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1 © 2007 Pearson Prentice Hall This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. The work and materials from it should never be made available to students except by instructors using the accompanying text in their classes. All recipients of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on these materials. Lecture Outlines Chapter 7 Physics, 3 rd Edition James S. Walker

2 Chapter 7 Work and Kinetic Energy

3 Units of Chapter 7 Work Done by a Constant Force Kinetic Energy and the Work-Energy Theorem Work Done by a Variable Force Power

4 7-1 Work Done by a Constant Force The definition of work, when the force is parallel to the displacement: (7-1) SI unit: newton-meter (N·m) = joule, J

5 7-1 Work Done by a Constant Force Om F = 82,0 N, d= 3,00 m (och eftersom kraften är parallell med förflyttningen) så blir det uträttade arbetet W = Fd = 82,0 N 3,00 m = 246 Nm = 246 J (7-1) Exercise 7-1 Darwin’s fink och hårt fröskal F=205 N, d=0,40 cm, ger W = 0,82 J

6 7-1 Work Done by a Constant Force

7 Example 7-1 (p.181) Heading for the ER (m p =72,0 kg, m gurney = 15 kg, if d=2,5 m, W =? Anta att vagnen rör sig friktionslöst)

8 7-1 Work Done by a Constant Force If the force is at an angle to the displacement: (7-3)

9 Figure 7-3 Force at an angle to direction of motion: another look

10 7-1 Work Done by a Constant Force The work can also be written as the dot product of the force and the displacement:

11 Example 7-2 (p.183) Gravity Escape System (d = 5,0 m, m = 4970 kg W g = ?) (Lägg märke till att θ är inte definierad på det vanliga sättet )

12 Conceptual Checkpoint 7-1(p.184) Path Dependence of Work (W 1 >W 2 ?, W 1 >W 2 ?, W 1 =W 2 ? (Lådan glider friktionlöst)

13 7-1 Work Done by a Constant Force The work done may be positive, zero, or negative, depending on the angle between the force and the displacement:

14 7-1 Work Done by a Constant Force If there is more than one force acting on an object, we can find the work done by each force, and also the work done by the net force: (7-5)

15 Example 7-3 A Coasting Car I

16 W N = Nd(cos90º) =0 W mg = mg(sinΦ)d W air = F air (cos180º)d = - F air d W total = W N + W mg + W air = 0 + mgsin(Φ) d - F air

17 Example 7-4 A Coasting Car II

18 Nu skall vi beräkna arbetet genom att beräkna skalärprodukten mellan F total och förflyttningen d. Då blir W total = (mgsin(Φ) - F air )d Eftersom F total är given på föregående bild. Vilket förstås är det uttryck som vi erhöll i Example 7-3

19 7-2 Kinetic Energy and the Work-Energy Theorem When positive work is done on an object, its speed increases; when negative work is done, its speed decreases.

20 7-2 Kinetic Energy and the Work-Energy Theorem Med hjälp av rörelseekvationen v f 2 =v i 2 +2ad och Newton’s lag a=F tot /m fås v f 2 =v i 2 +2(F tot /m)d Rörelseenergin definieras nedan och är ALLTID positiv (eller = 0) (7-6)

21 Table 7-2 Typical Kinetic Energies

22 Exercise 7-2 (p.188) A truck moving at 15 m/s has a kinetic energy of J a) What is the mass of the truck? b) If the speed is doubled? K → (factor)K? a)m = (2K/v 2 ) = J/(15 m/s) 2 = 3700 kg b) v → 2v då går K → 4 K (7-6)

23 7-2 Kinetic Energy and the Work-Energy Theorem Work-Energy Theorem: The total work done on an object is equal to its change in kinetic energy. (7-7)

24 Example 7-5(p.188) Hit the Books (m=4,10 kg, F app = 52,7 N) a) Hur stort arbete uträttar F app ? b) Hur stort arbete uträttar gravitationen? c) v f = ?

