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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: 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: 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? • 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 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 lines, quadruple the distance of cable would be used • MA = • 4 • Fe then is ? • 2.5 N • What about friction? 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 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 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