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Power and work and efficiency. In text Chapt er 10-12. Additional Note. In case you did read in your text, which I would prefer if you did you may have missed the following definition… Mechanical energy: Is the sum of potential and kinetic energy in a system.
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Power and work and efficiency In text Chapter 10-12
Additional Note In case you did read in your text, which I would prefer if you did you may have missed the following definition… Mechanical energy: Is the sum of potential and kinetic energy in a system. Same rules apply to any type of energy or system, that because the law of ______________ tells us that __________ before has to be the same as ___________ after
Pa-Pa-Pa Power Wheels!!! We all realize I think that it takes power to get stuff done, stuff like work, but what is power? Great question: Power is how much work we do, over how long we take to do it! P= W/t where w= work and t=time The unit of measure for power is the watt (named after James Watt) and given as a w. “What’s a watt Mr. Caddy?” Who said that? They get a cookie. A watt is 1J of energy transferred over 1 second. OR a watt is a J/s.
Example EZ How much power is needed to lift a cup 2m that has a mass of 3 kg in 2 seconds? P=w/t Work is F x d F is mg So P= fd/t or 3 x 9.8 x 2/2 P= 29.4 W or J/s As a side note, not that you don’t know this but a kilowatt is 1000w
Example harder An electric motor lifts an elevator 9m in 15 seconds but exerting an upward force of 12000N. What is the power required in kW? Known: 9m, t=15s, f=12000N Unknown: Power, work P= w/t , w=fd So again 12000x 9 / 15 = 7200W 7200W but I want kW so divide by 1000 and 7.2 kW
Efficiency Efficiency: The rate or ratio or percent of how effectively something or someone turns input energy into useful energy. Eff %= (useful energy/ input energy) x 100 This applies to anything…like work etc, output/input…..kinetic/potential…..
Example Ez Calculate the efficiency of a light bulb in converting energy if its given 1000W of energy and only 60W of that is used to illuminate the room? Eff= 60/1000 = 0.06x 100 = 6%
Example Harder • Suppose when you were at the top of the hill, your height was 25 m, and when you coasted down the hill, your 68kg body had a final speed of 13m/s. How efficient was I at converting types of energy? • Write out what you know • Given • m = 68kg • ∆h = 25 • v = 13m/s • g = 9.80m/s2 • Need: Efficiency = output x 100% input Efficiency = Ek x 100% Ep
Gravitational Potential Energy Ep = mg∆h = (68kg)(9.80m/s2)(25m) = 16677J • Kinetic Energy Ek = ½ mv2 = 0.5(68kg)(13m/s)2 = 5746J • Efficiency = output x 100% input • Efficiency = 5746J x 100% 16677J • Efficiency = 34.4564% • Efficiency = 34%
Problems in Text You can now do Page 262 4-8 work type problems Page 264 9-14 power problems Page 278-281 review questions on work, power and efficiency Page 291 4-8 potential energy questions Page 297 15 -15 kinetic and potential energy questions Page 300, 19-21 putting together mo, kinetic and conservation