1. Alternate Definition of Work

2. Suppose an object is moving in a direction given by its displacement as shown. Suppose the net force is acting as shown. What component of the net force does no work on the object? Δ d F net F net || F net 90° θ

3. Alternate Definition of Work Suppose an object is moving in a direction given by its displacement as shown. Suppose the net force is acting as shown. What component of the net force does no work on the object? Δ d F net F net || F net 90° θ F net 90° is perpendicular to Δ d. It does no work on the object. Depending on θ, F net || along Δ d can only speed up or slow down the object. Therefore, work done by the net force only changes the kinetic energy of the object.

4. Alternate Definition of Work Suppose an object is moving in a direction given by its displacement as shown. Suppose the net force is acting as shown. What component of the net force does no work on the object? Δ d F net F net || F net 90° θ F net 90° is perpendicular to Δ d. It does no work on the object. Depending on θ, F net || along Δ d can only speed up or slow down the object. Therefore, work done by the net force only changes the kinetic energy of the object. We can write... W net = Δ E K

5. Alternate Definition of Work Suppose an object is moving in a direction given by its displacement as shown. Suppose the net force is acting as shown. What component of the net force does no work on the object? Δ d F net F net || F net 90° θ F net 90° is perpendicular to Δ d. It does no work on the object. Depending on θ, F net || along Δ d can only speed up or slow down the object. Therefore, work done by the net force only changes the kinetic energy of the object. We can write... W net = Δ E K This is called the work-energy theorem.

6. More on the Work-Energy Theorem

7. W net = Δ E K

8. More on the Work-Energy Theorem W net = Δ E K What is Δ E K in terms of final kinetic energy E K2 and initial kinetic energy E K1 ?

9. More on the Work-Energy Theorem W net = Δ E K Δ E K = E K2 - E K1

10. More on the Work-Energy Theorem W net = Δ E K Δ E K = E K2 - E K1 So W net = ?

11. More on the Work-Energy Theorem W net = Δ E K Δ E K = E K2 - E K1 So W net = E K2 - E K1

12. More on the Work-Energy Theorem W net = Δ E K Δ E K = E K2 - E K1 So W net = E K2 - E K1 But what is W net in terms of F net, Δd, and θ ?

13. More on the Work-Energy Theorem W net = Δ E K Δ E K = E K2 - E K1 So W net = E K2 - E K1 Remember W net = | Δd | | F net | cosθ

14. More on the Work-Energy Theorem W net = Δ E K Δ E K = E K2 - E K1 So W net = E K2 - E K1 Remember W net = | Δd | | F net | cosθ Therefore we can create a new work-energy formula...

15. More on the Work-Energy Theorem W net = Δ E K Δ E K = E K2 - E K1 So W net = E K2 - E K1 Remember W net = | Δd | | F net | cosθ Therefore we can create a new work-energy formula... | Δd | | F net | cosθ = E K2 - E K1

16. Example: A 1260 kg car is moving at 72.0 km/h [E]. Suddenly the brakes are applied and a constant kinetic friction force of 5850 N [W] brings the car to a stop. Using the work-energy theorem, find the braking distance.

17. Given:

18. Example: A 1260 kg car is moving at 72.0 km/h [E]. Suddenly the brakes are applied and a constant kinetic friction force of 5850 N [W] brings the car to a stop. Using the work-energy theorem, find the braking distance. Given: m =1260 kg v 1 = 72.0 km/h = 20.0 m/s v 2 = 0 m/s f k = 5850 N θ = ?

19. Example: A 1260 kg car is moving at 72.0 km/h [E]. Suddenly the brakes are applied and a constant kinetic friction force of 5850 N [W] brings the car to a stop. Using the work-energy theorem, find the braking distance. Given: m =1260 kg v 1 = 72.0 km/h = 20.0 m/s v 2 = 0 m/s f k = 5850 N θ = 180°

20. Example: A 1260 kg car is moving at 72.0 km/h [E]. Suddenly the brakes are applied and a constant kinetic friction force of 5850 N [W] brings the car to a stop. Using the work-energy theorem, find the braking distance. Given: m =1260 kg v 1 = 72.0 km/h = 20.0 m/s v 2 = 0 m/s f k = 5850 N θ = 180° Unknown:

21. Example: A 1260 kg car is moving at 72.0 km/h [E]. Suddenly the brakes are applied and a constant kinetic friction force of 5850 N [W] brings the car to a stop. Using the work-energy theorem, find the braking distance. Given: m =1260 kg v 1 = 72.0 km/h = 20.0 m/s v 2 = 0 m/s f k = 5850 N θ = 180° Unknown: | Δd | = ? Formula: ?

