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- 1. Introduction to Work
- 2. Where we have been <ul><li>Previously we used Newton’s Laws to analyze motion of objects </li></ul><ul><li>Force and mass information were used to determine acceleration of an object (F=ma) </li></ul><ul><li>We could use the acceleration to determine information about velocity or displacement </li></ul><ul><ul><li>Did the object speed up or slow down? </li></ul></ul><ul><ul><li>How far did the object travel? </li></ul></ul>
- 3. Where we are going <ul><li>Now we will take a new approach to looking at motion </li></ul><ul><li>We will now look at work and power in relation to motion </li></ul><ul><li>Today we will focus on “work” </li></ul>
- 4. Definition of “work” <ul><li>The everyday definition of “work” and the one that we use in physics are quite different from each other </li></ul><ul><li>When most people think about “work”, they think of the job that they have </li></ul><ul><li>Although it is possible that you are doing the physics definition of work while at your job, it is not always the case </li></ul>
- 5. Physics Definition of “Work” <ul><li>Like so many other things in physics, we have to use an exact definition to really explain what “work” is </li></ul><ul><li>PHYSICS DEFINITION </li></ul><ul><ul><li>Work happens when a force causes an object to move through a displacement </li></ul></ul><ul><li>When a force acts upon an object to cause a displacement of the object, it is said that WORK has been done upon the object </li></ul>
- 6. Work <ul><li>There are three key ingredients to work </li></ul><ul><ul><li>Force </li></ul></ul><ul><ul><li>Displacement </li></ul></ul><ul><ul><li>Cause </li></ul></ul><ul><li>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 </li></ul>
- 7. Everyday Examples of “Work” <ul><li>There are several good examples of work which can be observed in everyday life </li></ul><ul><ul><li>A horse pulling a plow through a field </li></ul></ul><ul><ul><li>A person pushing a shopping cart </li></ul></ul><ul><ul><li>A student lifting a backpack onto her shoulder </li></ul></ul><ul><ul><li>A weightlifter lifting a barbell above his head </li></ul></ul><ul><li>In each case described here there is a force exerted upon an object to cause that object to be displaced </li></ul>
- 8. Work <ul><li>Work – Exerting force in a way that makes a change in the world. </li></ul><ul><ul><li>Throwing a rock is work : you’re exerting a force, and the rock’s location changes (i.e. “the world has been changed”) </li></ul></ul><ul><ul><li>Pushing on a brick wall is not work : you’re exerting a force, but “the world doesn’t change” (the wall’s position doesn’t change). </li></ul></ul>
- 9. Work <ul><li>So exerting force alone isn’t enough. You have to both exert a force , and make a change . </li></ul><ul><li>If you’re not exerting a force, you’re not doing work. </li></ul><ul><li>Example: Throwing a ball. </li></ul><ul><ul><li>While you are “throwing the ball” (as opposed to just holding it) you are exerting a force on the ball. And the ball is moving. So you’re doing work . </li></ul></ul><ul><ul><li>After the ball leaves your hand, you are no longer exerting force. The ball is still moving, but you’re no longer doing work. </li></ul></ul>
- 10. Work <ul><li>So, mathematically, we define work as “exerting a force that causes a displacement”: </li></ul><ul><li>(Work) = (Force exerted) (Displacement of object) (cos Θ ) </li></ul><ul><li>or </li></ul><ul><li>W = F*d*cos Θ </li></ul><ul><li>W = Work done (J) </li></ul><ul><li> F = Force exerted on object (N) </li></ul><ul><li>d = Displacement of object (m) </li></ul><ul><li>Θ = Angle between the force and the displacement </li></ul>
- 11. New Unit! <ul><li>The units for work are Nm (Newtons × meters). As we did with Newtons (which are kg m/s 2 ), we will “define” the Newton-meter to be a new unit. We’ll call this unit the Joule . </li></ul><ul><li>Abbreviation for Joule: J </li></ul><ul><li>So, 1 Nm = 1 J </li></ul><ul><li>Example: 1 joule = work done to lift a ¼ lb hamburger (1 N) 1 meter </li></ul>
- 12. Defining Θ – “the angle” <ul><li>This is a very specific angle </li></ul><ul><li>Not just “any” angle - It is the angle between the force and the displacement </li></ul><ul><li>Scenario A: A force acts rightward (@ 0 °) upon an object as it is displaced rightward (@ 0 °) . The force vector and the displacement vector are in the same direction, therefore the angle between F and d is 0 degrees </li></ul>F d Θ = 0 degrees 0° - 0° = 0° Subtract the smaller angle from the larger angle to determine the angle “between” the vectors
