE 1800J 4 c) What is the bullet s kinetic energy if the speed is doubled? KE (7200 J)(4) J 6

7 The Work-Energy Theorem The net work due to all forces equals the change in the kinetic energy of a system. W net = DK W net : work due to all forces acting on an object DK: change in kinetic energy (K f K i ) 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? 2 2 v vo 2gy 2 2 v m s m 0 2(10 / )(10.0 ) v 14.1 m / s b) Did air resistance do positive, negative or zero work on the acorn? Why? Negative because it made the actual speed less than the expected speed 7

8 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? W DKE 1 1 DKE (.008 kg)(11.0 m / s) (.008 kg)(14.1 m / s) 2 2 DKE.484 J.795J W 0.311J 2 2 d) What was the average force of air resistance? W Fd F( 10 m) 0.311J F.0311N Assignment pg #6,11,19,26 8

9 Work and Energy An Introduction Constant force and work The force shown is a constant force. W = FDr 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 a x b x 9

10 Variable force and work The force shown is a variable force. W = FDr 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 a x b x 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? W.25(.4).25(.8) 0.3J Figure from Physics, James S. Walker, Prentice-Hall

11 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? 1 W (2)(.6) 0.6 J 2 Figure from Physics, James S. Walker, Prentice-Hall 2002 Power Power is the rate of which work is done. P = W/Dt W: work in Joules Dt: 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. 11

12 Unit of Power SI unit for Power is the Watt. 1 Watt = 1 Joule/s Named after the Scottish engineer James Watt ( ) who perfected the steam engine. British system horsepower 1 hp = 746 W How We Buy Energy The kilowatt-hour is a commonly used unit by the electrical power company. Power companies charge you by the kilowatthour (kwh), but this not power, it is really energy consumed. 1 kw = 1000 W 1 h = 3600 s 1 kwh = 1000J/s 3600s = 3.6 x 106J 12

13 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 man s mass was 80.0 kg, what was his average power output during the climb? Give your answer in both watts and horsepower. h 1600(0.20 m) 320m W DE mgh kg m s m W J W J P t 1200s P 213Watts hp (80.0 )(10 / )(320 ) Potential energy Energy of position or configuration Stored energy For gravity: U g = mgh 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 13

14 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? 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! 14

15 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! Conservative forces and Potential energy W c = -DU 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 = -DU g = -(mgh f mgh i ) For springs W s = -DU s = -(½ k x f 2 ½ k x i2 ) 15

16 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 Sample problem A diver drops to the water from a height of 20.0m, his gravitational potential energy decreases by 12,500 J. How much does the diver weigh? U g mgh m(10 m / s )(20.0 m) 200m m 62.5kg 16

17 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. Law of Conservation of Mechanical Energy E = K + U = Constant K: Kinetic Energy (1/2 mv 2 ) U: Potential Energy (gravity or spring) DE = DU + DK = 0 DK: Change in kinetic energy DU: Change in gravitational or spring potential energy 17

18 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. Pendulum Energy h ½mv max 2 = mgh For minimum and maximum points of swing K 1 + U 1 = K 2 + U 2 For any points 1 and 2. 18

19 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. Spring Energy All U m -x 0 K 1 + U 1 = K 2 + U 2 = E For any two points 1 and 2 All K m x All U m ½kx max 2 = ½mv max 2 For maximum and minimum displacements from equilibrium 19

20 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? Work DEnergy k k(.04 m) k(.02 m) 30.0J k Work (50000)(.06 m) (50000)(.04 m) 2 2 Work 50.0J 2 2 Sample problem What is the speed of the pendulum bob at point B if it is released from rest at point A? v max max max 2gh o h cos m v m s m v 2 2(10 / )(.351 ) 2.65 m / s 40 o 1.5cos 40 h 1.5 m A B Max speed at point B 20

21 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 apple s 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. at 4.0 m energy is all potential 2 E U mgh (0.21 kg)(10 m / s )(4.0 m) 8.4J K 0J at 2.0m energy is both kinetic and potential but total is unchanged E 8.4J 2 U mgh (0.21 kg)(10 m / s )(2.0 m) 4.2J K E U J at 0m, right before stopping, total and kinetic energy are equal E K 8.4J U 0J Sample problem Problem copyright Physics, James S. Walker, Prentice-Hall 2002 A 1.60 kg block slides with a speed of 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. at 4.0 m energy is all potential (902)(.04) E U kx 2 2 J K 0J at 2.0m energy is both kinetic and potential but total is unchanged E J (902)(.02) U kx J 2 2 K E U J at 0m total and kinetic energy are equal and total is unchanged E K J U 0J 21

