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v = v o + at

v = v o + at. A) 47 = 5 + a(3.8) a = 11.05 m/s/s x = x o + v o t + ½ at 2 B) x = 0 + 5(3.8) + ½ (11.05)(3.8) 2 = 98.78 m C) v = v o + at v = 5 + (11.05)(2.7) = 34.84 m/s. A) -24.9 m/s B) v = v o + at -24.9 = 24.9 – 9.8t t = 5.08 sec C) v = v o + at v = 24.9 – 9.8(1.3) v = 12.16 m/s.

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v = v o + at

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  1. v = vo + at A) 47 = 5 + a(3.8) a = 11.05 m/s/s x = xo + vot + ½ at2 B)x =0 +5(3.8) + ½ (11.05)(3.8)2 = 98.78 m C) v = vo + at v = 5 + (11.05)(2.7) = 34.84 m/s

  2. A) -24.9 m/sB) v = vo + at-24.9 = 24.9 – 9.8tt = 5.08 secC) v = vo + atv = 24.9 – 9.8(1.3)v = 12.16 m/s

  3. D) v = vo + at -12.16 = 24.9 –9.8 tt = 3.78 secE) y = yo + vot + ½ at2 Part C+ D: y = 24.1 m

  4. F) y = yo + vot + ½ at2 Part C) = 24.1 m Part D) Goes from ground to ypeak = 31.63 m at t =2.54 sec then down to 24.1 m at 3.78 secDistance = 31.63 m+(31.63-24.1) = 39.17 m

  5. G) y = yo + vot + ½ at2 ypeak = 31.63 m at t =2.54 secH) 0 m/s (That’s why it’s at the peak!!)I & J) – 9.8 m/s/s (g for all locations!)

  6. 3. y = yo + voyt + ½ at20 =1.4–4.9t2t = 0.53 seconds to floorx = vxtx = 2.25(0.53)x = 1.20 m

  7. voy q vx STEP 0!4A) Vx= 43cos(30) = 37.24 m/s Voy= 43sin(30) = 21.50 m/s Determines the time in the air

  8. A) vy = voy + ayt-21.5 = 21.5 –9.8tt = 4.39 seconds total time in the air

  9. 4B) y = yo + vot + ½ at2y = 0 + 21.5(2.20) – 4.9(2.20) 2ypeak = 23.58 m at t =2.20 sec ypeak

  10. 4C) x =vxtx = 37.24(4.39)(Remember to use total time in the air to get Da’ Range)x = 163.48 m x

  11. Ff Fapp 5A) Fnet = ma(Fapp – Ff) = ma Fapp – 0 = 4(6.8) Fapp = 27.2 N net

  12. FN Ff Fapp 5B) Fnet = ma(Fapp – Ff) = ma Fapp – mk(FN) = 4(6.8) Fapp = mk(FN) + 27.2 Fapp = 0.45(39.2) + 27.2 = 44.84 N Fg

  13. Fapp 5C) Fnet = ma(Fapp – Fg) = ma Fapp – mg = 4(6.8) Fapp = mg + 27.2 Fapp = 39.2 + 27.2 = 66.40 N Fg

  14. 5D) Fnet = ma(Fapp – Ff) = ma Fapp – 0 = 4(0) Fapp = 0 NInertia keeps it going!!

  15. 5E) Fnet = ma(Fapp – Ff) = ma Fapp – mk(FN) = 4(0) Fapp = mk(FN) + 0 Fapp = 0.45(39.2) + 0 = 17.64 N

  16. 5F) Fnet = ma(Fapp – Fg) = ma Fapp – mg = 4(0) Fapp = mg + 0 Fapp = 39.2 + 0 = 39.20 N

  17. Phase 21A) Fg = mg = (12)(9.8) =117.6N1B) Fgperpendicular = mgcosq = 109.04 N Fgperp Fg q q = 22 degrees Fgparallel 1C) Fgparallel = mgsinq = 44.05 N

  18. Phase 21D) Fnet = ma FN Ff Fgparallel Fnet = Fgpar – Ff Fnet = 44.05 – 0 = 44.05 N a = Fnet/m = 44.05/12 a = 3.67 m/s/s Fgperp

  19. 1A) 1B) and 1C) do not change 1E) Fnet = ma FN Fnet = Fgpar – Ff Fnet = 44.05 – mkFN = 44.05 –(.07)(109.04) = 36.42 N a = Fnet/m = 36.42/12 a = 3.04 m/s/s Ff Fgparallel Fgperp

  20. 1F) Fnet = ma FN Fgparallel Fnet = Fgpar – Ff 0 = mgsinq – mk mgcosq mgsinq = mk mgcosq Tanq = mk q = Tan-1(mk) = 4.00 deg Ff Fgpar Fgperp

  21. Phase 2 #2 • 2. KEtrans =1/2 mv2 (Energy of Motion) KEtrans =1/2 (28)(25)2 = 8,750 Joules Holey Kinetic Energy Batman!

