2. 1 Conservation of Energy “with non-conservative forces” Some Key Terms Internal force: any force exerted on an object in the system due to another object in the system. External force: any force exerted by a object that is not part of the system on an object within the system.

3. 2 Closed system: a system that can exchange energy with its surroundings, but not with matter. Isolated system: a system that does not exchange either matter or energy with its surroundings. Open system: a system that can exchange both matter and energy with its surroundings.

4. 3 So far we have only looked at conservation of energy for conservative forces. The work done by non-conservative forces is equal to the difference of the final mechanical energy and the initial mechanical energy of a system. Now we will look at what happens the total energy of a system when a non-conservative force acts on the system.

5. In short, the Work/Energy theorem tells us that when you do work on a system you increase its final energy. If a force does work on a system so that the initial energy is different then the final energy, then the force must be a non conservative force. (A.K.A. dose not conserve energy)

6. If we now rearrange and solve for the initial velocity we get, If there is no work done by a no conservative force then, We are right back to conservation of energy.

7. We can now see that the Law of Conservation of Energy and the Work/Energy Theorem are in actuality the same thing. Where the work is done by a non conservative force.

8. Example: A 65.0 kg skydiver stepped of a hot air balloon that is 500 m above the ground. After freefalling a short distance, she deploys a parachute, finally reaching the ground of the velocity of 8.00 m/s a) Find the gravitational potential energy of the skydiver, relative to the ground, before she jumps. 3.19 x 10 5 J b) Find the kinetic energy of the skydiver just before she lands on the ground. 2.08 kJ c) How much work was done by air friction. -3.17 x 10 5 J

9. Example: A roller coaster car of mass of 200 kg is moving at a speed of 4.00 m/s at point A in the diagram. This point is 15 m above the ground. The car then heads down the slope toward point B, which is 6 m above the ground. If 3.40 x 10 3 J of work are done by friction between points A and B, determine the speed of the car at point B. 12.5 m/s

10. 9 Do Practice Problems Page 308 (pdf 41) #’s 18-23 (for #18 you will need to refer to the example problem on pg 305) End of Chapter Review Pg 327 (pdf 44) #’s 1, 2, 5, 22, 23, 26.