1. Chapter 5 Work and Energy

2. Work and Energy A force that causes a displacement of an object does work on the object. W = Fd *Force in the direction of displacement Perpendicular forces do not do work.

3. If an object is moved by a force other than horizontal….. θ F d Only the horizontal component causes displacement. W = Fdcosθ Units: N * m = joules (J)

4. * Sample 5A pg 169 Figure 5-3 Positive and Negative Work Pg 170

5. Kinetic Energy -> Energy of Motion - depends on speed & mass ½ mv 2 = KE Pg. 172 derivation of W net = F Δ x = maΔx W net = ½ mv f 2 – ½ mv i 2 x = d Sample 5B pg 173 Work – Kinetic Energy Theorem W net = Δ KE W net = KE f – KE i W net = ½ mv f 2 – ½ mv i 2 Net Work = Change in Kinetic Energy Sample 5C pg 175

6. Potential Energy -> stored energy - potential to move due to position Gravitational P.E. ->depends on height from a zero level Unit: Joule = J P.E. g = mgh Elastic P.E. -> depends on distance compressed or stretched P.E. elastic = ½ kx 2 k = spring constant x = distance compressed or stretched Figure 5-B pg 178 Sample 5D pg 179

7. Conservation of Energy Energy Remains constant ! Mechanical Energy -> is the sum of K.E. & all forms of P.E. associated with an object or group of objects (a system) ME = KE + PE = ½ mv 2 + mgh

8. Energy Non-Mechanical Mechanical KineticPotential Gravitational Elastic

9. Conservation of Mechanical Energy In a closed or isolated system the sum of the P.E. & K.E. remains constant; energy can change from P.E.  K.E., but the sum remains the same. (P.E. + K.E.) i = (P.E. + K.E.) f

10. Conservation of Energy Equation : ½ mv i 2 + mgh i = ½ mv f 2 + mgh f

11. MASS ON TABLE m h1h1 h2h2 0 m v2v2 m v3v3 h 3 = 0 PE 1 + KE 1 =PE 2 + KE 2 = PE 3 + KE 3 mgh 1 + ½ m(0) 2 = mgh 2 + ½ mv 2 2 = mg(0) + ½ mv 3 2 mgh 1 +0= mgh 2 + ½ mv 2 2 = 0 + ½ mv 3 2 v 1 = 0

12. MASS ON TABLE m h1h1 h2h2 0 m v2v2 m v3v3 h = 0 PE 1 + KE 1 =PE 2 + KE 2 = PE 3 + KE 3 mgh 1 + ½ m(0) 2 =mgh 2 + ½ mv 2 2 = mg(0) + 1/2mv 3 2 mgh 1 +0=mgh 2 + ½ mv 2 2 = 0 + ½ mv 3 2 mgh 1 = mgh 2 + ½ mv 2 2 = ½ mv 3 2 v 1 = 0

13. MASS ON TABLE m h1h1 h2h2 0 m v2v2 m v3v3 h = 0 PE 1 + KE 1 =PE 2 + KE 2 = PE 3 + KE 3 mgh 1 + ½ m(0) 2 =mgh 2 + ½ mv 2 2 = mg(0) + 1/2mv 3 2 mgh 1 +0=mgh 2 + ½ mv 2 2 = 0 + ½ mv 3 2 mgh 1 = mgh 2 + ½ mv 2 2 =½ mv 3 2 gh 1 = gh 2 + ½ v 2 2 = ½ v 3 2 v 1 = 0 * You would utilize any 2 equations depending on the situation

14. Sample Problem Pg. 184 ME i = ME f PE i + KE i = PE f + Ke f mgh 1 + ½ mv 1 2 = mgh 2 + ½ mv 2 2 736 J + 0 J = 0 J + (.5)(25kg)v f 2 736J/12.5kg=v f 2 v f = 7.67 m/s * ME not conserved in the Presence of Friction Fig. 5-12 pg 186

15. Rate of Work  POWER!!!!! P = W/Δt = (Fd)/Δt = F(d/t) = Fv = ma(d/t) Units for Power: Watt = J / sec or hp = 746 W * Sample Problem 5F pg. 188

16. Hmwk #1 Book Chp. 5 5A pg. 170 1,3,4 (d=1.1m) 5B pg. 174 1,3,5 5C pg. 176 1,5 5D pg.180 1,2 (PE=.031 J) 5E pg. 185 1,5 5F pg. 189 1,3

17. Hmwk #2 Workbook Chp. 5 5A 1. d= 195m5D 1. m= 159kg 4. d= 20m 3. m= 83.8 kg 6. F= 400N 5. h= 2.09m 10. W= 20364 J 7. PE= 1.6 x 10 5 J 5B 1. m= 66kg5E 3. h f =2970m 3. m= 67kg 5. 300 km/h (? m/s) 6. V= 1.084 m/s KE= 32J 8. KE= 82300 J 24:1 5C 2. KE f = 1760 J5F 1. W= 2 x10 11 J 4. 81m/s ? Km/hr 3. t= 48.5 s 6. m= 70 kg 5. P= 300W

18. Chapter Six Momentum

19. Momentum  p Vector quantity Product of an object’s mass & velocity p = mv Units: kg * m/s **Sample 6A pg 209  A change in Momentum takes force & time F = ma F = m (v/t) F = (mv)/t F = Δp/t

20. FΔtFΔt Impulse – Momentum Theorem Units: N*sec Impulse 

21. Examples Time >Force * Punch in a fight Time< Force * Karate Chop F t or f t = same p (change in momentum) Sample 6B pg 211 *How does stopping time Relate to p = FΔt? Sample 6C pg 212

22. Hmwk #3 BK & WKBK Chp. 6 Momentum A-C Book Pg. 209 6A 1, 2 #1. 2482 kg m/s right #2 a. 120 kg m/s NW b. 94 kg m/s NW c. 27 kg m/s NW Pg. 211 6B 1,3 #1 380N is to the left #3 -16 N sec (to the south) Pg. 213 6C 1,2 #2 a. 14m/s N b. 42m N c. 8 sec WKBK 6-A: 1. 59 kg 3. 83.2 m/s 6. 4.17 x 10 -2 kg m/s 6-B: 1. 1.8 sec 4. 11 m/s 8. 219 N up 6-C: 1. Δp= - 6 x 10 7 kg m/s Δx= 32m 3. Δp= 2.44 x 10 4 kg m/s Δx= 88.6 m east 7. 54 km

23. Einstein's General Relativity: compares measurements between two frames of reference moving relative to each other. Special Relativity: examines the measurements at relative speeds near the speed of light. E=mc 2 ie. Objects moving near the speed of light will be shorter in the direction of motion, be more massive and have slower clocks than measurements made by the moving object RELATIVE MOTION to different Frames of Reference: Motion with respect to high speed motion Time dilation Length contraction Mass change Book: pg. 66-67, 110-111,190-191

24. Test will cover : Relativity/Special relativity E=mc 2 Work: W=Fd Kinetic Energy: KE= ½ mv 2 Work-KE Theorem : W= ΔKE Potential Energy: PE g = mgh Elastic potential Energy: PE elastic = ½ kΔx 2 Conservation of Energy: KE i + PE i = KE f + PE f Power: P= W/t = Fd/t = Fv = mad/t = mgd/t Momentum: p = mv Impulse: Ft = Δp= mv f -mv i