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Lecture 16

Lecture 16. Agenda: Review for exam Assignment: For Tuesday, Read chapter 10.1-10.5 M. Tobin’s office hours today are from 3:05 to 3:55 PM I will have extra office hours tomorrow from 1 to 2:30 PM. m 2. T 1. m 1. m 3. Newton’s Laws. Three blocks are connected on the table as shown.

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Lecture 16

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  1. Lecture 16 Agenda: • Review for exam • Assignment: For Tuesday, Read chapter 10.1-10.5 • M. Tobin’s office hours today are from 3:05 to 3:55 PM • I will have extra office hours tomorrow from 1 to 2:30 PM

  2. m2 T1 m1 m3 Newton’s Laws Three 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. If m3 starts from rest how fast is it going after it goes up 2.0 m Use Work energy theorem DK= WConservative+ WNon-conservative

  3. Newton’s Laws 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. m2 T1 m1 m3 If m3 starts from rest how fast is it going after it goes up 2.0 m DK=½ m1v1f2 +½ m2v2f2 + ½ m3v3f2 -½ m1v1i2 -½ m2v2i2 - ½ m3v3i2 = ½ (m1+m2+m3) v2 WC. = Wg(mass 1) + Wg (mass 3) = + m1gh - m3gh WNC = Wfriction = FxDx = - m2gm h ½ (m1+m2+m3) v2 = + m1gh - m3gh - m2gm h v2 = 2gh(m1-m3-m2m)/(m1+m2+m3)

  4. Work, Energy & Circular Motion • A mass, 11 kg, slides down of a frictionless circular path of radius, 5.0 m, as shown in the figure. Initially it moves only vertically and, at the end, only horizontally (1/4 of a circle all told). Gravity, 10 m/s2, acts along the vertical. If the initial velocity is 2 m/s downward then(a) What is the work done by gravity on the mass? (b) What is the final speed of the mass when it reaches the bottom? (c) What is the normal force on the mass when it reaches the bottom (while still on the curved sections)?

  5. Work, Energy & Circular Motion • A mass, 11 kg, slides down of a frictionless circular path of radius, 5.0 m, as shown in the figure. Initially it moves only vertically and, at the end, only horizontally (1/4 of a circle all told). Gravity, 10 m/s2, acts along the vertical. If the initial velocity is 2 m/s downward then(a) What is the work done by gravity on the mass? W = mgR= 11 x 10 x 5 = 550 J • (b) What is the final speed of the mass when it reaches the bottom (while still on the curve)? ½ mvf2 = ½ m vi2 + mgR = 22 J + 550 J = 572 J vf = (1144 / 11) ½ m/s

  6. Work, Energy & Circular Motion • A mass, 11 kg, slides down of a frictionless circular path of radius, 5.0 m, as shown in the figure. Initially it moves only vertically and, at the end, only horizontally (1/4 of a circle all told). Gravity, 10 m/s2, acts along the vertical. If the initial velocity is 2 m/s downward then(c) What is the normal force on the mass when it reaches the bottom SFy = m ac = N – mg = m v2 /R N = mg + m v2 /R = (110 + 11 x 1144/11) N = 1254 N =1300 N

  7. h = 1 m sin 30° = 0.5 m 1 meter 30° Exercise Work/Energy for Non-Conservative Forces • 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 m • 2.5 J • 5.0 J • 10. J • -2.5 J • -5.0 J • -10. J

  8. Springs • 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 6.0 m more meters. • 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 J

  9. Chapter 5 & 6 (Forces and Newton’s Laws)

  10. Chapter 6

  11. Chapter 5 & 6

  12. Chapter 5 & 6 Note: Drag in air is proportional to v or v2 At terminal velocity, drag force cancels out other forces

  13. Chapter 5 & 6

  14. Chapter 7 & 8

  15. Chapter 7 & 8

  16. Chapter 7 & 8 Note: Ethermal is the same as Einternal (Serway) Note: Wdissapative is the same as Wnon-conservative

  17. Chapter 7 & 8 s here refers to the path

  18. Chapter 7 & 8

  19. Work and Energy • A block of mass m is connected by a spring to the ceiling. The block is held at a position where the spring is unstretched and then released. When released, the block (a) remains at rest. (b) oscillates about the unstretched position (c) oscillates about a position that is lower than the unstretched position (d) oscillates about a position that is higher than the unstretched position

