Energy: ability to do work

1. Energy:ability to do work Another way of analyzing motion…

2. Work: • In physics we say that work is done on an object if a force is applied to it and that force causes it to move a certain distance. Work = W = Fd N m WORK IS ENERGY!!! N•m = J = Joule

3. What is a Joule? N • m kg m• m s2 kg m2 s2 The energy required to lift a small apple one meter straight up.

4. Here’s the important thing about work…work is only done if the force has a component in the same direction as the displacement. m = 10 kg F = 25N d = 2m Is this guy doing work on the box? Yes. The force vector is in the same direction as the displacement. How much work is being done? W = Fd W =25N(2m) W = 50 Nm = 50J

5. Is work done when pulling this dog? Yes. The force vector has some component in the same direction as the displacement. How much work is being done? F Fy 30° Fx Fx = cosΘ = A H = 70N Fx = FcosΘ 30° Fx = 70Ncos30° d = 10m Fx = 61N W = Fd W =61N(10m) W =610J

6. Is work being done by this waiter? No. He could carry around that tray all day and according to physics he wouldn’t be doing any work. There is a force (the waiter pushes up on the tray) and there is a displacement (the tray is moved horizontally across the room). Yet the force does not cause the displacement. To cause a displacement, there must be a component of force in the direction of the displacement.

7. W = Fd • How much work is needed to lift at a constant speed a 15kg book 3m? W = mgd W = (15kg)(10m/s2)(3m) W = 450 J

8. Which path (incline vs. ladder) requires more work to get the box to the top? mbox = 10 kg 10 m 5 m 30° W = Fd W = Fd W = (50N)(10m) W = (100N)(5m) W = 500 J W = 500 J Same amount of work!

9. A particular task may require a certain amount of work but it might be done over different lengths of time… • This is known as Power (P). It measures the rate at which work is done. P = W t d = v t P = Fd t P = Fv J = watt = W s P = Fd t

10. Who has more power? • Dan Parker and Brad Bowen are in the weightlifting room. Dan lifts the 50 kg barbell over his head 10 times in one minute; Brad lifts the 50 kg barbell over his head 10 times in 10 seconds. Which student does the most work? Which student delivers the most power? Brad is more "power-full" since he does the same work in less time. Power and time are inversely proportional.

11. Try this… • A crane lifts a load with a mass of 1000kg a vertical distance of 25m in 9s at a constant velocity. How powerful is the crane? P = W = Fd t t = Fgd t = mgh t = (1000kg)(10m/s2)(25m) 9s = 27000 W

12. Try this… • A 45 kg bicyclist climbs a hill at a constant speed of 3 m/s by applying an average force of 80 N. How much power does the bicyclist develop? P = Fv P = (80 N)(3 m/s) P = 240 W

13. Power is a rate (ENERGY PER SECOND). • Your electric bill (power bill) is based on your rate of energy use. • A lightbulb with a 60 Watt power rating means that the bulb uses 60 joules of energy per second. W = ΔET Work = a change in total energy

14. Energy is the ability to do work! • Energy is measured by the amount of work it can do.

15. Energy comes in different forms… • Potential energy (PE): • Energy possessed by an object due to its position • Sometimes referred to as “stored energy”

16. Gravitational Potential Energy • If an object, originally at rest on Earth’s surface, is lifted to some height, work is done against the gravitational force. • The work done in lifting the object is equal to the objects gravitational potential energy.

17. work done = gravitational potential energy W = ΔPE W = Fd Fg w mg W = mgd h ΔPE = mgh

18. Knowing that the potential energy at the top of the tall platform is 50 J, what is the potential energy at the other positions shown on the stair steps and the incline?

19. Path doesn’t matter… Remember that the changes in an object's potential energy only depend on comparing its starting position and its ending position, not on whether it does or does not pass through various points in-between.

20. Try this… • How much potential energy is gained by an object with a mass of 2 kg that is lifted from the floor to the top of a .8 m high table? ΔPE = mgh ΔPE = (2kg)(10m/s2)(.8m) ΔPE =16 J

21. Try this… • King Kong is on top of the Empire State Building 426 m above the surface of the Earth. What is his gravitational potential energy relative to the ground? Let’s say his mass is 1000 kg (a metric ton). ΔPE = mgh ΔPE = (1000kg)(10m/s2)(426m) ΔPE = 4,260,000 J

22. Draw how the graph would look that represents this relationship- PE vs. h ΔPE = mgh PE = mg h PE What if m = .1 kg h

23. Elastic Potential Energy • Energy can be stored in a spring and is measured as the work required to stretch or compress it.

24. Remember Hooke’s Law… • The compression or elongation of a spring is directly proportional to the applied force. Fs = kx Spring constant The larger the k, the stiffer the spring.

25. x F What’s the spring constant of this spring? F = kx k = F x = 25 N .50 m = 50 N/m

26. Potential Energy of a Spring W = PEs W = Fd x ½ kx PEs = ½ kx2

27. What would the graph look like that shows this relationship- PE vs. x? PE PE PE PE x x x x PEs = ½ kx2 What if we made k = 2 N/m PEs = x2

28. Elastic potential energy can be stored in rubber bands, bungee chords, trampolines, springs, an arrow drawn into a bow, etc.

