2. Chapter 7.1 and 7.7 Work and Power

3. What is energy? Energy is defined as the ability to do work.Energy is defined as the ability to do work. But in some circumstances not all energy is available to do work.But in some circumstances not all energy is available to do work. Because of the association of energy with work, we begin the chapter with a discussion of work.Because of the association of energy with work, we begin the chapter with a discussion of work. What is a work?What is a work? What it means to DO Work?What it means to DO Work?

4. Objectives: Describe work in terms of force and displacement, using the definition of the scalar product.Describe work in terms of force and displacement, using the definition of the scalar product. Explain how relative directions of force and displacement of an object determine whether the work done on the object is positive, negative, or zero.Explain how relative directions of force and displacement of an object determine whether the work done on the object is positive, negative, or zero. Able to do graphical analysis on a force- distance curveAble to do graphical analysis on a force- distance curve

5. What it means to DO WORK The scientific definition of work differs in some ways from its everyday meaning.The scientific definition of work differs in some ways from its everyday meaning. The scientific definition of work reveals its relationship to energy —whenever work is done, energy is transferred.The scientific definition of work reveals its relationship to energy —whenever work is done, energy is transferred.

6. Three things are necessary for the performance of work: F  F  x There must be an applied force F.There must be an applied force F. There must be a displacement x.There must be a displacement x. The force must have a component along the displacement.The force must have a component along the displacement.

7. If a force does not affect displacement, it does no work. F W The force F exerted on the pot by the man does work. The earth exerts a force W on pot, but does no work even though there is displacement.

8. Definition of Work Work is a scalar quantity equal to the product of the displacement x and the component of the force F x in the direction of the displacement. Work = Force component X displacement Unit: [N∙m]=[J] (Joule)

9. Work can be (+), (-), and zero. If (+) work is done on object, that objects gains energy. (-) work being done removes energy from an object. Work is sometimes defined as the transfer of energy; Energy is defined as the ability to do work.

10. Positive Work F x Force F contributes to displacement x. Example: If F = 40 N and x = 4 m, then Work Work = (40 N)(4 m) = 160 N  m Work = 160 J 1 N  m = 1 Joule (J)

11. Negative Work f x The friction force f opposes the displacement. Example: If f = -10 N and x = 4 m, then Work = (-10 N)(4 m) = - 40 J Work = - 40 J

12. Resultant work is the algebraic sum of the individual works of each force. Example:F = 40 N, f = -10 N and x = 4 m Example: F = 40 N, f = -10 N and x = 4 m Work = (40 N)(4 m) + (-10 N)(4 m) Work = 120 J Resultant Work or Net Work F x f

13. Resultant work is also equal to the work of the RESULTANT force. Example: Example: Work = (F - f) x Work = (40 - 10 N)(4 m) Work = 120 J 40 N 4 m4 m -10 N Resultant Work (Cont.)

14. Work of a Force at an Angle x = 12 m F = 70 N 60 o Work = F x x Work = (F cos  ) x Work = (70 N) Cos 60 0 (12 m) = 420 J Work = 420 J Only the x-component of the force does work!

15. Work = Zero, when? Centripetal forces do no work, they are always perpendicular to the direction of motion. v FcFc Cos 90=zero

16. Quiz You lift a book with your hand in such a way that it moves up at constant speed. While it is moving, what is the total work done on the book? 1.m*g*d 2.F (hand)*d 3.(F+mg)*d 4.Zero 5.None of the avobe Answer: 4. The total work is zero since the net force acting on the book is zero. The work done by the hand is positive, while the work done by gravity is negative. The sum of the two is zero. Note that the KE of the book does not change either!

17. Example 1: A lawn mower is pushed a horizontal distance of 20 m by a force of 200 N directed at an angle of 30 0 with the ground. What is the work of this force? 30 0 x = 20 m F = 200 N Work = (F cos  ) x Work = (200 N)(20 m) Cos 30 0 Work = 3460 J Note: Work is positive since F x and x are in the same direction. F

18. Example 2: A 40-N force pulls a 4-kg block a horizontal distance of 8 m. The rope makes an angle of 35 0 with the floor and µ k = 0.2. What is the work done by each acting on block? 1. Draw sketch and given values 1. Draw sketch and given values. x P  P = 40 N; x = 8 m, µ k = 0.2;  = 35 0 ; m = 4 kg 2. Draw free-body diagram showing all forces. (Cont.) Work = (F cos  ) x +x 40 N   x n mg 8 m P fkfk

19. Example 3: What is the resultant work on a 4-kg block sliding from top to bottom of the 30 0 inclined plane? (h = 20 m and  k = 0.2) Work = (F cos  ) x h 30 0 n f mg x Net work =  (works) Find the work of 3 forces. First find magnitude of x from trigonometry: h x 30 0

