1. Aim: How can we explain the Law of Conservation of Energy? Do Now: Homework Review

2. Aim: How can we explain the Law of Conservation of Energy? Do Now: A 10 kg object free falls off the top of a 100 m tall building. At the top, calculate its potential and kinetic energies. PE = mghKE = ½mv 2 PE = (10 kg)(9.8 m/s 2 )(100 m)KE = ½(10 kg)(0 m/s) 2 PE = 9,800 JKE = 0 J

3. Conservation of Energy The total energy (E T ) of an object or system is constant The total energy (E T ) of an object or system is constant Energy cannot be created or destroyed Energy cannot be created or destroyed Energy can be changed from one form to another. Energy can be changed from one form to another.

4. A Dropped Sphere What energy does it have when held above the ground? What energy does it have when held above the ground? Potential Energy What happens to this PE as the ball drops? What happens to this PE as the ball drops? It becomes smaller and smaller Is the energy just disappearing? Is the energy just disappearing? No!! It is being converted into KE (the object is speeding up)

5. 100 m m = 10 kg At the top: PE = 9,800 J (PE = mgh) KE = 0 J (at rest) E T = 9,800 J (PE + KE = E T ) At the bottom: PE = 0 J (no height) E T = 9,800 J (total energy is constant!) KE = 9,800 J (PE + KE = E T ) If PE = 4,900 J, what is KE? KE = 4,900 J PE = 3,000 J, what is KE? KE = 6,800 J

6. As an object falls, PE is being converted into KE ΔPE = ΔKE mgΔh = ½mv 2 Problem A 2 kg mass is dropped from a height of 10 m. What is the KE as it strikes the ground? ΔKE = ΔPE ΔKE = mgΔh ΔKE = (2 kg)(9.8 m/s 2 )(10 m) ΔKE = 196 J

7. How fast is the object moving when it strikes the ground? KE = ½mv 2 196 J =.5(2kg)v 2 196 J/kg = v 2 v = 14 m/s

8. After the mass falls 5 m, what is its KE? ΔKE = ΔPE ΔKE = mgΔh ΔKE = (2 kg)(9.8 m/s 2 )(5 m) ΔKE = 98 J What is its PE? Solution 1 PE = mgh PE = (2 kg)(9.8 m/s 2 )(5 m) PE = 98 J Solution 2 PE + KE = E T PE + 98 J = 196 J PE = 98 J

9. After the mass falls 7 m, what is its KE? ΔKE = ΔPE ΔKE = mgΔh ΔKE = (2 kg)(9.8 m/s2)(7 m) ΔKE = 137.2 J What is its PE? PE + KE = E T PE + 137.2 J = 196 J PE = 58.8 J

10. How fast is the object moving when it strikes the ground? ΔPE = ΔKE mgΔh = ½mv 2 gΔh = ½v 2 2gΔh = v 2 (2) Remember mechanics? v f 2 = v i 2 + 2ad v i = 0 The two formulas are the same! v = 14 m/s

11. Sometimes energy is lost due to heat or friction When this happens: PE + KE + Q = E T Q = energy lost due to friction It is not used in every problem!

12. 10 m m = 2 kg KE = 190 J How much energy was lost due to friction? This means solve for Q At the top: No KE Has not moved, so no frictional loss The only energy is PE PE at the top equals E T

13. PE = mgh PE = (2 kg)(9.8 m/s 2 )(10 m) PE = 196 J At the bottom, there is no height (no PE) KE should equal PE at the top, unless there is frictional loss PE top = 196 J KE bottom = 190 J PE + KE + Q = E T 0 J + 190 J + Q = 196 J Q = 6 J