1. Energy Introduction Lesson 1

2. Definition of energy When energy is transferred from one form to another it may be transferred by doing work. For example, when you lift an object you do work by transferring chemical energy to kinetic energy and gravitational potential energy. This concept of work gives us a way of defining energy Energy is defined as the stored ability to do work

3. Work When a force acts upon an object to cause a displacement of the object, it is said that work was done upon the object. There are three key ingredients to work - force, displacement, and cause. In order for a force to qualify as having done work on an object, there must be a displacement and the force must cause the displacement.

4. Everyday Examples a horse pulling a plow through the field, a father pushing a grocery cart down the aisle of a grocery store, a student lifting a backpack full of books upon her shoulder, a weightlifter lifting a barbell above his head, an Olympian launching the shot-put, etc. In each case described here there is a force exerted upon an object to cause that object to be displaced.

5. Activity (in small groups) Read the five statements on the cards and determine whether or not they represent examples of work. Write down the reasons for your answers We will then discuss the answers as a class

6. A man applies a force to a wall and becomes exhausted No. This is not an example of work. The wall is not displaced. A force must cause a displacement in order for work to be done.

7. A book falls off a table and free falls to the ground Yes. This is an example of work. There is a force (gravity) which acts on the book which causes it to be displaced in a downward direction (i.e., "fall").

8. A waiter carries a tray full of meals above his head by one arm straight across the room at constant speed. No. This is not an example of work. There is a force (the waiter pushes up on the tray) and there is a displacement (the tray is moved horizontally across the room). Yet the force does not cause the displacement. To cause a displacement, there must be a component of force in the direction of the displacement.

9. A rocket accelerates through space. Yes. This is an example of work. There is a force (the expelled gases push on the rocket) which causes the rocket to be displaced through space.

10. Mathematically, work can be expressed by the following equation: W = F x d x Cos θ where F is the force, d is the displacement, and the angle (theta) is defined as the angle between the force and the displacement vector.

11. How do we find the angle, θ? The angle measure is defined as the angle between the force and the displacement. To gather an idea of it's meaning, consider the following three scenarios:

12. A force acts rightward upon an object as it is displaced rightward. In such an instance, the force vector and the displacement vector are in the same direction. Thus, the angle between F and d is 0 degrees.

13. A force acts upward on an object as it is displaced rightward. In such an instance, the force vector and the displacement vector are at right angles to each other. Thus, the angle between F and d is 90 degrees.

14. A force acts leftward upon an object that is displaced rightward. In such an instance, the force vector and the displacement vector are in the opposite direction. Thus, the angle between F and d is 180 degrees. An example of negative work

15. Negative Work Sometimes a force acts upon a moving object to hinder a displacement. Examples might include: – a car skidding to a stop on a roadway surface –baseball runner sliding to a stop on the infield dirt. In such instances, the force acts in the direction opposite the objects motion in order to slow it down. The force doesn't cause the displacement but rather hinders it. The negative refers to the numerical value that results when values of F, d and theta are substituted into the work equation. Since the force vector is directly opposite the displacement vector, theta is 180 degrees. The cosine(180 degrees) is -1 and so a negative value results for the amount of work done upon the object.

16. Which path needs the most energy? link to animation

17. Units of Work For work (and also energy), the standard metric unit is the Joule (abbreviated J). One Joule is equivalent to one Newton of force causing a displacement of one meter. In other words: 1 Joule = 1 Newton * 1 meter 1 J = 1 N * m

18. Consider an Arctic explorer dragging a sledge across a frozen lake. The explorer attaches the rope to his waist and the force of 200 N is applied at 30 o to the horizontal. We can model this situation in the diagram below. How much work is done by the explorer in dragging the sledge 150 metres across the ice at a steady speed? Worked Example

19. The difficulty here is that the force, F (200 N), and the displacement, s, are not in the samedirection. One solution is to resolve the 200 N force into its vertical and horizontal components as shown below. Provided the sledge does not rise above the ice, no work is done by the vertical force of 100 N. Work is done only by the horizontal component of the applied force (173.2 N). Work done = constant force × distance moved in direction of the force = 173.2 N × 150 m = 25 980 J

20. The other way of solving this is to use the general formula: W = Fd cos θ To apply this formula to the sledge example above we would write: W = Fd cos θ = 200 × 150 × cos 30 o = 200 × 129.9 = 25 980 J

21. Check Your Understanding Show your understanding of the concept and mathematics of work by answering the questions on the worksheet

22. Q1 Diagram A Answer: W = (100 N) * (5 m)* cos(0 degrees) = 500 J The force and the displacement are given in the problem statement. It is said (or shown or implied) that the force and the displacement are both rightward. Since F and d are in the same direction, the angle is 0 degrees.

23. Diagram B Answer: W = (100 N) * (5 m) * cos(30 degrees) = 433 J The force and the displacement are given in the problem statement. It is said that the displacement is rightward. It is shown that the force is 30 degrees above the horizontal. Thus, the angle between F and d is 30 degrees.

24. Diagram C Answer: W = (147 N) * (5 m) * cos(0 degrees) = 735 J The displacement is given in the problem statement. The applied force must be 147 N since the 15-kg mass (Fgrav=147 N) is lifted at constant speed. Since F and d are in the same direction, the angle is 0 degrees.

25. Answers to Q2 a) Only Fapp does work. Fgrav and Fnorm do not do work since a vertical force cannot cause a horizontal displacement. Wapp= (10 N) * (5 m) *cos (0 degrees) =+50 Joules b) Only Ffrict does work. Fgrav and Fnorm do not do work since a vertical force cannot cause a horizontal displacement Wfrict =(10 N) * (5 m) * cos (180 degrees) =-50 Joules c) Fapp and Ffrict do work. Fgravand Fnorm do not do work since a vertical force cannot cause a horizontal displacement. Wapp = (10 N) * (5 m) * cos (0 deg) =+50 Joules Wfrict = (10 N) * (5 m) * cos (180 deg) =-50 Joules

26. Answer to question 3 Be careful! If the F is parallel to the incline and the d is parallel to the incline, then the angle theta in the work equation is 0 degrees. In each case, the work is approximately1.18 x106 Joules. The angle does not affect the amount of work done on the roller coaster car.

27. Answer Q4 The motion has two parts: pulling vertically to displace the suitcase vertically (angle = 0 degrees) and pushing horizontally to displace the suitcase horizontally (angle = 0 degrees). For the vertical part, W = (200 N) * (10 m) * cos (0 deg) = 2000 J. For the horizontal part, W = (50 N) * (35 m) * cos (0 deg) = 1750 J. The total work done is 3750 J (the sum of the two parts).

28. Answers Q5: W = F * d * cos θ W = (50 N) * (3 m) * cos (30 degrees) =129.9 Joules Q6: To lift a 15-Newton block at constant speed, 15-N of force must be applied to it (Newton's laws). Thus, W = (15 N) * (3 m) * cos (0 degrees) =45 Joules