  20. Work and Energy • A block of mass m is connected by a spring to the ceiling. The block is held at a position where the spring is unstretched and then released. When released, the block (a) remains at rest. (b) oscillates about the unstretched position (c) oscillates about a position that is lower than the unstretched position (d) oscillates about a position that is higher than the unstretched position

  21. Work and Energy • A mass is attached to a Hooke’s law spring on a horizontal surface as shown in the diagram below. When the spring is at its equilibrium length, the block is at position Y. • When released from position X, how will the spring’s potential energy vary as the block moves from X to Y to Z ? (a) It will steadily increase from X to Z. (b) It will steadily decrease from X to Z. (c) It will increase from X to Y and decrease from Y to Z. (d) It will decrease from X to Y and increase from Y to Z.

  22. Work and Energy • A mass is attached to a Hooke’s law spring on a horizontal surface as shown in the diagram below. When the spring is at its equilibrium length, the block is at position Y. • When released from position X, how will the spring’s potential energy vary as the block moves from X to Y to Z ? (a) It will steadily increase from X to Z. (b) It will steadily decrease from X to Z. (c) It will increase from X to Y and decrease from Y to Z. (d) It will decrease from X to Y and increase from Y to Z.

  23. Work and Energy • An object moves along a line under the influence of a single force. The area under the force vs. position graph represents (a) the impulse delivered to the object (b) the work done on the object. (c) the change in the velocity of the object. (d) the momentum of the object.

  24. Work and Energy • An object moves along a line under the influence of a single force. The area under the force vs. position graph represents (a) the impulse delivered to the object (b) the work done on the object. (c) the change in the velocity of the object. (d) the momentum of the object.

  25. Concept problem • At high speeds the drag force of the air is found to be proportional to the square of a car’s speed. Assume that at 60 mph that 100% of car’s power is being used against wind resistance (i.e., there are no other non-conservative forces.) In terms of the ratio P(120 mph) / P (60 mph), how much more power will the car’s engine need to provide if this car is to travel at 120 mph? • 2 • 23/2 • 4 • 25/2 • 8 P = F v = v3 so ratio is 8

  26. Work and Energy • A block slides along a frictionless surface before colliding with a spring. The block is brought momentarily to rest by the spring after traveling some distance. The four scenarios shown in the diagrams below are labeled with the mass of the block, the initial speed of the block, and the spring constant. • Rank the scenarios in order of the distance the block travels, listing the largest distance first. (a) B , A , C = D (b) B , C , A , D (c) B , C = D , A (d) C = B, A , D (e) C = B = D , A

  27. Work and Energy • A block slides along a frictionless surface before colliding with a spring. The block is brought momentarily to rest by the spring after traveling some distance. The four scenarios shown in the diagrams below are labeled with the mass of the block, the initial speed of the block, and the spring constant. • Rank the scenarios in order of the distance the block travels, listing the largest distance first. (a) B , A , C = D (b) B , C , A , D (c) B , C = D , A (d) C = B, A , D (e) C = B = D , A

  28. Work and Forces • A 25.0 kg chair is pushed 2.00 m at constant speed along a horizontal surface with a constant force acting at 30.0 degrees below the horizontal. If the friction force between the chair and the surface is 55.4 N, what is the work done by the pushing force? (a) 85 J (b) 98 J (c) 111 J (d) 113 J (e) 128 J

  29. Work and Forces • A 25.0 kg chair is pushed 2.00 m at constant speed along a horizontal surface with a constant force acting at 30.0 degrees below the horizontal. If the friction force between the chair and the surface is 55.4 N, what is the work done by the pushing force? (a) 85 J (b) 98 J (c) 111 J (d) 113 J (e) 128 J

