1. Chapter 7 Work and Kinetic Energy

2. Units of Chapter 7 Work Done by a Constant Force Kinetic Energy and the Work-Energy Theorem Work Done by a Variable Force Power

3. 7-1 Work Done by a Constant Force The definition of work, when the force is in the direction of the displacement: (7-1) SI unit: newton-meter (N·m) = joule, J This is assuming the body can be treated as a particle.

4. 7-1 Work Done by a Constant Force British Unit is ft*lb 1 J = 0.7376 ft*lbs

5. 7-1 Work Done by a Constant Force If the force is at an angle to the displacement: (7-3) Only the force that is in the direction ( + or -) of the displacement is calculated Dot product!

6. The Sign of Work The work done may be positive, zero, or negative, depending on the angle between the force and the displacement:

8. Negative Work (example)

9. Total Work Sum of the works from each force Or: Work done by Net Force

11. Is it possible to do work on an object that remains at rest? 1) yes 2) no To Work or Not to Work

12. Is it possible to do work on an object that remains at rest? 1) yes 2) no force acts over a distance no displacementno work done Work requires that a force acts over a distance. If an object does not move at all, there is no displacement, and therefore no work done. To Work or Not to Work

13. Friction and Work 1) friction does no work at all 2) friction does negative work 3) friction does positive work A box is being pulled across a rough floor at a constant speed. What can you say about the work done by friction?

14. f N mg displacement Pull opposite negative W = F d cos   = 180 o W < 0 Friction acts in the opposite direction to the displacement, so the work is negative. Or using the definition of work (W = F d cos  ), since  = 180 o, then W < 0. Friction and Work 1) friction does no work at all 2) friction does negative work 3) friction does positive work A box is being pulled across a rough floor at a constant speed. What can you say about the work done by friction?

15. Normal Force and Work 1) Normal force does no work at all 2) Normal force does negative work 3) Normal force does positive work A box is being pulled across a rough floor at a constant speed. What can you say about the work done by the normal force?

16. f N mg displacement Pull zero W = F d cos  = 90 o W = 0 The normal force is perpendicular to the displacement, so the work is zero. Or using the definition of work (W = F d cos  ), since  = 90 o, then W = 0. Normal Force and Work 1) Normal force does no work at all 2) Normal force does negative work 3) Normal force does positive work A box is being pulled across a rough floor at a constant speed. What can you say about the work done by friction?

17. Force and Work 1) one force 2) two forces 3) three forces 4) four forces 5) no forces are doing work A box is being pulled up a rough incline by a rope connected to a pulley. How many forces are doing work on the box?

18. Force and Work N f T mg displacement Any force not perpendicular to the motion will do work: no work N does no work positive T does positive work f does negative work mg does negative work 1) one force 2) two forces 3) three forces 4) four forces 5) no forces are doing work A box is being pulled up a rough incline by a rope connected to a pulley. How many forces are doing work on the box?

19. Work Review Problems 1. Find the work needed by a weightlifter to lift a 30 kg barbell 1.5 meters upwards at a constant speed. 2. A 3000 kg car is moving across level ground at 5 m/s when it begins an acceleration that ends with the car moving at 15 m/s. Is work done in this situation? How do you know? 3. An Alaskan Huskie pulls a sled using a 500 N force across a 10 m wide street where the force of friction on the 90 kg sled is 200 N. How much work is done by friction? How much work is done by the huskie? How much by gravity? 4. A 60 kg man climbs a 3 m tall flight of stairs. What work was done on the man by the force of gravity? 5. Find the work done when a 100 N force pushes a car 10 m to the right as shown below. 25°

20. 7-2 Kinetic Energy and the Work-Energy Theorem When positive work is done on an object, its speed increases; when negative work is done, its speed decreases.

21. Kinetic Energy After algebraic manipulations of the equations of motion, we find (only in an inertial frame of reference): Therefore, we define the kinetic energy: (7-6) Based upon W = Fd = mad = ma  x and v 2 =v i 2 +2a  x (show me)

22. The Work-Energy Theorem Work-Energy Theorem: The total work done on an object is equal to its change in kinetic energy. (7-7)

23. Draw me an illustration that shows me the relationship between work, force, and speed

24. A child on a skateboard is moving at a speed of 2 m/s. After a force acts on the child, her speed is 3 m/s. What can you say about the work done by the external force on the child? 1) positive work was done 2) negative work was done 3) zero work was done ConcepTest 7.7 Work and KE

25. A child on a skateboard is moving at a speed of 2 m/s. After a force acts on the child, her speed is 3 m/s. What can you say about the work done by the external force on the child? 1) positive work was done 2) negative work was done 3) zero work was done The kinetic energy of the child increased because her speed increasedincrease in KE positive work being done KE f > KE i work W must be positive The kinetic energy of the child increased because her speed increased. This increase in KE was the result of positive work being done. Or, from the definition of work, since W =  KE = KE f – KE i and we know that KE f > KE i in this case, then the work W must be positive. ConcepTest 7.7 Work and KE Follow-up: What does it mean for negative work to be done on the child?

