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Ch6.1 – Work and Energy Work - force applied over a distance W = F net . s

Ch6.1 – Work and Energy Work - force applied over a distance W = F net . s If the force is applied at some angle : W = F net . s . cos θ - If the force and direction of motion are in the same direction (+) work - If force opposes motion (–) work

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Ch6.1 – Work and Energy Work - force applied over a distance W = F net . s

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  1. Ch6.1 – Work and Energy Work - force applied over a distance W = Fnet.s If the force is applied at some angle: W = Fnet.s.cosθ - If the force and direction of motion are in the same direction (+) work - If force opposes motion (–) work Ex1) An engine exerts a force of 400kN on a train, pulling it 500m, at constant speed. a) How much work is done by the engine on the train? b) The brakes apply a force of 100kN to bring the train to a stop in 1000m. How much work do they do on the train?

  2. Ex2) The Egyptians placed 500 kg stones on pyramids by making ramps out of dirt, and using round logs as wheels. If a ramp 100m long is used to place a stone 25m above the ground: a) How much work is done overcoming gravity? b) How much work is done overcoming friction? 100m 25m

  3. HW#11) A nurse pushes someone in a wheelchair , with her arms at 30°, 100m down a hallway, doing 400 J of work in the process. What average force did the nurse exert in the direction of motion? 16. A pickup truck is hauling a barge along a canal at a constant speed. The truck, driving parallel to the waterway at a 30º angle with the forward direction. If the truck exerts a force of 1000N on the cable, how much work is done in overcoming friction as the barge is moved 10 km? Ch6 HW#1 p205+ 1,5,11,12,14,16,19,20,22,23

  4. Ch6 HW#1 p205 1,5,11,12,14,16,19,20,22,23 • While floating out in space, a constant force of 500N is applied to a 542.3kg robot by a small rocket motor. The robot moves along a straight line in the direction of the thrust of the motor. How much work is done on it by the rocket for every 10.0m traveled? • 5. A book is slowly slid in a straight line, at a constant speed, 1.5m across a level table by a 15N horizontal force. How much work is done on the book-table system and what is it done against?

  5. Ch6 HW#1 p205 1,5,11,12,14,16,19,20,22,23 • While floating out in space, a constant force of 500N is applied to a 542.3kg robot by a small rocket motor. The robot moves along a straight line in the direction of the thrust of the motor. How much work is done on it by the rocket for every 10.0m traveled? • 5. A book is slowly slid in a straight line, at a constant speed, 1.5m across a level table by a 15N horizontal force. How much work is done on the book-table system and what is it done against?

  6. Ch6 HW#1 p205 1,5,11,12,14,16,19,20,22,23 • While floating out in space, a constant force of 500N is applied to a 542.3kg robot by a small rocket motor. The robot moves along a straight line in the direction of the thrust of the motor. How much work is done on it by the rocket for every 10.0m traveled? • 5. A book is slowly slid in a straight line, at a constant speed, 1.5m across a level table by a 15N horizontal force. How much work is done on the book-table system and what is it done against?

  7. (11,16 in class) 19. A 100N box filled with books is slid along a 13m long ramp up to a platform 5.0m above the ground. How much work is done if friction is negligible? How much work would have been done if the box were lifted straight up to the platform?

  8. 19. A 100N box filled with books is slid along a 13m long ramp up to a platform 5.0m above the ground. How much work is done if friction is negligible? How much work would have been done if the box were lifted straight up to the platform? Fg=100N 20. Several crates, having a total weight of 400N, are loaded into a 10.0kg wagon that is then pulled up a wooden ramp 10.0m long making an angle of 30.0°. If friction was negligible, how much work was done? Wg = Fg||.s.cos0 or Wg = Fg.sy

  9. 19. A 100 N box filled with books is slid along a 13 m long ramp up to a platform 5.0 m above the ground. How much work is done if friction is negligible? How much work would have been done if the box were lifted straight up to the platform? Fg=100N 20. Several crates, having a total weight of 400N, are loaded into a 10.0kg wagon that is then pulled up a wooden ramp 10.0m long making an angle of 30.0°. If friction was negligible, how much work was done? Wg = Fg||.s.cos0 or Wg = Fg.sy = Fgsinθ.s = Fg.(5m) = 500Nsin30°.10m = 500N.5m = 2500 J = 2500 Nm sy = s.sin30° = 10m.(.5) = 5m

