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(a) Find the useful power output of an elevator motor that lifts a 2500-kg load at a height of 35.0 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 2500 kg is raised in height, but the full 10,000 kg is accelerated.

(b) What does it cost, if electricity is $0.0900 per kW·h?

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Vikas
VikasLv10
4 Dec 2020

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Related textbook solutions

Physics: Principles with Applications
7th Edition, 2013
Giancoli
ISBN: 9780321625922

Related questions

(a) Find the useful power output of an elevator motor that lifts a 2500-kg load at a height of 35.0 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 2500 kg is raised in height, but the full 10 000 kg is accelerated. (b) What does it cost, if electricity is $0.900 per kWh?

Find the useful power output of an elevator motor that lifts a 2500 kg load a height of 35.0 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 system is balanced by 10000 kg so that only 2500 kg is raised in height, but the full 10000 kg is accelerated.

Consider the following system ashaving frictionless surfaces. A 20 kg block is connected by astring to a 30 kg block over a frictionless pulley. The 30 kg blockis also connected to a spring, which is currently at equilibrium 20cm above a table top. The 20 kg block, originally at a height of 45cm above the table top, is pulled down the incline a distance of 20cm - stretching the spring and raising the 30 kg mass to a heightof 40 cm (20 cm above its original position). The 20 kg block isthen released from rest.

If k=250 N/m, find the velocity of the 30 kg mass as it reachesits original height on its way back down.

[hint: Use conservation of energy, setting U = 0 at theequilibrium point of the spring. Also note that the 45 cm height ofthe 20 kg block is not its initial height.]

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