1. Work Kinetic Energy Work – Energy Theorem Gravitational Potential Energy Mechanical Energy Conservation of Energy Work Kinetic Energy Work – Energy Theorem Gravitational Potential Energy Mechanical Energy Conservation of Energy

2. The angle between two vectors is found by placing their tails together and picking the smallest angle between them. For example, the angle between B and D is 45 degrees. Find the angle between: 1.A and B 2.B and C 3. C and D 4. A and E 1. 180 o 2. 90 o 3. 135 o 4. 90 o

4. F = magnitude of the force r = magnitude of the displacement  = the angle between F and r

5. F = 100N r = 2m  = 30 degrees A 100N force acts on a box at 30 degrees with respect to the horizon, moving the box 2m to the right. How much work did the force do? Work = Frcos(  ) = 100(2)cos30 = 173J

6. Example Find the total work done on this box as it moves 3m to the right. Work done by P: W = 200(3)cos(0) = 600J Work done by N: W = Nrcos(90) = 0 Work done by f : W = 80(3) cos(180) = -240J Work done by Weight : W = 100(3)cos(90) = 0 Total Work = 600 + 0 + (-240) + 0 = 360J

7. KE = (1/2)mv 2 Example: Find the kinetic energy of a 20kg mass moving at 10 m/s. KE = (1/2)mv 2 = (1/2)(20)(10) 2 = 1000J Example: If you double your velocity, what happens to the KE? KE (new) = (1/2)m(2v) 2 = 4KE(original)

8. Work Energy Theorem Work Total = ½mv f 2 – ½mv i 2 or Work Total =  KE

9. Example If this box starts at rest, what is its velocity after it moves 3m? Previously we saw that the total work was 360J. Work Total = (1/2)mv f 2 – (1/2)mv i 2 360 = (1/2)(10)v f 2 v f = 8.48 m/s

10. Example If a mass is dropped from rest 20m above the ground, how fast is it moving when it hits the ground? Work Total = (1/2)mv f 2 – (1/2)mv i 2 = (1/2)mv f 2 Work = Frcos  = mgrcos(0) = mgr = (1/2)mv f 2 v f = [ 2gr ] ½ = [ 2(9.8)(20) ] ½ = 19.8 m/s Work Total = (1/2)mv f 2 – (1/2)mv i 2 = (1/2)mv f 2 Work = Frcos  = mgrcos(0) = mgr = (1/2)mv f 2 v f = [ 2gr ] ½ = [ 2(9.8)(20) ] ½ = 19.8 m/s

11. At the top of her swing, 2 meters from the ground, a child has a potential energy of 500 Joules. When she is 1m from the ground, her kinetic energy is At the top of her swing, 2 meters from the ground, a child has a potential energy of 500 Joules. When she is 1m from the ground, her kinetic energy is a. 500J b. 300J c. 250J d. 150J.

12. A golf ball moving at 40 m/s has a kinetic energy of 35 Joules. What is its kinetic energy if it instead moves at 120 m/s? A golf ball moving at 40 m/s has a kinetic energy of 35 Joules. What is its kinetic energy if it instead moves at 120 m/s? a. 75J b. 105J c. 315J d. None of these.

13. Energy of Motion – Kinetic Energy Stored Energy – Potential Energy

14. Gravitational Potential Energy The work you do to lift an object is equal to its increase in gravitational PE. PE = mgh h = vertical distance from ground level.

15. Example How much work does a 100kg person do walking up a flight of stairs that takes her 10m off of the ground? Work = increase in PE = mgh = 100(9.8)10 = 9800 Joules. (Note: 1 Calorie = 4128J, so this person burned a little over 2 calories. ) ( Note: 1 taco = 200 Calories = 825,600 Joules) Work = increase in PE = mgh = 100(9.8)10 = 9800 Joules. (Note: 1 Calorie = 4128J, so this person burned a little over 2 calories. ) ( Note: 1 taco = 200 Calories = 825,600 Joules)

16. E = KE + PE If there is no friction, the total mechanical energy is conserved. This means KE + PE is always the same number. So if KE drops, PE must rise. If PE drops, KE must rise. If there is no friction, the total mechanical energy is conserved. This means KE + PE is always the same number. So if KE drops, PE must rise. If PE drops, KE must rise. Another way to say this, is that KE is converted to PE, or PE is converted to KE.

17. Conservation of Energy

18. Example A 1 kg rock is dropped from rest from a building 20m high. Fill in the table. HeightKEPEE 20 10 5 0 PE at the top is mgh = 1(9.8)(20) = 196J 196J KE at the top is zero. 0 E at the top (and everywhere) is 0 + 196J = 196J 196J PE at the bottom is zero………… 0 196J 98J 49J 98J 147J

19. Work = Frcos(  ) KE = (1/2)mv 2 Work Total = ½mv f 2 – ½mv i 2 or Work Total =  KE Work Total = ½mv f 2 – ½mv i 2 or Work Total =  KE PE = mgh E = KE + PE

20. A 10kg rock is moving at 10m/s and is 10m above the ground. What is its total energy, E? A 10kg rock is moving at 10m/s and is 10m above the ground. What is its total energy, E? KE = (1/2)mv 2 = (.5)(10)(10 2 ) = 500J PE = mgh = 10(9.8)10 = 980J E = PE + KE = 1480J