  10. 22. A 50kg keg of beer slides upright down a 3.0m long plank leading from the back of a truck 1.5m high to the ground. Determine the amount of work done on the keg by gravity. 3m 1.5m Fg=500N

  11. 22. A 50kg keg of beer slides upright down a 3.0m long plank leading from the back of a truck 1.5m high to the ground. Determine the amount of work done on the keg by gravity. Wg = Fg.sy 3m 1.5m = 500N.1.5m = 750 J (usually considered negative work) Fg=500N

  12. 23. A newspaper delivery boy pulls horizontally on a rubber-wheeled cart • that has a coefficient of rolling friction of 0.02 and a mass of 25 kg. • It travels 10 km on level streets. • a) How much work does he do in overcoming friction? • b) If he pulls his cart 25m along a road inclined at 10°, • how much work does he do to overcome road friction? • c) How much work overcoming gravity? • a) Wf = Ff.s b) Wf = Ff.s • FN • Fg|| F c) Wg = Fg.sy • Ff • FgFg┴

  13. 23. A newspaper delivery boy pulls horizontally on a rubber-wheeled cart • that has a coefficient of rolling friction of 0.02 and a mass of 25 kg. • It travels 10 km on level streets. • a) How much work does he do in overcoming friction? • b) If he pulls his cart 25m along a road inclined at 10°, • how much work does he do to overcome road friction? • c) How much work overcoming gravity? • a) b) Wf = Ff.s • FN • Fg|| F c) Wg = Fg.sy • Ff • FgFg┴

  14. 23. A newspaper delivery boy pulls horizontally on a rubber-wheeled cart • that has a coefficient of rolling friction of 0.02 and a mass of 25 kg. • It travels 10 km on level streets. • a) How much work does he do in overcoming friction? • b) If he pulls his cart 25m along a road inclined at 10°, • how much work does he do to overcome road friction? • c) How much work overcoming gravity? • a) b) Wf = Ff.s • = µ.FN.25m • = (.02)(Fg┴)25m • = (.02)(Fgcos10°)25m • = 123 Nm • FN • Fg|| F c) Wg = Fg.sy • Ff = • = • FgFg┴

  15. 23. A newspaper delivery boy pulls horizontally on a rubber-wheeled cart • that has a coefficient of rolling friction of 0.02 and a mass of 25 kg. • It travels 10 km on level streets. • a) How much work does he do in overcoming friction? • b) If he pulls his cart 25m along a road inclined at 10°, • how much work does he do to overcome road friction? • c) How much work overcoming gravity? • a) b) Wf = Ff.s • = µ.FN.25m • = (.02)(Fg┴)25m • = (.02)(Fgcos10°)25m • = 123 Nm • FN • Fg|| F c) Wg = Fg.sy • Ff = 250N.(25m.sin10˚) • = • FgFg┴

  16. Ch6.2 - Work Graphs Ex1) A force of 5N is applied to an object, and is uniformly increased to 10N while the object moves a total distance of 8m, as shown. How much work is done on the object? 10 F (N) 5 5 10 s (m)

  17. If the force is not uniform finding the area is tricky. 10 F (N) 5 5 10 s (m)

  18. Ex2) A mass is attached to a spring, on a frictionless surface . The mass is initially at 0.3m, the spring’s natural length. The mass is pulled back to the 0.0m mark and released. a) How much work is done from 0.0-0.3? b) How much work is done from 0.0-0.6? 4 3 2 F (N) 1 0 .1 .2 .3 .4 .5 .6 (m) -1 -2 - 0.6 - - - 0.3 - - - 0.0 -3 -4 Ch6 HW#2 Practice Graphs

  19. Ch6 HW#2 Practice Graphs A force is gradually placed on an object, uniformly increased to 20 N over a distance of 5 m. It then stays constant for the next 5 m. How much work was done on this object? 20 F (N) 10 5 10 s (m)

  20. Ch6 Practice Graphs WS A force is gradually placed on an object, uniformly increased to 20 N over a distance of 5 m. It then stays constant for the next 5 m. How much work was done on this object? 20 A = ½.b.h = 50 Nm A = b.h = 100 Nm Atotal = 150 Nm F (N) 10 5 10 s (m)

