PHYSICS 231 Energy & work!

Transcription

1 PHYSICS 231 Energy & work! Remco Zegers 1

2 WORK Work: Transfer of energy Quantitatively: The work W done by a constant force on an object is the product of the force along the direction of displacement and the magnitude of displacement. W=(Fcos ) x Units: =Nm=Joule 1 calorie=4.184 J 1 Calorie=1000 cal Fcos F x 2

3 Fcos Fcos Non-constant force/angle W=(Fcos ) x: what if Fcos not constant while covering x? x F Example: what if or F changes while pulling the block? Area= A=(Fcos ) x W= ( A)=total area x x The work done is the same as the area under the graph of Fcos versus x 3 x

4 example force A person is dragging a block over a floor, with a force parallel to the floor. 4N 2N After 4 meter, the floor turns rough and instead of a force of m 2N a force of 4N must be applied. distance The force-distance diagram shows the situation. How much work did the person do over 8 meter? a) 0 J b) 16 J c) 20 J d) 24 J e) 32 J Work: area under F-x diagram: 4x2+4x4=24 J 4

5 Power: The rate of energy transfer Work (the amount of energy transfer) is independent of time. W=(Fcos ) x no time in here! To measure how fast we transfer the energy we define: Power(P)=W/ t (J/s=Watt) P =(Fcos ) x/ t=(fcos )v average 1 Watt = horsepower 5

6 A running person While running, a person dissipates about 0.60 J of mechanical energy per step per kg of body mass. If a 60 kg person develops a power of 70 Watt during a race, how fast is she running (1 step=1.5 m long) What is the force the person exerts on the road? W=F x P=W/ t=fv Work per step: 0.60 J/kg * 60 kg=36 J Work during race: 36*(racelength(L)/steplength)=24L Power= W/ t=24l/ t=24v average= 70 so v average =2.9 m/s F=P/v so F=24 N 6

7 Example A toy-rocket of 5.0 kg, after the initial acceleration stage, travels 100 m in 2 seconds. What is the work done by the engine? What is the power of the engine? h=100m W=(Fcos ) h=m rocket g h=4905 J (Force by engine must balance gravity!) P=W/ t=4905/2=2453 W (=3.3 horsepower) or P=(Fcos )v=mgv= /2=2453 W 7

8 Potential Energy Potential energy (PE): energy associated with the position of an object within some system. Gravitational potential energy: Consider the work done by the gravity in case of the rocket: W gravity =F g cos(180 o ) h=-mg h=-(mgh f -mgh i )=mgh i -mgh f =PE i -PE f The system is the gravitational field of the earth. PE=mgh Since we are usually interested in the change in gravitational potential energy, we can choose the ground level (h=0) in a convenient way. 8

9 Kinetic energy Consider object that changes speed only t=2s x=100m V=0 a) W=F x=(ma) x used Newton s second law b) v=v 0 +at so t=(v-v 0 )/a c) x=x 0 +v 0 t+0.5at 2 so x-x 0 = x=v 0 t+0.5at 2 Combine b) & c) d) a x=(v 2 -v 02 )/2 Combine a) & d) W=½m(v 2 -v 02 ) Kinetic energy: KE=½mv 2 When work is done on an object and the only change is its speed: The work done is equal to the change in KE: W=KE final -KE initial 9

10 Conservation of energy Mechanical energy = Potential Energy + Kinetic energy Mechanical energy is conserved if: the system is closed (no energy can enter or leave) the forces are conservative (see soon) Heat, chemical energy (e.g battery or fuel in an engine) Are sources or sinks of internal energy. 10

11 Conservation of mechanical energy Mechanical energy = potential energy + kinetic energy In a closed system, mechanical energy is conserved * V=100 m/s ME=mgh+½mv 2 =constant 5 kg What about the accelerating rocket? h=100m At launch: ME=5*9.81*0+½5*0 2 =0 At 100 m height: ME=5*9.81*100+½5*100 2 =29905 We did not consider a closed system! (Fuel burning) * There is an additional condition, see next lecture V=0 11

12 Example of closed system An object (0.2 kg) is dropped from a height of 35 m. Assuming no friction, what is the velocity when it reaches the ground? h=35 m v=0 m/s At launch: ME=mgh+½mv 2 0.2*9.81*35+½*0.2*0 2 =68.67 J At ground: ME=mgh+½mv 2 =0.2*9.81*0+½*0.2*v 2 =0.1v 2 J v at h=0? Conservation of ME: 68.67=0.1v 2 v=26.2 m/s 12

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14 Work and energy WORK POTENTIAL ENERGY KINETIC ENERGY 14

