1. Unit 6, Lesson 1: Work is Converted to Energy

2. POTENTIAL ENERGY “stored energy”
In physics, “work” is done when an applied force causes an object to move.
The potential energy stored is equal to the work done creating it.
5 min

3. Gravitational Potential Energy
The work done to lift (or lower) something is equal to its change in gravitational potential energy. m
EP = m·g·h h
5 min
h=0 (reference point)

4. relative to the floor in the room.
Try This: Find the potential energy of a 60 kg student sitting on an 80 cm high stool…
relative to the floor in the room.
relative to the floor in the basement below (3.0 m lower).
a) EP = mgh
= 60 x 9.8 x 0.8
= 470 J
b) EP = mgh
10 min
= 60 x 9.8 x 3.8
= 2200 J

5. KINETIC ENERGY “the energy of motion”
Kinetic energy = the amount of work needed to accelerate a mass “m” from rest to velocity “v”
5 min
EK = ½ mv2

6. Brain Break!

7. Heat Energy
Definition: The total kinetic energy of the molecules in a substance.
Temperature: A measure of the average kinetic energy per molecule.
Q - amount of heat energy (J)
C - specific heat capacity of the substance
(energy needed to raise 1 kg by 1°C)
m - mass (kg)
ΔT - change in
temperature (°C)
10 min
Q = mcΔT

8. Example 1: Find the heat energy needed to raise a cup of water (250 g) from room temperature (20°C) to the boiling point.
Q = mcΔT
= 0.25 x 4200 x (100-20)
= J
Example 2 (Try it!): Find the final temperature when J of heat is added to a 3.0 kg concrete block at 20°C. (c = 2900 J / kg°C for concrete)
20 min
Q = mcΔT
20000 = 3.0 x 2900 x ДT
ДT = 2.3°C
T = 22.3 °C

9. Unit 6, Lesson 2: Work and Power

10. Work work = force x distance (Joules (J) = N x m)
Example: How much work is done by a crane that lifts 120 kg up 75 m?
F = mg
W = F·d
= m·g·d
= 120 x 9.8 x 75
= J
10 min

11. Power power = rate of doing work power = rate of using energy
power = work
(Watts (W) = J/s)
time
Example: Find the power output of a 60 kg athlete who climbs the Grouse Grind in 40 minutes. (Elevation gain = 800 m)
10 min
power = work time
= (mg)d t
= 60 x 9.8 x
= 196 W

12. Brain Break!

13. Practice:
Pg. 139 #1-5 and 8-10
35 min

14. Lesson 3: Human Power Lab
Once you’ve gathered your materials, you may go make your measurements on any hill you choose – just let me know where you are going! *

15. Unit 6, Lesson 4: Work-Energy Theorem

16. Work – Energy Theorem W = ΔE
The work done on an object by a net force equals the change in mechanical energy of that object.
5 min

17. Example 1
A 1000 kg car going 20 m/s brakes and comes to a stop. How much work is done? What happens to the lost kinetic energy?
EK = ½mv2
= ½(1000)(20)2
= J
W = ΔEK
= 0 – J
= – J
(lost kinetic energy)
10 min
The lost kinetic energy is converted to heat due to work done by friction.

18. Example 2 (Try It!)
How much work is done when a car of mass 1200 kg speeds up from 10 m/s to 20 m/s?
If this is done in 15 seconds, find the power output.
W = Δ EK
W = ½mv2 - ½mvo2
W = ½(1200)(20)2 - ½(1200)(10)2
= J
15 min
P = W t =
= W

19. Brain Break!

20. Example 3 (Try It!)
A 22 kg sled at rest is pushed 12 m with a force of 30 N against a friction force of 10 N. Find:
a) the work done by the pusher
b) the work done by friction
c) the kinetic energy gained by the sled
a) W = F·d
b) W = F·d
= 30x12
= -10x12
15 min
= 360 J
= -120 J
c) ΔEK = W (done by net force)
= 20x12
= 240 J

21. Example 4 (Try It!)
How much work is done to slow down a 1200 kg vehicle from 80 km/h to 50 km/h?
W = ΔEK
W = ½mv2 - ½mvo2
10 min
W = ½(1200)(13.889)2 - ½(1200)(22.222)2
= J