  21. 2. A force of 10 N is applied to an object over a distance of 3 m, and then is uniformly increased to 15 N while the object moves a distance of 4 m. How much work is done on the object? 20 F (N) 10 5 10 s (m)

  22. 2. A force of 10 N is applied to an object over a distance of 3 m, and then is uniformly increased to 15 N while the object moves a distance of 4 m. How much work is done on the object? 20 A = ½.b.h = ½(4m)(5N) = 10 Nm A = b.h = (7m)(10N) = 70 Nm Atotal = 80 Nm F (N) 10 5 10 s (m)

  23. 3. A 0.20 kg object moves along a straight line. The net force acting on the object varies with the object’s displacement as shown. The object starts from rest at displacement x = 0 and time t = 0 and is displaced a distance of 20 m. Determine each of the following. a. The accl of the particle when its displacement x is 6 m b. The time taken for the object to be displaced the first 12 m c. The amount of work done displacing the object the first 12 m d. The speed of the object at displacement x = 12 m e. The final speed of the object at displacement x = 20 m 5 4 3 2 1 F (N) 2 4 6 8 10 12 14 16 18 20 s (m)

  24. 3. A 0.20 kg object moves along a straight line. The net force acting on the object varies with the object’s displacement as shown. The object starts from rest at displacement x = 0 and time t = 0 and is displaced a distance of 20 m. Determine each of the following. a. The accl of the particle when its displacement x is 6 m b. The time taken for the object to be displaced the first 12 m c. The amount of work done displacing the object the first 12 m d. The speed of the object at displacement x = 12 m e. The final speed of the object at displacement x = 20 m 5 4 3 2 1 a. F = ma a = 4N/.20kg = 20 m/s2 F (N) 2 4 6 8 10 12 14 16 18 20 s (m)

  25. 3. A 0.20 kg object moves along a straight line. The net force acting on the object varies with the object’s displacement as shown. The object starts from rest at displacement x = 0 and time t = 0 and is displaced a distance of 20 m. Determine each of the following. a. The accl of the particle when its displacement x is 6 m b. The time taken for the object to be displaced the first 12 m c. The amount of work done displacing the object the first 12 m d. The speed of the object at displacement x = 12 m e. The final speed of the object at displacement x = 20 m 5 4 3 2 1 a. F = ma a = 4N/.20kg = 20 m/s2 b. s = vit + ½at2 12 = 0 + ½(20)t2t = 1.1s F (N) 2 4 6 8 10 12 14 16 18 20 s (m)

  26. 3. A 0.20 kg object moves along a straight line. The net force acting on the object varies with the object’s displacement as shown. The object starts from rest at displacement x = 0 and time t = 0 and is displaced a distance of 20 m. Determine each of the following. a. The accl of the particle when its displacement x is 6 m b. The time taken for the object to be displaced the first 12 m c. The amount of work done displacing the object the first 12 m d. The speed of the object at displacement x = 12 m e. The final speed of the object at displacement x = 20 m 5 4 3 2 1 a. F = ma a = 4N/.20kg = 20 m/s2 b. s = vit + ½at2 12 = 0 + ½(20)t2t = 1.1s c. W = F.s W = (4N)(12m) = 48 J F (N) 2 4 6 8 10 12 14 16 18 20 s (m)

  27. 3. A 0.20 kg object moves along a straight line. The net force acting on the object varies with the object’s displacement as shown. The object starts from rest at displacement x = 0 and time t = 0 and is displaced a distance of 20 m. Determine each of the following. a. The accl of the particle when its displacement x is 6 m b. The time taken for the object to be displaced the first 12 m c. The amount of work done displacing the object the first 12 m d. The speed of the object at displacement x = 12 m e. The final speed of the object at displacement x = 20 m 5 4 3 2 1 a. F = ma a = 4N/.20kg = 20 m/s2 b. s = vit + ½at2 12 = 0 + ½(20)t2t = 1.1s c. W = F.s W = (4N)(12m) = 48 J d. vf2 = vi2 + 2as or vf = vi +at = 0 + (20)(1.1) = 21.9 m/s F (N) 2 4 6 8 10 12 14 16 18 20 s (m)