15 Mechanical Energy Mechanical energy Gravitational Potential Energy (mgh) Kinetic Energy ½mv 2 15

16 Work and Energy Work: W=Fcos( ) x Energy transfer The work done is the same as the area under the graph of Fcos versus x Power: P=W/ t Rate of energy transfer Potential energy (PE) Energy associated with position. Gravitational PE: mgh Energy associated with position in grav. field. Kinetic energy KE: ½mv 2 Energy associated with motion Elastic PE: ½kx 2 energy stored in stretched/compressed spring. Conservative force: Work done does not depend on path Non-conservative force: Work done does depend on path Mechanical energy ME: ME=KE+PE Conserved if only conservative forces are present KE i +PE i =KE f +PE f Not conserved in the presence of non-conservative forces (KE i +PE i )-(KE f +PE f )=W nc 16

17 question Old faithful geyser in Yellowstone park shoots water hourly to a height of 40 m. With what velocity does the water leave the ground? a) 7.0 m/s b) 14 m/s c) 20 m/s d) 28 m/s e) don t know KE i +PE i =KE f +PE f At ground level: ME=0.5mv 2 +mgh= 0.5mv 2 +0=0.5mv 2 At highest point: ME=0.5mv 2 +mgh= 0+m*9.8*40=392m Conservation of energy: 0.5mv 2 =392m so v=28 m/s 17

18 Extra credit quiz A ball (A) is dropped from a height h. A second ball (B) is rolled down a slope starting at height h. The initial speed of both is 0. Ignoring friction which one has the highest speed when reaching the ground? A B a) Ball A b) Ball B c) Same speed d) Cannot tell without more information Conservation of energy: mgh+0.5mv 2 =constant For both: initial mechanical energy: mgh final mechanical energy: 0.5mv 2 mgh=0.5mv 2 v= (2gh) 18

19 A swing 30 o L=5m If relieved from rest, what is the velocity of the ball at the lowest point? h (PE+KE)=constant PE release =mgh (h=5-5cos(30 o )) =6.57m J KE release =0 PE bottom =0 KE bottom =½mv 2 ½mv 2 =6.57m so v=3.6 m/s 19

20 Conservative forces A force is conservative if the work done by the force when Moving an object from A to B does not depend on the path taken from A to B. Example: work done by gravitational force Using the stairs: W g =mgh f -mgh i =mg(h f -h i ) h=10m Using the elevator: W g =mgh f -mgh i =mg(h f -h i ) The path taken (longer or shorter) does not matter: only the displacement does! 20

21 Non conservative forces A force is non-conservative if the work done by the force when moving an object from A to B depends on the path taken from A to B. object on rough surface top view Example: Friction You have to perform more work Against friction if you take the long path, compared to the short path. The friction force changes kinetic energy into heat. 21

22 Conservation of mechanical energy only holds if the system is closed AND all forces are conservative ME i -ME f =(PE+KE) i -(PE+KE) f =0 if all forces are conservative Example: throwing a snowball from a building neglecting air resistance ME i -ME f =(PE+KE) i -(PE+KE) f =W nc if some forces are nonconservative. W nc =work done by non-conservative forces. Example: throwing a snowball from a building taking into account air resistance 22

23 Overview Newton s second Law F=ma Work W=(Fcos ) x Conservation of mechanical energy W nc =0 Closed system Work-energy Theorem W nc =E f -E i Equations of kinematics X(t)=X(0)+V(0)t+½at 2 V(t)=V(0)+at 23

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25 Race track With friction KE PE TME NC KE PE TME NC KE PE TME NC KE PE TME NC KE PE TME NC 25

26 energy energy question A time B In the absence of friction, which energy-time diagram is correct? potential energy total energy kinetic energy C time time 26

27 A B h Ball on a track end h end In which case has the ball the highest velocity at the end? A) Case A B) Case B C) Same speed In which case does it take the longest time to get to the end? A) Case A B) Case B C) Same time 27

28 question A ball of 1 kg rolls up a ramp, with initial velocity of 6 m/s. It reaches a maximum height of 1 m (I.e. velocity 0 at at that point). How much work is done by friction? a) 0. b) 8.2 J c) 9.8 J d) 18 J e) 27.8 J (KE i +PE i )-(KE f +PE f )=W nc initial: E=0.5mv 2 (kinetic only)=18j final: E=mgh (potential only)=9.8 J W nc =18-9.8=8.2 J kinetic energy: 0.5mv 2 potential energy: mgh g=9.8 m/s 2 29

