Introductory Physics PHYS101

Transcription

1 Introductory Physics PHYS101

2 Dr Richard H. Cyburt Office Hours Assistant Professor of Physics My office: 402c in the Science Building My phone: (304) My TRF 9:30-11:00am F 12:30-2:00pm Meetings may also be arranged at other times, by appointment In person or is the best way to get a hold of me. Check my schedule on my office door. PHYS101

3 PHYS101: Introductory Physics 400 Lecture: 8:00-9:15am, TR Science Building Lab 1: 3:00-4:50pm, F Science Building 304 Lab 2: 1:30-3:20pm, M Science Building 304 Lab 3: 3:30-5:20pm, M Science Building 304 Lab 20: 6:00-7:50pm, M Science Building 304 PHYS101

4 Mastering Physics Online Go to HYPERLINK " Under Register Now, select Student. Confirm you have the information needed, then select OK! Register now. Enter your instructor s Course ID (RCYBURTPHYS101), and choose Continue. Enter your existing Pearson account username and password and select Sign in. You have an account if you have ever used a Pearson MyLab & Mastering product, such as MyMathLab, MyITLab, MySpanishLab, or MasteringChemistry. If you don t have an account, select Create and complete the required fields. Select an access option. Enter the access code that came with your textbook or was purchased separately from the bookstore. PHYS101

5 Midterm 2 There will be a sign out sheet in my office You must sign to get your exam PHYS101

6 Midterm 3 Thursday, Oct 27 8:00-9:15am S400 Chapters 8-11 Review Session: Wednesday, Oct 26 7:00-9:00pm S300 Bring questions!!! (HW, Lab, End of Chapter, Workbook) Allowed half sheet paper of formula PHYS101

7 Introductory Physics PHYS101

8 Douglas Adams Hitchhiker s Guide to the Galaxy PHYS101

9 You re already know physics! You just don t necessarily know the terminology and language we use!!! Physics of NASCAR Physics of Anger Birds PHYS101

10 Sweet Spot Just some Baseball statistics The New Science of Hitting By: Rob Arthur PHYS101

11 In class!! PHYS101

12 This lecture will help you understand: Thermal Energy Using the Conservation of Energy Energy in Collisions Power Transforming Energy PHYS101

13 Section 10.5 Thermal Energy

14 Thermal Energy Thermal energy is the sum of the kinetic energy of atoms and molecules in a substance and the elastic potential energy stored in the molecular bonds between atoms.

15 Creating Thermal Energy Friction on a moving object does work. That work creates thermal energy.

16 Example 10.9 Creating thermal energy by rubbing A 0.30 kg block of wood is rubbed back and forth against a wood table 30 times in each direction. The block is moved 8.0 cm during each stroke and pressed against the table with a force of 22 N. How much thermal energy is created in this process? PREPARE The hand holding the block does work to push the block back and forth. Work transfers energy into the block + table system, where it appears as thermal energy according to Equation The force of friction can be found from the model of kinetic friction introduced in Chapter 5, f k = µ k n; from Table 5.2 the coefficient of kinetic friction for wood sliding on wood is µ k = 0.20.

17 Example 10.9 Creating thermal energy by rubbing (cont.) To find the normal force n acting on the block, we draw the free-body diagram of the figure, which shows only the vertical forces acting on the block.

18 Example 10.9 Creating thermal energy by rubbing (cont.) SOLVE From Equation we have ΔE th = f k Δx, where f k = µ k n. The block is not accelerating in the y-direction, so from the free-body diagram Newton s second law gives or S Fy = n - w - F = ma y = 0 n = w + F = mg + F = (0.30 kg)(9.8 m/s 2 ) + 22 N = 24.9 N The friction force is then f k = µ k n = (0.20)(24.9 N) = 4.98 N. The total displacement of the block is cm = 4.8 m. Thus the thermal energy created is ΔE th = f k Δx = (4.98 N)(4.8 m) = 24 J ASSESS This modest amount of thermal energy seems reasonable for a person to create by rubbing.

