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h four fewer passengers than before) is 1600 kg, what should the spring constant be for a maximum compression of 3.00

h four fewer passengers than before) is 1600 kg, what should the spring constant be for a maximum compression of 3.00

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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If the mass of the elevator (with four fewer passengers than before) is 1600 kg, what should the spring constant be for a maximum compression of 3.00 m?

SOLUTION SET UP AND SOLVE We take the initial point as the elevator's position when it first contacts the spring (yi = 3.00 m) and the final point as its position when the spring is fully compressed (yf = 0). The elevator's initial speed is vi = 8.00 m/s, so K = mv;? =(2000 kg)(8.00 m/s)? = 64, 000 J The elevator stops at the final point; thus, Kf = 0. The elastic potential energy at the initial point is zero because the spring isn't yet compressed. The initial gravitational potential energy is Uj = mgyi = (2000 kg) (9.80 m/s²)(3.00 m) = 58, 800 J At the final point, the gravitational potential energy is zero (because yf = 0) and the elastic potential energy is Uf = ka? = k(-3.00 m)? where the force constant k is to be determined. Inserting all these values into Kf - K = Ugravi – Ugrav.f + Ueli - Uelf + Wother: we find that Ki + U = Kf + Ut 64, 000 J+ 58, 000 J = 0+ k(-3.00 m)? k = 2.73 x 104 N/m REFLECT You still have to explain to the client that the elevator won't stay at the bottom of the shaft, but will bounce back up and then return to hit the spring again and again until enough energy has been removed by friction for it to stop.
Transcribed Image Text:SOLUTION SET UP AND SOLVE We take the initial point as the elevator's position when it first contacts the spring (yi = 3.00 m) and the final point as its position when the spring is fully compressed (yf = 0). The elevator's initial speed is vi = 8.00 m/s, so K = mv;? =(2000 kg)(8.00 m/s)? = 64, 000 J The elevator stops at the final point; thus, Kf = 0. The elastic potential energy at the initial point is zero because the spring isn't yet compressed. The initial gravitational potential energy is Uj = mgyi = (2000 kg) (9.80 m/s²)(3.00 m) = 58, 800 J At the final point, the gravitational potential energy is zero (because yf = 0) and the elastic potential energy is Uf = ka? = k(-3.00 m)? where the force constant k is to be determined. Inserting all these values into Kf - K = Ugravi – Ugrav.f + Ueli - Uelf + Wother: we find that Ki + U = Kf + Ut 64, 000 J+ 58, 000 J = 0+ k(-3.00 m)? k = 2.73 x 104 N/m REFLECT You still have to explain to the client that the elevator won't stay at the bottom of the shaft, but will bounce back up and then return to hit the spring again and again until enough energy has been removed by friction for it to stop.
In this problem we have to account for both gravitational potential energy and elastic potential energy. In a "worst-case" design scenario, a 2000 kg elevator with broken cables is falling at 8.00 m/s when it first contacts a cushioning spring at the bottom of the shaft. The spring is supposed to stop the elevator, compressing 3.00 m as it does so (Figure 1). As an energy consultant, you are asked to determine what the force constant of the spring should be. Ignore air resistance and friction in the elevator guides. Figure 1 of 1 2000 kg 8.00 m/s Initial point 3.00 m mg Final point
Transcribed Image Text:In this problem we have to account for both gravitational potential energy and elastic potential energy. In a "worst-case" design scenario, a 2000 kg elevator with broken cables is falling at 8.00 m/s when it first contacts a cushioning spring at the bottom of the shaft. The spring is supposed to stop the elevator, compressing 3.00 m as it does so (Figure 1). As an energy consultant, you are asked to determine what the force constant of the spring should be. Ignore air resistance and friction in the elevator guides. Figure 1 of 1 2000 kg 8.00 m/s Initial point 3.00 m mg Final point
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