Work, Energy and Power AP style

1. Work, Energy and PowerAP style

2. Energy Energy: the currency of the universe. Everything has to be “paid for” with energy. Energy can’t be created or destroyed, but it can be transformed from one kind to another and it can be transferred from one object to another.

3. Doing WORK is one way to transfer energy from one object to another. Work = Force x displacement W = F∙d • Unit for work is Newton x meter. One Newton-meter is also called a Joule, J.

4. Work- the transfer of energy

5. Work = Force · displacement • Work is not done unless there is a displacement. • If you hold an object a long time, you may get tired, but NO work was done on the object. • If you push against a solid wall for hours, there is still NO work done on the wall.

6. For work to be done, the displacement of the object must be along the same direction as the applied force. They must be parallel. • If the force and the displacement are perpendicular to each other, NO work is done by the force.

7. F d • For example, in lifting a book, the force exerted by your hands is upward and the displacement is upward- work is done. • Similarly, in lowering a book, the force exerted by your hands is still upward, and the displacement is downward. • The force and the displacement are STILL parallel, so work is still done. • But since they are in opposite directions, now it is NEGATIVE work. d F

8. F • On the other hand, while carrying a book down the hallway, the force from your hands is vertical, and the displacement of the book is horizontal. • Therefore, NO work is done by your hands. • Since the book is obviously moving, what force IS doing work??? The static friction force between your hands and the book is acting parallel to the displacement and IS doing work! d

9. Both Force and displacement are vectors So, using vector multiplication, Work = This is a DOT product- only parallel components will yield a non-zero solution. In many university texts, and sometimes the AP test, the displacement, which is the change in position, is represented by “s” and not “d” W = The solution for dot products are NOT vectors. Work is not a vector. Let’s look an example of using a dot product to calculate work…

10. An object is subject to a force given by F = 6i – 8j as it moves from the position r = -4i + 3j to the position r = i + 7j (“r” is a position vector) What work was done by this force? First find the displacement, s s = Dr = rf – ro = (i + 7j) - (-4i + 3j) = 5i + 4j Then, do the dot product W = F · s (6i – 8j) · (5i + 4j) = 30 – 32 = -2 J of work

11. Example How much work is done to push a 5 kg cat with a force of 25 N to the top of a ramp that is 7 meters long and 3 meters tall at a constant velocity? W = Force · displacement Which measurement is parallel to the force- the length of the ramp or the height of the ramp? W = 25 N x 7 m W = 175 J 7 m F = 25 N 3 m

12. Example How much work is done to carry a 5 kg cat to the top of a ramp that is 7 meters long and 3 meters tall at a constant velocity? W = Force · displacement What force is required to carry the cat? Force = weight of the cat Which is parallel to the Force vector- the length of the ramp or the height? d = height NOT length W = mg x h W = 5 kg x 10 m/s2 x 3 m W = 150 J 7 m 3 m

13. If all four ramps are the same height, which ramp would require the greatest amount of work in order to carry an object to the top of the ramp? W = F∙Dr For lifting an object, the distance, d, is the height. Because they all have the same height, the work for each is the same!

14. Your Force • And,….while carrying yourself when climbing stairs or walking up an incline at a constant velocity, only the height is used to calculate the work you do to get yourself to the top! • The force required is your weight! Vertical component of d Horizontal component of d

15. How much work do you do on a 30 kg cat to carry it from one side of the room to the other if the room is 10 meters long? ZERO, because your Force is vertical, but the displacement is horizontal.