22. Example: A 1260 kg car is moving at 72.0 km/h [E]. Suddenly the brakes are applied and a constant kinetic friction force of 5850 N [W] brings the car to a stop. Using the work-energy theorem, find the braking distance. Given: m =1260 kg v 1 = 72.0 km/h = 20.0 m/s v 2 = 0 m/s f k = 5850 N θ = 180° Unknown: | Δd | = ? Formula: | Δd | | F net | cosθ = E K2 - E K1

23. Example: A 1260 kg car is moving at 72.0 km/h [E]. Suddenly the brakes are applied and a constant kinetic friction force of 5850 N [W] brings the car to a stop. Using the work-energy theorem, find the braking distance. Given: m =1260 kg v 1 = 72.0 km/h = 20.0 m/s v 2 = 0 m/s f k = 5850 N θ = 180° Unknown: | Δd | = ? Formula: | Δd | | F net | cosθ = E K2 - E K1 | Δd | | F net | cosθ = E K2 - mv 1 2 /2

24. Example: A 1260 kg car is moving at 72.0 km/h [E]. Suddenly the brakes are applied and a constant kinetic friction force of 5850 N [W] brings the car to a stop. Using the work-energy theorem, find the braking distance. Given: m =1260 kg v 1 = 72.0 km/h = 20.0 m/s v 2 = 0 m/s f k = 5850 N θ = 180° Unknown: | Δd | = ? Formula: | Δd | | F net | cosθ = E K2 - E K1 | Δd | | F net | cosθ = E K2 - mv 1 2 /2 Sub: ?

25. Example: A 1260 kg car is moving at 72.0 km/h [E]. Suddenly the brakes are applied and a constant kinetic friction force of 5850 N [W] brings the car to a stop. Using the work-energy theorem, find the braking distance. Given: m =1260 kg v 1 = 72.0 km/h = 20.0 m/s v 2 = 0 m/s f k = 5850 N θ = 180° Unknown: | Δd | = ? Formula: | Δd | | F net | cosθ = E K2 - E K1 | Δd | | F net | cosθ = E K2 - mv 1 2 /2 Sub: | Δd | (5850) cos180° = 0 – (1260) (20.0) 2 /2 Note f k supplies the net force in this case.

26. Example: A 1260 kg car is moving at 72.0 km/h [E]. Suddenly the brakes are applied and a constant kinetic friction force of 5850 N [W] brings the car to a stop. Using the work-energy theorem, find the braking distance. Given: m =1260 kg v 1 = 72.0 km/h = 20.0 m/s v 2 = 0 m/s f k = 5850 N θ = 180° Unknown: | Δd | = ? Formula: | Δd | | F net | cosθ = E K2 - E K1 | Δd | | F net | cosθ = E K2 - mv 1 2 /2 Sub: | Δd | (5850) cos180° = 0 – (1260) (20.0) 2 /2 - 5850 | Δd | = - 252000 Note f k supplies the net force in this case.

27. Example: A 1260 kg car is moving at 72.0 km/h [E]. Suddenly the brakes are applied and a constant kinetic friction force of 5850 N [W] brings the car to a stop. Using the work-energy theorem, find the braking distance. Given: m =1260 kg v 1 = 72.0 km/h = 20.0 m/s v 2 = 0 m/s f k = 5850 N θ = 180° Unknown: | Δd | = ? Formula: | Δd | | F net | cosθ = E K2 - E K1 | Δd | | F net | cosθ = E K2 - mv 1 2 /2 Sub: | Δd | (5850) cos180° = 0 – (1260) (20.0) 2 /2 - 5850 | Δd | = - 252000 | Δd | = 43.1 m Note f k supplies the net force in this case.

28. Mechanical Energy

29. In a mechanical “system”, the quantity “Mechanical Energy” is often used

30. Mechanical Energy In a mechanical “system”, the quantity “Mechanical Energy” is often used. It has the symbol _____.

31. Mechanical Energy In a mechanical “system”, the quantity “Mechanical Energy” is often used. It has the symbol E m.

32. Mechanical Energy In a mechanical “system”, the quantity “Mechanical Energy” is often used. It has the symbol E m. E m = _____ + _____ + _____

33. Mechanical Energy In a mechanical “system”, the quantity “Mechanical Energy” is often used. It has the symbol E m. E m = E K + E g + E e

34. Mechanical Energy In a mechanical “system”, the quantity “Mechanical Energy” is often used. It has the symbol E m. E m = E K + E g + E e If a mechanical “system” loses no energy to the surroundings in the form of heat, light, sound etc over time, the total mechanical energy remains __________ or is __________.