19 Try It Yourself: Agitating Atoms Vigorously rub a somewhat soft object such as a blackboard eraser on your desktop for about 10 seconds. If you then pass your fingers over the spot where you rubbed, you ll feel a distinct warm area. Congratulations: You ve just set some 100,000,000,000,000,000,000,000 atoms into motion!

20 Section 10.6 Using the Law of Conservation of Energy

21 Using the Law of Conservation of Energy We can use the law of conservation of energy to develop a before-and-after perspective for energy conservation: K + U g + U s + E th = W K f + (U g ) f + (U s ) f + E th = K i + (U g ) i + (U s ) i + W This is analogous to the before-and-after approach used with the law of conservation of momentum. In an isolated system, W = 0: K f + (U g ) f + (U s ) f + E th = K i + (U g ) i + (U s ) i

22 Choosing an Isolated System Text: p. 300

23 QuickCheck A spring-loaded gun shoots a plastic ball with a launch speed of 2.0 m/s. If the spring is compressed twice as far, the ball s launch speed will be 1.0 m/s 2.0 m/s 2.8 m/s 4.0 m/s 16.0 m/s

24 QuickCheck A spring-loaded gun shoots a plastic ball with a launch speed of 2.0 m/s. If the spring is compressed twice as far, the ball s launch speed will be 1.0 m/s 2.0 m/s 2.8 m/s 4.0 m/s 16.0 m/s Conservation of energy: Double Dx Þ double v

25 QuickCheck A spring-loaded gun shoots a plastic ball with a launch speed of 2.0 m/s. If the spring is replaced with a new spring having twice the spring constant (but still compressed the same distance), the ball s launch speed will be 1.0 m/s 2.0 m/s 2.8 m/s 4.0 m/s 16.0 m/s

26 QuickCheck A spring-loaded gun shoots a plastic ball with a launch speed of 2.0 m/s. If the spring is replaced with a new spring having twice the spring constant (but still compressed the same distance), the ball s launch speed will be 1.0 m/s 2.0 m/s 2.8 m/s 4.0 m/s 16.0 m/s Conservation of energy: Double k Þ increase v by square root of 2

27 Example Pulling a bike trailer Monica pulls her daughter Jessie in a bike trailer. The trailer and Jessie together have a mass of 25 kg. Monica starts up a 100-m-long slope that s 4.0 m high. On the slope, Monica s bike pulls on the trailer with a constant force of 8.0 N. They start out at the bottom of the slope with a speed of 5.3 m/s. What is their speed at the top of the slope?

28 Example Pulling a bike trailer (cont.) PREPARE Taking Jessie and the trailer as the system, we see that Monica s bike is applying a force to the system as it moves through a displacement; that is, Monica s bike is doing work on the system. Thus we ll need to use the full version of Equation 10.18, including the work term W.

29 Example Pulling a bike trailer (cont.) SOLVE If we assume there s no friction, so that ΔE th = 0, then Equation is or K f + (U g ) f = K i + (U g ) i + W

30 Example Pulling a bike trailer (cont.) Taking y i = 0 m and writing W = Fd, we can solve for the final speed: from which we find that v f = 3.7 m/s. Note that we took the work to be a positive quantity because the force is in the same direction as the displacement. ASSESS A speed of 3.7 m/s about 8 mph seems reasonable for a bicycle s speed. Jessie s final speed is less than her initial speed, indicating that the uphill force of Monica s bike on the trailer is less than the downhill component of gravity.

31 Energy and Its Conservation Text: p. 304

32 Energy and Its Conservation Text: p. 304

33 Section 10.7 Energy in Collisions

34 Energy in Collisions A collision in which the colliding objects stick together and then move with a common final velocity is a perfectly inelastic collision. A collision in which mechanical energy is conserved is called a perfectly elastic collision. While momentum is conserved in all collisions, mechanical energy is only conserved in a perfectly elastic collision. In an inelastic collision, some mechanical energy is converted to thermal energy.

