2. Work F The force, F, pushes the box for a short distance. This causes the box to start moving!!!!!! It gains energy!!!! I just don’t have any energy  Whoa!!! Now I do!!!

3. Work If a force, F, is used to move an object a displacement, d, then the force does work on the object. F d W = Fd

4. Units for Work  The units for force are Newtons (N) and the units for displacement are meters (m).  Since Work is equal to force times displacment. We could use a Newton-meter (N·m).  But the more common unit is the Joule (J)  1 J = 1 N·m  1 J is defined as the amount of work for 1- N of force to move an object 1-m

5. In order for work to be done:  There must be a force applied.  The object must have a displacement.  The force must be responsible for the displacement.

6. How much work is being done by this weightlifter? NONE!!!!! Since he doesn’t move the weights any distance (There is no displacement), the work done is 0 J.

7. How much work is being done in these cases? NONE!!!!! In each case the forces are perpendicular to the displacement, therefore they are not “responsible” for the motion. In general, when a force is perpendicular to the motion of an object, it DOES NO WORK!!!!

8. Watch Video The weight lifter in the video lifts 263.5-kg a distance of 2.43-m. How much work does he do during one lift? If he lifts it twice, how much work would he do?

9. F d f If more than one force is acting on an object, you can find the total work done on the object by adding the work done by each force together. If the force is in the same direction as the motion, then the work done is positive because it adds energy to the object. If the force is in the opposite direction as the motion, then the work done is negative because it removes energy from the object.

10. A box is being pushed with a force of 25-N. There is also friction between the box and the table of 12-N. The box is moved 3.0-m How much work is being done by the 25-N force? How much work is being done by the friction? How much total work is being done on the box? F = 25-N d = 3.0-m W 25 = Fd = (25)(3.0) = 75-J This work is positive because the force is in the direction of the motion and adds energy to the box. 25-N 3.0-m 12-N W 12 = -fd = -(12)(3.0) = -36-J This work is negative because the force is in the opposite direction of the motion and removes energy from the box. d = 3.0-m f = 12-N

11. A box is being pushed with a force of 25-N. There is also friction between the box and the table of 12-N. The box is moved 3.0-m. Find the total work done on the box. What is the net force on the box? How much total work is being done on the box? F = 25-N 25-N 3.0-m 12-N d = 3.0-m f = 12-N SAME PROBLEM, DIFFERENT METHOD F net = F – f = 25 – 12 = 13-N F net = 13-N Another way to find the total work done on the box is to find the net force on the box and multiply this by the displacement.