Lecture 18Newton’s LawsSlide 3Slide 4Exercise Work/Energy for Non-Conservative ForcesSlide 6Slide 7Momentum and collisionsSlide 9SpringsChapter 7 (Newton’s 3rd Law) & Chapter 8Chapter 8Chapter 9Slide 14Chapter 10Slide 16Slide 17Chapter 11Slide 19Chapter 12Slide 21Important ConceptsSlide 23Example problem: Going in circlesWork & FrictionSlide 26Spinning ball on inclineSlide 30Slide 31Slide 32Work and EnergySlide 34Momentum & ImpulseSlide 36Momentum, Work and EnergySlide 38Work and EnergySlide 40Slide 41Slide 42Momentum and ImpulseSlide 44Slide 45Slide 46Slide 47Slide 48Slide 49Slide 50Slide 51Slide 52Slide 53Work and ForcesSlide 55Work and PowerSlide 57Conservation of MomentumSlide 59Slide 60Slide 61Slide 62Slide 63Slide 64Slide 65Slide 66Circular MotionWork, Energy & Circular MotionSlide 69Slide 70Slide 71Slide 72Slide 73Slide 74Slide 75Slide 76Physics 207: Lecture 18, Pg 1Lecture 18Agenda: Agenda: Review for examReview for examAssignment: For Tuesday, Read chapter 14Assignment: For Tuesday, Read chapter 14Physics 207: Lecture 18, Pg 2Newton’s LawsThree blocks are connected on the table as shown. The table has a coefficient of kinetic friction of 0.350, the masses are m1 = 4.00 kg, m2 = 1.00kg and m3 = 2.00kg. a) If m3 starts from rest how fast is it going after it goes up 2.0 mm1T1m2m3Physics 207: Lecture 18, Pg 3Newton’s LawsThree blocks are connected on the table as shown. The table has a coefficient of kinetic friction of 0.350, the masses are m1 = 4.00 kg, m2 = 1.00kg and m3 = 2.00kg. a) If m3 starts from rest how fast is it going after it goes up 2.0 mUse Emech = K+U E= Efinal-Einit = Wn.c.m1T1m2m3Physics 207: Lecture 18, Pg 4Newton’s LawsThree blocks are connected on the table as shown. The table has a coefficient of kinetic friction of 0.350, the masses are m1 = 4.00 kg, m2 = 1.00kg and m3 = 2.00kg. a) If m3 starts from rest how fast is it going after it goes up 2.0 mUse Emech = K+U E= Efinal-Einit = Wn.c.E =½ m1v1f2 + m1gh1f +½ m2v2f2 +m2gh2f + ½ m3v3f2 + m3gh3f -(½ m1v1i2 + m1gh1i +½ m2v2i2 +m2gh2i + ½ m3v3i2 + m3gh3i) = ½ (m1+m2+m3) v2 - (0+0+0)+m1g(h1f-h1i)+m2g(0)+m3g(h3f-h3i) = ½ (m1+m2+m3) v2 -m1gh+m3gh = - m2g h (friction) v2= 2gh(m1-m3-m2)/(m1+m2+m3)m1T1m2m3Physics 207: Lecture 18, Pg 5Exercise Work/Energy for Non-Conservative ForcesA. 2.5 JB. 5.0 JC. 10. JD. -2.5 JE. -5.0 JF. -10. J1 meter30°An air track is at an angle of 30° with respect to horizontal. The cart (with mass 1.0 kg) is released 1.0 meter from the bottom and hits the bumper at a speed, v1. This time the vacuum/ air generator breaks half-way through and the air stops. The cart only bounces up half as high as where it started. How much work did friction do on the cart ? (g=10 m/s2) Notice the cart only bounces to a height of 0.25 mh = 1 m sin 30° = 0.5 mPhysics 207: Lecture 18, Pg 6Exercise Work/Energy for Non-Conservative ForcesHow much work did friction do on the cart ? (g=10 m/s2) W = F x = mg cos d is not easy to do, esp. if not givenWork done (W) is equal to the change in the mech. energy of the “system” (U+K). Efinal - Einitial and is < 0. (E = U+K)Here Ugravity is “in” the system and Kfinal = Kinitial = 0Use W = Ufinal - Uinit = mg ( hf - hi ) = - mg sin 30° 0.5 mW = -2.5 N m = -2.5 J or (D)1 meter30°(A) 2.5 J (B) 5 J (C) 10 J (D) –2.5 J (E) –5 J (F) –10 JhfhiPhysics 207: Lecture 18, Pg 7Exercise Work/Energy for Non-Conservative ForcesAlternatively we could look at Wnet=K Again Kfinal = Kinitial = 0Wnet=K = 0 = Wgravity + Wfriction = (mg sin ) (0.5 meter) + Wfriction Wfriction = -2.5 N m = -2.5 J or (D)And the result is the same1 meter30°(A) 2.5 J (B) 5 J (C) 10 J (D) –2.5 J (E) –5 J (F) –10 JhfhiPhysics 207: Lecture 18, Pg 8Momentum and collisionsRemember vector componentsA 5 kg cart rolling without friction to the right at 10 m/s collides and sticks to a 5 kg motionless block on a 30° frictionless incline. How far along the incline do the joined blocks slide?Physics 207: Lecture 18, Pg 9Momentum and collisionsRemember vector componentsA 5 kg cart rolling without friction to the right at 10 m/s collides and sticks to a 5 kg motionless block on a 30° frictionless incline. Momentum parallel to incline is conservedNormal force (by ground on cart) is to the incline mvi cos + m 0 = 2m vf vf = vi cos / 2 = 4.4 m/s Now use work-energy 2mgh + 0 = ½ 2mv2f d = h ½ v2f / (g sin ) mvimvi cos Physics 207: Lecture 18, Pg 10SpringsA Hooke’s Law spring with a spring constant of 200 N/m is first stretched 3.0 m past its equilibrium distance and then is stretched 9.0 m.How much work must be done to go from 3.0 m to 9.0 m?W = Ufinal-Uinitial= ½ k (x-xeq)final2 -½ k (x-xeq)init2 = 100 [(9)2 –(3)2] J= 100(72) J = 7200 JPhysics 207: Lecture 18, Pg 11Chapter 7 (Newton’s 3rd Law) & Chapter 8Physics 207: Lecture 18, Pg 12Chapter 8Physics 207: Lecture 18, Pg 13Chapter 9Physics 207: Lecture 18, Pg 14Chapter 9Physics 207: Lecture 18, Pg 15Chapter 10Physics 207: Lecture 18, Pg 16Chapter 10Physics 207: Lecture 18, Pg 17Chapter 10Physics 207: Lecture 18, Pg 18Chapter 11Physics 207: Lecture 18, Pg 19Chapter 11Physics 207: Lecture 18, Pg 20Chapter 12and Center of MassPhysics 207: Lecture 18, Pg 21Chapter 12Physics 207: Lecture 18, Pg 22Important ConceptsPhysics 207: Lecture 18, Pg 23Chapter 12Physics 207: Lecture 18, Pg 24 Example problem: Going in circles A 2.0 kg disk tied to a 0.50 m string undergoes circular motion on a rough but horizontal table top. The kinetic coefficient of friction is 0.25. If the disk starts out at 5.0 rev/sec how many revolutions does it make before it comes to rest?Work-energy theoremW = F d = 0 – ½ mv2 F = -mg d = - ½ mv2d= v2 /(2g)=(5.0 x 2 x 0.50)2/ (0.50 x 10) m = 5 2 mRev = d / 2r = 16 revolutionsWhat if the disk were tilted by 60° ? Top viewPhysics 207: Lecture 18, Pg 25Work & FrictionA.A.The same distance. Not so exciting.The same distance. Not so exciting.B. 2 times as far (only ~7/10 of the way up the driveway) C.C.Twice as far, right to the door. Whoopee!Twice as far, right to the door. Whoopee! D.D.Four times as far crashing into the house. (Oops.)Four times as far crashing into the house. (Oops.)You like to drive home fast, slam on your brakes at the start of the driveway, and
View Full Document
Unlocking...