12/12 do now

1. 12/12 do now • An 8. kg ball is fired horizontally from a 2.0 x 103 kg cannon initially at rest. After having been fired, the momentum of the ball is 2.40 x 103 kg·m/s east (neglect friction.]. Calculate the magnitude of the cannon’s velocity after the ball is fired. [show all work]

2. Work, Energy, and Power - Chapter Outline Lesson 1: Basic Terminology and Concepts Work Definition and Mathematics of Work Calculating the Amount of Work Done by Forces Potential Energy Kinetic Energy Mechanical Energy Power Lesson 2: The Work-Energy Theorem Internal vs. External Forces The Work-Energy Connection Analysis of Situations Involving External Forces Analysis of Situations in Which Mechanical Energy is Conserved Application and Practice Questions Bar Chart Illustrations

3. Objective - Work • Definition and Mathematics of Work • Calculating the Amount of Work Done by Forces • Homework – castle learning

4. Definition and Mathematics of Work force • In physics, work is defined as a _________ acting upon an object to ____________ a __________________. • In order for a force to qualify as having done work on an object, there must be a displacement and the force must ___________ the displacement cause displacement cause

5. Work is done

6. Work is not done

7. Let’s practice – work or no work • A student applies a force to a wall and becomes exhausted. • A calculator falls off a table and free falls to the ground. • A waiter carries a tray full of beverages above his head by one arm across the room • A rocket accelerates through space. no work work no work work

8. F θ d Calculating the Amount of Work Done by Forces • F - is the force in Newton, which causes the displacement of the object. • d - is the displacement in meters • θ = angle between force and displacement • W - is work in N∙m or Joule (J). 1 J = 1 N∙m = 1 kg∙m2/s2 • Work is a _____________ quantity • Work is independent of time the force acts on the object. W = F∙d∙cosθ scalar

9. W = F∙d∙cosθ Work done – positive, negative or zero work Positive work negative work - force acts in the direction opposite the objects motion in order to slow it down. no work

10. To Do Work, Forces Must Cause Displacements W = F∙d∙cosθ = 0

11. Only the horizontal component of the force (Fcosθ) causes a horizontal displacement.

12. The angle in work equation • It is important to recognize that the angle has a precise definition. It is the angle between the force and the displacement vectors. F & d are in the same direction, θ is 0o. d F

13. Each path up to the seat top requires the same amount of work. The amount of work done by a force on any object is given by the equation W = F∙d∙cosθ where F is the force, d is the displacement and θ is the angle between the force and the displacement vector. In all three cases, θ equals to 0o

14. example • A 20.0 N force is used to push a 2.00 kg cart a distance of 5.00 meters. Determine the amount of work done on the cart by the force. Given: F = 20.0 N; d = 5.00 meters; m = 2.00 kg; θ = 0 Unknown: W = ? W = F∙dcosθ W = (20.0 N)(5.00 m)cos0o W = 100. J 20.0 N

15. example • How much work is done in lifting a 5.0 kg box from the floor to a height of 1.2 m above the floor? Given: d = h = 1.2 meters; m = 5.0 kg; θ = 0 Unknown: W = ? W = F∙dcosθ F = mg = (5.0 kg)(9.81 m/s2) cos0o = 49 N W = F∙d = (49 N) (1.2 m) = 59 J

16. example • A 2.3 kg block rests on a horizontal surface. A constant force of 5.0 N is applied to the block at an angle of 30.o to the horizontal; determine the work done on the block a distance of 2.0 meters along the surface. • Given: F = 5.0 N; m = 2.3 kg d = 2.0 m θ= 30o 5.0 N 30o 2.3 kg • unknown: • W = ? J • Solve: • W = F∙d∙cosθ • W = (5.0 N)(2.0 m)(cos30o) = 8.7 J

17. practice • Matt pulls block along a horizontal surface at constant velocity. The diagram show the components of the force exerted on the block by Matt. Determine how much work is done against friction. • Given: Fx = 8.0 N Fy = 6.0 N dx = 3.0 m 6.0 N • unknown: W = ? J W = Fxdx W = (8.0 N)(3.0 m) = 24 J 8.0 N 3.0 m

18. example • A neighbor pushes a lawnmower four times as far as you do but exert only half the force, which one of you does more work and by how much? Wyou = Fd Wneighbor = (½ F)(4d) = 2 Fd = 2 Wyou The neighbor, twice as much