22. Unit 6, Lesson 5: Efficiency

23. Efficiency
Power Output
Energy Output
Efficiency =
x 100% =
x 100%
Power Input
Energy Input
Example 1: Find the efficiency of a 1200 W kettle that heats 1 cup (250 g) of water from 20 to 70°C in 1 minute.
Output: Q = mcДT
= (0.25)(4200)(70-20)
= J
15 min
Input: E = Pt
= (1200)(60)
= J
Eff = (52500/72000)x100
= 73%

24. Example 2 (Try it!): A crane operates at 15 kW and lifts a 700 kg mass 50 m in 2 minutes. Find the crane’s efficiency.
Input: E = Pt
= (15000)(120)
= J
Output: E = EP
= mgh
= (700)(9.8)(50)
= J
15 min
Eff = (343000) x100
= 19%

25. Brain Break!

26. Example 3 (Try it!): A 6.5 hp forklift is 15% efficient when lifting a 220 kg box. How long does it take to lift the box 1.5 m?
(1 hp = 746 W)
Output
Efficiency = 0.15 =
Input
Pin = 6.5 hp x 746 W/hp = 4849 W
15 min
EP/t
= mgh
0.15 = Pout/Pin =
= (220)(9.8)(1.5)
Pin
t Pin
t (4849)
t = 4.5 s

27. Practice:
Work and Power Problems #1
10 min

28. Unit 6, Lesson 6: Efficiency Practice

29. Example 4 (Try it!): An 800 kg 100-hp car is 22% efficient at transferring energy into forward motion. How long does it take to go from 0 to 108 km/h? (1 hp = 746 W)
Pout = W
t = E/P = ½ mv2 / Pout = ½ (800)(30)2 / = 22 s
Output
Efficiency = 0.22 =
Input
Pout_______
0.22 = Pout/Pin =
(100 hp)(746 W/hp)
15 min

30. Brain Break!

31. Practice:
Work and Power Problems #2-5
40 min

32. Unit 6, Lesson 7: Conservation of Energy

33. Conservation of Mechanical Energy
Total Mechanical Energy of an object moving in a gravitational field:
E = EP + EK
E = mgh + ½mv2
Conservation of Energy Law: If no other force (besides gravity) does work on an object, the total mechanical energy is conserved.
5 min
Eo = Efinal

34. Example 1 (Try it!):
Matt kicks a soccer ball of mass 0.6 kg with an initial speed of 15 m/s.
Find its
total mechanical energy
kinetic energy at height 4.0 m
speed when at 4.0 m
height when the speed is 5.0 m/s. v h
15m/s
15 min a)
E = mgh + ½mv2
= 0 + ½ (0.6)(15)2
= 67.5 J

35. b) c) EK = ½mv2 E = EP + EK E = mgh + EK EK = ½(0.6)v2
b) kinetic energy at height 4.0 m
c) speed when at 4.0 m
d) height when the speed is 5.0 m/s
m = 0.6 kg
ET = 67.5 J b)
E = EP + EK c)
EK = ½mv2
E = mgh + EK
EK = ½(0.6)v2
67.5 = (0.6)(9.8)(4) + EK
v = 12 m/s
EK = 44 J d)
10 min
E = mgh + ½mv2
67.5 = (0.6)(9.8) h + ½ (0.6)(5)2
67.5 = 5.88h + 7.5
h = 10.2 m

36. Brain Break!

37. Example 2 (Try it!): Christine (75 kg including her gear) skis straight down an icy slope, from rest. After she has lost 250 m of elevation,
a) find her final velocity assuming no effects from air-resistance or friction.
b) if her final velocity is 210 km/h, find the total amount of energy lost due to friction forces. What happens to the “lost” energy? a)
Etop = Ebottom
0m/s
mgh = ½ mv2
15 min
gh = ½ v2
2gh = v2
2(9.8)(250) = v2
250m
v = 70 m/s (252 km/h)
h = 0

