1. Work When a force acts on an object and the object moves a parallel distance.

2. Mechanical Energy Kinetic energy is the energy of an object because of speed. Potential energy is energy that is stored up. Work is a transfer of energy. A pitcher gives a ball energy by doing work on it.

3. Work The product of the displacement and the force parallel to the displacement. W = Fd Since the force must be parallel to the displacement, W = Fdcosϑ ϑ

4. Work 1. A person is horizontally pushing a cart 30.m along the floor with 50.N. How much work does the person do on the cart? W = Fdcosϑ W = (50.)(30.)(cos 0 o ) W = 1500N.m W = 1500J Work is NOT a vector. It has no direction.

5. Not Work A force can be exerted on an object and yet do no work. Ex. If you hold a bag of heavy groceries it may feel like work, but if there is no displacement, work is equal to zero.

6. Not Work If you held up a 25N bag of groceries as you walk 3.0m at a constant speed across the kitchen you do no work because W = Fdcosϑ W = (25)(3.0)(cos 90 o ) W = (25)(3.0)(0) W = 0 J

7. Work Be clear for what you are calculating work. You can calculate work done on an object or person. You can calculate work done by an object or person. You can calculate the net work done.

8. Work 2a. Determine the work a hiker does on a 15.0kg backpack to carry it up a hill of height 10.0m. F g = mg F g =(15)(9.8) F g =147N W = Fdcosϑ W = (147)(10)(cos 0 o ) W = 1470 J Work done on the backpack: 1470N

9. Work 2b. Determine the work gravity does when a hiker carries a 15.0kg backpack up a hill of height 10.0m. W = Fdcosϑ W = (147)(10)(cos 180 o ) W = (147)(10)(1) W = - 1470 J Gravity does negative work because it is in the direction opposite to the displacement.

10. Work 2c. Determine the net work done on a 15.0kg backpack to carry it up a hill of height 10.0m. You can do this two ways. Find the net force and multiply by displacement, or find the work done by each force and add them. W net = W gravity + W hiker W net = -1470J + 1470J W net = 0 J

11. A 12.0kg box is being pulled across the floor. The person is exerting a force of 45N, and μ = 0.220. The person pulls the box a distance of 7.20 m. a)How much work was done by the person? b)How much work was done by friction? c)Calculate the net work done on the box.

12. How much work did a horse do that dragged an 80 kg cart along the ground 70 km a) Without acceleration if μ = 0.060? b) With an acceleration of 2.2 m/s 2 if μ = 0.060? c) How much work is done by friction in (a) and (b)?