Title: Physical Science
1Physical Science
- Work and Energy
- Slides subject to change
2Work and Energy
- Work refers to an activity involving force and
motion. - The force is in the direction of the motion.
3What Is Work?
- Work done by a constant force F equals the
product of the force and parallel distance d
through which the object moves while the force is
applied. - Units are N-m or joules (J).
force
distance
W Fd
4Example
- Given Formula
- F1 800 N W Fd
- d 20 m
F1
d 20 meters
0
5Example
- Given Formula
- F1 800 N W Fd
- d 20 m
- W (800)(20) 16,000 J
F1
d 20 meters
0
6Luggage
- Given Formula
- F2 400 N W Fd
- Motion in direction of F2 0 m.
- d0.
luggage F2
0
20 meters
7Luggage
- Given Formula
- F2 400 N W Fd
- Motion in direction of F2 0 m.
- d0.
- W (400)(0) 0 J
- F2 (luggage weight) does NO work.
0
20 meters
8Weightlifting
- Lift m 23 kg, d 2 meters.
- F mg 23(9.8) 225 N
- W Fd (225)(2) 450 J
- Hold it there for 300 seconds.
- How much additional work does it take?
- d 0
- W (1000)(0) 0 J
9Kinetic Energy
- Energy associated with motion.
- Motion Gr. kinesis
- KE ½ mv2
- Energy units in SI system are joules (J)
v
10Kinetic Energy
- Energy associated with motion.
- Given Formula
- m 1,000 kg KE ½ mv2
- v 13 m/s
- KE (½)(1000)(13)2 84,500 J
Energy associated with motion
11Kinetic Energy
Mass Speed KE ½ mv2
1,000 kg 13 m/s 84,500 J
1,000 kg 26 m/s 338,000 J
- As speed goes up, the kinetic energy goes up as
the speed squared! - Double the speed from 13 m/s to 26 m/s, energy
increases four-fold.
12Energy
- When positive work is done, there is an increase
of kinetic energy. - When negative work is done, there is an decrease
of kinetic energy. - ?W ?KE , or amount of work done equals the
change in kinetic energy.
? means change in
13Kingda Ka
- How much work does is take to accelerate Kingda
Ka from 0 to 128 mi/h (57 m/s)? Assume that all
work goes into kinetic energy.
- Calculate how much kinetic energy it gains.
Mass Speed KE ½ mv2
Initial 9,000 kg 0 m/s 0 J
Final 9,000 kg 57 m/s 14.6x106 J
14Kingda Ka
- Initial KE is 0 J.
- Final KE is 14.6x106 J.
- ?E KEf - KEi 14.6x106 - 0
- 14.6x106 J
- Since W ?KE, it must take 14.6x106 J of
- work to accelerate the Kingda Ka car.
15Potential Energy
- Work can also be done to change the position of
an object and create potential energy (PE). - Example work done against gravity to lift an
object and give it potential energy. - Gravitational potential energy
-
- mg weight of an object
- h height raised
PE mgh
16Other PE Examples
17Potential Energy
- Gravitational PE comes from raising the height of
mass m. - Given Formula
- m 80 kg PE mgh
- h 50 m
- PE mg h
- (80)(9.8)(50)
- 38,200 J
70
20
18Kingda Ka
139
- How much gravitational potential energy does
Kingda Ka gain by climbing from from 0 m to 139
m? (h 139 m)
Mass Height PE mgh
9,000 kg 139 m 12.3x106 J
0
19Mechanical Energy
- Total mechanical energy E is the total of kinetic
energy and potential energy. -
- E KE PE
20Conservation of Energy
- The total mechanical energy of an isolated system
remains constant. - Initial total energy equals final total energy.
- Ebefore Eafter
- (KE PE)before (KE PE)after
- Isolated means no work is done from the outside
that affects the system.
21An Example
- Slowly lift a 10-kg anvil to h 5 meters and
hold it there. What is its total energy at h? - Initally
- KE 0
- PE mgh 10(9.8)(5) 490 J
- Einitial KE PE 0 490 490 J
5 m
0 m
22An Example
- Release it, it falls to h 0. Without your hands
in the way, the anvil and gravity form an
isolated system. - Finally
- KE ½ mv2
- PE 0
- Efinal (KE PE)final ½ mv2
5 m
0 m
23An Example
- As it falls, by conservation of energy,
- Ebefore Eafter
- (KE PE)before (KE PE)after
- (490) (½ mv2)
- Solve for v,
- v2 2(490)/10 98
- v 9.9 m/s
24A Pendulum
- At h 3 m above equilibrium (the lowest point),
and at rest. - m 5 kg
- KE 0
- PE mgh 5(9.8)(3) 147 J
- Einitial 147 J
3 m
0 m
Equilibrium
25A Pendulum
- At lowest point h 0 m
- m 5 kg
- KE ½ mv2
- PE mgh 0 J
- Efinal
- ½ (5)v2 2.5v2
- Conservation of Energy Einitial Efinal
- 147 2.5 v2
- v 6.8 m/s
26Look at Energy
KE 147 J
PE 147 J
PE
PE
KE
KE
KE 0
PE 0
in-between, a mix of both
initial
final
27(No Transcript)
28Power
- WORK Done on an object, and results in a
transfer of energy. - W Fd
- POWER The rate of this energy transfer.
