A it depends how fast you do it f f f du w f

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Unformatted text preview: W= F (x power dx by a dU = a Mechanical)dx = − delivered = − force on Ui − Uf dx i i i moving particle: Work = W = Kf − Ki = −Uf + Ui Thus : Ki the i = and the f P = power, depends on+ UforceKf + Uvelocity. A simple formula: (W = work) E = K + U is conserved ￿r dW F · d￿ ￿v P= = = F ·￿ dt dt Example: lift 100 kg a height of 1 meter in 1 second: P = (100 kg ) (g) (1m/s) = 31000 Watts = 10 light bulbs Demo: bike generator Q: How many horses does it take to replace a light bulb? A: 0.13 horses 1 horsepower = 746 Watt. 1 lightbulb/1horse = 100 W/746W = 0.13 Automotive horsepower... How much power should a car engine provide in order for the car to be capable of going up a 30 degree hill at a speed of 100 km/hr? (=28 m/s). Assume the car’s mass is 1000kg. v 30° Fg = mg sin 30° P= mg sin30°v = (1000)(9.8) sin 30°(28)/746 = 184 hp Hydropower — Gravitational Potential Energy Hoover Dam — Colorado River (Nevada/Arizona) Hydropower — Gravitational Potential Energy Niagara Falls — NY/Canada 57 m drop 2800 m3/s flow rate Niagara Falls Horseshoe Falls, Canada Robert Moses Hydroelectric Power Plant & Reservoir Lewiston, NY Hydropower In time Δt: — Gravitational Potential Energy ΔV H (V= volume) ΔV ΔUg = Δm g H = ΔV ρwater g H Available power: P = g ΔU g Δt = Δm Δt g H = ΔV Δt ρ water gH P = (flow rate) (density of water) g H Ex. Niagara Falls H = 57 m flow rate P = g ΔV Δt ρ water ΔV Δt ~ 2800 m3 / s gH ~ (2.8 x 103m3/s)(1.0 x 103 kg/m3)(9.8 N/kg)(57 m) = 1600 megaWatts Aside: ¿What’s wrong with this picture‽ No “adhesion”, i.e. no glue, only static friction what’s wrong with this picture...... Recall if held by static friction, tan θmax = μs But based on the picture θmax > 80° Thus, μs > 5.7 ! But I said, “for all practical purposes” μs < 1 hmmmm........ Gecko inspired super-friction: Cartoon of friction synthetic nanofiber (what is nano? A: the diameter of the fiber, 600nm) Gecko’s feet Center of mass and Momentum * So far: we’ve treated objects as point particles. * We now consider co!ections of particles or extended objects. * We wi! see: there is a special point ca!ed the CENTER OF MASS that behaves like a point particle in response to external forces. * This leads to MOMENTUM CONSERV ATION * A! these new concepts are derived "om Newton’s Laws. Center of Mass Motivating demo: projectile motion of extended objects. Observe: there is a special point on the object that has a perfect parabolic trajectory as for projectile motion. The object does not have to be rigid: De...
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This note was uploaded on 09/29/2012 for the course MATH 1920 taught by Professor Pantano during the Spring '06 term at Cornell.

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