Discover a world of knowledge at Westonci.ca, where experts and enthusiasts come together to answer your questions. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
Find how much work ∆W is done by the motor in lifting the elevator:
P = ∆W / ∆t
where
• P = 45.0 kW = power provided by the motor
• ∆W = work done
• ∆t = 20.0 s = duration of time
Solve for ∆W :
∆W = P ∆t = (45.0 kW) (20.0 s) = 900 kJ
In other words, it requires 900 kJ of energy to lift the elevator and its passengers. The combined mass of the system is M = (m + 490.0) kg, where m is the mass of the elevator alone. Then
∆W = M g h
where
• g = 9.80 m/s² = acceleration due to gravity
• h = 35.0 m = distance covered by the elevator
Solve for M, then for m :
M = ∆W / (g h) = (900 kJ) / ((9.80 m/s²) (35.0 m)) ≈ 2623.91 kg
m = M - 490.0 kg ≈ 2133.91 kg ≈ 2130 kg
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.