35. Mechanical Energy In a mechanical “system”, the quantity “Mechanical Energy” is often used. It has the symbol E m. E m = E K + E g + E e If a mechanical “system” loses no energy to the surroundings in the form of heat, light, sound etc over time, the total mechanical energy remains constant or is conserved.

36. Mechanical Energy In a mechanical “system”, the quantity “Mechanical Energy” is often used. It has the symbol E m. E m = E K + E g + E e If a mechanical “system” loses no energy to the surroundings in the form of heat, light, sound etc over time, the total mechanical energy remains constant or is conserved. This is called the conservation of mechanical energy.

37. Example: A 72.0 N applied force compresses a vertical spring 12.0 cm. A small 21.0 g marble is placed at rest on the compressed spring. The compressed spring is then used to project the marble straight up into the air. If friction is negligible, to what maximum height does the marble rise? F s = 72 N X = 0.120 m h max = ? ground m = 0.021 kg

38. Example: A 72.0 N applied force compresses a vertical spring 12.0 cm. A small 21.0 g marble is placed at rest on the compressed spring. The compressed spring is then used to project the marble straight up into the air. If friction is negligible, to what maximum height does the marble rise? F s = 72 N X = 0.120 m h max = ? ground m = 0.021 kg Find K first: K = ?

39. Example: A 72.0 N applied force compresses a vertical spring 12.0 cm. A small 21.0 g marble is placed at rest on the compressed spring. The compressed spring is then used to project the marble straight up into the air. If friction is negligible, to what maximum height does the marble rise? F s = 72 N X = 0.120 m h max = ? ground m = 0.021 kg Find K first: K = F s / X = ?

40. Example: A 72.0 N applied force compresses a vertical spring 12.0 cm. A small 21.0 g marble is placed at rest on the compressed spring. The compressed spring is then used to project the marble straight up into the air. If friction is negligible, to what maximum height does the marble rise? F s = 72 N X = 0.120 m h max = ? ground m = 0.021 kg Find K first: K = F s / X = K = 72 / 0.120 K = ?

41. Example: A 72.0 N applied force compresses a vertical spring 12.0 cm. A small 21.0 g marble is placed at rest on the compressed spring. The compressed spring is then used to project the marble straight up into the air. If friction is negligible, to what maximum height does the marble rise? F s = 72 N X = 0.120 m h max = ? ground m = 0.021 kg Find K first: K = F s / X = K = 72 / 0.120 K = 600 N/m

42. Example: A 72.0 N applied force compresses a vertical spring 12.0 cm. A small 21.0 g marble is placed at rest on the compressed spring. The compressed spring is then used to project the marble straight up into the air. If friction is negligible, to what maximum height does the marble rise? F s = 72 N X = 0.120 m h max = ? ground m = 0.021 kg Find K first: K = F s / X = K = 72 / 0.120 K = 600 N/m Choose a zero reference level for gravitational potential energy. Why can't it be ground?

43. Example: A 72.0 N applied force compresses a vertical spring 12.0 cm. A small 21.0 g marble is placed at rest on the compressed spring. The compressed spring is then used to project the marble straight up into the air. If friction is negligible, to what maximum height does the marble rise? F s = 72 N X = 0.120 m h max = ? ground m = 0.021 kg K = 600 N/m Choose a zero reference level for gravitational potential energy. Why can't it be ground? Apply conservation of mechanical energy:

44. Example: A 72.0 N applied force compresses a vertical spring 12.0 cm. A small 21.0 g marble is placed at rest on the compressed spring. The compressed spring is then used to project the marble straight up into the air. If friction is negligible, to what maximum height does the marble rise? F s = 72 N X = 0.120 m h max = ? ground m = 0.021 kg K = 600 N/m Choose a zero reference level for gravitational potential energy. Why can't it be ground? Apply conservation of mechanical energy: E m (start) = E m (max height)