16. Example Displacement = 20 m A boy pushes a lawnmower 20 meters across the yard. If he pushed with a force of 200 N and the angle between the handle and the ground was 50 degrees, how much work did he do? F cos q q F W = (F cosq )d W = (200 N cos50˚) 20 m W = 2571 J

17. NOTE: If while pushing an object, it is moving at a constant velocity, the NET force must be zero. So….. Your applied force must be exactly equal to any resistant forces like friction. S F = ma FA – f = ma = 0

18. q A 5.0 kg box is pulled 6 m across a rough horizontal floor (m = 0.4) with a force of 80 N at an angle of 35 degrees above the horizontal. What is the work done by EACH force exerted on it? What is the NET work done? ●Does the gravitational force do any work? NO! It is perpendicular to the displacement. ● Does the Normal force do any work? No! It is perpendicular to the displacement. ● Does the applied Force do any work? Yes, but ONLY its horizontal component! WF = FAcosq x d = 80cos 35˚ x 6 m = 393.19 J ● Does friction do any work? Yes, but first, what is the normal force? It’s NOT mg! Normal = mg – FAsinq Wf = -f x d = -mN∙d= -m(mg – FAsinq)∙d = -7.47 J ● What is the NET work done? 393.19 J – 7.47 J = 385.72 J Normal FA f mg

19. For work to be done, the displacement of the object must be in the same direction, as the applied force. They must be parallel. • If the force and the displacement are perpendicular to each other, NO work is done by the force. So, using vector multiplication, W = F • d (this is a DOT product!!) (In many university texts, as well as the AP test, the displacement is given by d = Dr, where r is the position vector: displacement = change in position W = F • Dr If the motion is in only one direction, W = F • Dx Or W = F • Dy

20. An object is subject to a force given by F = 6i – 8j as it moves from the position r = -4i + 3j to the position r = i + 7j (“r” is a position vector) What work was done by this force? First find the displacement, Dr rf – ro = (i + 7j) - (-4i + 3j) = Dr= 5i + 4j Then, complete the dot product W = F ·Dr (6i – 8j) · (5i + 4j) = 30 – 32 = -2J of work

21. Varying Forces The rule is… “If the Force varies, you must integrate!” If the force varies with displacement- in other words, Force is a function of displacement, you must integrate to find the work done.

22. Examples of Integration An object of mass m is subject to a force given by F(x) = 3x3 N. What is the work done by the force?

23. Examples of Definite Integration To stretch a NON linear spring a distance x requires a force given by F(x) = 4x2 – x. What work is done to stretch the spring 2 meters beyond its equilibrium position?

24. Force, N Position, m Graphing Force vs. postion • If you graph the applied force vs. the position, you can find how much work was done by the force. Work = F·d = “area under the curve”. Total Work = 2 N x 2 m + 3N x 4m = 16 J Area UNDER the x-axis is NEGATIVE work = - 1N x 2m F Net work = 16 J – 2 J = 14 J d

25. The Integral = the area under the curve! If y = f(x), then the area under the f(x) curve from x = a to x = b is equal to

26. Therefore, given a graph of Force vs. position, the work done is the area under the curve.

28. Work and Energy

29. Work and Energy Often, some force must do work to give an object potential or kinetic energy. You push a wagon and it starts moving. You do work to stretch a spring and you transform your work energy into spring potential energy. Or, you lift an object to a certain height- you transfer your energy into the object in the form of gravitational potential energy. Work = Force x distance = transfer of energy

30. Kinetic Energy the energy of motion K = ½ mv2

31. Where does K = ½ mv2 come from?? Did your teacher just amazingly, miraculously make that equation up?? Hmmm…

32. The “Work- Kinetic Energy Theorem” SWork = DK SF • d = DK = ½ mvf2 - ½ mvo2 College Board AP Objective: You are supposed to be able to derive the work-kinetic energy theorem….. (this does not mean to take a derivative- it means to start with basic principles-S SF= ma, S W = S F •d, and kinematics equations and develop the equation for kinetic energy) So…. do it!! Then you’ll see where the equation we call “kinetic energy” comes from. (Hint: start with S F = ma and use the kinematics equation that doesn’t involve time.)

33. NET Work by all forces = D Kinetic EnergySF·d = D½ mv2 How much more distance is required to stop if a car is going twice as fast (all other things remaining the same)? The work done by the forces stopping the car = the change in the kinetic energy SF·d = D½ mv2 With TWICE the speed, the car has FOUR times the kinetic energy. Therefore it takes FOUR times the stopping distance.

