2. Work and Energy

3. Energy KINETIC ENERGY (KE): energy associated with motion of an object POTENTIAL ENERGY (PE): energy associated with position of an object

4. WORK WORK: the product of the component of the force along the direction of displacement and the magnitude of the displacement. W=F dcos  where d = magnitude of displacement When the force is not in the direction of the displacement, then either sin or cos is needed. Units of work: 1 Joule = 1 Nm

5. The con may expend energy when he pushes on the wall, but if it doesn’t move, no work is performed on the wall. In order to do work, you must exert a force over a distance.

6. W = FD cos  The force must be in the same direction as the movement. If not, you have to use a vector component of the force that is in the same direction. Hence the cosine. A lot of the time, you will see the work equation written like this: Why?

7. Example: A man cleaning his apartment pulls a vacuum cleaner with a force of 50. N at an angle of 30.0 . A frictional force of magnitude 40. N dampens the motion, and the vacuum is pulled a distance of 3.0 m. Calculate: a)the work done by the 50. N pull; b)the work done by the frictional force; c)the net work done on the vacuum by all forces acting on it.

8. Example: A man cleaning his apartment pulls a vacuum cleaner with a force of 50. N at an angle of 30.0 . A frictional force of magnitude 40. N dampens the motion, and the vacuum is pulled a distance of 3.0 m. Calculate: a)the work done by the 50 N pull; b)the work done by the frictional force; c)the net work done on the vacuum by all forces acting on it. W F = (Fcos  ) d W F = (50. N)(cos 30  )(3.0 m) W F = 130 J 30  direction of displacement 50 N 40 N

9. Example: A man cleaning his apartment pulls a vacuum cleaner with a force of 50. N at an angle of 30.0 . A frictional force of magnitude 40. N dampens the motion, and the vacuum is pulled a distance of 3.0 m. Calculate: a)the work done by the 50. N pull; b)the work done by the frictional force; c)the net work done on the vacuum by all forces acting on it. W f = (Fcos  ) d W f = (40. N)(cos 180  )(3.0 m) W f = -120 J 30  direction of displacement 50 N 40 N

10. Example: A man cleaning his apartment pulls a vacuum cleaner with a force of 50 N at an angle of 30 . A frictional force of magnitude 40 N dampens the motion, and the vacuum is pulled a distance of 3 m. Calculate a)the work done by the 50 N pull; b)the work done by the frictional force; c)the net work done on the vacuum by all forces acting on it. W net = W F + W f W net = 130 J – 120 J W net = 10 J 30  direction of displacement 50 N 40 N *normal force, weight, & upward component of pull do no work because they are perpendicular to the displacement.

11. Work and Kinetic Energy

12. This is known as the work-energy theorem. Roughly speaking, energy is the ability to do work.

13. Work-Energy Theorem The net work done on an object by the force or forces acting on it is equal to the change in the kinetic energy of the object. –The speed of an object will increase if the net work done on it is positive; –The speed of an object will decrease if the net work done on it is negative.

14. KE i + PE i + Pe i(s) + Work added = KE f + PE f + Pe f(s) + Work lost This is where we are going with this unit. We will now break this equation down piece by piece. Note: Energy can be in all four or any combination of the four types on either side. You need to decide which pieces you will use.

15. Example: A 1400-kg car has a net forward force of 4500 N applied to it. The car starts from rest and travels down a horizontal highway. What are its kinetic energy and speed after it has traveled 100.m? (ignore friction and air resistance) 4500 N 100 m m = 1400 kg F = 4500 N V o = 0 d = 100. m KE f = ? KE i = 0 V f = ?

16. Example: A 1400-kg car has a net forward force of 4500 N applied to it. The car starts from rest and travels down a horizontal highway. What are its kinetic energy and speed after it has traveled 100.0m? (ignore friction and air resistance) 4500 N 100 m

17. Potential Energy Gravitational potential energy: the energy an object has due to its position in space. PE = mgh In working problems, choose the zero level for PE (the pt. at which the grav. PE=0) so that you can easily calculate the difference in PE

18. Example: A 60.0-kg skier is at the top of a slope. At the initial point A, the skier is 10.0 m vertically above point B. Find the gravitational PE of the skier at A and B, and the difference in PE between the 2 pts. choose B = 0 PE i = mgh i PE i = (60.0kg)(9.8m/s2)(10.0m) PE i =5880 J PE f = mgh f PE f = (60.0kg)(9.8m/s2)(0m) PE f =0 J A B