22 Law of Conservation of Energy E = U + K + E int = Constant E int is thermal energy. DU + DK + D E int = 0 Mechanical energy may be converted to and from heat. Work done by non-conservative forces W net = W c + W nc Net work is done by conservative and non-conservative forces W c = -DU Potential energy is related to conservative forces only! W net = DK Kinetic energy is related to net force (work-energy theorem) DK = -DU + W nc From substitution W nc = DU + DK = DE Nonconservative forces change mechanical energy. If nonconservative work is negative, as it often is, the mechanical energy of the system will drop. 22

23 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? 1 2 Eo Uo Ko mgho mvo W DE J 2 2 Eo (72 kg)(10 m / s )(1.75 m) (72 kg)(1.3 m / s) 1321J 2 E f U f K f 0 (72 kg)(8.2 m / s) 2421J NC Assignment pg #32,36,41,44,49,51,56,62,68 23

24 Momentum and Momentum Change Which do you think has more momentum? 24

25 Momentum Momentum is a measure of how hard it is to stop or turn a moving object. What characteristics of an object would make it hard to stop or turn? Sample Problem Calculate the momentum of a 65-kg sprinter running east at 10 m/s. p mv p (10 m / s)(65 kg) 650 kg m / s 25

26 Sample Problem Calculate the momentum of a system composed of a 65-kg sprinter running east at 10 m/s and a 75-kg sprinter running north at 9.5 m/s. Sample Problem Calculate the momentum of a system composed of a 65-kg sprinter running east at 10 m/s and a 75-kg sprinter running north at 9.5 m/s. 26

27 Change in momentum Like any change, change in momentum is calculated by looking at final and initial momentums. Dp = p f p i Dp: change in momentum p f : final momentum p i : initial momentum Momentum change demonstration Using only a meter stick, find the momentum change of each ball when it strikes the desk from a height of exactly one meter. Which ball, Bouncy or Lazy, has the greatest change in momentum? 27

28 Wording dilemma In which case is the magnitude of the momentum change greatest? In which case is the change in the magnitude of the momentum greatest? Impulse (J) Impulse is the product of an external force and time, which results in a change in momentum of a particle or system. J = F t and J = DP Therefore Ft = DP Units: N s or kg m/s (same as momentum) 28

29 Impulsive Forces Usually high magnitude, short duration. Suppose the ball hits the bat at 90 mph and leaves the bat at 90 mph, what is the magnitude of the momentum change? What is the change in the magnitude of the momentum? F(N) 3000 Impulse (J) on a graph area under curve t (ms) 29

30 Sample Problem Suppose a 1.5-kg brick is dropped on a glass table top from a height of 20 cm. a) What is the magnitude and direction of the impulse necessary to stop the brick? b) If the table top doesn t shatter, and stops the brick in 0.01 s, what is the average force it exerts on the brick? c) What is the average force that the brick exerts on the table top during this period? Sample Problem Suppose a 1.5-kg brick is dropped on a glass table top from a height of 20 cm. a) What is the magnitude and direction of the impulse necessary to stop the brick? b) If the table top doesn t shatter, and stops the brick in 0.01 s, what is the average force it exerts on the brick? c) What is the average force that the brick exerts on the table top during this period? 30

31 Sample Problem F(N) 2,000 1, t(s) This force acts on a 1.2 kg object moving at m/s. The direction of the force is aligned with the velocity. What is the new velocity of the object? Law of Conservation of Momentum If the resultant external force on a system is zero, then the vector sum of the momentums of the objects will remain constant. SP before = SP after 31

32 Sample problem A 75-kg man sits in the back of a 120-kg canoe that is at rest in a still pond. If the man begins to move forward in the canoe at 0.50 m/s relative to the shore, what happens to the canoe? Sample problem A 75-kg man sits in the back of a 120-kg canoe that is at rest in a still pond. If the man begins to move forward in the canoe at 0.50 m/s relative to the shore, what happens to the canoe? 32

33 External versus internal forces External forces: forces coming from outside the system of particles whose momentum is being considered. External forces change the momentum of the system. Internal forces: forces arising from interaction of particles within a system. Internal forces cannot change momentum of the system. An external force in golf The club head exerts an external impulsive force on the ball and changes its momentum. The acceleration of the ball is greater because its mass is smaller. The System 33

34 An internal force in pool The System The forces the balls exert on each other are internal and do not change the momentum of the system. Since the balls have equal masses, the magnitude of their accelerations is equal. Explosions When an object separates suddenly, as in an explosion, all forces are internal. Momentum is therefore conserved in an explosion. There is also an increase in kinetic energy in an explosion. This comes from a potential energy decrease due to chemical combustion. 34