35 Example Energy transformations in a perfectly inelastic collision The figure shows two train cars that move toward each other, collide, and couple together. In Example 9.8, we used conservation of momentum to find the final velocity shown in the figure from the given initial velocities. How much thermal energy is created in this collision?

36 Example Energy transformations in a perfectly inelastic collision (cont.) PREPARE We ll choose our system to be the two cars. Because the track is horizontal, there is no change in potential energy. Thus the law of conservation of energy, Equation 10.18a, is K f + ΔE th = K i. The total energy before the collision must equal the total energy afterward, but the mechanical energies need not be equal.

37 Example Energy transformations in a perfectly inelastic collision (cont.) SOLVE The initial kinetic energy is Because the cars stick together and move as a single object with mass m 1 + m 2, the final kinetic energy is

38 Example Energy transformations in a perfectly inelastic collision (cont.) From the conservation of energy equation on the previous slide, we find that the thermal energy increases by ΔE th = K i - K f = J J = J This amount of the initial kinetic energy is transformed into thermal energy during the impact of the collision. ASSESS About 96% of the initial kinetic energy is transformed into thermal energy. This is typical of many real-world collisions.

39 Elastic Collisions Elastic collisions obey conservation of momentum and conservation of mechanical energy. Momentum: Energy: m ( v ) = m ( v ) + m ( v ) 1 1x i 1 1x f 2 2x f m ( v ) = m ( v ) + m ( v ) x i 1 1x f 2 2x f

40 Example Velocities in an air hockey collision On an air hockey table, a moving puck, traveling to the right at 2.3 m/s, makes a head-on collision with an identical puck at rest. What is the final velocity of each puck?

41 Example Velocities in an air hockey collision (cont.) PREPARE The before-and-after visual overview is shown in the figure. We ve shown the final velocities in the picture, but we don t really know yet which way the pucks will move. Because one puck was initially at rest, we can use Equations to find the final velocities of the pucks. The pucks are identical, so we have m 1 = m 2 = m.

42 Example Velocities in an air hockey collision (cont.) SOLVE We use Equations with m 1 = m 2 = m to get The incoming puck stops dead, and the initially stationary puck goes off with the same velocity that the incoming one had.

43 Example Protecting your head A bike helmet basically a shell of hard, crushable foam is tested by being strapped onto a 5.0 kg headform and dropped from a height of 2.0 m onto a hard anvil. What force is encountered by the headform if the impact crushes the foam by 3.0 cm? PREPARE We can use the work-energy equation, Equation 10.18, to calculate the force on the headform. We ll choose the headform and the earth to be the system; the foam in the helmet is part of the environment. We make this choice so that the force on the headform due to the foam is an external force that does work W on the headform.

44 Example Protecting your head (cont.) Before-and-after visual overview of the bike helmet test

45 Example Protecting your head (cont.) SOLVE The headform starts at rest, speeds up as it falls, then returns to rest during the impact. Overall, then, K f = K i. Furthermore, ΔE th = 0 because there s no friction to increase the thermal energy. Only the gravitational potential energy changes, so the work-energy equation is (U g ) f - (U g ) i = W The upward force of the foam on the headform is opposite the downward displacement of the headform. Referring to Tactics Box 10.1, we see that the work done is negative: W = -Fd, where we ve assumed that the force is constant. Using this result in the work-energy equation and solving for F, we find

46 Example Protecting your head (cont.) Taking our reference height to be y = 0 m at the anvil, we have (U g ) f = 0. We re left with (U g ) i = mgy i, so This is the force that acts on the head to bring it to a halt in 3.0 cm. More important from the perspective of possible brain injury is the head s acceleration: ASSESS The accepted threshold for serious brain injury is around 300g, so this helmet would protect the rider in all but the most serious accidents. Without the helmet, the rider s head would come to a stop in a much shorter distance and thus be subjected to a much larger acceleration.

