2. Chapter 8A. Work A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University © 2007

3. 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.

4. 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.

5. 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 Work = F x x

6. 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)

7. 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

8. 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

9. 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.)

10. 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!

11. 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.

12. 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

13. 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 u k = 0.2. What is the work done by each acting on block? 1. Draw sketch and find given values 1. Draw sketch and find given values. x P  P = 40 N; x = 8 m, u 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

14. Example 2 (Cont.): Find Work Done by Each Force. +x 40 N   x n W = mg 8 m P fkfk Work = (P cos  ) x P = 40 N; x = 8 m, u k = 0.2;  = 35 0 ; m = 4 kg 4. First find work of P. Work P = (40 N) cos 35 0 (8 m) = 262 J 5. Next consider normal force n and weight W. Each makes a 90 0 angle with x, so that the works are zero. (cos 90 0 =0): Work P = 0 Work n = 0

15. Example 2 (Cont.): 6. Next find work of friction. +x 40 N   x n W = mg 8 m P fkfk P = 40 N; x = 8 m, u k = 0.2;  = 35 0 ; m = 4 kg Work P = 262 J Work n = Work W = 0 Recall: f k =  k n n + P cos 35 0 – mg = 0; n = mg – P cos 35 0 n = (4 kg)(9.8 m/s 2 ) – (40 N)sin 35 0 = 16.3 N f k =  k n = (0.2)(16.3 N); f k = 3.25 N

16. Example 2 (Cont.): +x 40 N   x n W = mg 8 m P fkfk Work P = 262 J Work n Work n = Work W = 0 6. Work of friction (Cont.) f k = 3.25 N; x = 8 m Work f = (3.25 N) cos 180 0 (8 m) = -26.0 J Note work of friction is negative cos 180 0 = -1 7. The resultant work is the sum of all works: 262 J + 0 + 0 – 26 J (Work) R = 236 J

17. 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

18. 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 )(40 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 

19. 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

20. 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)(20 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.

21. 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.

22. 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

23. 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

24. 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

25. 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

26. 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

27. 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

28. Example 4: A 4-kg mass suspended from a spring produces a displacement of 20 cm. What is the spring constant? F 20 cm m The stretching force is the weight (W = mg) of the 4-kg mass: F = (4 kg)(9.8 m/s 2 ) = 39.2 N Now, from Hooke’s law, the force constant k of the spring is: k = = FxFx  0.2 m k = 196 N/m

29. Example 5: What work is required to stretch this spring (k = 196 N/m) from x = 0 to x = 30 cm? Work = ½(196 N/m)(0.30 m) 2 Work = 8.82 J F 30 cm Note: The work to stretch an additional 30 cm is greater due to a greater average force.

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. 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.

32. 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.

33. 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.

34. 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.

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

36. 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).

37. CONCLUSION: Chapter 8A - Work