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```Energy + Power
Energy of Motion and Simple
Machines
Energy: Comes in many forms
among these:
• Kinetic Energy
(Energy of moving Objects)
KE = 1/2 mv2
• Gravitational Potential Energy
(Energy of falling objects)
PE = mgh
Kinetic Energy of a two liter bottle of water moving at 1 m/s
KE = 1/2 mv2
KE = 1/2 2kg (1m/s)2
KE = 1 kg m 2 /s 2 = 1 Joule (J)
1 m/s
Gravitational Potential Energy is simply energy afforded by
relative height, the potential to fall. This energy can convert to
Kinetic Energy.
So, lets say a 100 kg person stands on a 10 m cliff near sea level
GPE = mgh
GPE = (100kg)(9.8m/s2)(10m)
GPE = 9800 kgm2/s2
James Prescott Joule:
British largely
self-trained physicist
1818-1889
Fascinated by electricity,
he and his brother used
to experiment by giving
shocks to each other
and the family servants
Credited with the
First law of
Thermodynamics
Mechanical Equivalent of Heat
The falling mass yields
energy according to
GPE = mgh
The water gains heat
Energy by being stirred…
Kinetic energy, temperature
A visit with Tarzan and family:
50 m above the forest floor, the Tarzans wait to swing.
100 kg Tarzan swings
and grabs a banana from
the forest floor. How fast
Is he going?
Why can’t he make it
back to the branch he
started on?
Tarzan’s Swing
• GPE = mgh
• GPE = (100kg)(9.8 m/s2)(50m)
• GPE = 49,000 Joules
• If all GPE becomes KE at the bottom of the Swing
GPE = KE = 1/2 mv2
• 49,000 kgm2/s2 = 1/2 100kg v2
• v = 31.3 m/s
~ 112 km/h
Why doesn’t Tarzan make it back to the branch? Energy is lost…
to the system:
Drag in air…air and vine get warmer as Tarzan swings.
Energy is the ability to do Work (W)
W = Fd
Lets say I push
with a 500 N
Force for 100 m
W = 50,000 N-m
W= 50,000 J
Force exerted at an angle to the direction of effective work is
proportional to the cosine of the angle
W= F d cos Q
So lets say I push the mower for 60 m with a force of 200 N at a
50 degree angle to the horizontal. How much work gets done on
my lawn mower.
W = F d cos Q
W = (200N)(60m)(cos 50o)
W = 7700 J
o
30
Lift a wheelbarrow at
above
the horizontal, push for 75 m,
with a force of 500 N
• W = F d cos Q
• W = (500N)(75m) (cos 30o)
• W = 32,500 J
Backing a wheelbarrow up a stair makes the pulling vector
more in alignment with the direction of desired work
Power is the rate at which work or energy can be produced
P = W/t
Power is the rate at which work or energy can be produced
P = W/t
Power is measured in J/s = Watts
Power Plant
Atlantic City wind turbines
Wind Power, rate produced
•
•
•
•
•
•
•
(3) 50m x 1 m blades in a 10 m/s wind.
Air has a density of about 1.2 kg/m3
150 m2 hit w/ 1.2 kg/m3 x 10 s= 1800 kg/s
KE/s = (1/2 mv2 ) /s
KE/s = (1/2 1800 kg (10 m/s)2 )/s
KE/s = (90,000 kg m 2 /s2 )/s
KE/s = 90,000 J/s = 90 kW
Atlantifc city wind turbines each produces 1.5 MW
http://www.njwind.com/webcam.html
Hoover Dam, CA NV
A 50 m Hydroelectric Dam 1 m3
water passes the turbine in 0.6 s
• What power is produced?
• PE/s = 1/2 mgh/t
• PE/s
=1/2(1000kg)(9.8m/s2)(50m)
/(0.6s)
PE/s = 408,000 W = Power of
that turbine
(avg PE for water column =1/2mgh)
Simple Machines:
Lever terminology
are proportional to lever arms
• MA = Fr/Fe
• MA = mechanical
• Fr = resistance force
(exerted by the
machine)
• Fe = exertion force
(exerted by you)
What force is necessary in the gluteus maximus to lift 115 lbs
if the torso is 10x the length of the pelvis?
A Third class lever
A machine can increase force but it can’t increase energy…
Great Moments in Physics:
2006
Jake Wulff invents
“Privy Prop”
What kind of lever is Homer?
• Homer has a fulcrum at
his waist, so 1st class…
• If you push his head down
5cm with a force of 40 N,
how much force is applied
to lift the edge of the cap
rising 0.5 cm?
• Wi = Wo, Fed = Frd
• (40N)( .05m) = Fr(.005m)
• 400N = Fr
What kind of lever is this?
resistance, so 2nd class…
• If you push the handle
down
4 cm with a force
of 40 N, how much force is
applied to lift the edge of
the cap rising 0.5 cm?
