2. 1 7/8/04 Midterm 1: July 9 Will cover material from Chapters 1-6 Go to the room where you usually have recitation Practice exam available on-line and in the library

3. 7/8/04 2 Chapter 7 Kinetic Energy and Work

4. 3 7/8/04 Energy A property of the state of an object Scalar quantity – no direction Conserved – cannot be created or destroyed, but it can change from one form to another or be exchanged from one object to another Units: Joule = kg m 2 /s 2

5. 4 7/8/04 Kinds of Energy Kinetic (movement) Potential  Gravitational  Spring  Chemical bonds Mass (E=mc 2 …) Thermal And more…

6. 5 7/8/04 Kinetic Energy Kinetic Energy  Energy of motion For an object of mass m moving with speed v: Also kinetic energy associated with rotation, vibration, etc. More on that later…

7. 6 7/8/04 Kinetic Energy: Orders of Magnitude Earth orbiting sun: 2x10 29 J Car at 60 mph: 100,000 J Baseball pitch: 300 J K = ½mv 2 for some common objects: A man walking: 40 J Angry bee: 0.005 J

8. 7 7/8/04 Work W > 0  energy added W < 0  energy taken away Work  Energy transferred by a force Work done on an object is the energy transferred to/from it Work done on an object by a constant force F while moving through a displacement r  W = F r

9. 8 7/8/04  is the angle between the vectors if you put their tails together Dot or Scalar Product B  A Measures “how much” one vector lies along another

10. 9 7/8/04 What Does It Mean Physically? W > 0 if  < 90° force is adding energy to object W 90° force is reducing energy of object W = 0 if F  r r F  r F

11. 10 7/8/04 Work Examples F tot = 0  W tot = 0 Note: Kinetic Energy is constant… Push on a wall  r = 0, so no work is done (W = 0) d Lift a weight against gravity at constant speed…

12. 11 7/8/04 Work Against Gravity A weightlifter does work when lifting a weight W lift =mgh

13. 12 7/8/04 Work for Tension, Gravity, Friction What is the work done by gravity, tension, and friction? Iguazu Falls: 269 ft = 82 m Guess that the weight of the pack is 250 lb. = 113 kg Vertical ascent: W gravity = -mgh = -(113 kg)(9.8 m/s 2 )(82 m) = -90,800 J W tension = -W gravity = 90,800 J

14. 13 7/8/04 Horizontal pull (at the end): use d~50 ft=15.2 m, μ k = 1.0 W friction = F f d = -μ k mg d = -(1.0)(113 kg)(9.8 m/s 2 )(15.2 m) = -16,800 J W tension = -W friction = 16,800 J What is the work done by gravity, tension, and friction? Work for Tension, Gravity, Friction

15. 14 7/8/04 Work Due to Friction v F fr The frictional force always opposes the motion: Moving to the right:Moving to the left: W negative in both cases

16. 15 7/8/04 Another Work Example Consider a pig sliding down frictionless ramp:  N FgFg d θ d h Work done by the normal force: Work from the gravitational force:

17. 16 7/8/04 Work-Kinetic Energy Theorem The total work done on an object is the change in its kinetic energy! Easy to extrapolate to three dimensions For constant acceleration: Since F tot = ma: Multiply by ½m: But ½mv 2 = K:

18. 17 7/8/04 Total Work in One Dimension For a small segment  x,  W  F(x)  x From x 1 to x 2 : W tot  F(x)  x = area under curve  x 1 < x < x 2 To be exact: x1x1 x2x2 xx F x

19. 18 7/8/04 Work in Two Dimensions In one dimension motion and force are always in the same direction (or opposite directions) This is not true in two dimensions F dr How do we generalize work and kinetic energy to motion in two dimensions?

20. 19 7/8/04 Work Along an Arbitrary Path F ΔxΔx ΔyΔy ll Over the whole path: Looking at a small patch of the path:

21. 20 7/8/04 Restating the Work-Kinetic Energy Theorem Constant force: Variable force:

22. 21 7/8/04 A Formal Derivation Newton’s 2 nd Law Using the chain rule:

23. 22 7/8/04 An Example v i =60 mph =26.8 m/s 150 ft μ k =0.9 45.7 m x Will the car be able to stop before hitting the moose?