45. Example: A 72.0 N applied force compresses a vertical spring 12.0 cm. A small 21.0 g marble is placed at rest on the compressed spring. The compressed spring is then used to project the marble straight up into the air. If friction is negligible, to what maximum height does the marble rise? F s = 72 N X = 0.120 m h max = ? ground m = 0.021 kg K = 600 N/m Choose a zero reference level for gravitational potential energy. Why can't it be ground? Apply conservation of mechanical energy: E m (start) = E m (max height) E K1 + E g1 + E e1 = E K2 + E g2 + E e2

46. Example: A 72.0 N applied force compresses a vertical spring 12.0 cm. A small 21.0 g marble is placed at rest on the compressed spring. The compressed spring is then used to project the marble straight up into the air. If friction is negligible, to what maximum height does the marble rise? F s = 72 N X = 0.120 m h max = ? ground m = 0.021 kg K = 600 N/m Choose a zero reference level for gravitational potential energy. Why can't it be ground? Apply conservation of mechanical energy: E m (start) = E m (max height) E K1 + E g1 + E e1 = E K2 + E g2 + E e2 ? + ? + ? = ? + ? + ?

47. Example: A 72.0 N applied force compresses a vertical spring 12.0 cm. A small 21.0 g marble is placed at rest on the compressed spring. The compressed spring is then used to project the marble straight up into the air. If friction is negligible, to what maximum height does the marble rise? F s = 72 N X = 0.120 m h max = ? ground m = 0.021 kg K = 600 N/m Choose a zero reference level for gravitational potential energy. Why can't it be ground? Apply conservation of mechanical energy: E m (start) = E m (max height) E K1 + E g1 + E e1 = E K2 + E g2 + E e2 0 + ? + ? = ? + ? + ?

48. Example: A 72.0 N applied force compresses a vertical spring 12.0 cm. A small 21.0 g marble is placed at rest on the compressed spring. The compressed spring is then used to project the marble straight up into the air. If friction is negligible, to what maximum height does the marble rise? F s = 72 N X = 0.120 m h max = ? ground m = 0.021 kg K = 600 N/m Choose a zero reference level for gravitational potential energy. Why can't it be ground? Apply conservation of mechanical energy: E m (start) = E m (max height) E K1 + E g1 + E e1 = E K2 + E g2 + E e2 0 + 0 + ? = ? + ? + ?

49. Example: A 72.0 N applied force compresses a vertical spring 12.0 cm. A small 21.0 g marble is placed at rest on the compressed spring. The compressed spring is then used to project the marble straight up into the air. If friction is negligible, to what maximum height does the marble rise? F s = 72 N X = 0.120 m h max = ? ground m = 0.021 kg K = 600 N/m Choose a zero reference level for gravitational potential energy. Why can't it be ground? Apply conservation of mechanical energy: E m (start) = E m (max height) E K1 + E g1 + E e1 = E K2 + E g2 + E e2 0 + 0 + KX 2 /2 = ? + ? + ?

50. Example: A 72.0 N applied force compresses a vertical spring 12.0 cm. A small 21.0 g marble is placed at rest on the compressed spring. The compressed spring is then used to project the marble straight up into the air. If friction is negligible, to what maximum height does the marble rise? F s = 72 N X = 0.120 m h max = ? ground m = 0.021 kg K = 600 N/m Choose a zero reference level for gravitational potential energy. Why can't it be ground? Apply conservation of mechanical energy: E m (start) = E m (max height) E K1 + E g1 + E e1 = E K2 + E g2 + E e2 0 + 0 + KX 2 /2 = 0 + ? + ?

51. Example: A 72.0 N applied force compresses a vertical spring 12.0 cm. A small 21.0 g marble is placed at rest on the compressed spring. The compressed spring is then used to project the marble straight up into the air. If friction is negligible, to what maximum height does the marble rise? F s = 72 N X = 0.120 m h max = ? ground m = 0.021 kg K = 600 N/m Choose a zero reference level for gravitational potential energy. Why can't it be ground? Apply conservation of mechanical energy: E m (start) = E m (max height) E K1 + E g1 + E e1 = E K2 + E g2 + E e2 0 + 0 + KX 2 /2 = 0 + mgh + ?

