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Peter[Deuce]
United States93 Posts
A 1.0 x 10^3 kg elevator carries a maximum load of 800 kg. A constant frictional force of 4.0 x 10^3 N retards the elevator's motion upward. What minimum power, in kilowatts, must the motor deliver to lift the fully loaded elevator at a constant speed of 3.00 m/s. + Show Spoiler + 66kW TY   | ||
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Kula
United States140 Posts
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Luddite
United States2315 Posts
so, recall that potential energy from height = MgH. Moving the elevator upwards at 3 m/s increases height at 3.00 m/s, and Mg is a force. force * m/s = power, and you can add the frictional force to the force of gravity. That should give you the correct answer. | ||
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evanthebouncy!
United States12816 Posts
Power = ForceNeeded * Speed so just add up the force, friction and gravity, times that to speed... Don't see it as too difficult, maybe review your section abit? :D | ||
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Kula
United States140 Posts
Without adding gravity it becomes 64981.44=65 kw How'd you guys do it? | ||
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Peter[Deuce]
United States93 Posts
I did P=(1800x9.81+4000)3 P=64974 It is only 65kW =[ | ||
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magicbullet
Singapore163 Posts
Power = work done/ time = (work done against friction + work done to pull the elevator up) / time taken = (1800KG*10 m/s^2*3m + 4000N*3m) / 1s = 66 kw | ||
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Luddite
United States2315 Posts
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Peter[Deuce]
United States93 Posts
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