  28. 3. A 0.20 kg object moves along a straight line. The net force acting on the object varies with the object’s displacement as shown. The object starts from rest at displacement x = 0 and time t = 0 and is displaced a distance of 20 m. Determine each of the following. a. The accl of the particle when its displacement x is 6 m b. The time taken for the object to be displaced the first 12 m c. The amount of work done displacing the object the first 12 m d. The speed of the object at displacement x = 12 m e. The final speed of the object at displacement x = 20 m f. The change in momentum of the object as it is displaced from x = 12m to x = 20m. a. F = ma a = 4N/.20kg = 20 m/s2 b. s = vit + ½at2 12 = 0 + ½(20)t2t = 1.1s c. W = F.s W = (4N)(12m) = 48 J d. vf2 = vi2 + 2as or vf = vi +at = 0 + (20)(1.1) = 21.9 m/s e. Work = ∆KE (Area) = KEf – KEi += ½mvf2 – 0 48J+ 16J = ½(.2kg)vf2vf2= 25.3 m/s 5 4 3 2 1 F (N) 2 4 6 8 10 12 14 16 18 20 s (m)

  29. 3. A 0.20 kg object moves along a straight line. The net force acting on the object varies with the object’s displacement as shown. The object starts from rest at displacement x = 0 and time t = 0 and is displaced a distance of 20 m. Determine each of the following. a. The accl of the particle when its displacement x is 6 m b. The time taken for the object to be displaced the first 12 m c. The amount of work done displacing the object the first 12 m d. The speed of the object at displacement x = 12 m e. The final speed of the object at displacement x = 20 m f. The change in momentum of the object as it is displaced from x = 12m to x = 20m. a. F = ma a = 4N/.20kg = 20 m/s2 b. s = vit + ½at2 12 = 0 + ½(20)t2t = 1.1s c. W = F.s W = (4N)(12m) = 48 J d. vf2 = vi2 + 2as or vf = vi +at = 0 + (20)(1.1) = 21.9 m/s e. Work = ∆KE (Area) = KEf – KEi += ½mvf2 – 0 48J+ 16J = ½(.2kg)vf2vf2= 25.3 m/s 5 4 3 2 1 F (N) 2 4 6 8 10 12 14 16 18 20 s (m) f. ∆p = mvf– mvi = = (.2kg)(25.3m/s) – (.2kg)(21.9m/s) = 0.68 kg.m/s

  30. Ch6.3 – Energy Kinetic Energy: Potential Energy: Work Energy Theorem: Ex1) A 2,200,000N jet starts at rest and takes off at 268 m/s. a) What is it’s initial KE? b) What is it’s final KE? c) How much work was done to bring it up to speed?

  31. Ex2) Spiderman (60 kg) is on top of a 30 m building when he decides to scale a 10m tall flagpole, at constant speed. a) How much work did he do? b) What was his initial PE? c) What was his final PE?

  32. HW #34) Suppose that a 0.149 kg baseball is traveling at 40m/s/ a) How much work is done to stop it? b) If it is brought to rest in .02m, what average force is acted on it? Ch6 HW#3 p209 28,34,38,41,44,47,55

  33. Ch6 HW#3 p209 28,34,38,41,44,47,55 28. The record average speed for the men’s 10km walk is 4.4m/s. How much KE would a 70kg athlete have at that speed? 38. A video of a 70kg male sprinter shows him going from 0m/s to 3.0 m/s in the 1st step, reaching 4.2m/s on 2nd step, and 5.1m/s on 3rd. Compare speed to KE in each step. 41. A 0.046kg golf ball is driven from rest to 70m/s in about 1 ms. Determ KE and approx what dist the club in contact with the ball.

  34. The energy content of beer is about 1.8x106 J/kg. If that energy could be turned completely into PEG, how much beer is needed to raise a 1kg mass 1km into the air? 47. A car with a mass of 1000kg …

  35. Ch6.2 - Conservation of Energy - The total amount of mechanical energy stays constant Ex1) A 5kg mass is dropped from a 2 meter height. What speed right before it hits the ground? Ex2) A 5kg mass slides down an incline plane, as shown. How fast is it going at bottom? 10 m

  36. Ex3) A cannon is fired at 50 m/s at 40° degrees to horizon. What height does it reach? How fast is it going when it comes back down? Ch6 HW#4 p209 59,60,61,62,63

  37. Ch6 HW#4 p209 59,60,61,62,63 59. What is the kinetic energy of a 10.0kg piece of concrete after it has fallen for 2.00sec from the side of an old building? 60. A 60kg stuntperson runs off a cliff at 5.0m/s and lands safely in the river 10.0m below. What is the splashdown speed?