- P W/t
- Example A 60 watt light bulb transforms 60
joules of electric energy into thermal energy and
light in 1 second.
29Power Units
- MKS unit for power is watt (W).
- 1 joule per second 1 watt (W)
- In the U.S. system, also measure power in
horsepower (hp). - 1 hp 746 W
30Example
- A 0.0020 N force pushes a piece of broccoli 2.1
cm in a time of 0.40 seconds. Calculate the power
output. - F 0.0020 N
- d 2.1 cm or 0.021 m
- Work Fd (0.0020)(0.021) 4.2x105 J
- P Work/?t 4.2x105 /0.40 1.1x104 W
31Example
- John runs up 20 stairs in 5.0 seconds. He has a
mass of 80 kg. What amount of power has John
generated? Each step is 19 cm. - Step height 19 cm, or 0.19 meter.
- Height raised h 20 (0.19) 3.8 m
- Work Fd mg ?h 80(9.8)(3.8) 3,000 J
- P Work/?t 3,000/5.0 600 watts
32Kingda Ka
- Recall work to accelerate Kingda Ka from 0 to 128
mi/h (57 m/s) - Work ?KE 14.6x106 joules
- Recall, it does this in t 3.5 seconds.
- What is the minimum power required of the
hydraulic motors? - P Work/t 14.6x106 J/3.5 s (at least).
- 4.9x106 watts
33Kinda Ka
- Convert 4.9x106 W to horsepower (hp).
- 1 hp 746 W
- P 6,500 hp
- Kinda Kas hydraulic motors are actually rated at
7,400 hp.
34Top Fuel Dragster
- Mass 1,000 kg, v 148 m/s in 4.5 s
- ½ mv2 ½ (1000)(148)2 11.0 x 106 J
- Work ?KE 11.0 x 106 joules
- P Work/t 11.0 x 106/4.5 2.43 x 106 W
- Convert to hp
- 3,300 hp
35Energy Delivered
- Power P W/t
- Rearrange
- W P t
- The energy delivered, or work done P t
- How much work or energy (in joules) is sent to a
light bulb in 60 seconds? - W P t (100)(60) 6,000 J
36Measure Energy Delivered
- Power companies bill customers according to
energy or work done in kilowatt-hours. - Price in LA County about 17 /kW-hr (1 kW (1,000
watts) consumed over 1 hour). - Units are kW and hours.
power
time
37Measure Energy Delivered
- In Power Company units
- How much energy is sent to a 1,300-W hair dryer
in 15 minutes? In kW-hrs. - P 1,300 W 1.3 kW
- t 15 min 0.25 hr
- W P t (1300)(0.25) 0.325 kW-hr
- At 17 / kW-hr, what is the cost?
- Cost (0.325 kW-hr)(17/kW-hr) 6
38Measure Energy Delivered
- Power of 100-W bulb 100 W 0.10 kW
- In one hour it uses
- W Pt (0.10)(1) 0.10 kW-hr
- In 24 hours it uses
- W Pt (0.10)(24) 2.4 kW-hr
- _at_17 / kW-hr 41
- In one month (30 days) 12.30
39U.S. Sources of Energy
40Utility-Scale Energy Sources
- Fossil energy - burn fossil fuels, heat water,
and create steam, turn turbine. - Nuclear heat water, create steam, turn turbine.
- Geothermal steam from deep in Earth (6000 feet
Hawaii), turn turbine. - Gas turbine hot gases direct into turbine.
- Wind wind turns turbine.
- Hydroelectric water pressure turns turbine.
41Coal Power
NPR Visualizing the U.S. Grid
42Longwall Issues
Longwall
43Gas Power
NPR Visualizing the U.S. Grid
44Nuclear Power
NPR Visualizing the U.S. Grid
45Hydroelectric Power
NPR Visualizing the U.S. Grid
46Solar Power
NPR Visualizing the U.S. Grid
47Historical View
Energy Information Administration / Annual Energy
Review 2007
http//www.eia.doe.gov/cneaf/electricity/epa/epa_s
um.html