52. Example: A 72.0 N applied force compresses a vertical spring 12.0 cm. A small 21.0 g marble is placed at rest on the compressed spring. The compressed spring is then used to project the marble straight up into the air. If friction is negligible, to what maximum height does the marble rise? F s = 72 N X = 0.120 m h max = ? ground m = 0.021 kg K = 600 N/m Choose a zero reference level for gravitational potential energy. Why can't it be ground? Apply conservation of mechanical energy: E m (start) = E m (max height) E K1 + E g1 + E e1 = E K2 + E g2 + E e2 0 + 0 + KX 2 /2 = 0 + mgh + 0

53. Example: A 72.0 N applied force compresses a vertical spring 12.0 cm. A small 21.0 g marble is placed at rest on the compressed spring. The compressed spring is then used to project the marble straight up into the air. If friction is negligible, to what maximum height does the marble rise? F s = 72 N X = 0.120 m h max = ? ground m = 0.021 kg K = 600 N/m Choose a zero reference level for gravitational potential energy. Why can't it be ground? Apply conservation of mechanical energy: E m (start) = E m (max height) E K1 + E g1 + E e1 = E K2 + E g2 + E e2 0 + 0 + KX 2 /2 = 0 + mgh + 0 KX 2 /2 = mgh

54. Example: A 72.0 N applied force compresses a vertical spring 12.0 cm. A small 21.0 g marble is placed at rest on the compressed spring. The compressed spring is then used to project the marble straight up into the air. If friction is negligible, to what maximum height does the marble rise? F s = 72 N X = 0.120 m h max = ? ground m = 0.021 kg K = 600 N/m Choose a zero reference level for gravitational potential energy. Why can't it be ground? Apply conservation of mechanical energy: E m (start) = E m (max height) E K1 + E g1 + E e1 = E K2 + E g2 + E e2 0 + 0 + KX 2 /2 = 0 + mgh + 0 KX 2 /2 = mgh 600(.12) 2 /2 = 0.021(10)h

55. Example: A 72.0 N applied force compresses a vertical spring 12.0 cm. A small 21.0 g marble is placed at rest on the compressed spring. The compressed spring is then used to project the marble straight up into the air. If friction is negligible, to what maximum height does the marble rise? F s = 72 N X = 0.120 m h max = ? ground m = 0.021 kg K = 600 N/m Choose a zero reference level for gravitational potential energy. Why can't it be ground? Apply conservation of mechanical energy: E m (start) = E m (max height) E K1 + E g1 + E e1 = E K2 + E g2 + E e2 0 + 0 + KX 2 /2 = 0 + mgh + 0 KX 2 /2 = mgh 600(.12) 2 /2 = 0.021(10)h 4.32 = 0.21h

56. Example: A 72.0 N applied force compresses a vertical spring 12.0 cm. A small 21.0 g marble is placed at rest on the compressed spring. The compressed spring is then used to project the marble straight up into the air. If friction is negligible, to what maximum height does the marble rise? F s = 72 N X = 0.120 m h max = ? ground m = 0.021 kg K = 600 N/m Choose a zero reference level for gravitational potential energy. Why can't it be ground? Apply conservation of mechanical energy: E m (start) = E m (max height) E K1 + E g1 + E e1 = E K2 + E g2 + E e2 0 + 0 + KX 2 /2 = 0 + mgh + 0 KX 2 /2 = mgh 600(.12) 2 /2 = 0.021(10)h 4.32 = 0.21h h max = 20.6 m

57. Systems With Friction

58. Most mechanical systems have kinetic friction. The short-cut formula for the work done by the kinetic friction force is W = ?

59. Systems With Friction Most mechanical systems have kinetic friction. The short-cut formula for the work done by the kinetic friction force is W = - f k d

60. Systems With Friction Most mechanical systems have kinetic friction. The short-cut formula for the work done by the kinetic friction force is W = - f k d The negative sign represents a _______ of heat energy from the system to the surroundings

61. Systems With Friction Most mechanical systems have kinetic friction. The short-cut formula for the work done by the kinetic friction force is W = - f k d The negative sign represents a loss of heat energy from the system to the surroundings

62. Systems With Friction Most mechanical systems have kinetic friction. The short-cut formula for the work done by the kinetic friction force is W = - f k d The negative sign represents a loss of heat energy from the system to the surroundings How can we modify the conservation of mechanical energy formula to take friction into account? E m (start) = E m (later) ??????

63. Systems With Friction Most mechanical systems have kinetic friction. The short-cut formula for the work done by the kinetic friction force is W = - f k d The negative sign represents a loss of heat energy from the system to the surroundings How can we modify the conservation of mechanical energy formula to take friction into account? E m (start) - f k d = E m (later)

64. Work done by an applied force in a system with friction

65. Critical thinking: Can you explain why this equation is valid to find the work done by an applied force introduced to a system where energy is lost to the surroundings due to friction? W applied = ΔE m + f k d