(a) Find the useful power output (in W) of an elevator motor that lifts a 2300 kg load a height of 25.0 m in 12.0 s, if it also increases the speed from rest to 4.00 m/s. Note that the total mass of the counterbalanced system is 10,000 kg-so that only 2300 kg is raised in height, but the full 10,000 kg is accelerated. (Enter a number.) 210833.33 (b) What does it cost (in cents), if electricity is $0.0900 per kW · h? (Enter a number.) 1897.5 X cents

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(a) Find the useful power output (in W) of an elevator motor that lifts a 2300 kg load a height of 25.0 m in 12.0 s, if it also increases the speed from rest to 4.00 m/s. Note that the total mass of the
counterbalanced system is 10,000 kg-so that only 2300 kg is raised in height, but the full 10,000 kg is accelerated. (Enter a number.)
210833.33
X W
(b) What does it cost (in cents), if electricity is $0.0900 per kW · h? (Enter a number.)
1897.5
cents
Transcribed Image Text:(a) Find the useful power output (in W) of an elevator motor that lifts a 2300 kg load a height of 25.0 m in 12.0 s, if it also increases the speed from rest to 4.00 m/s. Note that the total mass of the counterbalanced system is 10,000 kg-so that only 2300 kg is raised in height, but the full 10,000 kg is accelerated. (Enter a number.) 210833.33 X W (b) What does it cost (in cents), if electricity is $0.0900 per kW · h? (Enter a number.) 1897.5 cents
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