Page 1Physics 207 – Lecture 18Physics 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 m3starts from rest how fast is it going after it goes up 2.0 mm1T1m2m3Page 2Physics 207 – Lecture 18Physics 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 m3starts 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 m3starts 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 = - m2gµ h (friction)v2= 2gh(m1-m3-m2µ)/(m1+m2+m3)m1T1m2m3Page 3Physics 207 – Lecture 18Physics 207: Lecture 18, Pg 5Newton’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 Now the pulleys are uniform disks of 4.00 kg mass and 0.50 m radius. The rope does not slip.a) If m3starts from rest how fast is it going after it goes up 2.0 mm1T1m2m3Physics 207: Lecture 18, Pg 6Exercise 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 hori zontal. 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 mPage 4Physics 207 – Lecture 18Physics 207: Lecture 18, Pg 7Exercise Work/Energy for Non-Conservative Forces How much work did friction do on the cart ? (g=10 m/s2) W = F ∆x = µ mg cos θ d is not easy to do, esp. if µ not given Work done (W) is equal to the change in the mech. energy of the “system” (U+K). Efinal- Einitialand is < 0. (E = U+K) Here Ugravityis “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 8Exercise Work/Energy for Non-Conservative Forces Alternatively we could look at Wnet=∆K Again Kfinal= Kinitial= 0 Wnet=∆K = 0 = Wgravity+ Wfriction= (mg sin θ ) (0.5 meter) + WfrictionWfriction= -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 JhfhiPage 5Physics 207 – Lecture 18Physics 207: Lecture 18, Pg 9A problem that almost made it to the exam Another kind of collision A 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 10Springs A 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 JPage 6Physics 207 – Lecture 18Physics 207: Lecture 18, Pg 11Chapter 7 (Newton’s 3rdLaw) & Chapter 8Physics 207: Lecture 18, Pg 12Chapter 8Page 7Physics 207 – Lecture 18Physics 207: Lecture 18, Pg 13Chapter 9Physics 207: Lecture 18, Pg 14Chapter 9Page 8Physics 207 – Lecture 18Physics 207: Lecture 18, Pg 15Chapter 10Physics 207: Lecture 18, Pg 16Chapter 10Page 9Physics 207 – Lecture 18Physics 207: Lecture 18, Pg 17Chapter 10Physics 207: Lecture 18, Pg 18Chapter 11Page 10Physics 207 – Lecture 18Physics 207: Lecture 18, Pg 19Chapter 11Physics 207: Lecture 18, Pg 20Chapter 12and Center of MassPage 11Physics 207 – Lecture 18Physics 207: Lecture 18, Pg 21Chapter 12Physics 207: Lecture 18, Pg 22Important ConceptsPage 12Physics 207 – Lecture 18Physics 207: Lecture 18, Pg 23Chapter 12Physics 207: Lecture 18, Pg 24Example 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 theorem W = F d = 0 – ½ mv2 F = -µmg d = - ½ mv2 d= v2/(2µg)=(5.0 x 2π x 0.50)2/ (0.50 x 10) m = 5 π2m Rev = d / 2πr = 16 revolutions What if the disk were tilted by 60° ?Page 13Physics 207 – Lecture 18Physics 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 ofthe driveway, and screech to a stop “laying rubber” all the way. It’s particularly fun when your mother is in the car with you. You practice this trick driving at 20 mph and with some groceries inyour car with the same mass as your mother. You find that you only travel half way up the driveway. Thus when your mom joins you in the car, you try it driving twice as fast. How far will you go this time ?Physics 207: Lecture 18, Pg 26Work & Friction W = F d = - µ N d = - µ mg d = ∆K = 0 – ½ mv2W1= - µ mg d1= ∆K1= 0 – ½ mv12W2= - µ mg d2= ∆K2= 0 – ½ m(2v1)2 = – 4 (½ mv12)- µ
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