PS 11 GeneralPhysics I for the Life Sciences

Transcription

1 PS GeneralPhysics I for the Life Sciences W O R K N D E N E R G Y D R. E N J M I N C H N S S O C I T E P R O F E S S O R P H Y S I C S D E P R T M E N T J N U R Y 0 4

2 Questions and Probles for Conteplation Chapter 6 Questions:, 7, 8, 0,4, 8,, 5 Probles:, 5, 7, 5, 8,, 6, 3, 33, 37, 40, 4, 43, 47, 5, 57, 58, 59, 60, 66 General Probles: 74, 8, 83, 88, 9, 93

3 Making Objects Moe pply force on the oing body Finite tie of application Push, pull, throw, kick Hit with racket, bat, golf club, stick Engine, otors, achines How about Graity? No one can escape graity! (infinite tie of application!) Work and Energy Considerations Force Displaceent Tie

4 Dragging a Crate on the Floor

5 Doing Work Work Quantitatie entity (scalar) Product of the displaceent and the coponent of the force parallel to the displaceent W = 0 W F d Fd cos = 90 applied force is perpendicular to displaceent cos 90 = 0 d = 0 ( no atter how hard you push or pull!) F = 0 (Now, you re not really exerting any effort!)

6 Force Diagras pplied Force F Useful coponent of F Coponent parallel to displaceent Does work! Free ody Diagra No friction Object pulled at an angle with force F

7 re You Doing Work?

8 Doing Work Work done by an object/force Work done by graity Work done on a specific object Work done on a crate Units of work N = J (joule) Sae as energy

9 Maxiize Work F parallel to d W Fd cos0 Fd Works well for net force Frictional Forces ay coplicate the picture

10 Exaple 6- person pulls a 50-kg crate 40 along a horizontal floor with a constant force of 00 N at an angle of 37. The frictional force between the floor and the crate is 50 N. Deterine a) the work done by each force acting on the crate and b) the net work done on the crate.

11 Solution Work done by graitational and noral force is zero (perpendicular to the displaceent) Work done by the applied force F P WP FP xcos (00N)(40)cos37 300J Work done by frictional force F fr WP Ffrxcos80 (50N)(40)( ) 000J Work done by friction on the crate is negatie!

12 Solution Net work = su of work done by each force W net W G W N W P J 00J W fr 000J lternatie solution Deterine the work done by the net force W net F net x ( F P cos F ) x (00 N cos37 50N)(40) 00J fr

13 What is Energy? Duhhh.. It is easured in joules There are two types: kinetic and potential KE = ½ (translational: along a straight line) Moing object PE = gh (graitational PE) Position with respect to ground It is related to work

14 Work and Energy For a constant acceleration a Work done by the net force F net is d d ad d F W net net d a ad

15 Work-Energy Principle The net work done on an object is equal to the change in the object s kinetic energy W net KE KE Doing work on an object changes its kinetic energy

16 Work-Energy Theore Exaple Deterine the elocity of the crate at the 40 ark W net KE f i f where i =0 00J f (50kg) f (00 N) /(50kg) 6.9 / s

17 Work-Energy Theore pplied gain If the person stops pulling the box at the 40 ark, how far will the box slide on the floor before it stops? W fr KE f i i where f = 0 ( 50N) x s ) (50kg)(6.98 / 00N x 4 50N

18 aseball Exaple 45-g baseball is thrown so that it acquires a speed of 5 /s. a) What is its kinetic energy? b) What was the net work done on the ball to ake it reach this speed assuing it started fro rest? Solution a) KE (0.45kg)(5 / s) 45J b) W net = KE = 45 J 0 = 45 J Using the Work-Energy Principle

19 aseball Exaple If the throwing action extended oer.0, what was the aerage force exerted on the baseball? Wnet Fae x KE 45J 45N F ae 45N.0

20 Exercise: Stopping a Car car traeling at 50 kph can brake to a stop in 0. If the car is going twice as fast, what will be its stopping distance? ssue the axiu braking force is independent of speed.

21 Possible Solution Work done fro initial braking exercise W brakes KE f i i Work done by second braking exercise W ( ) 4( ) 4W brakes ssuing braking force F brakes to be constant W brakes raking distance is 40! i i brakes 4W 4F (0) F (40) brakes brakes brakes

22 Energy Kinetic energy Due to otion Potential Energy Due to how an object is positioned Not necessarily oing Soe exaples Object aboe the ground Copressed/stretched spring Copressed/stretched aterials

23 Graitational Potential Energy Lifting objects aboe ground leel Do work against graity Lift an object fro y to y (h = y y ) Wext Fext ( y y ) cos 0 gh Work done by graity WG FG ( y y )cos 80 gh Graity does work against the lifting otion!