26. ConcepTest 7.8b Speeding Up I 1) 0  30 mph 2) 30  60 mph 3) both the same A car starts from rest and accelerates to 30 mph. Later, it gets on a highway and accelerates to 60 mph. Which takes more energy, the 0  30 mph, or the 30  60 mph?

27. 1/2 mv 2 velocity squared The change in KE (1/2 mv 2 ) involves the velocity squared. 1/2 m (30 2 - 0 2 ) = 1/2 m (900) So in the first case, we have: 1/2 m (30 2 - 0 2 ) = 1/2 m (900) 1/2 m (60 2 - 30 2 ) = 1/2 m (2700) In the second case, we have: 1/2 m (60 2 - 30 2 ) = 1/2 m (2700) bigger energy changesecond case Thus, the bigger energy change occurs in the second case. ConcepTest 7.8b Speeding Up I 1) 0  30 mph 2) 30  60 mph 3) both the same A car starts from rest and accelerates to 30 mph. Later, it gets on a highway and accelerates to 60 mph. Which takes more energy, the 0  30 mph, or the 30  60 mph?

29. Example A 1300 kg car coasts on a horizontal road with a velocity of 18 m/s, E. After crossing an unpaved, sandy stretch of road 30.0 m long, it’s velocity decreases to 15 m/s, E. –Was the net work done on the car positive, negative, or zero? –Find the net work done on the car. –What is the magnitude and direction of the average net force on the car in the sandy section?

30. Physics Try these kinetic speed problems to get you ready for class Agenda – –Bellringer –Finish Kinetic Energy –Practice –Introduce Power –More practice. 1.A 10.0 N (weight) lightsaber is accelerated from rest at a rate of 2.5 m/s 2. What is the kinetic energy of the lightsaber after it has accelerated over a distance of 15.0 m. 2.A 1200.0 N Wookie jumps off a cliff on Earth. What is its kinetic energy after it falls for 4.50 s?

32. Exceptions In this situation, the person can not be treated like a point!

33. 7-3 Work Done by a Variable Force If the force is constant, we can interpret the work done graphically: (cross product)

34. Multiple Rectangles (calculus) If the force takes on several successive constant values: (Sum of individual works)

35. Varying Force (calculus) We can then approximate a continuously varying force by a succession of constant values. (area under the curve (or line as the case may be)

36. Work done by a spring The force needed to stretch a spring an amount x is F = kx. (k is the force or spring constant) Therefore, the work done in stretching the spring is (calculus) (7-8)

37. Hooke’s Law Force is directly proportional to elongation of a spring –F=kx (where k is the force or spring constant) –Example. If it takes 300 N to stretch a spring 2 m, what is the force constant

38. Example A 1.2 kg block is held against a spring of force constant 1.0x10 4 N/m, compressing it a distance of 0.15 m. How fast is the block moving after it has been released and the spring pushes it away (in other words, just after the spring stops pushing on it)?

39. 7-4 Power Power is a measure of the rate at which work is done: (scalar) (7-10) SI unit: J/s = watt, W 1 horsepower = 1 hp = 550 ft*lbs/s = 746 W kilowatt-hour (kW*h) = (1 kW/s) for an hour (work or energy, not power)

40. 7-4 Power

41. Engine #1 produces twice the power of engine #2. Can we conclude that engine #1 does twice as much work as engine #2? 1) yes 2) no ConcepTest 7.11c Power

42. Engine #1 produces twice the power of engine #2. Can we conclude that engine #1 does twice as much work as engine #2? 1) yes 2) no No!! We cannot conclude anything about how much work each engine does. work will depend upon how much time is used No!! We cannot conclude anything about how much work each engine does. Given the power output, the work will depend upon how much time is used. For example, engine #1 may do the same amount of work as engine #2, but in half the time. ConcepTest 7.11c Power

43. ConcepTest 7.12b Energy Consumption Which contributes more to the cost of your electric bill each month, a 1500-Watt hair dryer or a 600-Watt microwave oven? 1) hair dryer 2) microwave oven 3) both contribute equally 4) depends upon what you cook in the oven 5) depends upon how long each one is on 1500 W 600 W

44. energy you have to know how long it was running We already saw that what you actually pay for is energy. To find the energy consumption of an appliance, you must know more than just the power rating — you have to know how long it was running. ConcepTest 7.12b Energy Consumption Which contributes more to the cost of your electric bill each month, a 1500-Watt hair dryer or a 600-Watt microwave oven? 1) hair dryer 2) microwave oven 3) both contribute equally 4) depends upon what you cook in the oven 5) depends upon how long each one is on 1500 W 600 W

45. 7-4 Power If an object is moving at a constant speed in the face of friction, gravity, air resistance, and so forth, the power exerted by the driving force can be written: (7-13) Efficiency = Power output/Power input *100%

47. Summary of Chapter 7 If the force is constant and parallel to the displacement, work is force times distance If the force is not parallel to the displacement, The total work is the work done by the net force:

48. Summary of Chapter 7 SI unit of work: the joule, J Total work is equal to the change in kinetic energy: where

49. Summary of Chapter 7 Work done by a spring force: Power is the rate at which work is done: SI unit of power: the watt, W