20. Example 3(Cont.): What is the resultant work on 4-kg block? (h = 20 m and  k = 0.2) h 30 0 n f mg x = 40 m 1. First find work of mg. Work = (4 kg)(9.8 m/s 2 )(20 m) Cos 60 0 Work = 784 J Positive Work Work = mg(cos  ) x 60 0 mg x 2. Draw free- body diagram Work done by weight mg mg cos 

21. Example 3 (Cont.): What is the resultant work on 4-kg block? (h = 20 m and  k = 0.2) h 30 0 n f mg r 3. Next find work of friction force f which requires us to find n. 4. Free-body diagram: nf mg mg cos 30 0 30 0 n = mg cos 30 0 = (4)(9.8)(0.866) n = 33.9 N f =  k n f = (0.2)(33.9 N) = 6.79 N

22. Example 3 (Cont.): What is the resultant work on 4-kg block? (h = 20 m and  k = 0.2) 5. Find work of friction force f using free-body diagram Work = (6.79 N)(40 m)(cos 180 0 ) Work = (f cos  ) x Work = (272 J)(-1) = -272 J Note: Work of friction is Negative. f x 180 0 What work is done by the normal force n ? h 30 0 n f mg r Work of n is 0 since it is at right angles to x.

23. Example 3 (Cont.): What is the resultant work on 4-kg block? (h = 20 m and  k = 0.2) Net work =  (works) Weight: Work = + 784 J Force n : Work = 0 J Friction: Work = - 272 J Resultant Work = 512 J h 30 0 n f mg r Note: Resultant work could have been found by multiplying the resultant force by the net displacement down the plane.

24. Graph of Force vs. Displacement Assume that a constant force F acts through a parallel displacement  x. Force, F Displacement, x F x1x1x1x1 x2x2x2x2 The area under the curve is equal to the work done. Work = F(x 2 - x 1 ) Area

25. Example for Constant Force What work is done by a constant force of 40 N moving a block from x = 1 m to x = 4 m? Work = F(x 2 - x 1 ) 40 N Force, F Displacement, x 1 m 4 m Area Work = (40 N)(4 m - 1 m) Work = 120 J

26. Work of a Varying Force Our definition of work applies only for a constant force or an average force. What if the force varies with displacement as with stretching a spring or rubber band? F x Fx

27. Hooke’s Law When a spring is stretched, there is a restoring force that is proportional to the displacement. F = -kx The spring constant k is a property of the spring given by: K = FxFx F x m

28. Work Done in Stretching a Spring F x m Work done ON the spring is positive; work BY the spring is negative. From Hooke’s law: F = kx x F Work = Area of Triangle Area = ½ (base)(height) = ½ (x)(F avg ) = ½ x(kx) Work = ½ kx 2

29. Compressing or Stretching a Spring Initially at Rest: Two forces are always present: the outside force F out ON spring and the reaction force F s BY the spring. Compression: F out does positive work and F s does negative work (see figure). Stretching: F out does positive work and F s does negative work (see figure). x m x m Compressing Stretching

30. General Case for Springs: If the initial displacement is not zero, the work done is given by: x1x1 x2x2 F x1x1 m x2x2 m

31. Power Power is defined as the rate at which work is done: (P = dW/dt )

32. Units of Power 1 W = 1 J/s and 1 kW = 1000 W One watt (W) is work done at the rate of one joule per second. One horsepower is work done at the rate of 550 ft lb/s. ( 1 hp = 550 ft lb/s = 746 W)

33. Example 4: A 100-kg cheetah moves from rest to 30 m/s in 4 s. What is the power? Recognize that work is equal to the change in kinetic energy: m = 100 kg Power Consumed: P = 1.22 kW

34. Summary x xx x F 60 o Work = F x x Work = (F cos  ) x Work is a scalar quantity equal to the product of the displacement x and the component of the force F x in the direction of the displacement.

35. 1. Draw sketch and establish what is given and what is to be found. Procedure for Calculating Work 2. Draw free-body diagram choosing positive x-axis along displacement. Work = (F cos  ) x +F x n mg 3. Find work of a single force from formula. 4. Resultant work is work of resultant force.

36. 1. Always draw a free-body diagram, choosing the positive x-axis in the same direction as the displacement. Important Points for Work Problems: 2. Work is negative if a component of the force is opposite displacement direction 3. Work done by any force that is at right angles with displacement will be zero (0). 4. For resultant work, you can add the works of each force, or multiply the resultant force times the net displacement.

37. Summary For Springs F x m Hooke’s Law: F = -kx Spring Constant: The spring constant is the force exerted BY the spring per unit change in its displacement. The spring force always opposes displacement. This explains the negative sign in Hooke’s law.

38. Summary (Cont.) Work = ½ kx 2 Work to Stretch a Spring: x1x1 x2x2 F x1x1 m x2x2 m

39. Springs: Positive/Negative Work x m x m + Compressing Stretching Two forces are always present: the outside force F out ON spring and the reaction force F s BY the spring. Compression: F out does positive work and F s does negative work (see figure). Stretching: F out does positive work and F s does negative work (see figure).