- 13. Defining Θ – “the angle” <ul><li>Scenario B: A force acts leftward (@ 180 °) upon an object which is displaced rightward (@ 0 °) . The force vector and the displacement vector are in the opposite direction, therefore the angle between F and d is 180 degrees </li></ul>F d Θ = 180 degrees 180° - 0° = 180° Subtract the smaller angle from the larger angle to determine the angle “between” the vectors
- 14. Defining Θ – “the angle” <ul><li>Scenario C: A force acts upward (@ 90 °) upon an object as it is displaced rightward (@ 0 °) . The force vector and the displacement vector are at a right angle to each other, therefore the angle between F and d is 90 degrees </li></ul>F d Θ = 90 degrees 90° - 0° = 90° Subtract the smaller angle from the larger angle to determine the angle “between” the vectors
- 15. To Do Work, Forces must CAUSE Displacement <ul><li>Consider scenario C from the previous slide </li></ul><ul><li>The situation is similar to a waiter who carried a tray full of meals with one arm (F=20N) straight across a room (d=10m) at constant speed </li></ul><ul><li>W = F*d*cos Θ </li></ul><ul><li>W = (20N)(10m)(cos 90°) </li></ul><ul><li>W = 0J </li></ul><ul><li>The waiter does not do work </li></ul><ul><li>upon the tray as he carries it </li></ul><ul><li>across the room </li></ul>
- 16. The Meaning of Negative Work <ul><li>On occasion, a force acts upon a moving object to hinder a displacement </li></ul><ul><ul><li>A car skidding to a stop on a roadway surface </li></ul></ul><ul><ul><li>A baseball player sliding to a stop on the infield dirt </li></ul></ul><ul><li>In such cases the force acts in the direction opposite the objects motion in order to slow it down </li></ul><ul><li>The force doesn’t cause the displacement, but it hinders the displacement </li></ul><ul><li>This is commonly called negative work </li></ul>
- 17. The Meaning of Negative Work <ul><li>If you substitute the numerical values into the work equation, you are left with a negative answer </li></ul><ul><li>W = F*d*cos Θ </li></ul><ul><li>W = (40 N)(10 m)(cos 180°) </li></ul><ul><li>W = (40 N)(10 m)(-1) </li></ul><ul><li>W = -400 J </li></ul><ul><li>The 10 m displacement is hindered by a 40 N force causing -400 J worth of work </li></ul><ul><li>This will be an important concept a little later </li></ul>
- 18. Example of Work <ul><li>You are pushing a very heavy stone block (200 kg) across the floor. You are exerting 620 N of force on the stone, and push it a total distance of 20 m in 1 direction before you get tired and stop. </li></ul><ul><li>How much work did you just do? </li></ul><ul><ul><li>W = (620 N)(20 m) = 12,400 J </li></ul></ul>
- 19. Work Problems <ul><li>Austin lifts a 200 N box 4 meters. How much work did he do? </li></ul><ul><li>W = (200N)(4m)(cos 0 °) </li></ul><ul><li>W = (200N)(4m)(1) </li></ul><ul><li>W = 800 J </li></ul>
- 20. Work Problems <ul><li>Caitlin pushes and pushes on a loaded shopping cart for 2 hours with 100 N of force. The shopping cart does not move. How much work did Caitlin do? </li></ul><ul><li>Chase lifts a 100 kg (220 lbs) barbell 2 meters. How much work did he do? </li></ul>
- 21. Work Done By “Lifting” Something <ul><li>Notice that when we were pushing something along the ground, the work done didn’t depend on the mass. </li></ul><ul><li>Lifting up something does do work that depends on mass. </li></ul><ul><li>Because of gravity: </li></ul><ul><ul><li>Gravity always pulls down with a force equal to m*a g , where m is the mass, and a g = 9.8 m/s 2 . </li></ul></ul><ul><ul><li>So we must exert at least that much force to lift something. </li></ul></ul><ul><ul><li>The more mass something has, </li></ul></ul><ul><ul><li>the more work required to lift it. </li></ul></ul>
- 22. Work Done By “Lifting” Something <ul><li>Example: A weightlifter lifts a barbell with a mass of 280 kg a total of 2 meters off the floor. What is the minimum amount of work the weightlifter did? </li></ul><ul><ul><li>The barbell is “pulled” down by gravity with a force of (280 kg)(9.8 m/s 2 ) = 2,744 N </li></ul></ul><ul><ul><li>So the weightlifter must exert at least 2,744 N of force to lift the barbell at all. </li></ul></ul><ul><ul><li>If that minimum force is used, the work done will be: </li></ul></ul><ul><ul><li>W = (2,744 N)(2 m) = 5,488 J </li></ul></ul>
- 23. Questions???

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