24. 23 7/8/04 An Example (continued) Friction from the tires on the road will slow the car: The work done by friction will be: From the work-energy theorem: Just miss the moose!

25. 24 7/8/04 Example: Pile Driver d Big Mass Drop a big mass to drive a nail into a board

26. 25 7/8/04 The Pile Driver h x m How much force is exerted on the nail? 1) Work done by gravity on freely falling pile W 1 = (-mg)(-h-x) = mg(h+x) 2) Work done on the nail W 2 = F(-x) = -Fx 3) Total Work = 0 W = W 1 +W 2 = mg(h+x) – Fx = 0 F= mg(h+x) x

27. 26 7/8/04 Example: Slowing a Bus v i =75 mph=33.5 m/s How much work is done slowing down a bus? 5.45 x 10 3 kg v f =65 mph=29.0 m/s

28. 27 7/8/04 Example: (continued) F brake = 2.75 x 10 4 N How far does the bus travel while slowing? Area under F(x) curve: x F xixi xfxf F brake

29. 28 7/8/04 Example: (continued)

30. 29 7/8/04 The Spring: A Variable Force Springs exert force when stretched or compressed positive x x = 0 Defines equilibrium position F = - kx k = "spring constant" big k  stiff spring x = 0  no force x 0 spring pushes out x > 0  F < 0 spring pulls in Hooke’s Law:

31. 30 7/8/04 Work Done By a Spring How much work by spring in moving from x i to x f ? xfxf If |x i | < |x f | then spring takes energy away xixi x=0 F spring x=0

32. 31 7/8/04 Example: You stretch a spring 10 cm and must apply a 10 N force to hold the spring in place. What is the spring constant, and how much work did you do on the spring to stretch it? 10 cm F pull F spring

33. 32 7/8/04 Example: 10 cm F pull F spring The work done by the spring is: So the work I do on the spring is

34. 33 7/8/04 Power Work doesn’t depend on the time interval Work to climb a flight of stairs~3000 J 10 s 1 min 1 hour Power is work done per unit time Average Power: Instantaneous Power: 1 hp = 746 W 1 J 1 s = 1 Watt Units Work time

35. 34 7/8/04 Power in 3-D For a constant force: So power will be:

36. 35 7/8/04 Express Elevator Say a 900 kg elevator travels 135 floors (400 m) in 40 s. It accelerates at 2.4 m/s 2 until it reaches a velocity of 12 m/s. Find the average power (going up) W = mgh = (900 kg)(9.8 m/s 2 )(400 m) = 3.5 x 10 6 J m = 900 kg a = 2.4 m/s 2 v max = 12 m/s

37. 36 7/8/04 Maximum Instantaneous Power: P = Fv Maximum just when elevator reaches cruising speed going up P = (900 kg)(9.8 m/s 2 +2.4 m/s 2 )(12 m/s) = 130,000 W Express Elevator (continued) m = 900 kg a = 2.4 m/s 2 v max = 12 m/s F – mg = ma a F F = m(g+a)

38. 37 7/8/04 An Application A car engine can supply some maximum amount of power. How will the car’s velocity change when it accelerates at constant power?

39. 38 7/8/04 Acceleration at Constant Power  Start from v = 0 and use work-energy theorem: Acceleration is not constant!

40. 39 7/8/04 Why Cars Need Gears There is some maximum power at which the engine can operate. Shifting to a higher gear reduces the force applied to the wheels, allowing for a higher top speed. F must be larger than the air resistance for the car to continue accelerating.

41. 40 7/8/04 Accelerating from 0 to 60 How much average horsepower is needed to accelerate a 1100 kg car from 0 to 60 mph in 3.1 seconds? In reality, there are transmission losses, etc., so you need much more horsepower to achieve this level of performance.

42. 41 7/8/04 Example: (Problem 7.37) The force (but not the power) to tow a boat at constant velocity is proportional to the speed. If a speed of 4.0 km/h requires 7.5 kW, how much power does a speed of 12 km/hr require?