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Chapter 5

Chapter 5 . 5-1 work . Objectives. Recognize the difference between the scientific and ordinary definitions of work. Define work by relating it to force and displacement. Identify where work is being performed in a variety of situations.

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Chapter 5

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  1. Chapter 5 5-1 work

  2. Objectives • Recognize the difference between the scientific and ordinary definitions of work. • Define work by relating it to force and displacement. • Identify where work is being performed in a variety of situations. • Calculate the net work done when many forces are applied to an object.

  3. What is 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. There are several good examples of work that can be observed in everyday life - a horse pulling a plow through the field, a father pushing a grocery cart down the aisle of a grocery store, a freshman 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.

  4. Examples of work • Read the following five statements and determine whether or not they represent examples of work. • 1. • Solution : NO, no displacement • 2 .A book falls off a table and free falls to the ground. • Solution: yes, there is gravity acting on the object which causes displacement • 3. A waiter carries a tray full of meals above his head by one arm straight across the room at constant speed. (Careful! This is a very difficult question that will be discussed in more detail later.) • Solution: No, no displacement • 4. rocket accelerates through space. • solution: yes, displacement A teacher applies a force to a wall and becomes exhausted.

  5. Work • Work is equal to the magnitude of the force f, times the magnitude of the displacement d. • W=f*d

  6. Definition of work

  7. Work • 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. Perhaps the most difficult aspect of the above equation is the angle "theta." The angle is not just any 'ole angle, but rather a very specific angle. 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. • Scenario A: 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.   • Scenario B: 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.   • Scenario C: 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.

  8. Work

  9. Units of work • Whenever a new quantity is introduced in physics, the standard metric units associated with that quantity are discussed. In the case of 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, • The Joule is the unit of work. • 1 Joule = 1 Newton * 1 meter • 1 J = 1 N * m • In fact, any unit of force times any unit of displacement is equivalent to a unit of work. Some nonstandard units for work are shown below. Notice that when analyzed, each set of units is equivalent to a force unit times a displacement unit.

  10. The meaning of negative work • On occasion, a force acts upon a moving object to hinder a displacement. Examples might include a car skidding to a stop on a roadway surface or a 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. These situations involve what is commonly called negative work. The negative of negative work 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. Negative work will become important (and more meaningful) in Lesson 2 as we begin to discuss the relationship between work and energy.

  11. Work • In summary, work is done when a force acts upon an object to cause a displacement. Three quantities must be known in order to calculate the amount of work. Those three quantities are force, displacement and the angle between the force and the displacement. • lets Investigate

  12. Example#1 • How much work is done in a vacuum cleaner pulled 3m by a force of 50 N at an angle 30 degree above the horizontal?

  13. Example#2 • Ben Travlun carries a 200-N suitcase up three flights of stairs (a height of 10.0 m) and then pushes it with a horizontal force of 50.0 N at a constant speed of 0.5 m/s for a horizontal distance of 35.0 meters. How much work does Ben do on his suitcase during this entire motion? • Solution • 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).

  14. Example#3 • A force of 50 N acts on the block at the angle shown in the diagram. The block moves a horizontal distance of 3.0 m. How much work is done by the applied force? • Solution • W = F * d * cos(Theta) W = (50 N) * (3 m) * cos (30 degrees) = 129.9 Joules

  15. Example#4 • How much work is done by an applied force to lift a 15-Newton block 3.0 meters vertically at a constant speed? • solution 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

  16. Example#5 • A tired squirrel (mass of 1 kg) does push-ups by applying a force to elevate its center-of-mass by 5 cm. Estimate the number of push-ups that a tired squirrel must do in order to do a approximately 5.0 Joules of work. The squirrel applies a force of approximately 10 N (9.8 N to be exact) to raise its body at constant speed. The displacement is 0.05 meters. The angle between the upward force and the upward displacement is 0 degrees. The work for 1 push-up is approximately W = 10 N * 0.05 m * cos 0 degrees = 0.5 Joules If the squirrel does a total of 5.0 Joules of work, then it must have done about 10 push-ups.

  17. Homework • Do problems 1-4 in your book page 156.

  18. Closure • Today we learned about work • Next we are going to learned about energy

  19. Have a great day

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