29 Conservation of mechanical energy A) what is the speed of m 1 and m 2 when they pass each other? 5 kg 4.0 m 3 kg (PE 1 +PE 2 +KE 1 +KE 2 )=constant At time of release: PE 1 =m 1 gh 1 =5.00*9.81*4.00 =196. J PE 2 =m 2 gh 2 =3.00*9.81*0.00 =0.00 J KE 1 =0.5*m 1 *v 2 =0.5*5.00*(0.) 2 =0.00 J KE 2 =0.5*m 1 *v 2 =0.5*3.00*(0.) 2 =0.00 J Total =196. J At time of passing: PE 1 =m 1 gh 1 =5.00*9.81*2.00 =98.0 J PE 2 =m 2 gh 2 =3.00*9.81*0.00 =58.8 J KE 1 =0.5*m 1 *v 2 =0.5*5.00*(v) 2 =2.5v 2 J KE 2 =0.5*m 1 *v 2 =0.5*3.00*(v) 2 =1.5v 2 J Total = v 2 196= v 2 so v=3.13 m/s 30

30 work How much work is done by the gravitational force when the masses pass each other? W=F x=m 1 g2.00+m 2 g(-2.00)=39.2 J 5 kg Pe start - Pe passing =( )= 39.2 J The work done by F g is the same as the change in potential energy 3 kg 4.0 m 31

31 Friction (non-conservative) The pulley is not completely frictionless. The friction force equals 5 N. What is the speed of the objects when they pass? ( PE+ KE) start -( PE+ KE) passing =W nc W nc =F friction x=5.00*2.00=10.0 J 5 kg 4.0 m 3 kg (196.)-( KE)=10 J KE=29.2 J=0.5*(m 1 +m 2 )v 2 =4v 2 v=2.7 m/s 32

32 5 kg 4.0 m 3 kg questions In the absence of friction, when m 1 starts to move down: 1) potential energy is transferred from m 1 to m 2 2) potential energy is transformed into kinetic energy 3) m 1 and m 2 have the same kinetic energy a) 1) and 2) and 3) are true b) only 2) and 3) are true c) only 1) and 2) are true d) only 2 is true. 33

33 x=-a x=0 A spring F s =-kx Hooke s law: k: spring constant (N/m) F s (x=0)=0 N F s (x=-a)=ka average: F s =(0+ka)/2=ka/2 W s =F s x=(ka/2)*(a)=ka 2 /2 The energy stored in a spring depends on the location of the endpoint: elastic potential energy. E potential,spring =½kx 2 34

34 PINBALL The ball-launcher spring has a constant k=120 N/m. A player pulls the handle 0.05 m. The mass of the ball is 0.1 kg. What is the launching speed? (PE gravity +PE spring +KE ball ) pull =(PE gravity +PE spring +KE ball ) launch mgh pull +½kx pull2 +½mv pull 2 = mgh launch +½kx launch2 +½mv launch 2 0.1*9.81*0+½120(0.05) 2 +½0.1(0) 2 = 0.1*9.81*(0.05*sin(10 o ))+½120*(0) 2 +½0.1v launch =8.5E v 2 v=1.7 m/s 35

35 Where is the kinetic energy 1) highest? 2) lowest? Assume height of catapult is negligible to the maximum height of the stone. And what about potential energy? Parabolic motion t=0 t=1 t=2 t=3 A B C D E t=5 36

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37 question An outfielder throws a baseball of 0.15 kg at a speed of 40 m/s and angle of 30 degrees with the field. What is the kinetic energy of the baseball at the highest point, ignoring friction? a) 0 J b) 30 J c) 90 J d) 120 J e) don t know Horizontal component of velocity at start: v x =v o cos30=34.65 m/s At highest point: only horizontal velocity v x,highest =34.65 m/s kinetic energy: 0.5mv 2 =90 J 38

38 friction A box of 2 kg slides (from rest) down a slope. At the bottom, it has a velocity of 4 m/s. If h=2 m, how much work was done by the frictional force? If F friction =2 N, over what length did the box slide. ME i -ME f =(PE+KE) i -(PE+KE) f =W nc PE i =mgh i =2x9.8x2=39.2 J PE f =mgh f =2x9.8x0=0 J combine: W nc =W friction = KE i =0.5mv 2 =0.5*2*0 2 =0 J =23.2 J KE f =0.5mv 2 =0.5*2*4 2 =16 J W=F x cos :angle between direction of force and direction of motion=0 in this case: W friction =F x 23.2=2 x, so x=16.6 m. 39

39 question A worker pushes a wheelbarrow with a force of 50 N over a distance of 10 m. A frictional force acts on the wheelbarrow in the opposite direction, with a magnitude of 30 N. What net work is done on the wheelbarrow? a) don t know b) 100 J c) 200 J W net =F x=(50-30)10=200 J d) 300 J e) 500 J 40