34. A car going 20 km/h will skid to a stop over a distance of 7 meters. If the same car was moving at 50 km/h, how many meters would be required for it to come to a stop? The velocity changed by a factor of 2.5, therefore the stopping distance is 2.52 times the original distance: 7 meters x 6.25 = 43.75 meters

35. Example, Wnet = SF∙d = DK A 500 kg car moving at 15 m/s skids 20 m to a stop. How much kinetic energy did the car lose? DK = ½ mvf2 DK = -½ (500 kg)(15 m/s) 2 DK = -56250 J What is the magnitude of the net force applied to stop the car? SF·d = DK SF = DK / d SF = 56250 J / 20 m F = 2812.5 N

36. SF∙d = D ½ mv2 A 0.02 kg bullet moving at 90 m/s strikes a block of wood. If the bullet comes to a stop at a depth of 2.5 cm inside the wood, how much force did the wood exert on the bullet? F = 3240 N

37. Example Wnet = SF∙d = DK A 500 kg car moving on a flat road at 15 m/s skids to a stop. How much kinetic energy did the car lose? DK = D½ mvf2 DK = -½ (500 kg)(15 m/s)2 DK = -56250 J How far did the car skid if the effective coefficient of friction was m = 0.6? Stopping force = friction = mN = mmg SF·d = DK -(mmg)·d = DK d = DK / (-mmg) *be careful to group in the denominator! d = 56250 J / (0.6 · 500 kg · 9.8 m/s2) = 19.13 m

39. First, a review of Conservative Forces and Potential Energy from the notes… What we have covered so far- the College Board Objectives state that the student should know: The two definitions of conservative forces and why they are equivalent. Two examples each of conservative and non-conservative forces. The general relationship between a conservative force and potential energy. Calculate the potential energy if given the force. Calculate the force if given the potential energy.

40. Conservation of Mechanical Energy

41. Mechanical Energy Mechanical Energy = Kinetic Energy + Potential Energy E = ½ mv2 + U

42. Potential Energy Stored energy It is called potential energy because it has the potential to do work.

43. Example 1: Spring potential energy in the stretched string of a bow or spring or rubber band. Us= ½ kx2 • Example 2: Chemical potential energy in fuels- gasoline, propane, batteries, food! • Example 3: Gravitational potential energy- stored in an object due to its position from a chosen reference point.

44. Gravitational potential energy Ug= weight x height Ug= mgh

45. The Ugmay benegative. For example, if your reference point is the top of a cliff and the object is at its base, its “height” would be negative, so mgh would also be negative. • The Ugonly depends on the weight and the height, not on the path that it took to get to that height.

46. “Conservative” forces - mechanical energy is conserved if these are the only forces acting on an object. The two main conservative forces are: Gravity, spring forces “Non-conservative” forces - mechanical energy is NOT conserved if these forces are acting on an object. Forces like kinetic friction, air resistance (which is really friction!)

47. Conservation of Mechanical Energy If there is no kinetic friction or air resistance, then the total mechanical energy of an object remains the same. If the object loses kinetic energy, it gains potential energy. If it loses potential energy, it gains kinetic energy. For example: tossing a ball upward

48. Conservation of Mechanical Energy The ball starts with kinetic energy… Which changes to potential energy…. Which changes back to kinetic energy PE = mgh What about the energy when it is not at the top or bottom? E = ½ mv2 + mgh Energybottom = Energytop ½ mvb2 = mght K = ½ mv2 K = ½ mv2

49. Examples • dropping an object • box sliding down an incline • tossing a ball upwards • a pendulum swinging back and forth • A block attached to a spring oscillating back and forth First, let’s look at examples where there is NO friction and NO air resistance…..

50. Conservation of Mechanical Energy If there is no friction or air resistance, set the mechanical energies at each location equal. Remember, there may be BOTH kinds of energy at any location. And there may be more than one form of potential energy, U! E1 = E2 U1 + ½mv12 = U2 + ½ mv22