24 The Ground Ground = zero potential! Suitable reference point Let y coincide with the ground h = height of the object aboe the ground PE G gy Note W ext gh PE PE PE W G gh PE PE PE

25 Notes on Potential Energy elongs to a syste Not to a single object Change in PE is the physically eaningful quantity PE not useful on its own

26 Loop-the-loop Exaple: Roller Coaster

27 Conseratie Forces Work done depends only on the initial and final positions of the object Work done by Graity Work done around a closed loop is zero! Initial and final points coincide h = 0

28 Mechanical Energy Mechanical Energy = Kinetic E + Potential E If only conseratie forces are acting on the syste KE KE PE PE 0 KE PE KE KE PE 0 PE ME KE ME PE

29 Conseration of Mechanical Energy If only conseratie forces are acting, the total echanical energy of a syste neither decreases or increases in any process it stays constant. Change in the kinetic energy coes fro a change in potential energy

30 Proble Soling Using Conseration of Energy Free Fall Deterine the elocity of the rock as it hits the ground (h =.0 ) Deterine the tie it takes the rock to reach the ground

31 Throw all Up Into the ir o = 5 /s How high will the ball go up? Do you need the ass of the ball? What will its elocity be when it falls? How uch tie will the ball stay up in the air?

32 Slide Deterine the elocity of the child as he hits the water

33 Slide Who will be oing with a larger elocity? Who will reach the botto of the slide first?

34 Springs F = -kx (restoring force) U = ½ kx (energy stored)

35 Springs Elastic potential energy =½ k x =.60 kg, h = 55.0 c, x = 5.0 c (copression) Deterine k for the spring.

36 Solution Conseration of energy ME ME ME 3 gh g( Y) k( Y) ME =ME 3 gh g( h k gy Y) g( h Y k 590 Y) ky N ky (.60kg)(9.80 / s (0.50) )( )

37 Trapoline 65-kg trapoline artist o = 5.0 /s upward Trapoline 3.0 below Trapoline behaes like a spring with k = 6. x 0 4 N/ How far will the trapoline be depressed? How high will the trapoline artist go up after bouncing off the trapoline?

38 Solution Conseration of energy ME =ME 3 ME ME ME 3 gh i gh f i g( x) k( x) gh i g( x) ) k( x =65 kg, h i =3.0, i =5.0 /s, g=9.80 /s, k=6.x0 4 N/ ( 637N)(3.0) x Sole for x! ( ) (65kg)(5.0 / s) 637Nx (3. 0 N / N / ) x 637Nx 73.5N 0

39 Solution How high? ME =ME gh h f i gh i g i i gh h i f i g (5.0 / s) h f / s

40 Power Rate of doing work erage power Units: J/s = W (Watt) Horsepower (hp) = 746 W P Energy consued = P t W t

41 Stair-Clibing Power 60-kg jogger runs up a 4.5 high staircase in 4.0 s Estiate the power output of the jogger W gy (60kg)(9.8 / s )(4.5) P 660W t t 4.0s How uch energy did she consue? J E P t s 600J s

42 Conseration of Energy Total Energy (including dissipatie ones like heat/friction) neither increases nor decreases in any process Energy can be transfored fro one for to another and transferred fro one object to another but the total aount reains constant W NC gy KE PE gy F fr d

43 Graitational Potential Energy Work done by graitational force in oing a ass towards the center of the earth W GME r GM r E Graitational PE, U U GM r E Where is U = 0? When r =!