47 Section 10.8 Power

48 Power Power is the rate at which energy is transformed or transferred. The unit of power is the watt: 1 watt = 1 W = 1 J/s

49 Output Power of a Force A force doing work transfers energy. The rate that this force transfers energy is the output power of that force:

50 QuickCheck Four students run up the stairs in the time shown. Which student has the largest power output?

51 QuickCheck Four students run up the stairs in the time shown. Which student has the largest power output? B.

52 QuickCheck Four toy cars accelerate from rest to their top speed in a certain amount of time. The masses of the cars, the final speeds, and the time to reach this speed are noted in the table. Which car has the greatest power?

53 QuickCheck Four toy cars accelerate from rest to their top speed in a certain amount of time. The masses of the cars, the final speeds, and the time to reach this speed are noted in the table. Which car has the greatest power?

54 Example Power to pass a truck Your 1500 kg car is behind a truck traveling at 60 mph (27 m/s). To pass the truck, you speed up to 75 mph (34 m/s) in 6.0 s. What engine power is required to do this? PREPARE Your engine is transforming the chemical energy of its fuel into the kinetic energy of the car. We can calculate the rate of transformation by finding the change ΔK in the kinetic energy and using the known time interval.

55 Example Power to pass a truck (cont.) SOLVE We have so that ΔK = K f - K i = ( J) - ( J) = J To transform this amount of energy in 6 s, the power required is This is about 71 hp. This power is in addition to the power needed to overcome drag and friction and cruise at 60 mph, so the total power required from the engine will be even greater than this.

56 Section 11.1 Transforming Energy

57 Transforming Energy Energy cannot be created or destroyed, but it can be transformed. The energy in these transformations is not lost; it is converted to other forms that are less useful to us. Text: p. 319

58 Transforming Energy Energy cannot be created or destroyed, but it can be transformed. The energy in these transformations is not lost; it is converted to other forms that are less useful to us. Text: p. 319

59 Transforming Energy Energy cannot be created or destroyed, but it can be transformed. The energy in these transformations is not lost; it is converted to other forms that are less useful to us. Text: p. 319

60 QuickCheck 11.1 When you walk at a constant speed on level ground, what energy transformation is taking place? E chem U g U g E th E chem K E chem E th K E th

61 QuickCheck 11.1 When you walk at a constant speed on level ground, what energy transformation is taking place? E chem U g U g E th E chem K E chem E th K E th

62 Transforming Energy The work-energy equation includes work, an energy transfer. We now include electric and radiant energy in our definition of work: Work is positive when energy is transferred into the system and negative when energy is transferred out of the system. When other forms of energy are transformed into thermal energy the change is irreversible. The energy isn t lost, but it is lost to our use.

63 Efficiency The larger the energy losses in a system, the lower its efficiency. Reductions in efficiency are caused by process limitations and fundamental limitations. Process limitations cause an energy loss due to practical details. Fundamental limitations cause an energy loss due to physical laws.

64 Efficiency 35% efficiency is close to the theoretical maximum for power plants due to fundamental limitations.

65 Example 11.1 Lightbulb efficiency A 15 W compact fluorescent bulb and a 75 W incandescent bulb each produce 3.0 W of visible-light energy. What are the efficiencies of these two types of bulbs for converting electric energy into light?

66 Example 11.1 Lightbulb efficiency (cont.) PREPARE The problem statement doesn t give us values for energy; we are given values for power. But 15 W is 15 J/s, so we will consider the value for the power to be the energy in 1 second. For each of the bulbs, what you get is the visible-light output 3.0 J of light every second for each bulb. What you had to pay is the electric energy to run the bulb. This is how the bulbs are rated. A bulb labeled 15 W uses 15 J of electric energy each second. A 75 W bulb uses 75 J each second.

