Transcript Work and Energy

Physical Science
Work and Energy
Slides subject to change
Work and Energy
Work refers to an activity involving force
and motion.
 The force is in the direction of the motion.

What 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
Example
Given
 F1 = 800 N
 d = 20 m
Formula
W = Fd
F1
0
d = 20 meters
Example
Given
Formula
 F1 = 800 N
W = Fd
 d = 20 m
 W = (800)(20) = 16,000 J
F1
0
d = 20 meters
Luggage
Given
Formula
 F2 = 400 N
W = Fd
 Motion in direction of F2 = 0 m.
 d=0.
luggage
F2
0
20 meters
Luggage
Given
Formula
 F2 = 400 N
W = Fd
 Motion in direction of F2 = 0 m.
 d=0.
 W = (400)(0) = 0 J
 F2 (luggage weight)
does NO work.
0
20 meters
Weightlifting
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

Kinetic Energy
Energy associated with motion.
 Motion = Gr. kinesis

KE = ½ mv2

Energy units in SI system are joules (J)
v
Kinetic Energy
Energy associated with motion.
Given
Formula

m = 1,000 kg
 v = 13 m/s


KE = ½ mv2
KE = (½)(1000)(13)2 = 84,500 J
Energy associated with motion
Kinetic Energy
Mass
Speed
KE = ½ mv2
13 m/s
84,500 J
26 m/s
338,000 J
1,000 kg
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.

Energy
When positive work is done, there is an
increase of kinetic energy.
 When negative work is done, there is an
decrease of kinetic energy.


ΔW = ΔKE , or amount of work done
equals the change in kinetic energy.
“Δ” means “change in …”
Kingda Ka
 How much work does is take
 Calculate how much kinetic energy
to accelerate Kingda Ka from
gains.
0 to 128 mi/h (57 m/s)?
Assume that all work goes
into kinetic energy.
it
Mass
Speed
KE = ½ mv2
Initial
9,000 kg
0 m/s
0J
Final
9,000 kg
57 m/s
14.6x106 J
Kingda Ka
Initial KE is 0 J.
 Final KE is 14.6x106 J.
ΔE = KEf − KEi = 14.6x106 − 0
= 14.6x106 J


Since W = ΔKE, it must take 14.6x106 J of
work to accelerate the Kingda Ka car.
Potential 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:

PE = mgh
mg = weight of an object
 h = height raised

Other PE Examples

Springs

Archery Bow
Potential 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)
70

= 38,200 J

20
139
Kingda Ka

0
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
Mechanical Energy

Total mechanical energy E is the total of
kinetic energy and potential energy.
E = KE + PE
Conservation 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.”
An Example

Slowly lift a 10-kg anvil to h = 5 meters
and hold it there. What is its total energy at
h?
 Initally
5m
 KE = 0
 PE = mgh = 10(9.8)(5) = 490 J
 Einitial = KE + PE = 0 + 490 = 490 J
0m
An 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

5m
0m
An 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

A Pendulum
At h = 3 m above equilibrium (the lowest
point), and at rest.
 m = 5 kg
3m
 KE = 0
 PE = mgh = 5(9.8)(3) = 147 J
 Einitial = 147 J

Equilibrium
0m
A 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

Look at Energy

Total is always the same …
KE = 147 J
PE = 147 J
PE
KE
PE
KE
PE = 0
KE = 0
initial
in-between, a
mix of both
final
Power

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.
Power 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

Example

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) = 4.2x10–5 J


P = Work/Δt = 4.2x10–5 /0.40 = 1.1x10–4 W
Example
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 Δh = 80(9.8)(3.8) = 3,000 J


P = Work/Δt = 3,000/5.0 = 600 watts
Kingda Ka
Recall work to accelerate Kingda Ka from
0 to 128 mi/h (57 m/s)
 Work = ΔKE = 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

Kinda Ka
Convert 4.9x106 W to
horsepower (hp).
 1 hp = 746 W


P = 6,500 hp

Kinda Ka’s hydraulic motors are actually
rated at 7,400 hp.
Top 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 = ΔKE = 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

Energy Delivered
Power P = W/t
 Rearrange
W=Pt
 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

Measure 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
Measure 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¢

Measure 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

 @17¢
 In
/ kW-hr = 41¢
one month (30 days) ~ $12.30
U.S. Sources of Energy
US DOE, 2009
Utility-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.

Coal Power

Longwall
Coal
NPR: Visualizing
theMining
U.S. Grid
Longwall
Longwall Issues
Gas Power
NPR: Visualizing the U.S. Grid
Nuclear Power
NPR: Visualizing the U.S. Grid
Hydroelectric Power
NPR: Visualizing the U.S. Grid
Solar Power
NPR: Visualizing the U.S. Grid
Historical View
Energy Information Administration /
Annual Energy Review 2007
http://www.eia.doe.gov/cneaf/electricity/epa/epa_sum.html