29. Try this… • A force of 50 N is needed to compress a spring a distance of 1 m. What is the potential energy stored in the compressed spring? PEs = ½ kx2 Fs = kx k = 50 N 1 m PEs = ½ (50)(1m)2 Fs = kx x x PEs = 25 J k = 50 N/m k = Fs x

30. Try this… • When a spring is stretched .2 m from its equilibrium position, it possesses a potential energy of 10 J. What is the spring constant for the spring? PEs = ½ kx2 k = 2PE x2 k = 500 N/m

31. Kinetic Energy • When a moving object strikes another object and displaces it, the moving object exerts a force on the second object and does work on it.

32. Kinetic Energy- the energy an object possesses due to its motion. W = ΔKE W = Fd vt from rest- v 2 ma v t W = ΔKE = mvvt t 2 ΔKE = ½ mv2

33. Try this… • What is the kinetic energy of a 980 kg race car traveling at 90 m/s? ΔKE = ½ mv2 ΔKE = ½ (980kg)(90m/s)2 ΔKE = 3,969,000 J

34. Try this… • Determine the kinetic energy of a 625-kg roller coaster car that is moving with a speed of 18.3 m/s. ΔKE = ½ mv2 ΔKE = ½ (625kg)(18.3m/s)2 ΔKE = 104,653 J

35. Try this… • A platform diver for the Circus has a kinetic energy of 12 000 J just prior to hitting the bucket of water . If the divers mass is 40 kg, then what is her speed? ΔKE = ½ mv2 v2 = 2KE m v= 25 m/s

36. Conservation of Energy • Energy can neither be created nor destroyed. But it can be transferred from one type to another (i.e. potential to kinetic) in a closed system.

37. Examples… The ball is losing height (falling – h is decreasing) and gaining speed (v is increasing). Energy is transformed from PE (height) to KE (speed). Motion - A ball falls from a height of 2 meters in the absence of air resistance.

38. Examples… The skier is losing height (the final location is lower than the starting location) and gaining speed (the skier is faster at B than at A). Energy is transformed from PE (height) to KE (speed). Motion - A skier glides from location A to location B across a friction free ice.

39. Examples… The ball is gaining height (rising) and losing speed (slowing down). Energy is transformed from KE (speed) to PE (height). Motion - A baseball is traveling upward toward a man in the bleachers.

40. Examples… The jumper is losing speed (slowing down) and the bungee cord is stretching. Energy is transformed from KE (speed) to PE (a stretched "spring"). Motion - A bungee cord begins to exert an upward force upon a falling bungee jumper.

41. Examples… The spring changes from a compressed state to a relaxed state and the dart starts moving. Energy is transformed from PEs (a compressed spring) to KE (speed). Motion - The spring of a dart gun exerts a force on a dart as it is launched from an initial rest position.

42. Closed System? • A closed system is one in which there are no external forces doing work on the system, and no transfer of energy into or out of the system. • External Forces- FA, Ff, FT, Fair, FN • The total energy (ET) of a closed system ALWAYS remains the same.

43. Total Mechanical Energy • In a closed (ideal) system… ΔPE + ΔKE = TME and ΔPE + ΔKE = 0 ΔKE = - ΔPE

44. In a non-ideal system there is an external force acting on the system and the total energy is… internal energy- influenced by heat ET = PE + KE + Q Here’s the equation we’re going to use: W = ΔPE + ΔKE + Wf

45. There are 3 different approaches we can take to solve this problem: v = 0 A 10 kg 1. PE = KE’ mgh = 1mv2 2 2mgh = 1mv22 2 20 m 2mgh = mv2 m m v2 = 2gh B v2 = 2(10m/s2)(20m) v = ? 20 m/s v2 = 400 m/s

46. 0 0 2. v = 0 A 10 kg W = ΔKE + ΔPE + Wf Any work being done on the system (W = Fd)? Is anyone pulling or pushing on the block? Any friction? No – neglecting air friction No 0 = ΔKE + ΔPE 20 m 0 0 0 = KEf – KEi + PEf – PEi 0 = KEf – PEi KEf = PEi B v = ?

47. v = 0 A 10 kg 3. A B 0 ½ mv2 = 2000 J 0 mgh= 2000 J 20 m 0 0 0 0 0 v2 = 2gh = 20 m/s 2000 J 2000 J B v = ?

48. v = 0 10 kg How could we of figured this out without energy? vi = 0 m/s a = 10 m/s2 d = 20 m 20 m vf = ? vf2 = vi2 + 2ad vf2 = 2ad vf2 = 2(10m/s2)(20m) v = ? 20 m/s vf2 = 400 m/s

49. A B C ½ mv2 = 1920 J KE = TE – PE = 1320 J 0 h = ? v = 0 m/s h = PE/mg = 3.2 m 0 mgh = 600 J 0 0 0 C 0 0 0 v2 = 2KE/m = 6.6 m/s 0 m/s 8 m/s 1920 J 1920 J 1920 J B A Remember energy is another way of analyzing motion.