1. Work and energy

2. Objectives 1.Recognize the difference between the scientific and the ordinary definitions of work. 2.Define work, relating it to force and displacement. 3.Identify where work is being performed in a variety of situations. 4.Calculate the net work done when many forces are applied to an object. Homework:

3. In physics, work is defined as a force acting upon an object to cause a displacement. Definition and Mathematics of Work Work is being done Work is not being done

4. Let’s practice – work or no work 1.A student applies a force to a wall and becomes exhausted. 2.A calculator falls off a table and free falls to the ground. 3.A waiter carries a tray full of beverages above his head by one arm across the room 4.A rocket accelerates through space. no work work no work work

5. Calculating the Amount of Work Done by Forces θ F d F - is the force in Newton, which causes the displacement of the object. d - is the displacement in meters θ = angle between force and displacement W - is work in N∙m or Joule (J). 1 J = 1 N∙m = 1 kg∙m 2 /s 2 Work is a scalar quantity Work is independent of time the force acts on the object. θ F d FxFx FyFy Only the horizontal component of the force (Fcosθ) causes a horizontal displacement.

6. Example 1 How much work is done on a vacuum cleaner pulled 3.0 m by a force of 50.0 N at an angle of 30 o above the horizontal?.

7. Example 2 How much work is done in lifting a 5.0 kg box from the floor to a height of 1.2 m above the floor? W = F∙dcosθ F = mg = (5.0 kg)(9.81 m/s 2 ) cos0 o = 49 N W = F∙d = (49 N) (1.2 m) = 59 J Given: d = h = 1.2 meters; m = 5.0 kg; θ = 0 Unknown: W = ?

8. Example 3 A 2.3 kg block rests on a horizontal surface. A constant force of 5.0 N is applied to the block at an angle of 30. o to the horizontal; determine the work done on the block a distance of 2.0 meters along the surface. Given: F = 5.0 N; Given: F = 5.0 N; m = 2.3 kg d = 2.0 m θ= 30 o 30 o 5.0 N 2.3 kg unknown: unknown: W = ? J W = ? J Solve: Solve: W = F∙d∙cos W = F∙d∙cosθ W = (5.0 N)(2.0 m)(cos30 o ) = 8.7 J W = (5.0 N)(2.0 m)(cos30 o ) = 8.7 J

9. Example 4 Matt pulls block along a horizontal surface at constant velocity. The diagram show the components of the force exerted on the block by Matt. Determine how much work is done against friction. 8.0 N 6.0 N 3.0 m W = F x d x W = (8.0 N)(3.0 m) = 24 J Given: F x = 8.0 N Given: F x = 8.0 N F y = 6.0 N d x = 3.0 m unknown: W = ? J unknown: W = ? J F

10. Class work Page 170 practice #1-4 1.1.50 x 10 7 J 2.7.0 x 10 2 J 3.1.6 x 10 3 J 4.1.1 m

11. The sign of work

12. When No work is done

13. Force vs. displacement graph The area under a force versus displacement graph is the work done by the force. Displacement (m) Force (N) work Example: a block is pulled along a table with 10. N over a distance of 1.0 m. W = Fd = (10. N)(1.0 m) = 10. J height basearea

14. The angle in work equation The angle in the equation is the angle between the force and the displacement vectors. F & d are in the same direction, θ is 0 o. F d

15. What is θ in each case?

16. Class work Page 171 #1-6