  30. Work and Power • A 100 kg elevator is carrying 6 people, each weighing 70 kg. They all want to travel to the top floor, 75 m from the floor they entered at. How much power will the elevator motor supply to lift this in 45 seconds at constant speed? (a) 1.2 · 102 W (b) 7.0 · 102 W (c) 8.7 · 102 W (d) 6.9 · 103 W (e) 8.5 · 103 W

  31. Work and Power • A 100 kg elevator is carrying 6 people, each weighing 70 kg. They all want to travel to the top floor, 75 m from the floor they entered at. How much power will the elevator motor supply to lift this in 45 seconds at constant speed? (a) 1.2 · 102 W (b) 7.0 · 102 W (c) 8.7 · 102 W (d) 6.9 · 103 W (e) 8.5 · 103 W

  32. Newton’s Laws • Two sleds are hooked together in tandem. The front sled is twice as massive as the rear sled. The sleds are pulled along a frictionless surface by a forceF, applied to the more massive sled. The tension in the rope between the sleds is T. Determine the ratio of the magnitudes of the two forces, T/F. (a) 0.33 (b) 0.50 (c) 0.67 (d) 1.5 (e) 2.0 (f) 3.0

  33. Newton’s Laws • A factory worker raises a 100. kg crate at a constant rate using a frictionless pulley system, as shown in the diagram. The mass of the pulleys and rope are negligible. (assume g= 9.8 m/s2) With what force is the worker pulling down on the rope? (a) 245 N (b) 327 N (c) 490 N (d) 980 N (e) 1960 N

  34. Work and Energy • A 6.0 kg block is pushed up against an ideal Hooke’s law spring (of spring constant 3750 N/m ) until the spring is compressed a distance x. When it is released, the block travels along a track from one level to a higher level, by moving through an intermediate valley (as shown in the diagram). The track is frictionless until the block reaches the higher level. There is a frictional force stops the block in a distance of 1.2 m. If the coefficient of friction between the block and the surface is 0.60, what is x ? (Let g = 9.81 m/s2 ) (a) 0.11 m (b) 0.24 m (c) 0.39 m (d) 0.48 m (e) 0.56 m

  35. Work and Energy • A 6.0 kg block is pushed up against an ideal Hooke’s law spring (of spring constant 3750 N/m ) until the spring is compressed a distance x. When it is released, the block travels along a track from one level to a higher level, by moving through an intermediate valley (as shown in the diagram). The track is frictionless until the block reaches the higher level. There is a frictional force stops the block in a distance of 1.2 m. If the coefficient of friction between the block and the surface is 0.60, what is x ? (Let g = 9.81 m/s2 ) (a) 0.11 m (b) 0.24 m (c) 0.39 m (d) 0.48 m (e) 0.56 m

  36. Conceptual Problem A bird sits in a birdfeeder suspended from a tree by a wire, as shown in the diagram at left. Let WB and WF be the weight of the bird and the feeder respectively. Let T be the tension in the wire and N be the normal force of the feeder on the bird. Which of the following free-body diagrams best represents the birdfeeder? (The force vectors are not drawn to scale and are only meant to show the direction, not the magnitude, of each force.)

  37. Conceptual Problem A block is pushed up a 20º ramp by a 15 N force which may be applied either horizontally (P1) or parallel to the ramp (P2). How does the magnitude of the normal force N depend on the direction of P? (A) N will be smaller if P is horizontal than if it is parallel the ramp. (B) N will be larger if P is horizontal than if it is parallel to the ramp. (C) N will be the same in both cases. (D) The answer will depend on the coefficient of friction. 20°

  38. Conceptual Problem A cart on a roller-coaster rolls down the track shown below. As the cart rolls beyond the point shown, what happens to its speed and acceleration in the direction of motion? A. Both decrease. B. The speed decreases, but the acceleration increases. C. Both remain constant. D. The speed increases, but acceleration decreases. E. Both increase. F. Other

