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Midterm 1: July 9

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. Chapter 7. Kinetic Energy and Work. Energy. A property of the state of an object Scalar quantity – no direction

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Midterm 1: July 9

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  1. 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

  2. Chapter 7 Kinetic Energy and Work

  3. 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 m2/s2

  4. Kinds of Energy • Kinetic (movement) • Potential • Gravitational • Spring • Chemical bonds • Mass (E=mc2…) • Thermal • And more…

  5. 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…

  6. Kinetic Energy: Orders of Magnitude K = ½mv2 for some common objects: Earth orbiting sun: 2x1029 J Car at 60 mph: 100,000 J Baseball pitch: 300 J A man walking: 40 J Angry bee: 0.005 J

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

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

  9. F F q r r W > 0 if q < 90° force is adding energy to object force is reducing energy of object W < 0 if q > 90° W = 0 if F  r What Does It Mean Physically?

  10. d Work Examples • Push on a wall • r = 0, so no work is done (W = 0) • Lift a weight against gravity at constant speed… Ftot = 0  Wtot = 0 Note: Kinetic Energy is constant…

  11. Work Against Gravity Wlift=mgh A weightlifter does work when lifting a weight

  12. Work for Tension, Gravity, Friction What is the work done by gravity, tension, and friction? Guess that the weight of the pack is 250 lb. = 113 kg Iguazu Falls: 269 ft = 82 m Vertical ascent: Wgravity = -mgh = -(113 kg)(9.8 m/s2)(82 m) = -90,800 J Wtension = -Wgravity = 90,800 J

  13. Work for Tension, Gravity, Friction What is the work done by gravity, tension, and friction? Horizontal pull (at the end): use d~50 ft=15.2 m, μk = 1.0 Wfriction = Ff d = -μkmg d = -(1.0)(113 kg)(9.8 m/s2)(15.2 m) = -16,800 J Wtension = -Wfriction = 16,800 J

  14. v Ffr Work Due to Friction The frictional force always opposes the motion: Moving to the right: Moving to the left: W negative in both cases

  15. N d Fg q d h θ Another Work Example Consider a pig sliding down frictionless ramp: Work done by the normal force: Work from the gravitational force:

  16. Work-Kinetic Energy Theorem For constant acceleration: Since Ftot = ma: Multiply by ½m: But ½mv2 = K: Easy to extrapolate to three dimensions The total work done on an object is the change in its kinetic energy!

  17. F x Dx x1 x2 S Wtot  F(x)Dx = area under curve x1 < x < x2 Total Work in One Dimension For a small segment Dx, DW  F(x)Dx From x1 to x2: To be exact:

  18. F dr 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 How do we generalize work and kinetic energy to motion in two dimensions?

  19. F Dl Δy Δx Work Along an Arbitrary Path Looking at a small patch of the path: Over the whole path:

  20. Restating the Work-Kinetic Energy Theorem Constant force: Variable force:

  21. A Formal Derivation Newton’s 2nd Law Using the chain rule:

  22. vi=60 mph =26.8 m/s x 150 ft An Example μk=0.9 45.7 m Will the car be able to stop before hitting the moose?

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

  24. Example: Pile Driver Drop a big mass to drive a nail into a board Big Mass d

  25. mg(h+x) F= x The Pile Driver How much force is exerted on the nail? m 1) Work done by gravity on freely falling pile W1 = (-mg)(-h-x) = mg(h+x) 2) Work done on the nail h W2 = F(-x) = -Fx 3) Total Work = 0 x W = W1+W2 = mg(h+x) – Fx = 0

  26. Example: Slowing a Bus 5.45 x 103 kg vi=75 mph=33.5 m/s vf=65 mph=29.0 m/s How much work is done slowing down a bus?

  27. F xi xf x Fbrake Example: (continued) How far does the bus travel while slowing? Fbrake = 2.75 x 104 N Area under F(x) curve:

  28. Example: (continued)

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

  30. Fspring Fspring Work Done By a Spring How much work by spring in moving from xi to xf? x=0 xi x=0 xf If |xi| < |xf| then spring takes energy away

  31. 10 cm Example: Fspring Fpull 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?

  32. 10 cm Fspring Fpull Example: The work done by the spring is: So the work I do on the spring is

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

  34. Power in 3-D For a constant force: So power will be:

  35. m = 900 kg a = 2.4 m/s2 vmax = 12 m/s Express Elevator Say a 900 kg elevator travels 135 floors (400 m) in 40 s. It accelerates at 2.4 m/s2 until it reaches a velocity of 12 m/s. Find the average power (going up) W = mgh = (900 kg)(9.8 m/s2)(400 m) = 3.5 x 106 J

  36. m = 900 kg F a = 2.4 m/s2 a vmax = 12 m/s Express Elevator(continued) Maximum Instantaneous Power: Maximum just when elevator reaches cruising speed going up P = Fv F – mg = ma F = m(g+a) P = (900 kg)(9.8 m/s2+2.4 m/s2)(12 m/s) = 130,000 W

  37. 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?

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

  39. 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.

  40. 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.

  41. 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?

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