44 Escape Velocity To escape the earth s graitational field, an object s kinetic energy ust atch its graitational PE For the earth, KE i U i GM R GM R E 0 E E E 0 ( N (6.38 / kg 0 6 )(5.97 ) 0 4 kg).k/ s

45 lack Holes Could we just set the escape elocity to the speed of light? Not this siple: we need to consider relatiistic effects Nonetheless the siple substitution works apparently Schwarzchild radius R S : the radius below which light cannot escape! c GM R S For a 3 solar ass star: R S GM (6.67 c 0 N (3.00 / kg 0 8 )(6.0 / s) 0 30 kg) 3 R S k

46 Questions and Probles for Conteplation Chapter 7 Questions:, 3, 6, 8,, 5 Probles:, 4, 5, 9,, 6,, 7, 30, 3, 40, 4, 53, 54, 55, 56 General Probles: 65, 67, 70, 73 Long Test 3 February 7, Thursday

47 Moentu Newton s second law p F t Ipulse Ipulse and Moentu p t Ipulse F t p a t Ipulse changes the oentu of an object!

48 Ipulse Deliering Ipulse F t F t: sall interaction tie, large force required F t: large interaction tie, sall force required

49 Exaple: end your knees when landing 70-kg person jups fro a height of 3.0 Stiff-legged landing:.0 c displaceent during ipact ending the knees: 50 c displaceent Calculate the ipulse experienced by the person Estiate the aerage force exerted by the person

50 Solution Ipulse changes oentu of the person Velocity of the person as he hits the ground KE PE gh (9.8 / s ) / s Ipulse as the person is brought to a stop I p p p 0 (70kg)( Ns o / s) Negatie sign eans force is opposite the original oentu

51 Solution (continued) Force ody decelerates fro 7.7 /s to 0 /s Stiff-legged a f i 0 (7.7 / s) 3000 d (0.00) s ent knees a f i 0 (7.7 / s) 59 d (0.50) s

52 Solution Force Stiff-legged force is around fifty ties bigger! So, bend your knees when landing on the ground!!

53 Exaple: Tennis First serice = 55 /s (0 ph) ball = kg Contact tie = 4 s Force exerted on the ball F p t t (0.060kg)( s / s) 0 800N

54 Waiter s Trick Pull paper slowly Static friction has a chance to act on glass and change its oentu (significant ipulse) Pull paper fast Static friction has too little tie to act on glass to change its oentu (negligible ipulse) Inertia law - Newton s First Law is satisfied

55 Conseration of Moentu In the absence of external forces, net ipulse deliered to the syste is zero Moentu is consered! p F f p p f t p i p 0 0 i 0

56 Collisions Moentu is consered Two types Elastic collision: kinetic energy of the syste is consered Inelastic collision: part of the kinetic energy is transfored into other fors (deforation/ heat energy)

57 Elastic Collisions Colliding objects bounce off each other in an elastic fashion

58 Inelastic collision Objects stick together or defored after the collision Syste loses kinetic energy

59 D Head-On Elastic Collision > 0, < 0 ( ) Conseration of oentu No kinetic energy loss f i p p

60 Special Case = 0 Rearranging ters Diide by the preious equation to get ) )( ( ) ( ) (

61 Special Case Substitute back and sole for and Special situations (Collision Deo) = >> >>

62 Newton s Cradle Swing,, 3, or 4 balls How any balls swing at the other end?

63 allistics Inelastic collision What is the elocity of the bullet? Conseration of oentu ( M ) = ass of bullet M = ass of block = initial elocity of bullet = final elocity of block and bullet ( M )

64 Deterining ullet Velocity Conseration of energy KE is conerted into GPE (pendulu) ( M )( ) ( M gh ) gh Substituting into oentu result, ( M ) gh

65 Exaple = 0.00 g M =.00 kg h = 0.0 c ( ) g ( g / s ) / s

66 Kinetic Energy ccounting Initial KE = ½ Final KE = ½ (M+) KE i = KE f? KE KE f i M ( M ) M M

67 Recoil Recoil elocity R of gun as it is fired 0 R M R M

68 Glancing Collision Reinds you of billiards! General equations for elastic collision ) ( ) ( ) ( ) (

69 Exaple ssuing = and = 3.0 /s, deterine the elocities of and after the collision. Solution pply conseration of oentu in the x and y direction x-direction y-direction cos( 45 ) cos( 45 ) 0 sin( 45 ) sin( 45 )

70 Exaple (continued) fter canceling out in the equations, the y equation yields The x equation yields Soling for yields Then sin( 45 ) sin( 45 ) 3.0 / s cos(45 ) 3.0 / s cos(45 ) cos(. / s cos( 45 ). / s 45 )

71 PS Experient 3D collision reduced to D z-coponent the sae for both balls = 0, = Split oentu into x and y coponents ) ( ) ( ) ( sin sin 0 cos cos

72 Siplified Equations 0 ( ) cos sin ( ) sin ( cos )

73 Third Long Exa Chapter 6 and 7 February 7, Thursday 6:00-7:30 p