Question

Conservation of Energy/Power Problem: The Sears tower has elevators that weigh 1000kg each. If the elevator is lifted 400m to one of the upper floors in a total of 15s. (You will need to use the kinematic equations for some parts of this problem) Part 1. How much Power must the engine doing the lifting be rated for? Give your answer to the nearest kilo-Watt (1 kW = 1000 W). kW Part 2. What is the Potential Energy of the Elevator setting at the top? Give your answer to the nearest kilo-Joule (1 kJ = 1000 J). kJ Part 3. If the cable snaps and the Elevator begins to freely fall, what is the Kinetic and Potential energy of the Elevator: At 200m above the ground? Potential Energy: 2 Kinetic Energy: J

Answer 1

11 PE w t mgh t 1000 X 9.6900 261333.33 W 15 P= 261.33 kW 2) U= mgh = 1000 x 9.8 x 400 = 3920 x103 J = 3900 kJ 3 Total energy = 3920 kJ AL 200m above gocund polential energy 1000x 9. 8x 200 = 1960 kJ - lanetic energy 3920 - 1960 = 1960 kj After 2 seconds U= Omls, a = g = 9.8 m/s² v= utat = 2x 9.8 = 19.6 m/sa kinetic energy = mv amv? x 1000 x(19-672= 192.08 kj Potential energy 3920 - 192.08 = 3727.92 kJ Right before striking ground 0 kj poterstal energy kinetic energy 3920 kJ 4) mu? 3920x10] => v? 263920 => V = 88.59 m/s 88.59 Ć - V-u 9.035 o 9.8