38. v = 210 = 58.3 m/s 3.6 ΔE = Ebottom – Etop = ½ mv2 - mgh
b) if her final velocity is 210 km/h, find the total amount of energy lost due to friction forces. What happens to the “lost” energy?
v = = 58.3 m/s
3.6
ΔE = Ebottom – Etop
= ½ mv2
- mgh
= ½ (75)(58.3)2 - (75)(9.8)(250)
5 min
= J (converts to heat)

39. Practice:
Conservation of Energy Problems #1
5 min

40. Unit 6, Lesson 8: Practice with Conservation Law

41. Find his velocity when he swings past ground level.
Practice #1: Tarzan, a 75 kg ape-man, swings from a branch 3.0 m above the ground with an initial speed of 5.0 m/s.
Find his velocity when he swings past ground level.
Find the maximum height he swings to.
What should his initial speed be if he is to just reach a branch 4.0 m high?
b) & c)
20 min
3.0 m a)
h = 0
Note: In the physics jungle there is no air resistance and the vines have no mass and are unstretchable!

42. a) mgho + ½mvo2 = mgh + ½mv2 m cancels out; h = 0 gho + ½vo2 = 0 + ½v2
v = √(vo2 + 2gho)
v = √((5.0)2 + 2(9.8)(3.0))
v = 9.2 m/s horizontal b)
gho + ½vo2 = 0 + gh
v = 0
h = vo2 + ho 2g
5 min
h = 4.3 m c)
gho + ½vo2 = 0 + gh
v = 0
vo = √(2gh - 2gho)
vo = 4.4 m/s

43. Brain Break!

44. Practice #2: George of the Jungle swings from the same 3
Practice #2: George of the Jungle swings from the same 3.0 m high branch with an initial speed of 3.5 m/s. Unfortunately, he hits a tree when he is1.5 m above the ground. Find his speed just before impact and the work done by the tree. His mass is 90 kg.
15 min
3.0 m
1.5 m

45. mgho + ½mvo2 = mgh + ½mv2 m cancels out gho + ½vo2 = gh + ½v2
v = √(vo2 + 2gho – 2gh)
v = √((3.5)2 + 2(9.8)(3.0) – 2(9.8)(1.5))
v = 6.5 m/s
W = ΔKE = KE - KEo
W = ½mv2 – ½mvo2
5 min
W = 0 – ½(90)(6.5)2
W = J

46. Practice:
Conservation of Energy Problems #2 and 3
10 min

47. Unit 6, Lesson 9: More Practice with Conservation Law
Jacob
Clare
Travis
McKenna
Chase
Balance
Carli
Nathan
Taylor
Anna
Aidan
Finley
Amira
Ashley
Kally
Annie

48. Heat = Work done by friction W = Ef - Eo W = ½ mv2 - mgh
Practice #3: How much heat energy is produced when a speed skier loses 300 m in elevation if he starts from rest and finishes going 40 m/s? His mass (including gear) is 90 kg.
Heat = Work done by friction
0m/s
W = Ef - Eo
W = ½ mv2 - mgh
W = ½(90)(40)2 – (90)(9.8)(300)
W = -1.9 x 105 J
300m
15 min
40m/s
J of heat are produced
h = 0

49. Heat energy gained by the pool = KE lost by the meteor
Practice #4:
A 5 kg meteorite lands in a swimming pool with L of water. The temperature rises 5°C. Find the velocity of the meteor right before it hits the water.
Q = mcΔT
= x 4200 x 5
= 2.1 x 109 J
Heat energy gained by the pool = KE lost by the meteor
Q = KE = ½mv2
2.1 x 109 = ½(5)v2
v = m/s
15 min

50. Brain Break!

51. Practice #5:
A 20 kg meteor crashes into a 1.5m-deep pool of dimensions 4m X 6m at 17 km/s. Find the final temperature if the initial temp is 22C and all of the meteor’s energy heats the water. (Density of H2O = 1000 kg/m3.)
Q = KE
= (0.5)(20)(17 000)2
= 2.89x109 J
mc∆T = Q
[(1.5 x 4 x 6) x 1000] x 4200 x ∆T = 2.89x109
36000 x 4200 x ∆T = 2.89x109
∆T = 19C
∆T = = 41C
20 min

52. Practice:
Conservation of Energy Problems #4 and 5
10 min