  39. Sample Problem • A 200 kg wood crate sits in the back of a truck. The coefficients of friction between the crate and the truck are μs = 0.9 and μk = 0.5. The truck starts moving up a 20° slope. What is the maximum acceleration the truck can have without the crate slipping out the back? • Solving: • Visualize the problem, Draw a picture if necessary • Identify the system and make a Free Body Diagram • Choose an appropriate coordinate system • Apply Newton’s Laws with conditional constraints (friction) • Solve

  40. Sample Problem • You have been hired to measure the coefficients of friction for the newly discovered substance jelloium. Today you will measure the coefficient of kinetic friction for jelloium sliding on steel. To do so, you pull a 200 g chunk of jelloium across a horizontal steel table with a constant string tension of 1.00 N. A motion detector records the motion and displays the graph shown. • What is the value of μk for jelloium on steel?

  41. Sample Problem S Fx =ma = F - ff = F - mk N = F - mk mg S Fy = 0 = N – mg mk = (F - ma) / mg & x = ½ a t2  0.80 m = ½ a 4 s2 a = 0.40 m/s2 mk = (1.00 - 0.20 · 0.40 ) / (0.20 ·10.) = 0.46

  42. Force Cart AirTrack Exercise: Newton’s 2nd Law A force of 2 Newtons acts on a cart that is initially at rest on an air track with no air and pushed for 1 second. Because there is friction (no air), the cart stops immediately after I finish pushing. It has traveled a distance, D. • 8 x as far • 4 x as far • 2 x as far • 1/4 x as far Next, the force of 2 Newtons acts again but is applied for 2 seconds. The new distance the cart moves relative to D is:

  43. Force Cart AirTrack Exercise: Solution We know that under constant acceleration, Dx = a (Dt)2 /2 (when v0=0) Here Dt2=2Dt1, F2 = F1a2 = a1 (B) 4 x as long

  44. Sample exam problem A small block moves along a frictionless incline which is 45° from horizontal. Gravity acts down at 10 m/s2. There is a massless cord pulling on the block. The cord runs parallel to the incline over a pulley and then straight down. There is tension, T1, in the cord which accelerates the block at 2.0 m/s2 up the incline. The pulley is suspended with a second cord with tension, T2. A. What is the tension magnitude, T1, in the 1st cord? B. What is the tension magnitude,T2, in the 2nd cord?

  45. Sample exam problem a = 2.0 m/s2 up the incline. What is the tension magnitude, T1, in the 1st cord? Use a FBD! Along the block surface S Fx = m ax = -mg sin q + T T = 5 x 2 N + 5 x 10 x 0.7071 N = (10 + 35) N = 45 N

  46. Sample exam problem a = 0.0 m/s2 at the pulley. What is the tension magnitude,T2, in the 2nd cord? Use a FBD!

  47. Sample exam problem • You have a 2.0 kg block that moves on a linear path on a horizontal surface. The coefficient of kinetic friction between the block and the path is μk. Attached to the block is a horizontally mounted massless string as shown in the figure below. The block includes an accelerometer which records acceleration vs. time. As you increase the tension in the rope the block experiences an increasingly positive acceleration. At some point in time the rope snaps and then the block slides to a stop (at a time of 10 seconds). Gravity, with g = 10 m/s2, acts downward.

  48. Sample exam problem A. At what time does the string break and, in one sentence, explain your reasoning? B. What speed did the block have when the string broke? C. What is the value of μk? D. Using μk above (or a value of 0.25 if you don’t have one), what was the tension in the string at t = 2 seconds?

  49. Sample exam problem B. What speed did the block have when the string broke? Don’t know initial v (t=0) so can’t integrate area at t < 4 sec. vf = 0 m/s and from t = 4 to 10 sec (6 second) a = - 2 m/s2 0 = vi + a t = vi – 2 x 6 m/s  vi = 12 m/s

  50. Sample exam problem C. What is the value of μk? Use a FBD! S Fx = m ax = - fk = - μk N S Fy = 0 = mg – N  N = mg So m ax = - fk = - μk mg  μk = - ax / g = - (-2)/10 = 0.20

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