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

2. Physics and Work
In this module, we will learn a measurable definition of work as the product of force and distance.

3. Objectives: After completing this module, you should be able to:
Describe work in terms of force and displacement, using the definition of the scalar product.
Solve problems involving concept of work.
Distinguish between the resultant work and the work of a single force.
Define the spring constant and calculate the work done by a varying spring force.

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

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

6. Work = Force component X displacement
Definition of Work
Work is a scalar quantity equal to the product of the displacement x and the component of the force Fx in the direction of the displacement.
Work = Force component X displacement
Work = Fx x

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

8. Negative Work x f 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

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

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

11. Work of a Force at an Angle
x = 12 m
F = 70 N
60o
Work = Fx x
Work = (F cos ) x
Work = (70 N) Cos 600 (12 m) = 420 J
Only the x-component of the force does work!
Work = 420 J

12. Procedure for Calculating Work
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 q x n mg x
3. Find work of a single force from formula.
4. Resultant work is work of resultant force.

13. Note: Work is positive since Fx and x are in the same direction.
Example 1: A lawn mower is pushed a horizontal distance of 20 m by a force of 200 N directed at an angle of 300 with the ground. What is the work of this force?
300
x = 20 m
F = 200 N F
Work = (F cos q ) x
Note: Work is positive since Fx and x are in the same direction.
Work = (200 N)(20 m) Cos 300
Work = 3460 J

14. n x P 1. Draw sketch and find given values. +x x mg 8 m P fk
Example 2: A 40-N force pulls a 4-kg block a horizontal distance of 8 m. The rope makes an angle of 350 with the floor and uk = 0.2. What is the work done by each acting on block? x P q
1. Draw sketch and find given values.
P = 40 N; x = 8 m, uk = 0.2; q = 350; m = 4 kg +x
40 N
350 x n mg
8 m P fk
2. Draw free-body diagram showing all forces. (Cont.)
Work = (F cos ) x

15. Example 2 (Cont.): Find Work Done by Each Force.
350 x n
W = mg
8 m P fk
P = 40 N; x = 8 m, uk = 0.2; q = 350; m = 4 kg
4. First find work of P.
Work = (P cos ) x
WorkP = (40 N) cos 350 (8 m) = 262 J
5. Next consider normal force n and weight W.
Each makes a 900 angle with x, so that the works are zero. (cos 900=0):
WorkP = 0 Workn = 0

16. 6. Next find work of friction.
Example 2 (Cont.): +x
40 N
350 x n
W = mg
8 m P fk
P = 40 N; x = 8 m, uk = 0.2; q = 350; m = 4 kg
WorkP = 262 J
Workn = WorkW = 0
Recall: fk = mk n
6. Next find work of friction.
n + P cos 350 – mg = 0;
n = mg – P cos 350
n = (4 kg)(9.8 m/s2) – (40 N)sin 350 = 16.3 N
fk = mk n = (0.2)(16.3 N);
fk = 3.25 N

17. n +x x 8 m P fk WorkP = 262 J Workn = WorkW = 0
Example 2 (Cont.): +x
40 N
350 x n
W = mg
8 m P fk
WorkP = 262 J
Workn = WorkW = 0
6. Work of friction (Cont.)
fk = 3.25 N; x = 8 m
Workf = (3.25 N) cos 1800 (8 m) = J
Note work of friction is negative cos 1800 = -1
7. The resultant work is the sum of all works:
262 J – 26 J
(Work)R = 236 J

18. First find magnitude of x from trigonometry:
Example 3: What is the resultant work on a 4-kg block sliding from top to bottom of the 300 inclined plane? (h = 20 m and mk = 0.2) h
300 n f mg x
Net work = S (works)
Find the work of 3 forces.
Work = (F cos q) x
First find magnitude of x from trigonometry: h x
300

19. Example 3(Cont. ): What is the resultant work on 4-kg block
Example 3(Cont.): What is the resultant work on 4-kg block? (h = 20 m and mk = 0.2)
600 mg x h
300 n f mg
x = 40 m
1. First find work of mg.
mg cos q
2. Draw free-body diagram
Work = mg(cos q) x
Work = (4 kg)(9.8 m/s2)(40 m) Cos 600
Work done by weight mg
Positive Work
Work = 784 J

20. Example 3 (Cont. ): What is the resultant work on 4-kg block
Example 3 (Cont.): What is the resultant work on 4-kg block? (h = 20 m and mk = 0.2) h
300 n f mg r
3. Next find work of friction force f which requires us to find n.
4. Free-body diagram: n f mg
mg cos 300
300
n = mg cos 300= (4)(9.8)(0.866)
n = 33.9 N
f = mk n
f = (0.2)(33.9 N) = 6.79 N

21. Example 3 (Cont. ): What is the resultant work on 4-kg block
Example 3 (Cont.): What is the resultant work on 4-kg block? (h = 20 m and mk = 0.2) h
300 n f mg r
5. Find work of friction force f using free-body diagram
Work = (f cos q) x
Work = (6.79 N)(20 m)(cos 1800) f x
1800
Work = (272 J)(-1) = -272 J
Note: Work of friction is Negative.
What work is done by the normal force n?
Work of n is 0 since it is at right angles to x.

22. Example 3 (Cont. ): What is the resultant work on 4-kg block
Example 3 (Cont.): What is the resultant work on 4-kg block? (h = 20 m and mk = 0.2) h
300 n f mg r
Net work = S (works)
Weight: Work = J
Friction: Work = J
Force n: Work = 0 J
Resultant Work = 512 J
Note: Resultant work could have been found by multiplying the resultant force by the net displacement down the plane.

23. Graph of Force vs. Displacement
Assume that a constant force F acts through a parallel displacement Dx.
Force, F
The area under the curve is equal to the work done. F x1 x2
Area
Work = F(x2 - x1)
Displacement, x

24. 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?
40 N
Force, F
Displacement, x
1 m
4 m
Area
Work = F(x2 - x1)
Work = (40 N)(4 m - 1 m)
Work = 120 J

25. 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 F x

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

27. Work Done in Stretching a Spring
Work done ON the spring is positive; work BY the spring is negative. F x m
From Hooke’s law: F = kx
Work = Area of Triangle x F
Area = ½ (base)(height) = ½ (x)(Favg ) = ½ x(kx)
Work = ½ kx2

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

29. 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/s2) = 39.2 N
Now, from Hooke’s law, the force constant k of the spring is:
k = = DF Dx
39.2 N
0.2 m
k = 196 N/m

30. 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
Note: The work to stretch an additional 30 cm is greater due to a greater average force.

31. General Case for Springs:
If the initial displacement is not zero, the work done is given by: x1 x2 F x1 m x2

32. Summary F Work = Fx 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 Fx in the direction of the displacement.

33. Procedure for Calculating Work
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 q x n mg
3. Find work of a single force from formula.
4. Resultant work is work of resultant force.

34. Important Points for Work Problems:
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.

35. Summary For Springs Hooke’s Law: x F = -kx Spring Constant: F m
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.

36. Work to Stretch a Spring:
Summary (Cont.) x1 x2 F x1 m x2
Work = ½ kx2
Work to Stretch a Spring:

37. Springs: Positive/Negative Workx m +
Compressing
Stretching
Two forces are always present: the outside force Fout ON spring and the reaction force Fs BY the spring.
Compression: Fout does positive work and Fs does negative work (see figure).
Stretching: Fout does positive work and Fs does negative work (see figure).

38. CONCLUSION: Chapter 8A - Work