19. Example: A 60.0-kg skier is at the top of a slope. At the initial point A, the skier is 10.0 m vertically above point B. Find the gravitational PE of the skier at A and B, and the difference in PE between the 2 pts. PE i =5880 J PE f =0 J  PE = PE f - PE i  PE = - 5880 J A B

20. Law of Conservation of Energy: Energy can not be created or destroyed. Actually, It’s conservation of mass - energy

21. Remember E=mc 2 That is to say, energy and mass are interchangeable.

22. Energy RIP Although energy cannot be destroyed, it can go places where we can never recover it. Kinetic Potential Chemical Heat Electrical Sound And many more…

23. Conservation of Mechanical Energy Although energy may be changed into a different form, the final value will be the same as the initial value (energy is not lost). Ei = Ef KE i + PE i = KE f + PE f ½ mv i 2 + mgh i = ½ mv f 2 + mgh f

24. Kinetic vs. Potential Energy Calculate the KE and PE energy at each of the 5 spots on the ski run. Assume the skier starts from rest. (m = 51 kg)

25. Kinetic vs. Potential Energy Calculate the KE and PE energy at each of the 5 spots on the roller coaster. Assume the skater starts from rest.

26. Example: A diver weighing 750. N (mass = 77.0 kg) drops from a board 10.0 m above the surface of the pool of water. 10 m A) Use the conservation of mechanical energy to find her speed at a point 5.00 m above the water surface. B) Find the speed of the diver just before she strikes the water.

27. Example: A diver weighing 750. N (mass = 77.0 kg) drops from a board 10.0 m above the surface of the pool of water. A) Use the conservation of mechanical energy to find her speed at a point 5 m above the water surface. B) Find the speed of the diver just before she strikes the water. h f = 5 m ½ mv i 2 + mgh i = ½ mv f 2 + mgh f ½(77)(0)+(77)(9.8)(10)=½(77)(v f 2 )+(77)(9.8)(5) V f = 9.90 m/s

28. Example: A diver weighing 750 N (mass = 77.0 kg) drops from a board 10 m above the surface of the pool of water. A) Use the conservation of mechanical energy to find her speed at a point 5 m above the water surface. B) Find the speed of the diver just before she strikes the water. h f =0 ½ mv i 2 + mgh i = ½ mv f 2 + mgh f ½(77)(0)+(77)(9.8)(10)=½(77)(v f 2 )+(77)(9.8)(0) V f = 13.9 m/s

29. Example: A sled and its rider together weigh 800. N. They move down a frictionless hill through a vertical distance of 10.0 m. Use conservation of energy to find the speed of the sled-rider system at the bottom of the hill, assuming the rider pushes off with an initial speed of 5.00 m/s. ½ v i 2 + gy i = ½ v f 2 + gy f ½(5) 2 + (9.8)(10)=½(v f 2 )+ (9.8)(0) v f = 14.9 m/s ½ mv i 2 + mgy i = ½ mv f 2 + mgy f 10 m

30. Conservative / Nonconservative Forces A force is CONSERVATIVE if the work it does on an object between 2 points is independent of the path the object takes between the 2 points. (The work done depends only on the initial and final positions.) (example: Work done by gravity)

31. Conservative / Nonconservative Forces A force is NONCONSERVATIVE if the work it does on an object moving between 2 points depends on the path taken. (i.e. Work done by sliding friction)

32. Conservation of Energy with Nonconservative Forces. A 15 kg kid, initially at rest, slides down a 8.0 m high slide. Ideally, what is his final velocity at the bottom of the slide? KE i + PE i = KE f + PE f 0 + mgh = 1/2 mv 2 + 0 (15kg)(9.8m/s2)(8.0m) = 1/2 (15kg)v 2 v f = 12.5m/s ~13m/s

33. In reality, the final velocity of the kid is only 10 m/s. Where did the extra energy go? Friction is an nonconservative force which used up some of the initial potential energy. To account for friction, we need to add back in the work done by (or energy lost by) friction.

34. KE i + PE i + Work added = KE f + PE f + Work lost Example: Work of someone pushing Example: Work done by friction Note: Usually work done by friction is negative (because it is in the opposite direction). But since we are ADDING BACK IN the work done by friction it is positive.