  38. Ch6 HW#4 p209 59,60,61,62,63 59. What is the kinetic energy of a 10.0kg piece of concrete after it has fallen for 2.00sec from the side of an old building? vf = vi + at = 19.6 m/s vfKEf = ½mvf2 = ½(10)(19.6) = 98 J 60. A 60kg stuntperson runs off a cliff at 5.0m/s and lands safely in the river 10.0m below. What is the splashdown speed?

  39. Ch6 HW#4 p209 59,60,61,62,63 59. What is the kinetic energy of a 10.0kg piece of concrete after it has fallen for 2.00sec from the side of an old building? vf = vi + at = 19.6 m/s vfKEf = ½mvf2 = ½(10)(19.6) = 98 J 60. A 60kg stuntperson runs off a cliff at 5.0m/s and lands safely in the river 10.0m below. What is the splashdown speed? KEi + PEi = KEf + PEf ½mvi2 + mgh = ½mvf2 + 0 ½(5)2 + (9.8)(10) = ½(vf)2 vf = 14.9 m/s

  40. 61. Two automobiles of weight 7.12 kN and 14.24 kN are traveling along horizontally at 96 km/h (26.7 m/s) when they both run out of gas. Luckily, there is a town in a valley not far off, but it’s just beyond a 33.5 mhigh hill. Neglect friction, which of the cars will make it to town? Is KEi greater than PEf ? 63. A kid in a wagon is traveling at 10 m/s just as she reaches the bottom of a hill and begins to climb a second hill. How high up it will she get before the wagon stops, assuming negligible friction losses?

  41. Ch6.5 Conservation of Energy with Friction Ex1) If the 5kg is actually going 5 m/s when it reaches the bottom, what is the coefficient of friction? 10 m

  42. Ch6 HW#5 p209 74 + 2 Bonus questions Bonus #1. A 5kg object starts at rest, slides down an incline plane with a kinetic coefficient of friction of 0.10, as shown. What is the speed at the bottom of the incline? 2.9m 1m

  43. Ch6 HW#5 p209 74 + 2 Bonus questions Bonus #1. A 5kg object starts at rest, slides down an incline plane with a kinetic coefficient of friction of 0.10, as shown. What is the speed at the bottom of the incline? 2.9m 1m PEi = KEf + Wfric mgh = ½mv2 + Ff.s : mgh = ½mv2 + µmgcosθ.s gh = ½v2 + µgcosθ.s (9.8)(1) = ½ v2 + (.1)(9.8)cos20°(2.9) v = 3.8 m/s

  44. Bonus #2. A 2kg object starts at rest, slides down an incline plane, as shown. If its speed at the bottom is 2.78 m/s, what is the kinetic coefficient of friction? 0.87m 0.5m

  45. Bonus #2. A 2kg object starts at rest, slides down an incline plane, as shown. If its speed at the bottom is 2.78 m/s, what is the kinetic coefficient of friction? 0.87m 0.5m PEi = KEf + Wfric mgh = ½mv2 + Ff.s : mgh = ½mv2 + µmgcosθ.s gh = ½v2 + µgcosθ.s (9.8)(0.5) = ½(2.78)2 + µ(9.8)cos35°(0.87) µ = 0.15

  46. 74. An inclined plane at 30.0º is 6.40 m long. A book, which has a kinetic coefficient of friction with the incline of 0.20, is placed at the top and immediately begins to slide. Using energy considerations, how long will it take for the book to reach the bottom of the incline? 6.4 m

  47. 74. An inclined plane at 30.0º is 6.40 m long. A book, which has a kinetic coefficient of friction with the incline of 0.20, is placed at the top and immediately begins to slide. Using energy considerations, how long will it take for the book to reach the bottom of the incline? MEi = MEf PEi = KEf + Wfric mgh = ½mv2 + Ff.s mgh = ½mv2 + µFN.s mgh = ½mv2 + µFg┴.s mgh = ½mv2 + µmgcosθ.s gh = ½v2 + µgcosθ.s v = 6.4 m/s s = ½(vi + vf)t t = 2 sec 6.4 m

  48. Ch6.4 – Power - rate a which work is done Units: Common substitutions:

  49. Common substitutions: 1.

  50. Common substitutions: 2. Overcoming gravity: 1.

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