25 Example 7-5 Hit the books a) W app = F app Δy = 52,7 N 1,60 m = 84,3 J b) W g = mg(cos180)Δy = 4,10 kg 9,81 m/s 2 (-1) 1,60 m = - 64,4 J c) ma y = F app - mg = 52,7 N - 4,10 kg 9,81 m/s 2 = 12,5 N v f 2 = v i 2 + 2Δy a y = ,60 m 12,5N/(4,10 kg) v f = 3,12 m/s

26 Example 7-6 (p.189) Pulling a Sled (m = 6,40 kg, v i = 0,50 m/s, ingen friktion, W = ?, v f = ?)

27 Example 7-6 Pulling a sled a) W boy = F(cos 29,0º)d = 11,0 N (cos 29,0º) 2,00 m = 19,2 J b) ΔK =m(v f 2 - v i 2 )/2 = 19,2 J v f 2 = v i ,2J/(6,40 kg) = (0,5 m/s) 2 + 6,00 m 2 /s 2 v f = 2,50 m/s

28 Conceptual Checkpoint 7-2 (p.190) Compare the Work a) W 2 = W 1 b) W 2 = 2W 1 c) W 2 = 3W 1 d) W 2 = 4W 1 ?

29 7-3 Work Done by a Variable Force If the force is constant, we can interpret the work done graphically:

30 7-3 Work Done by a Variable Force If the force takes on several successive constant values:

31 7-3 Work Done by a Variable Force We can then approximate a continuously varying force by a succession of constant values.

32 Figure 7-9 Stretching a spring

33 7-3 Work Done by a Variable Force The force needed to stretch a spring an amount x is F = kx. Therefore, the work done in stretching the spring is (7-8)

34 Example 7-7a+b (p.192-3) Flexing an AFM (atomic-force microscopy) Cantilever (kiselkonsol, 250 μm lång) Till att förlänga fjädern 0,10 nm krävs arbetet 1, J. a) Hur stor är fjäderkonstanten k? b) Hur stort arbete krävs för att öka x till 0,20 nm från 0,10 nm? a) k = 2W/x 2 = 2 1, J/(0, m) 2 = 2,4 N/m b) W(0 → 0,2 nm) = kx 2 /2 = 4 W(0 → 0,1 nm) + W(0,1 nm → 0,2 nm) = = 4 W(0 → 0,1 nm) = 4, J dvs W(0,1 nm → 0,2 nm)= 3, J

35 Figure 7-11 Work done in stretching a spring: average force

36 Example 7-7b Flexing an AFM Cantilever Fixeringsbild! Allt OK??

37 Figure 7-12 The work done by a spring can be positive or negative

38 Active example 7-1 (p.194) A block compresses a string. (m=1,5 kg, v 0 =2,2 m/s, k = 475 N/m Vad blir x (<0) då v=0? K i = mv 0 2 /2 = 3,6 J och K f = 0 Dvs fjädern utför ett negativt arbete på blocket eftersom ΔK < 0 Då blir - kx 2 /2 = - 3,6 J x = ((23,6 J)/(475 N/m)) 1/2 = 0,12 m

39 7-4 Power Power is a measure of the rate at which work is done: (7-10) SI unit: J/s = watt, W 1 (engelsk) horsepower = 1 hp = 746 W 1 (svensk) hästkraft = 75 kp m/s = 736 W

40 7-4 Power

41 Example 7-8 Passing Fancy

42 Example 7-8 (p.196) Passing Fancy m car = 1300 kg, v i = 13,4 m/s till v f = 17,9 m/s på t = 3,00 s. P=? W = ΔK = mv f 2 /2 – mv i 2 /2 = 91,6 kJ P = W/t = 91,6 kJ/3,00 s = 30,5 kW

43 Figure 7-13 Driving up a hill

44 Active Example 7-2 (p.197) Find the maximum (constant) speed. F = 1280 N (m car = 1500 kg, θ = 5,00°) P = 50,0 hp = 50,0 746 W = 37,3 kW P = W/t = F d /t = F v (om hastigheten är konstant) v = P/F = 37,3 kW/1280 N = 29,1 m/s

45 Summary of Chapter 7 If the force is constant and parallel to the displacement, work is force times distance If the force is not parallel to the displacement, The total work is the work done by the net force:

46 Summary of Chapter 7 SI unit of work: the joule, J Total work is equal to the change in kinetic energy: where

47 Summary of Chapter 7 Work done by a spring force: Power is the rate at which work is done: SI unit of power: the watt, W


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