  22. Phase 2 #3 • 3. PE= mgh (Energy of Position) PE =mgh= 4,116 Joules

  23. Phase 2 #4 • 4. PE1 + KE1= PE2 + KE2 (Conservation of Energy) mgh1 + 1/2mv12= mgh2 + 1/2mv22 gh1 + 1/2v12= gh2 + 1/2v22 9.8(25) + ½(5)2= 9.8(17) + 1/2v22 v2 = 13.48 m/s

  24. Phase 2 #5 5A) Work = (Average Force) times Displacement This word is key when the force is not constant Here force goes from 0 N to 250 N which is an average force of 125 N W = Fd = (125)(.26) =32.5 Joules 5B) W = DPEelastic = 32.5 Joules stored (Work-Energy Theorem)

  25. Phase 2 #6A 6A) Momentum, p = mv p = (95)(15) = + 1,425 kgm/s + momentum is momentum to the right

  26. Phase 2 #6B 6B) Impulse (FDt) equals Change in Momentum (Dp)

  27. Phase 2 #6B 6B) FDt = Dp m Dv Dp = (95)(4-15) = - 1,045 kgm/s = Impulse Momentum was lost so Impulse is negative

  28. Phase 2 #6C 6C) Impulse (FDt) = -1045 kg m/s Magnitude: FDt = 1045 kg m/s F(7.3) = 1045 Magnitude of the Force, F = 143.15 N

  29. Phase 2 #7A First part 7A) q = qo +wot + ½ at2 wo = 33rev1 min2p rad = 3.46rad/sec 1 min 60 sec 1 rev Then:a = Dw/Dt = (0-3.46)/13 = -0.27 rad/s/s

  30. Phase 2 #7A Second PArt 7A) q = 0 + 3.46(13) + ½ (-0.27)(13)2 q= 22.17 rad = 3.53 revolutions

  31. Phase 2 #7B & 7C 7B) a = Dw/Dt = (0-3.46)/13 = -0.27 rad/s/s 7C) w = wo + at = 3.46 + (-0.27)(9.7) w = 0.84 rad/sec v = rw

  32. Phase 2 #7D 7D) q = qo +wot + ½ at2 q = 0 +3.46(9.7) + ½ (-0.27)(9.7)2 q = 20.86 rad = 3.32 revolutions

  33. Phase 2 #8 a A 12 kg a 3 kg B

  34. T Phase 2 #8 A 12 kg a 117.6 N Fnet = ma 117.6 – T = 12a Equation #1

  35. T Phase 2 #8 a B 3 kg 29.4 N Fnet = ma T – 29.4 = 3a or T = (29.4 + 3a) Equation #2

  36. Phase 2 #8A) & 8B) Fnet = ma 117.6 – (29.4 + 3a) = 12a 88.2 = 15a 8A) a = 5.88 m/s/s [Down] 8B) a = 5.88 m/s/s [Up]

  37. T Phase 2 #8 a B 3 kg 29.4 N Back to Equation #2 T = (29.4 + 3a) = 29.4 + 3(5.88) = 47.04 N

  38. #1 and #2A-B • Electrons are mobile (protons are NOT!) Phase 3 2A) Force on charge A due to charge B 2B) FAB = kq1q2 (Coulomb’s Law) r2 Force is Repulsive for 2 negative charges: FAB = (9 x 109)(17 x 10-6)(3 x 10-6) = 93.67 N (.07)2

  39. #2C-E Phase 3 FAB =- FBA (Repulsive Forces are equal in magnitude but Opposite in Direction by Newton’s 3rd Law of Motion (Action/Reaction)

  40. Phase 3 #3 E = F/q = kQ/r2 E = (9 x 109)(3 x 10-6) (0.30)2 E = 300,000 N/C [radially inward]

  41. Phase 3 #4 Number of field lines a charge

  42. B +5mC Phase 3 #5 A -7mC -9mC C

  43. B +5mC Phase 3 #5 FAB A FAC -7mC -9mC C

  44. B +5mC Phase 3 #5 Using Coulomb’s Law FAB = 28.125 N FAC =157.5 N -7mC -9mC C

  45. Resultant Vector: FAnet =159.99 N[169.88 deg] Phase 3 #5 FAB = 28.125 N FAC =157.5 N Tan-1 (28.125/157.5) = 10.12 degrees

  46. Phase 3 #6A A) Current is the rate of flow of charge in Amperes Current, I = Coulombs/second I = 357/76.2 = 4.69 Amps

  47. Phase 3 #6B B) Key: Each electron carries 1.6 x 10-19Coulombs (See Equation Sheet) 4.69 Coulombs1 electron = 2.93 x 1019 1 sec 1.6 x 10-19Coulombselectrons

  48. Phase 3 #7: Very Important Rule V = IR V = (2.3)(9) = 20.7 volts P = VI = 47.61 Watts Or P = I2R since V=IR

  49. 1A) Series Resistors- Up and Add Them! Req = 50 W Phase 4 1A Series Resistors: Same Current Different Voltage Series Resistors

  50. 1B) Parallel Resistors- Add Reciprocals and Flip IT! Req = (10-1 + 15-1 + 25-1)-1 = 4.84 W Phase 4 1B Parallel Resistors: Different Current Same Voltage

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