WORK

1. WORK

2. Definition We’ll say that a force applied to an object produces work when the object, due to the applied force, undergoes a displacement. In symbols: 1) W = F · d [work done by a constant force equals F dot d] // [work equals F times d] (work is force multiplied by displacement) s F This formula is used only when the force “F” and the displacement “d” have the same direction. The symbol used to indicate work in Italian is L while in English it is W(for work). F d If we measure the force in Newton and the displacement in metres, work is measured in Joule. The joule is according SI the unit of work. It is equal to the work done in applying a force of one Newton (N) and it produces a distance of one meter (m). 1 J = 1 N · 1 m Pagina 2

3. As we know, force and displacement are vectorial quantities or vectors, while work is a physical scalar quantity that can be described by a magnitude or numeric value] What is work produced by? It is a force that applied to an object produces a displacement. Example 1 What is the work produced by a force of 20 N which applied to an object, in the same direction of displacement, produces a displacement of 8 m? In symbols: W = F · d = 20 N · 8 m = 160 J Pagina 3

4. When the force applied to an object forms an angle “α” with the displacement, F work is calculated in the following way: 2) W = F1 · d where F1 is the projection of F along d. Where: F1= d· cosα So: W = F1 · d In general: W = F · d· cosα Pagina 4

5. Specific case: See these pictures: Pagina 5

6. As you can see, as the α increases the force component along the direction of the displacement, F’ decreases; also the work done, consequently, decreases to become, even, zero when alpha = 90°. i.e. when the force is perpendicular to the displacement. Therefore the work is maximum when the applied force is in the same direction of the shift (displacement) However, if the applied force forms an angle α with the direction of movement, only a part of it produces work (and the other part is wasted): it means, as the next example makes it clear, if we apply a force of 20 N on an object so as to form with the direction of the displacement an angle of 6o °, it is like we have enforced to the body only a force of 10 N (in direction of the displacement). (α = 0°). Pagina 6

7. Example 2 A force of 20 N is applied to an object and it produces a displacement of 8 meters. If the direction of the force applied forms an angle of 60° with the direction of the displacement, calculate the work done.. In this case, as you can see, work done is less than previous example. Pagina 7

8. How many kinds of works are there ? Force can facilitate motion (angle between force and displacement is: 0°≤ α < 90° (in this case we have positive work). Work ispositive or put up resistance (hindering it) when force and displacement are as picture: in this case the angle between force and displacement is: 90°< α ≤ 180° Work isnegative (in this case we have negativework). The force is hindering the displacement. Work done can be positive or negative If α = 90° Work iszero Pagina 8

9. When force and displacement are perpendicular, work done is zero; the force doesn’t produce work, in this case! W = F1 · d = 0 · d = 0 So, the term (word) “work”, that in everyday life means “fatigue”/”effort”, in the daily life of a porter or waiter , etc. that moves horizontally carryng a weight, would mean in physics no work; in this case, we can say that, in phisycs, the work hasn’t been done. In this case, the term “work” in physics has a different meaning from the term “work” in everyday life. • So, Work is 0 , when/if • F = 0 • d = 0 • F isperpendiculartodisplacement Pagina 9

10. Look at these pictures: 1) Force and displacement have the same direction (vertical): The work done by the force F to lift an object is positive (force and displacement have the same direction), while the weight force P, which is opposed to the displacement (gravity and upward displacement have opposite directions), performs a negative work . 2) The direction of the force forms an angle “α” with the direction of the displacement: positive work. 3) Force a displacement have the same direction (horizontal): positive work. 4) When an object falls from the table reaching the floor, the object falls owing to its weight force as it has the same direction of the movement (downward), and therefore the weight force does a positive work. W = P · d Pagina 10

11. Example P h If an object, having mass m, falls from a height h, the angle between the gravity force and the displacement is zero, so the work done by this force is: W = P ∙ h = P ∙ h ∙ cos 0° = mgh Pagina 11

12. Is it Work? • 1. A teacher applies a force to a wall and becomes exhausted. Answer: No, it isn’t ! This isn’t an example of work. The wall isn’t displaced. A force MUST cause a displacement in order for work to be done. • 2. A book falls off a table and free falls to the ground. Answer: Yes, it is! 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"). Pagina 12

13. 3. A waiter carries a tray full of meals above his head by one arm straight across the room at constant speed. Answer: No, it isn’t! This isn’t 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. • 4. A rocket accelerates through space. Answer: Yes, it is! 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. Pagina 13

14. How can you sum works? A boy drags a box, for a distance of two meters, applying a force of 20 N, which has the same direction of the displacement. If on the box acts a friction force of 20 N , which is the total work? • We can solve the problem in two ways: • First method or way: calculating the resultant force: • and then calculating the work produced by the resistive force: W = R ∙ s = 40N ∙ 2m = 80 J • Second method: calculating singly the work produced by each force F and Fa: W1 = F ∙ s = 60N ∙ 2m = 120 J W2 = Fa ∙ s = -20N ∙ 2m = -40 J Then we calculate the total work summing all the works that the single forces have done: W = W1 + W2 = 120J + (-40J) = 120J – 40J = 80J N.B. Pay attention to the sign : motive work= positive, resistive= negative. Page 14

15. 1. What is the overall work which is done to lift along the vertical a book per 30 cm, with a force of 10 N, when we know the weight is of 8 N ? The (motive) work done by the force F is: W1 = F ∙ s = 10N ∙ 0,3m = 3 J The (resistive) work done by the weight force P is: W2 = P ∙ s = -8N ∙ 0,3m = -2,4 J The total work is: I lavoro totale è : W = W1 + W2 =3J + (-2,4J) = 3J – 2,4J = 0,6J 2. What is the work done lifting along the vertical a book of mass 1 Kg per 1 m, to costant speed? If the book is lifted to a costant speed, it means, according to the first principle of the dynamics, the resultant of the forces is zero, that is the applied force F is equal to the weight P of the body, that is: P = m ∙ g = 1Kg ∙ 9,8 m/s2 = 9,8 N F = P = 9,8 N W2 = F ∙ d = 9,8N ∙ 1m = 9,8J W 1 = P ∙ d = 9,8N ∙ 1m = -9,8J If we want to calculate, in this case, the overall work done by all the present forces, it would be : W = R ∙ d = 0 ∙ 1m = 0 J W = W1 + W2 = -9,8 J +(9,8 J) = -9,8 J + 9,8 J = 0 Page 15

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