• Wi = Wo, Fed = Frd
• (40N)( .04m) = Fr(.005m)
• 320 N = Fr
The ratio of lengths of lever to
fulcrum are the same as de:dr
• Longer lever, more
mechanicaal
• MA = Fr/Fe
Let’s say the rock is 1500 N the
man weighs 1000. N, the lever is
3 meters long and the fulcrum is
placed 1 meter from the end
• Whats the AMA
• MA = Fr/Fe
• MA = 1500N/1000N
• AMA = 1.5
How efficient is this system?
• AMA/ IMA = Efficiency
• 1.5/2 = 75%
• Or Wout/Win = Efficiency
• Eff = (1500 N) 0.1m
/(1000N) 0.2
• = 75%
What does a bat do?
• Lets say I can swing
with a force of 200N,
my second hand is the
fulcrum, 5 cm away.
The bat is 60 cm to the
“sweet spot”…How
much force do I apply
there?
Third class lever: a bat
• Fe de= Fr dr
• (200N)(.05m) = Fr(0.6m)
• 16.7 N…so why does the bat work?
• The distance traveled by a mass at the end is much
greater, so much faster
• 3 vertical lines raising
object…
• MA = 3
• Fr = 600N
• What’s Fe ?
• Fe = 200N
Block fixed to ceiling, single line
through block attached to mass
Fe
10 N
• Observe number of
lines lifting object
• MA =
• 1
• Fe then is ?
• 10 N
• This pulley just
changes direction of
effort…
Cable fixed to ceiling, single
block attached to mass
• Observe number of
lines, twice the
distance of cable
would be used
• MA =
• 2
• Fe then is ?
• 5N
10 N
Block and tackle fixed to ceiling,
10 N
• Observe number of
distance of cable
would be used
• MA =
• 4
• Fe then is ?
• 2.5 N
MA for inclined planes
slope
rise
MA = slope/rise
The Inclined Plane at Ronquieres,Belgium: a moving boat lock.
Slope is 1432 m long
Rise is 68 m high
MA = ?
91m x 12m x 3.5m of water in a caisson with a mass of 10 tones
Assuming 80% efficiency what is the force necessary to raise to lock?
Effort force at the Roquières
Inclined Plane
• (91m x 12 x 3.5m) + 10 tones = 3833 tones
• 3,833,000 kg x 9.8 m/s2 = Fr
• Fr = 3.75634 x 107 N
• MA = s/r = 1432m / 68m
• MA = 21.06
• MA = Fr/Fe
• Fe = 3.75634 x 107 N / 21.06
• Fe = 1.78 x 106N (for ideal mechanical advantage)
Efficiency of the Roquières
Inclined Plane
• Fe = 1.78 x 106N (for ideal mechanical advantage)
• But friction causes reduced mechanical advantage or
• Efficiency = AMA/IMA= Fo/Fi = Wo/ Wi
• Efficiency = 80% = 0.8 = 1.78 x 106N / Fi
• It will require 2.23 x 106 N to move the caisson
Work Input in the Roquières
Inclined Plane
• It will require 2.23 x 106 N to move the caisson.
• Its got to move 1432m…W = Fd
• W = 2.23 x 106 N x 1432m
• W = 3.192 x 109 N-m
Power Input in the Roquières
Inclined Plane
• It takes 45 minutes total, 20 minutes to rise, at a speed of
1.2 m/s.
• How much power is required to move the caisson?
• 2.23 x 106 N to move the caisson.
• P = Fv
• P = 2.23 x 106 N x1.2 m/s
• P = 2.68 x 106 N-m/s = 2.68 MW
• P = W/t
• P = (3.192 x 109N-m) / (20min)(60s/min)
• P = 2.66 x 106 W = 2.66 MW
Another type
Elastic Potential Energy (U)
• U = average F X d
Henry V
Agincourt
Falkirk
What stretches?
If the spring constant (k) is known
U = Elastic potential energy
X = the draw length
A bow stores potential energy
U = 1/2 Ffd d
U = 1/2 (45lbs)( 4.45N/lb) (0.38m)
U = (100N) (0.38m)
U = 38 Joules
The bow converts the elastic
potential to Kinetic Energy
KE = 1/2mv2
38 Joules = (1/2) 0.035 kg)(v2)
v = 46.6 m/s
If the arrow is fired at 45o what is
the range?
R = vo2 sin2q
g
R = (46.6m/s)2 sin2(45o)
9.8 m/s2
R = 220 m
OK so let’s fire an arrow into the air.
If our t is really half the flight
v = vo + at, so
vo = 9.8 m/s2 (t/2)
Range = R = vo sin2q/g
R = vo sin2(45o)/g
OK so let’s fire an arrow into the air.
If our t is really half the flight
v = vo + at, so
vo = 9.8 m/s2 (t/2)
Range = R = vo sin2q/g
R = vo sin2(45o)/g
A compound bow
cam
4186.8 J/Kcal
```
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