1. 1 Work, Power, Energy Glencoe Chapters 9,10,11

2. 2 Ch 9 assignments In class samples: 1,2,4,13,15 Assigned problems 7-9,17,20

3. 3 Chapter 9 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

4. 4 Conservation of Momentum This is used in collisions where Newton’s third law is related to conservation of momentum. See figure 9-6 on page 236  i =  f Ex. prob 2 on p. 237

5. #2, p237 5

6. 6 Recoil occurs when objects are at rest initially. Example problem 3 on page 240 General formula  i =  f but  i = 0  Ci +  Di =  Cf +  Df 0 =  Cf +  Df So:  Cf = -  Df

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9. 9 Chapter 10 Energy, Work, and Simple Machines Assigned problems: 29,52,57,60,66,81

10. 10 Work is the transfer of energy by mechanical means. In this section you will calculate work and power used. Terms: work energy kinetic energy work-energy theorem Joulepowerwatt

11. 11 Work = Force x distance cos  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

12. 12 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.

13. 13 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?

14. 14 W = Fdcos  W = 15 N (12 m) (cos 20) W = 169  170 J 20 o 15 N

15. 15 Power Power = Work / time j/s = watt or w Power is “The rate of doing work.”

16. 16 Sample: #2 on page 261 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) =

17. 17 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 does the climber expend in this effort? P = W = 603 J = 0.34 watt t 1800 s

18. 18 Machines

19. 19 Mechanical Advantage MA = resistance (or load) force effort force

20. 20 Calculated Mechanical Advantage: Fr / Fe = 200nm / 300 nm =.67

21. 21 Ideal Mechanical Advantage, IMA = d effort / d resistance IMA = 2.4 m / 1.2 m = 2

22. 22 Efficiency Efficiency, e = work out = MA x 100 work in IMA

23. 23 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

24. 24 Ch11 Assignment 11/6,55,59,73,74.79

25. 25 The total energy of a closed, isolated system is constant. The energy can change form. 20.0J

26. 26 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)

27. 27 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 73c. KE bottom = PE top KE bottom = 4.90 x 10 3 J