19. Force vs. displacement graph • The area under a force versus displacement graph is the work done by the force. Example: a block is pulled along a table with 10. N over a distance of 1.0 m. W = Fd = (10. N)(1.0 m) = 10. J work Force (N) Displacement (m) height base area

20. Class work • Work practice

21. 12/13 do now • Work, energy, and power packet – • page 1-4

22. objectives • Describe potential energy • gravitational potential energy • Hooke’s Law • Elastic potential energy • Homework – castle learning • Reminder: the momentum and collisions packet was due last Friday. 80% credit if hand in today. Last day of collection - Friday 12/16 for 50% credit • Food drive ends 12/16 – every two items for one extra credit on your average

23. Potential energy • An object can store energy as the result of its position. ________________________ is the stored energy of position possessed by an object. • Two form: • Gravitational • Elastic Potential energy

24. Gravitational potential energy vertical position (height). • Gravitational potential energy is the energy stored in an object as the result of its _________________________ • The energy is stored as the result of the _____________ attraction of the Earth for the object. • The work done in raising an object must result in an increase in the object's _______________________ • The gravitational potential energy of the box is dependent on two variables: • The mass of the object • The height of the object • Equation: ______________________ • m: mass, in kilograms • h: height, in meters • g: acceleration of gravity = 9.81 m/s2 gravitational gravitational potential energy PEgrav = m∙g∙h

25. GPE • GPE = mgh • The equation shows that . . . • . . . the more gravitational potential energy it’s got. • the more mass a body has • or the stronger the gravitational field it’s in • or the higher up it is

26. Unit of energy • The unit of energy is the same as work: _______ • 1 joule = 1 (kg)∙(m/s2)∙(m) = 1 Newton ∙ meter • 1 joule = 1 (kg)∙(m2/s2) Joules Work and energy has the same unit

27. Gravitational potential energy is relative • To determine the gravitational potential energy of an object, a _______ height position must first be assigned. • Typically, the ___________ is considered to be a position of zero height. • But, it doesn’t have to be: • It could be relative to the height above the lab table. • It could be relative to the bottom of a mountain • It could be the lowest position on a roller coaster zero ground

28. example • How much potential energy is gained by an object with a mass of 2.00 kg that is lifted from the floor to the top of 0.92 m high table? • Known: • m = 2.00 kg • h = 0.92 m • g = 9.81 m/s2 Solve: ∆PE = mg∆h ∆PE = (2.00 kg)(9.81m/s2)(0.92 m) = 18 J • unknown: • PE = ? J

29. The graph of gravitational potential energy vs. vertical height for an object near Earth's surface gives the weight of the object. The weight of the object is the slope of the line. Weight = 25 J/1.0 m = 25 N m = weight / g = 2.5 kg

30. Elastic potential energy • Elastic potential energy is the energy stored in ______________ materials as the result of their stretching or compressing. • Elastic potential energy can be stored in • Rubber bands • Bungee cores • Springs • trampolines elastic

31. Springs are a special instance of a device that can store elastic potential energy due to either compression or stretching. • A force is required to compress or stretch a spring; the more compression/stretch there is, the more force that is required to compress it further. • For certain springs, the amount of force is directly proportional to the amount of stretch or compression (x); the constant of proportionality is known as the spring constant (k).

32. Hooke’s Law F = kx Spring force = spring constant x displacement • F in the force needed to displace (by stretching or compressing) a spring x meters from the equilibrium (relaxed) position. The SI unit of F is Newton. • k is spring constant. It is a measure of stiffness of the spring. The greater value of k means a stiffer spring because more force is needed to stretch or compress it that spring. The Si units of k are N/m. depends on the material made up of the spring. k is in N/m • x the distance difference between the length of stretched/compressed spring and its relaxed (equilibrium) spring.

33. example • Determine the x in F = kx

34. force elongation F = kx • Spring force is directly proportional to the elongation of the spring (displacement) The slope represents spring constant: k = F / x

35. elongation force caution • Sometimes, we might see a graph such as this: The slope represents the inverse of spring constant: Slope = 1/k = x / F

36. example • Given the following data table and corresponding graph, calculate the spring constant of this spring.