35. How much work was done by friction as the kid slid down the slide ? (m = 15 kg, y i = 8.0m v f = 10 m/s) KE i + PE i + Work added = KE f + PE f + Work lost 0 + mgh + 0 = 1 / 2 mv f 2 + 0 + W friction (15kg)(9.8m/s 2 )(8.0m) = 1 / 2 (15kg)(10.m/s) 2 + W f 1176 J = 750 J +W f W f = 426 J ~ 430J

36. Potential Energy Stored in a Spring PE s = ½ kx 2 wherek = spring constant x = distance spring is compressed or stretched (KE + PE g + PE s ) i = (KE + PE g + PE s ) f

37. Example: A block of mass 0.500 kg rests on a horizontal, frictionless surface. The block is pressed lightly against a spring, having a spring constant k=80.0 N/m. The spring is compressed a distance of 2.00 cm and released. Find the speed of the block at the instant it loses contact with the spring at the x = 0 position. (PE s ) i = (KE) f ½kx i 2 = ½mv f 2 ½(80)(0.02) 2 = ½(0.5) v f 2 v = 0.253 m/s (KE + PE g + PE s ) i = (KE + PE g + PE s ) f 0.500 kg

38. Example: The launching mechanism of a toy gun consists of a spring of unknown spring constant. By compressing the spring a distance of 0.120 m, the gun is able to launch a 20.0-g projectile to a maximum height of 20.0 m when fired vertically from rest. Determine the value of the spring constant. (PE s ) i = (PE g ) f ½kx i 2 = mgh ½(k)(0.12) 2 = (0.02)(9.8)(20) k = 544 N/m (KE + PE g + PE s ) i = (KE + PE g + PE s ) f

39. Machines make work easier Work In = Work Out + Heat In a real machine the efficiency must always be less than 100%

40. Types of Simple Machines: Lever Pulley Ramp

41. Fulcrum Work In Work Out Distance In Distance Out Work In = Work Out Force x Distance = Force x Distance

42. 50 lbs 25 lbs 50 lbs Machines make work easier.

43. But the work stays the same. 4 ft 8 ft 4 ft Remember…W = F x d He lifts the package up 4 ft…

44. Power Power: the rate at which work is done

45. A bulldozer and a small boy can do the same amount of work. It’s just that the bulldozer can do it faster.

46. Units in the SI system….

47. Units in the English system 1 HP = 746 Watts

48. A new conveyor system at the local packaging plan will utilize a motor- powered mechanical arm to exert an average force of 890 N to push large crates a distance of 12 meters in 22 seconds. Determine the power output required of such a motor. 490 W (rounded from 485 W)

49. The Taipei 101 in Taiwan is a 1667-foot tall, 101-story skyscraper. The skyscraper is the home of the world’s fastest elevator. The elevators transport visitors from the ground floor to the Observation Deck on the 89th floor at speeds up to 16.8 m/s. Determine the power delivered by the motor to lift the 10 passengers at this speed. The combined mass of the passengers and cabin is 1250 kg. 2.06 x 10 5 W

50. Example: A 2.00x10 3 kg car starts from rest and accelerates to a final velocity of +20.0 m/s in a time of 15.0 s. Assume that the force of air resistance remains constant at a value of –500. N during this time A) Find the average power developed by the engine. B) Find the instantaneous power when the car reaches its final velocity

51. Example: A 2000 kg car starts from rest and accelerates to a final velocity of +20.0 m/s in a time of 15.0 s. The force of the engine is 3160 N. A) Find the average power developed by the engine (average power uses average v).

52. Example: A 2000 kg car starts from rest and accelerates to a final velocity of +20 m/s in a time of 15 s. Assume that the force of air resistance remains constant at a value of –500 N during this time. A) Find the average power developed by the engine (average power uses average v).

53. Example: A 2000 kg car starts from rest and accelerates to a final velocity of +20 m/s in a time of 15 s. Assume that the force of air resistance remains constant at a value of –500 N during this time. B) Find the instantaneous power when the car reaches its final velocity (use velocity at instant you are solving for).

54. Example: An elevator has a mass of 1000. kg and carries a maximum load of 800. kg. A constant frictional force of 4000. N retards its motion upward. What minimum power must the motor deliver to lift the fully-loaded elevator at a constant speed of 3.00 m/s? (In watts and horsepower) v is constant, so a=0 F T – F f – mg = 0 F T = F f + mg = 4000 N + (1800 kg)(9.8m/s 2 ) F T = 2.16 x 10 4 N P = F T v P = (2.16 x 10 4 N)(3.00 m/s) = 6.48 x 10 4 W P = (6.48 x 10 4 W)(1 hp / 746 W) = 86.9 hp