35 Recoil Guns and cannons recoil when fired. This means the gun or cannon must move backward as it propels the projectile forward. The recoil is the result of action-reaction force pairs, and is entirely due to internal forces. As the gases from the gunpowder explosion expand, they push the projectile forwards and the gun or cannon backwards. Sample problem Suppose a 5.0-kg projectile launcher shoots a 209 gram projectile at 350 m/s. What is the recoil velocity of the projectile launcher? 35

36 Sample problem Suppose a 5.0-kg projectile launcher shoots a 209 gram projectile at 350 m/s. What is the recoil velocity of the projectile launcher? Sample Problem An exploding object breaks into three fragments. A 2.0 kg fragment travels north at 200 m/s. A 4.0 kg fragment travels east at 100 m/s. The third fragment has mass 3.0 kg. What is the magnitude and direction of its velocity? 36

37 Sample Problem An exploding object breaks into three fragments. A 2.0 kg fragment travels north at 200 m/s. A 4.0 kg fragment travels east at 100 m/s. The third fragment has mass 3.0 kg. What is the magnitude and direction of its velocity? Collisions When two moving objects make contact with each other, they undergo a collision. Conservation of momentum is used to analyze all collisions. Newton s Third Law is also useful. It tells us that the force exerted by body A on body B in a collision is equal and opposite to the force exerted on body B by body A. 37

38 Collisions During a collision, external forces are ignored. The time frame of the collision is very short. The forces are impulsive forces (high force, short duration). Collision Types Elastic collisions Also called hard collisions No deformation occurs, no kinetic energy lost Inelastic collisions Deformation occurs, kinetic energy is lost Perfectly Inelastic (stick together) Objects stick together and become one object Deformation occurs, kinetic energy is lost 38

39 (Perfectly) Inelastic Collisions Simplest type of collisions. After the collision, there is only one velocity, since there is only one object. Kinetic energy is lost. Explosions are the reverse of perfectly inelastic collisions in which kinetic energy is gained! An 80-kg roller skating grandma collides inelastically with a 40- kg kid. What is their velocity after the collision? How much kinetic energy is lost? Sample Problem 39

40 An 80-kg roller skating grandma collides inelastically with a 40- kg kid. What is their velocity after the collision? How much kinetic energy is lost? Sample Problem Sample Problem A fish moving at 2 m/s swallows a stationary fish which is 1/3 its mass. What is the velocity of the big fish after dinner? 40

41 Sample Problem A fish moving at 2 m/s swallows a stationary fish which is 1/3 its mass. What is the velocity of the big fish after dinner? Sample problem A car with a mass of 950 kg and a speed of 16 m/s to the east approaches an intersection. A 1300-kg minivan traveling north at 21 m/s approaches the same intersection. The vehicles collide and stick together. What is the resulting velocity of the vehicles after the collision? 41

42 Sample problem A car with a mass of 950 kg and a speed of 16 m/s to the east approaches an intersection. A 1300-kg minivan traveling north at 21 m/s approaches the same intersection. The vehicles collide and stick together. What is the resulting velocity of the vehicles after the collision? Elastic Collision In elastic collisions, there is no deformation of colliding objects, and no change in kinetic energy of the system. Therefore, two basic equations must hold for all elastic collisions Sp b = Sp a (momentum conservation) SK b = SK a (kinetic energy conservation) 42

43 Sample Problem A 500-g cart moving at 2.0 m/s on an air track elastically strikes a 1,000-g cart at rest. What are the resulting velocities of the two carts? Sample Problem A 500-g cart moving at 2.0 m/s on an air track elastically strikes a 1,000-g cart at rest. What are the resulting velocities of the two carts? 43

44 Sample Problem Suppose three equally strong, equally massive astronauts decide to play a game as follows: The first astronaut throws the second astronaut towards the third astronaut and the game begins. Describe the motion of the astronauts as the game proceeds. Assume each toss results from the same-sized "push." How long will the game last? 2D-Collisions Momentum in the x-direction is conserved. SP x (before) = SP x (after) Momentum in the y-direction is conserved. SP y (before) = SP y (after) Treat x and y coordinates independently. Ignore x when calculating y Ignore y when calculating x Let s look at a simulation: 44

45 Sample problem Calculate velocity of 8-kg ball after the collision. y y 2 m/s 2 kg 3 m/s 2 kg 8 kg x 50 o x 0 m/s 8 kg v Before After Assignment pg #4,11,13,17,22,28,34,40,45 45

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