67 Example 11.1 Lightbulb efficiency (cont.) SOLVE The efficiencies of the two bulbs are computed using the energy in 1 second: ASSESS Both bulbs produce the same visible-light output, but the compact fluorescent bulb does so with a significantly lower energy input, so it is more efficient. Compact fluorescent bulbs are more efficient than incandescent bulbs, but their efficiency is still relatively low only 20%.

68 Section 11.2 Energy in the Body

69 Getting Energy from Food: Energy Inputs The chemical energy in food provides the necessary energy input for your body to function.

70 Getting Energy from Food: Energy Inputs Chemical energy in food is broken down into simpler molecules such as glucose and glycogen. They are metabolized in the cells by combining with oxygen. Oxidation reactions burn the fuel you obtain by eating.

71 Getting Energy from Food: Energy Inputs Burning food transforms all of its chemical energy into thermal energy. Thermal energy is measured in units of calories (cal) calorie = 4.19 joules

72 Getting Energy from Food: Energy Inputs

73 Getting Energy from Food: Energy Inputs

74 Example 11.2 Energy in Food A 12 oz can of soda contains approximately 40 g (or a bit less than 1/4 cup) of sugar, a simple carbohydrate. What is the chemical energy in joules? How many Calories is this?

75 Example 11.2 Energy in Food (cont.) SOLVE From Table 11.1, 1 g of sugar contains 17 kj of energy, so 40 g contains Converting to Calories, we get ASSESS 160 Calories is a typical value for the energy content of a 12 oz can of soda (check the nutrition label on one to see), so this result seems reasonable.

76 Using Energy in the Body: Energy Outputs At rest, your body uses energy to build and repair tissue, digest food and keep warm. Your body uses approximately 100 W of energy at rest. The energy is converted almost entirely into thermal energy, which is transferred as heat to the environment.

77 Using Energy in the Body: Energy Outputs

78 Using Energy in the Body: Energy Outputs When you engage in an activity, your cells require oxygen to metabolize carbohydrates. You can measure your body s energy use by measuring how much oxygen it is using. Total metabolic energy use is all of the energy used by the body while performing an activity.

79 Using Energy in the Body: Energy Outputs

80 Efficiency of the Human Body As you climb stairs, you change your potential energy. If you climb 2.7 m and your mass is 68 kg then your potential energy is On average, you use 7200 J to climb the stairs:

81 Efficiency of the Human Body We compute the efficiency of climbing the stairs: We typically consider the body s efficiency to be 25%.

82 Energy Storage Energy from food that is not used will be stored. If energy input from food continuously exceeds the energy outputs of the body, the energy is stored as fat under the skin and around the organs.

83 Example 11.6 Running out of fuel The body stores about 400 g of carbohydrates. Approximately how far could a 68 kg runner travel on this stored energy? PREPARE Table 11.1 gives a value of 17 kj per g of carbohydrate. The 400 g of carbohydrates in the body contain an energy of

84 Example 11.6 Running out of fuel (cont.) SOLVE Table 11.4 gives the power used in running at 15 km/h as 1150 W. The time that the stored chemical energy will last at this rate is And the distance that can be covered during this time at 15 km/h is to two significant figures.

85 Example 11.6 Running out of fuel (cont.) ASSESS A marathon is longer than this just over 42 km. Even with carbo loading before the event (eating high-carbohydrate meals), many marathon runners hit the wall before the end of the race as they reach the point where they have exhausted their store of carbohydrates. The body has other energy stores (in fats, for instance), but the rate that they can be drawn on is much lower.

86 Energy and Locomotion When you walk at a constant speed on level ground, your kinetic and potential energy are constant. You need energy to walk because the kinetic energy of your leg and foot is transformed into thermal energy in your muscles and shoes, which is lost.