1. 1 Newton’s 3 rd Law Momentum Prentice Hall Chapters 7 & 8

2. 2 Ch 7 & 8 assignments –Complete Packets for the chapters. On the problem sections, you must follow the problem solving method: write givens with symbols and units, working equation, work with units and answer with units. Use more paper if there is not enough room.

3. Newton’s 3 rd Law For every action there is an opposite and equal reaction. This is applied to momentum and collisions. 3

4. 4 Chapter 8 Momentum and Its Conservation Momentum,  = mass of an object times its velocity  = mv units: kgm/s Impulse is an object’s difference in final momentum and initial momentum F  t = m  v =  f –  i units: ns, kgm/s

5. 5 Conservation of Momentum This is used in collisions where Newton’s third law is related to conservation of momentum. See figure 8.11 on page 132  i =  f

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7. 7 Recoil occurs when objects are at rest initially. Example problem on page 135 General formula  i =  f but  I (small fish) = 0

8. Collision Types Elastic – when objects collide and rebound from each other without a loss in energy (no permanent change in shape and no heat change). Fig 8.11 (pg. 132) Inelastic – when objects collide and become distorted. Vehicle crashes. Perfectly inelastic – when objects collide and stick together after impact. Freight cars coupling on train track. 8

9. Asgn: 142/55,57,58due tomorrow. 9

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11. Now we begin: 11 Work Power Energy

12. Objectives *Differentiate among the various forms of energy and how they can be transformed from one into another. *Explore the Law of Conservation of Energy by differentiating among open, closed, and isolated systems and explain that the total energy in an isolated system is a conserved quantity. *Compare and contrast work and power both qualitatively and quantitatively. 12

13. 13 Work is the transfer of energy by mechanical means. In this section you will calculate work and power used. Terms/Units: Work/joule,j energy/joule,j kinetic energy/joule,j work-energy theorem Power/joule/sec or watt 746 w = 1 hp

14. 14 Work = Force x distance W = Fd units: nm or joule, j Work is equal to a constant force exerted on an object in the direction of motion, times the object’s displacement

15. 15 Kinetic Energy – the energy of motion KE = ½ m v 2 units: kg (m/s) 2 = (kgm/s 2 ) m = nm = joule Work Energy Theorem: W =  KE Work = the change in kinetic energy of a system.

16. 16 Work at an angle W = Fd cos  Work = the product of force and displacement, times the cosine of the angle between the force and the direction of the displacement. How much work is done pulling with a 15 N force applied at 20. o over a distance of 12 m?

17. 17 How much work is done pulling with a 15 N force applied at 20. o over a distance of 12 m? W = Fdcos  W = 15 N (12 m) (cos 20) W = 169  170 J 20 o 15 N

18. 18 Power Power = Work / time j/s = watt or w Power is “The rate of doing work.”

19. 19 Sample Problem: Together, two students exert a force of 825 n in pushing a car a distance of 35 m. A. How much work do they do on the car? W = Fd = 825n (35m) = B. If the force was doubled, how much work would they do pushing the car the same distance? W = 2Fd = 2(825n)(35m) =

20. 20 A rock climber wears a 7.5 kg backpack while scaling a cliff. After 30.0 min, the climber is 8.2 m above the starting point. a.How much work does the climber do on the backpack? W = Fd = mgd = 7.5kg(9.8m/s 2 )(8.2m) b. How much power did the climber expend? P = W = 6.0x10 2 j x 1min = 0.33 = 3.3x10 -1 w t 30.0min 60sec

21. 21 Machines

22. 22 Actual Mechanical Advantage – comparing torques or forces MA = resistance (or load) force effort force

23. 23 Calculated Actual Mechanical Advantage: Fr / Fe = 200nm / 300 nm =.67

24. 24 Ideal Mechanical Advantage, comparing distances: IMA = d effort / d resistance IMA = 2.4 m / 1.2 m = 2

25. 25 Efficiency Efficiency, e = work out = AMA x 100 work in IMA Sample problem on next slide.

26. Two people use a wheel and axle to raise a mass of 750. kg. The radius of the wheel is 0.50 m and the radius of the axle is 0.040 m. If the efficiency of the machine is 62% and each person exerts and equal force, how much force must each apply? Efficiency, e = work out = AMA x 100 work in IMA e = 0.62 IMA = R =0.50m = r 0.040m AMA = f effort  f effort = f resist AMA  f effort = ? f resistance 26

27. Efficiency, e = work out = AMA x 100 work in IMA e = 0.62 IMA = R =0.50m = r 0.040m AMA = f effort  f effort = f resist AMA  f effort = ? f resistance f effort = f resis r = 750. kg x 10.nkg-1 x 0.040m e x R 0.62 x 0.50 m f effort = 968 n Per person: 968 = 484 n 2 27

28. 28 Chapter 11 Energy & Its Conservation Kinetic Energy: KE = ½ m v 2 KE for a Spring: KE = ½ k (  d) 2 k = F / d Potential Energy: PE = mgh Rest Energy: E = mc 2 Mechanical Energy, E = KE + PE Conservation of Energy (collisions, etc.) KE i + PE i = KE f + PE f

29. 29 The total energy of a closed, isolated system is constant. The energy can change form. 20.0J

30. 30 Examples: Collisions Elastic Collisions: ones in which initial KE = final KE Inelastic Collisions: ones in which energy is changed into another form (i.e., KE into heat or sound)

31. A 98.0 –N sack of grain is hoisted to a storage room 50.0 m above the ground floor of a grain elevator. a.How much work was done? W = Fd W = 98.0 – N ( 50.0m) W = 4900 = 4.90 x 10 3 j 31

32. A 98.0 –N sack of grain is hoisted to a storage room 50.0 m above the ground floor of a grain elevator. b. What is the increase in potential energy of the sack of grain at this height? PE = mgh = Fd PE = 98.0 N (50.0 m) PE = 4900 j = 4.90 x 10 3 j 32

33. A 98.0 –N sack of grain is hoisted to a storage room 50.0 m above the ground floor of a grain elevator. c.The rope being used to lift the sack of grain breaks just as the sack rached the storage room. What kinetic energy does the sack have just before it strikes the ground floor? Ans.: KE bottom = PE top KE bottom = 4.90 x 10 3 J 33

34. 34 11/73 73a. W = Fd = 98.0 N (50.0 m) W = 4.90 x 10 3 J 73b. PE = mgh = Fd  PE = W = 4.90 x 10 3 J