1. Ch. 8 Energy

2. Learning Intention Understand how to describe, discuss, and quantify the energy of a system Journal: Why do you think this concept is important?

3. Work Application of a force in order to move something. W = (F)(d) Units: N  m Also called a Joule (J)

4. Power The rate at which work is done P = W/t Unit: J/s Also called a watt (W)

5. Time to practice: Calculate the work done when lifting a 5N load a distance of 7 m?

6. Time to practice: Calculate the work done when lifting a 5N load a distance of 7 m? W = Fd = (5N)(7m) = 35N  m or 35 J

7. Calculate the distance an object was moved when 5 J where expended using a force of 2 N.

8. Rearrange W = Fd d = W/F d = 5J/2N = 2.5 m

9. Calculate the power expended when 70 J of work is completed in 4 s.

10. P = W t P = 70 J / 4s P = 17.5 J/ s

11. Energy Becomes evident when it changes from one form to another

12. Energy is measured in Joules!!! Mechanical Energy – Potential Energy – Kinetic Energy

13. Potential Energy Stored energy Has the potential to do work PE = w  h PE = m  g  h

14. Kinetic Energy Energy of motion Depends on mass and speed of an object KE = ½ mv 2

15. Example Problems Calculate the energy of a 4 kg object that travels at 6 m/s. Given: Equation: Work: Answer: m = 4 kg V = 6 m/s

16. Example Problems Calculate the energy of a 4 kg object that travels at 6 m/s. Given: Equation: Work: Answer: m = 4 kg KE = ½ mv 2 V = 6 m/s

17. Example Problems Calculate the energy of a 4 kg object that travels at 6 m/s. Given: Equation: Work: Answer: m = 4 kg KE = ½ mv 2 KE= ½ (4 kg)(6m/s) 2 V = 6 m/s KE = 72 J

18. Calculate the energy stored in a 50 kg object that is 127 m above the ground. Given:Equation:Work & Answer: m = 50 kg h = 127 m

19. Calculate the energy stored in a 50 kg object that is 127 m above the ground. Given: Equation:Work & Answer: m = 50 kg PE = mgh h = 127 m g= 9.8 m/s 2

20. Calculate the energy stored in a 50 kg object that is 127 m above the ground. Given: Equation:Work & Answer: m = 50 kg PE = mgh PE = (50 kg)(9.8m/s 2 )(127m) h = 127 m PE = 62230 J g= 9.8 m/s 2

21. Mechanical Advantage

22. Work input = Work output (F  d) in = (F  d) out Levers are used to obtain a mechanical advantage in that they are force multipliers

23. Pulleys Change direction of force Multiply force Or Both!

24. Work input = Work output (F  d) in = (F  d) out A lever is used to lift a heavy load. When a 36 N force pushes one end of the lever down 1 m, the load rises 0.45 m. Calculate the weight, in Newtons, of the load. (F  d) in = (F  d) out (36N)(1m) = F (0.45 m) F = 36N  m 0.45 m F = 80 N

25. Your turn In raising a 3050 N fridge with a pulley system, it is noted that for every 1.5 m of rope pulled, the fridge rises 0.5 m. How much force is used to pull up the fridge? (F  d) in = (F  d) out (F  1.5m) = (3050 N)(0.5m) F = (3050 N)(0.5m) 1.5 m F = 1016.67 N

26. Efficiency Ideal machines have 100% of work input converted to output…. In practice, is this true??? Not typically, hence Ideal machines are “ideal” Efficiency = work ouput total work input

27. Efficiency = work ouput total work input Calculate the efficiency of a machine that requires 230 J of input energy to do 40 J of work. Efficiency = 40 J 230 J Efficiency = 0.17  17%

28. In class Practice