1. Gravitational Potential Energy p. 191 1-6extra p. 194 1-5,7 Conservation of Energy p. 197 1-6, 7(tricky)extra p. 201 1-11 Chapter Review for above p. 226 14-21,37-39 Elastic Potential Energy p. 206 1-5extra p. 219 5-14 p. 211 8-11, 12,13 tough Chapter Review for Elastic Potential p. 227 22,23,35

2. GRAVITATIONAL POTENTIAL ENERGY Conservative and Non-Conservative Forces A ball is thrown upwards and returns to the thrower with the same speed it departed with. A block slides into a spring, compresses it and leaves the spring with the same speed it first contacted it with.

3. A force is conservative if the kinetic energy of a particle returns to its initial value after a round trip (during the trip the E k may vary). A force is non- conservative if the kinetic energy of the particle changes after the round trip (Assume only one force does work on the object). Gravitational, electrostatic and spring forces are conservative forces. Friction is an example of a non-conservative force. For a round trip the frictional force generally opposes motion and only leads to a decrease in kinetic energy.

4. We must introduce the concept of potential energy. This is energy of configuration or position. As kinetic energy decreases the energy of configuration increases and vice versa.  E p Change in potential energy  E g Change in gravitational potential energy.  E e Change in elastic potential energy.

6. define E g as mgh h is height relative to a reference point Gravity does work on an object as its height changes. As an object increases its height gravity does negative work on the object and the object’s kinetic energy decreases. This loss of kinetic energy is a gain of potential energy.

7. define E m as the mechanical energy Mechanical Energy is conserved when an object is acted upon by conservative forces.

8. LAW OF CONSERVATION OF ENERGY Energy may be transformed from one kind to another, but it cannot be created or destroyed: the total energy is constant. There are many forms of energy such as electromagnetic, electrical, chemical, nuclear, and thermal.

9. A ball is launched from a height of 2 m with an initial velocity of 25 m/s [35 o ath]. What is the speed of the ball when its height is 7.5 m?

10. The speed of the ball is 22.73 m/s The energy approach doesn’t calculate the velocity but it is a quicker method. The kine- matic approach is longer but more precise. let’s try it for exam review!!!

11. ELASTIC POTENTIAL ENERGY Hooke’s Law (Robert Hooke 1678) The magnitude of the force exerted by a spring is directly proportional to the distance the spring has moved from its equilibrium position. An ideal spring obeys Hooke’s Law because it experiences no internal or external friction.

12. +F x = force exerted by hand on spring

13. x (m) F (N) slope = k elastic limit non-elastic region breaking point The linear region is sometimes called Hooke’s Law region. It applies to many elastic devices.

14. Hooke’s Law F = force exerted on the spring (N) k = force constant of spring (N/m) x = position of spring relative to the equilibrium (deformation) (m) The direction of compression on the spring is negative while the direction of elongation is positive (for F and x). The spring exerts an equal and opposite force on the object.

15. Derivation of Elastic Potential Energy A spring exerts a conservative force on a object. An object will have the same kinetic energy after a round trip with a spring. The spring will begin at its equilibrium position with zero potential energy.

16. zero since at equilibrium the force on the object and its  d have opp- osite directions

17. For an object interacting with an ideal spring. The potential energy of this object must be considered in the mechanical energy.

18. Remember a reference height is needed for height. The direction of x is not important unless solving for x. If it is known that the answer is compression then –x is correct. If the answer is elongation then +x is correct. If one form of energy is not present then it need not be included in the equation. try p.206 1-5 p.211 8-10 (they are quick)

19. A 2 kg ball is dropped from a height of 10 m onto a spring that is 0.75 m in length and has a spring constant of 1000 N/m. How far will the ball compress the spring? What force is exerted on the ball at its lowest point? 10 m 0.75 m x2x2 0.75+x 2 initial v 1 =0 h 1 =10 m x 1 =0 final v 2 =0 h 2 =0.75+x 2

20. Use the quadratic formula to solve.