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

1 m/s

KE = 1/2 mv2

KE = 1/2 2kg (1m/s)2

KE = 1 kg m 2 /s 2 = 1 Joule (J)

Kinetic Energy of a two liter bottle of water moving at 1 m/s

Gravitational Potential Energy is simply energy afforded by relative height, the potential to fall. This energy can convert toKinetic Energy.

So, lets say a 100 kg person stands on a 10 m cliff near sea levelGPE = mgh

GPE = (100kg)(9.8m/s2)(10m)

GPE = 9800 kgm2/s2

James Prescott Joule:

British largely self-trained physicist1818-1889

Fascinated by electricity, he and his brother usedto experiment by giving shocks to each other and the family servants

Credited with the First law ofThermodynamics

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 fastIs he going?

Why can’t he make it back to the branch hestarted 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

So lets say I push the mower for 60 m with a force of 200 N at a50 degree angle to the horizontal. How much work gets done on my lawn mower.

W = F d cos

W = (200N)(60m)(cos

W = 7700 J

Lift a wheelbarrow at 30o above the

horizontal, push for 75 m, with a force of

500 N • W = F d cos • W = (500N)(75m) (cos

• 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

http://www.njwind.com/webcam.html

Atlantifc city wind turbines each produces 1.5 MW

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:mechanical advantage

Lever terminology

Advantage of lever: Force ratios are

proportional to lever arms• MA = Fr/Fe

• MA = mechanical advantage

• 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:

2006Jake 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?

• Fulcrum ahead of the 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 advantage

• 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 • Observe number

of lines lifting object

• MA =• 1• Fe then is ?• 10 N• This pulley just changes direction of effort…

10 N

Fe

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 ?• 5 N

10 N

Block and tackle fixed to ceiling,

• Observe number of lines, quadruple the distance of cable would be used

• MA =• 4• Fe then is ?• 2.5 N• What about friction?10 N

MA for inclined planes

slope

rise

MA = slope/rise

The Inclined Plane at Ronquieres,Belgium: a moving boat lock.

Slope is 1432 m longRise 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 Actual Mechanical Advantage (AMA)

• 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 knownU = Elastic potential energyX = the draw length

A bow stores potential energy

U = 1/2 Ffd dU = 1/2 (45lbs)( 4.45N/lb) (0.38m)U = (100N) (0.38m)U = 38 Joules

The bow converts the elastic potential to

Kinetic EnergyKE = 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 sin2

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 sin2/g

R = vo sin2(/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 sin2/g

R = vo sin2(/g

A compound bow

cam

4186.8 J/Kcal