37. example • A 20.-newton weight is attached to a spring, causing it to stretch, as shown in the diagram. What is the spring constant of this spring?

38. 12/14 do now • Which of the following statements are true about work? Include all that apply. • Work is a form of energy. • Units of work would be equivalent to a Newton times a meter. • A kg•m2/s2 would be a unit of work. • Work is a time-based quantity; it is dependent upon how fast a force displaces an object. • Superman applies a force on a truck to prevent it from moving down a hill. This is an example of work being done. • An upward force is applied to a bucket as it is carried 20 m across the yard. This is an example of work being done. • A force is applied by a chain to a roller coaster car to carry it up the hill of the first drop of the Shockwave ride. This is an example of work being done.

39. objectives • Hooke’s law • Potential energy • Kinetic energy • Lab – determine spring constant using Hooke’s law • Homework – castle leaning • Reminder: the momentum and collisions packet was due last Friday. 70% credit if hand in today. Last day of collection - Friday 12/16 for 50% credit • Food drive ends 12/16 – every two items for one extra credit on your average • Quiz on Friday – work, potential and kinetic energy, Hooke’s Law – questions are from castle learning assignments

40. example • A spring has a spring constant of 25 N/m.  What is the minimum force required to stretch the spring 0.25 meter from its equilibrium position?

41. example • The graph below shows elongation as a function of the applied force for two springs, A and B. Compared to the spring constant for spring A, the spring constant for spring B is • smaller • larger • the same

42. Elastic potential energy in a spring • Elastic potential energy is the Work done on the spring. PEs = Favg∙d = Favg∙x = (½ k∙x)∙x = ½ kx2 Note: F is the average force • k: spring constant • x: amount of compression or extension relative to equilibrium position PEs = ½ k∙x2

43. Elastic potential energy elongation Elastic potential energy is directly proportional to x2

44. example • A spring has a spring constant of 120 N/m.  How much potential energy is stored in the spring as it is stretched 0.20 meter? Given: k = 120 N/m x = 0.20 m Unknown: PEs = ? J Solve: PEs = ½ kx2 PEs = ½ (120 N/m)(0.20 m)2 PEs = 2.4 J

45. example • The unstretched spring in the diagram has a length of 0.40 meter and a spring constant k.  A weight is hung from the spring, causing it to stretch to a length of 0.60 meter.  In terms of k, how many joules of elastic potential energy are stored in this stretched spring? PEs = ½ kx2 PEs = ½ k(0.20 m)2 PEs = (0.020 k) J

46. example • Determine the potential energy stored in the spring with a spring constant of 25.0 N/m when a force of 2.50 N is applied to it. Solve: PEs = ½ kx2 To find x, use Fs = kx, (2.50 N) = (25.0 N/m)(x) x = 0.100 m PEs = ½ (25.0 N/m)(0.100 m)2 PEs = 0.125 J Given: Fs = 2.50 N k = 25.0 N/m Unknown: PEs = ? J

47. example • As shown in the diagram, a 0.50-meter-long spring is stretched from its equilibrium position to a length of 1.00 meter by a weight. If 15 joules of energy are stored in the stretched spring, what is the value of the spring constant? PE = ½ kx2 15 J = ½ k (0.50 m)2 k = 120 N/m

48. example • A 10.-newton force is required to hold a stretched spring 0.20 meter from its rest position. What is the potential energy stored in the stretched spring? Given: F = 10. N x = 0.20 m Unknown: PEs = ? J PEs = ½ kx2 = ½ (kx)(x) PEs = ½ Fx PEs = ½ (10. N)(.20 m) PEs = 1.0 J

49. A force of 0.2 N is needed to compress a spring a distance of 0.02 meter. What is the potential energy stored in this compressed spring? Given: F = 0.2 N x = 0.02 m Unknown: PEs = ? J PEs = ½ kx2 = ½ (kx)(x) PEs = ½ Fx PEs = ½ (0.2 N)(0.02 m) PEs = 0.002 J

50. KE = ½ mv2 Kinetic energy motion • Kinetic energy is the energy of _______. • An object which has motion - whether it be vertical or horizontal motion - has kinetic energy. • The equation for kinetic energy is: __________________ • Where KE is kinetic energy, in joules • v is the speed of the object, in m/s • m is the mass of the object, in kg